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Chemical BondingIsomerism

Chemical Equilibrium Solid State

CHEMISTRY

1. CHEMICAL BONDING

Introduction 1.1Chemical Bond 1.1Cause of Chemical Combination 1.1Octet Rule (Lewis-Kossel Rule) 1.1Electron-Dot (Lewis) Structure 1.1Exceptions of Octet Rule 1.2Ionic Bond 1.3Covalent Character In Ionic Compounds 1.5Covalent Bond 1.6Valence Bond Theory 1.7Gilesspie and Nyhom Theory or Vsepr Theory 1.9Resonance 1.9Hybridisation 1.11Some Special Bonding Situations 1.17Bond Parameters 1.18Factors Affecting Bond Length 1.19Molecular Orbital Theory (MOT): 1.21Dipole Moment 1.25Hydrogen Bond 1.27Vander Waal’s Forces 1.30Exercise–0 (Recall Your Understanding) 1.31Exercise–1 (Check Your Understanding) 1.32Exercise–2 (Challenge Your Understanding) 1.39Exercise–3 (AIIMS Special) 1.47Exercis–4 (Archives From Various Medical Entrance) 1.50Answer Key 1.59

2. ISOMERISMIntroduction 2.1Structural Isomerism 2.1Chain Isomerism 2.1Position Isomerism 2.1Ring Chain Isomerism (RCI) 2.3Functional Isomerism 2.4

Metamerism 2.5Tautomerism Or Desmotropism 2.6Stereo Isomerism 2.9Configurational Isomerism 2.9Exercise–0 (Recall Your Understanding) 2.27Exercise–1 (Check Your Understanding) 2.28Exercise–2 (Challenge Your Understanding) 2.33Exercise–3 (AIIMS Special) 2.40Exercis–4 (Archives From Various Medical Entrance) 2.41Answer Key 2.44

3. CHEMICAL EQUILIBRIUM

Introduction 3.1

Chemical Reaction 3.1

Rate of Reaction 3.2

Law of Mass Action 3.2

Equilibrium Constant (K) 3.4

Applications of Equilibrium Constant 3.6

Characteristics of Equilibrium Constant & Factors Affecting It 3.8

Homogeneous Liquid System 3.9

Heterogenous Equilibrium 3.10

Degree of Dissociation (A) 3.10

External Factors Affecting Equilibrium 3.11

Application of Le- Chatelier’s Principle 3.14

Thermodynamics of Equilibrium 3.15

Exercise–0 (Recall Your Understanding) 3.17

Exercise–1 (Check Your Understanding) 3.18

Exercise–2 (Challenge Your Understanding) 3.27

Exercise–3 (AIIMS Special) 3.33

Exercis–4 (Archives From Various Medical Entrance) 3.34

Answer Key 3.40

CONTENTS

4. SOLID STATE Introduction 4.1

Properties of Solids 4.1

Types of Solids 4.1

Study of Crystals 4.3

Types of Unit Cell 4.4

Different Classes of Crystals 4.4

Mathematical Analysis of Cubic System (Types and Analysis) 4.5

Arrangement of the Atom / Particles of the Solids in Three Dimensions 4.7

Types of Voids Found in Close Packings 4.11

Ionic Solids 4.13

Crystal Defect (Point Defects) 4.16

Properties of Solids 4.19

Exercise–0 (Recall Your Understanding) 4.21

Exercise–1 (Check Your Understanding) 4.22

Exercise–2 (Challenge Your Understanding) 4.27

Exercise–3 (AIIMS Special) 4.32

Exercis–4 (Archives From Various Medical Entrance) 4.33

Answer Key 4.37

CHEMICAL BONDING (CHEMISTRY) 1.1

CHEMICAL BONDING

Chapter

1

INTRODUCTION (i) Most of the elements exist as molecules which

are cluster of atoms. How do atoms combine toform molecules and why do atoms form bonds.Such doubts will be discussed in this chapter.

(ii) A molecule will only be formed if it is more stableand has a lower energy, than the individual atoms.

CHEMICAL BOND (i) A force that acts between two or more atoms to

hold them together as a stable molecule.(ii) It is union of two or more atoms involving

redistribution of electron among them.(iii) This process accompanied by decrease in energy.(iv) Decrease in energy Strength of the bond.(v) Therefore molecules are more stable than atoms.

1. CAUSE OF CHEMICAL COMBINATION a. Tendency to acquire noble gas configuration :(classical concept)

(i) Atom combines to acquire noble gas configuration.(ii) Only outermost electron i.e. ns, np and (n-1)d

electrons participate in bond formation.(iii) Inert gas elements do not participate, as they have

stable electronic configuration and henceminimum energy. (Stable electronic configuration: 1s2 or ns2np6)

CHEMICAL BONDS

(A)Ionic bond

(Energy released >42 KJ/mole)

(B)Covalent bond

(C)Co-ordinate bond

(D)Metalic bond

STRONG BOND (Inter atomic)

(E)Hydrogen bond

(Energy 2 - 42 KJ/mole)

(F)Vander waal's bond

WEAK BOND(Inter Molecular)

(8 - 42 KJ/mole) (2 - 8 KJ/mole)

b. Tendency to acquire minimum energy :(modernconcept)

(i) When two atoms approaches to each other.Nucleus of one atom attracts the electron ofanother atom.

(ii) Two nuclei and electron of both the atoms repellseach other.

(iii) If net result is attraction, the total energy of thesystem (molecule) decreases and a chemicalbond forms.

(iv) So, Attraction 1/energy Stability..

2. OCTET RULE (LEWIS-KOSSEL RULE)

Noble gases have 8 electron in their outermost shell(complete octet) and outermost configuration is ns2p6.Every atom has a tendency to complete its octet by losingor gaining or by sharing electron.

Three different types of bonds—ionic, covalent,and metallic—are formed depending on the electropositiveor electronegative character of the atoms (of an element)involved in a bond formation. Atoms of various elementsmay be electropositive or electronegative dependingon the type of element. Elements may be classifiedas follows:

(i) Electropositive elements. These elements readilylose one or more electrons.

(ii) Electronegative elements. These elements acceptelectrons and achieve a stable electronicconfiguration.

(iii) Elements that have a little tendency to lose or gainelectrons.

ELECTRON-DOT (LEWIS) STRUCTUREFor single central atom Lewis structure drawing

1.2 CHEMICAL BONDING (CHEMISTRY)

Rules for Lewis Structure Drawing :(i) Calculate n1 = no. of valence shell electrons of all

atoms

+ no. of negative charge (if any)

– no. of positive charge (if any)

(ii) Calculate n2 = (no. of H-atom × 2) + (no. of atomsother than H-atoms × 8)

(iii) Calculate n3 = n2–n1 no. of shared electrons

i.e. No. of bonds = 2n3 ......(a)

(iv) Calculate n4 = n1 – n3 no. of unshared electrons

i.e. no. of lone pairs = 2n4 ......(b)

Using the informations (a) & (b) the structure isto be assigned as follows:

(i) Find out the central atom first (i.e. either least innumber or more electro positive)

(ii) Allocate the surrounding atoms around the centralatom with the help of bonds available in (a).

(iii) To fulfill the octet of each atom. utilise the lonepairs available in (b).

(iv) Finally calculate the formal charge for each atomand assign on the atoms according the formulagiven.

F.C. (of an atom) = Valence shell e– of that atom –no. of bonds associated with it – no. of unshared e–s on it.

In order to write the Lewis structure of a molecule,the points given below need to be kept in mind.

(i) Add the valence electrons of all atoms. In case ofan ion, add electrons equal to negative charge (fornegative ion) or subtract the number of electronsequal to the positive charge (for a positive ion).

(ii) Write the formula of the compound, and join atomsby single bonds.

(iii) Complete the octet of all the atoms (or twoelectrons for hydrogen) which are bonded to thecentral atom.

(iv) Fix up the remaining electrons, if any, on thecentral atom.

(v) In case the central atom does not have the requisitenumber of electrons to complete its octet, attachit by multiple bonds.

S.NO. CO2 NH3 PCl4+ NO2–

1. n1 16 8 32 182. n2 24 14 40 243. n2 – n1= n3 8 6 8 64. n3/2 4 3 4 35. n3 – n1= n4 8 2 24 126. n4/2 4 1 12 6

O = C = O

::

::

NH H H

:

Cl

PCl

::: :: :

Cl

:: :

Cl

: ::+

ON

:

::: O: :

(1) (2)

Formal changeC = zero N = zero P = +1 N = zeroO = zero H = zero Cl = zero O(1) = zero

O(2) = –1

3. EXCEPTIONS OF OCTET RULE(a) Incomplete octet molecules : or (electrondefficient molecules) or Hypovalent molecules

Compound in which octet is not complete in outermost orbit of central atom.

Examples: Halides of IIIA groups, BF3, AlCl3, BCl3,hydride of III A/13th group etc.

B× ×

×

Cl

Cl

Cl

selectron6onlyhasBoron

BClIn 3

Other examples - BeCl2 (4 electron), Ga(CH3)3 (6electron)(b) Expansion of octet or (electron efficientmolecules) or Hypervalent molecules

Compound in which central atom has more than 8electron in outermost orbits.

Examples: PCl5, SF6, IF7, the central atom P, Sand I contain 10, 12, and 14 electrons respectively.

P Cl ××

×

×Cl ×

Cl

Cl Cl

Electron dot formula of PCl5

Cl

Cl

(c) Pseudo inert gas configuration : Cations, whichcontains 18 electrons in outermost orbit

Examples : Ga+3, Cu+, Ag+, Zn+2, Cd+2, Sn+4, Pb+4

etc.


Electronic configuration of Ga+3 - 1s2, 2s2 2p6,3s23p63d10

18 electron(d) Odd electron molecules : Central atom have

an unpaired electron or odd number (7 electron) ofelectrons in their outer most shell.

Examples : NO, NO2 ClO2, ClO3 etc.

4. IONIC BOND(i) Atoms get established by ion formation in

formation of an ionic bond. One atom forms acation by losing electrons and other forms anionby accepting electrons.

(ii) The octets of the atoms are completed by transferof electrons.

(iii) Cations and anions are bonded by electrostaticforce of attraction in ionic bond.

(iv) Ionic bond is formed between metals (cations)and nonmetals (anions).

(v) During the formation of an ionic bond, cation canattain one of the following configurations.(a) Inert Configuration : I A and II A group

metals (octet configuration) Na+1, Mg+2,K+1, Rb+1 attain ns2 np6 inert configuration.

(b) Pseudo Inert Configuration : Metals ofI B and II B groups – Cu+, Zn+2, Ag+1, Au+1,Cd+2, Hg+2.

(c) Transition metals : e.g. Ti+3 (9), Ti+2 (10).(vi) Energies involved in ionic bond formation :

(a) Ionisation Energy – The energy requiredto remove an electron from the outermostshell of metal in gaseous state.X(g) X+(g) + 1e– (Ionisation Energy)

(b) Electron Affinity / Electron Gain Enthalpy(Heg) – The energy required duringformation of an anion by addition of anelectron to a nonmetal in gaseous state.Y(g) + 1e– Y–(g)(Electron Affinity)

(c) Lattice Energy –(i) The energy released during the formation of

ionic bond.(ii) Cations and anions form crystal lattice of ionic

crystal in space by electrostatic force ofattraction.

(iii) Lattice Energy –1

r r? ? a charge on

Cation/ AnionX+(g) + Y–(g) XY(s) (Lattice Energy)

(d) Ionic compound is formed when the energyrequired ( Ionisation energy) is less than energy released(Electron affinity + Lattice energy)

Example : (1) Lesser Ionisatoin energy Greatertendency to form cation.Na+ > Mg+2 > Al+3

Cs+ > Rb+ > K+ > Na+ > Li+

(2) Higher electron affinity Greater tendency toform anion

Cl– > F– > Br– > I–

F– > O–2 > N–3

(3) Energy released during the formation of one molecrystal lattice is called lattice energy.

Na+ + Cl– NaCl + 94.5 K. cal of energy (Latticeenegry)

(4) Magnitude of charge U z+ z– (Ionic charge)Lattice energy a Magnitude of chargeNaCl MgCl2 AlCl3Na+ Mg+2 Al+3

– Lattice energy increases– Size of cation decreases.

(5) Size of Cation :- Lattice energy 1

r r? ???

LiCl NaCl KCl RbCl CsCl– Size of cation increasing– Size of anion is constant– Lattice energy decreases.Characteristic of Ionic Compounds

(i) Solid and Crystalline Structure :Ionic compounds do not show molecular structure.Ionic compounds have definite crystal structures.

CsCl has body centered cubic (BCC) structure.(ii) Melting and Boiling Point :

Ionic compounds have high melting and boiling points.(iii) Hardness :

Ionic compounds are solid with brittle nature.

Cation formation tendency???


Brittleness

+ + +

+ +

+ + +

+ +

+ + +

+ +

+ + +

+ +

Attraction Repulsion

{Same charged ions comes nearer. So they repell eachother.}(iv) Conductivity :

Ions are not free in solid state so ionic compoundsare bad conductor of electric current in solid state but insolution and fused state electric current passes throughand the ionic compound becomes good conductor ofelectricity(v) Solubility of Ionic Compounds :

Ionic compounds are soluble in solvents which havehigh dielectric constant like water.

Highly soluble in water (Polar solvents)Example: NaCl in water(I) The Na+ ions get associates with - vely charged

'O' of water(II) And Cl– ions associates with +vely charged 'H' of

water.

+

�

HO2

+

�

H O2

+

�H

O2

+

�H O2+

�

HO2

+

�

HO2

+�

HO

2

Na+

+

�

HO2

+

�

H O2

+

�H

O2

+

�H O2

+

�

HO2

+

�

HO2

+�

HO

2

Cl�

Oxygen atom of H2O H atom of H2Oattracted to Na+ attracted to Cl–

(III) Thus charge on Na+ and Cl– decreases andelectrostatic force of attraction also decreaseswhich leads to free ion.

(IV) The energy released due to interaction betweensolvent and solute is called solvation energy. Ifwater is used as solvent it is called hydrationenergy.

(V) For an ionic compound to be soluble in water –Hydration energy > Lattice energy

Lattice energy 1

Solubility

Hydration energy Solubility.

Hydration energy (HE) 1 1r r? ?? { r+ & r– are

radius of cation and anion}(VI) Hydration energy mainly depends on the cation

radius because the value of 1r? is negligible in

comparision to 1r? .

(VII) The capacity of solvent to neutralise the chargeof ionic compounds is called Dielectric constant.It is represented by .

Water has maximum dielectric constant (= 80)(CH3OH = 35), (Acetone = 21), (C2H5OH = 27),

(Ether = 4.1), (Benzene = 2.3)Note: H2SO4 and H2O2 have high dielectric constant

but these are not a good solvent due to oxidising nature.Example: Keeping size of cation constant, the lattice

energy decreases with the increase of anionic radius.Hence order of solubility of LiX in water is LiF < LiCl

< LiBr < LiI As solubility 1

lattice energy

Note: In LiI covalent nature is more according toFajan's rule but HE > LE therefore LiI is more soluble inwater.

Example: Keeping size of anion constant, thehydration energy decreases with the increase of cationicradius. Hence order of solubility of MSO4 will be –

BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4(Exception of Fajan's rule)

Example: Solubility decreases in a period (as ionicnature decreases and covalent nature increases)

NaCl > MgCl2 > AlCl3Note: (I) Ionic compounds shows ionic reaction andcovalent compounds shows - molecular reaction.

(II) Ionic reactions are faster than molecular reactionbecause of free ions.

e.g. When NaCl is added in AgNO3 solution, whiteppt of AgCl is formed at once.

Ag+ NO3–+ Na+Cl– Na+ NO3

–+ AgCl white ppt.(vi) Isomorphism –

(I) Two compounds are said to be isomorphous ifthey have similar no. of electrons i.e. similarconfiguration of their cation and anion.

(II) They have similar crystal structure.


Example:Na+ F– Mg+2 O–2

Valency +1, –1 +2, –2electronic configuraton 2, 8 2, 8 2, 8 2, 8similarly Ca+2 2Cl–1 2K+1 S–2

2, 8, 8 2,8,8

2,8,8

2,8,8

2,8,8

2, 8, 8

5. COVALENT CHARACTER IN IONIC COMPOUNDS(FAJAN’S RULE) :When anion and cation approach each other, the valenceshell of anion is

pulled towards cation nucleus and thus shape of anion isdeformed. This phenomenon of deformation of anion bya cation is known as polarisation and the ability of cationto polarize a near by anion is called as polarizing power ofcation.

Fajan’s pointed out More distortion of anion, morewill be polarisation then covalent character increases.

Fajan’s gives some rules which govern the covalentcharacter in the ionic compounds, which are as follows:

(i) Size of cation : Smaller is the cation more is itspolarizing power and thus more will be the polarisation ofanion. Hence more will be covalent character incompound.

Size of cation 1

Polarisation

Example: BeCl2 MgCl2 CaCl2 SrCl2 BaCl2LiCl > NaCl > KCl > RbCl > CsCl

Size of cation increasesPolarisation decreasesCovalent character decreases

(ii) Size of anion : Larger is the anion, greater is itspolarisability and, therefore, more will be the polarisation.Thus more will be covalent character in compound.

Size of anion a polarisationExample: LiF LiCl LiBr Lil

– Size of anion increases and Polarisation increases– Covalent character increases(iii) Charge on cation : Higher is the oxidation state

of cation, more will be the deformation of anion and thus,more will be covalent character in compound.

Charge on cation a polarisation.Example: NaCl MgCl2 AlCl3

Na+ Mg2+ Al3+

– Charge of cation increases and Polarisation increases– Covalent character increases

(iv) Charge on anion : Higher is the charge on anionmore will be the polarisation of anion and thus more willbe covalent character in the compound.Charge on anion a polarisation.Example: AlF3 Al2O3 AlN

F– , O2– , N3–

– Charge on anion increases and Polarisation increases– Covalent character increases

(v) Pseudo inert gas configuration of cation :Cation having pseudo inert gas configuration has morepolarizing power than the cation that has inert gasconfiguration. Thus NaCl having inert gas configurationwill be more ionic whereas CuCl having pseudo inert gasconfiguration will be more covalent in nature.Cu+ = [Ne] 3s2 p6 d10 Na+ = 1s2 2s2 p6

18e– 8e–

Pseudo inert gas Inert gas configurationconfiguration (more shielding(poor shielding of of s and p electrons)d-electrons)

Application & Exceptions of Fajan’s Rules :(i) Ag2S is less soluble than Ag2O in H2O because

Ag2 S is more covalent due to bigger S2– ion.(ii) Fe(OH)3 is less soluble than Fe(OH)2 in water

because Fe3+ is smaller than Fe2+ and thus charge is more.Therefore, Fe(OH)3 is more covalent than Fe(OH)2 .

(iii) The colour of some compounds can be explainedon the basis of polarisation of their bigger negativeions.For example :

AgCl is white AgBr, AgI, Ag2CO3 are yellow. Similarly,SnCl2 is white but SnI2 is black. PbCl2 is white but PbI2 isyellow.

The bigger anions are more polarised and hence theirelectrons get excited by partial absorption of visible light.

(iv) Variation of melting point [melting point ofcovalent compound < melting point of ionic compound]


BeCl2 , MgCl2 , CaCl2, SrCl2, BaCl2––––––––––––––>–––––––––––––––––––––>–Ionic charater increases, melting point increases ; since

size of cation increases & size of anions is constant.CaF2, CaCl2, CaBr2, CaI2

–––––––––––––––>–––––––––––––––––––––>–Covalent character increase, melting point decrease ;

since size of cations increase & size of anions is constant.

EXAMPLE 1AgCl is colourless whereas AgI is yellow, because of :

(A) Ag+ have 18 electron shell to screen the nuclearcharge.

(B) Ag+ shows pseudo inert gas configuration.(C) distortion of I– is more pronounced than Cl– ion.(D) existence of d – d transition.

SOLUTION:(C) the bigger anions are more polarised and hence theirelectrons get excited by partial absorption of visible light.

EXAMPLE 2Write the increasing order of M.P. & B.P. of followingcompounds.(1) SnCl4, SnCl2 (2) FeSO4, Fe2(SO4)3

(3) PbCl4, PbCl2

SOLUTION:(1) Sn+4 < Sn+2 (2) Fe+3 < Fe+2 (3) Pb+4 < Pb+2

(Charge on cation polarisation power covalent

character1

M.P.? )

EXAMPLE 3Write increasing order of M.P. & B.P. of -

CaSO4 , Ca3(PO4)2 and ClO4– , SO4

–2 , PO4–3

SOLUTION:CaSO4 > Ca3(PO4)2 ClO4

– > SO4–2 > PO4

–3

Charge on anion polarisation covalent

nature1

M.P.?

6. COVALENT BOND(I) A covalent bond is formed by the mutual sharing ofelectrons between two atoms of electro negative elements

to complete their octet.(Except H which completes itsduplet)

..H H

H molecule2

H—H ..O

..: .. O

..:

O2

O O

..N.

. .... N..

N2

N N

(II) The shared pair of electrons should have oppositespins, and are localised between two atoms concerned.

(III) Shairing of electrons may occurs in threeways –

No. of electrons Electron pair Bondshared between

two atoms2 1 Single bond (—)4 2 Double bond ( )6 3 Triple bond ( )

Examples –|

H N H

H

? ? { Three single bonds (not triple

bond)N N Triple bond. (not three single bond) O O

Double bond (Not two single bond) H—O—H (Two singlebonds.)Lewis structure and covalent bond :

(I) Electron dot structures, also known as Lewisstructures of covalent molecules, are written inaccordance with octet rule.

(II) All atoms in a formula will have a total of eightelectrons by sharing in the valence shell exceptthe H-atom which forms the largest number ofbonds with other atoms placed in the centre ofskeleton structure. Other atoms surround it tocomplete the octet.

(III) Structure in which valence electrons arerepresented by dots are called lewis structures.

(IV) Structures represented by line (–) or dashes areknown as couper structure.

(V) Lewis dot formulae show only the number ofvalency electron, the number and kinds of bonds,but do not depict the three dimensional shapes ofmolecules and polyatomic ions.

Orbital concept of covalent bond :(I) One orbital can accomodate at the most 2 electrons

with opposite spins


(II) Half filled orbital or unpaired electron orbitalaccepts one electron from another atom, tocomplete its orbital.

(III) Tendency to complete orbital or to pair the electronis an essential condition of covalent bond.Completion of octet is not the essential conditionof covalent bond.

(IV) Covalency : It is defined as the number ofelectrons contributed by an atom of the elementfor shairing with other atoms to achieve noblegas configuration.

(V) If the outermost orbit has empty orbitals thencovalent bonds are formed in excited state.

Variable valency in covalent bonds :(I) Variable valencies are shown by those elements

which have empty orbitals in outermost shell.(II) Lone pair electrons get excited in the subshell of

the same shell to form the maximum number ofunpaired electrons. Maximum covalency isshown in excited state.

(III) The energy required for excitation of electrons iscalled promotion energy.

(IV) Promotion rule – Excitation of electrons in thesame orbit.

Example –(1) Phosphorus Ground state

3s 3p

Covalency 3 (PCl3)

Phosphorus Excited state

3s 3p 3d Covalency – 5 (PCl5)

(2) Sulphur Ground state.

3s 3p 3d

Covalency - 2 (SF2)

Sulphur Excited state

Ist excited state3s 3p 3d

Covalency - 4 (SF4)

2nd excited state3s 3p 3d

Covalency - 6 (SF6)So variable covalency of S is 2, 4, & 6.

(3) Iodine has three lone pair of electrons(Ground state)

5s 5p 5d

So it shows three excited states – Maximum number ofunpaired electrons = 7Variable Valencies are 1, 3, 5, 7

APPLICATIONS OF VARIABLE VALENCY(1) To explain existence of molecules –NCl3 — existsNCl5 — doesn't exists (due to absence of d-orbitals in

Nitrogen.) While PCl3 and PCl5 both existsbecause 3d orbitals are present in phosphorus.

OF2 — exists, but OF4 and OF6 doesn't exists due toabsence of d-orbitsls While SF4 and SF6 existsdue to presence of d-ortbital, present in itsvalence shell.

Wave mechanical concept of chemical bonding –(Overlapping)

To explain the nature of covalent bond two theoriesbased on quantum mechanics have been proposed.

(1) Valence bond theory (VBT)(2) Molecular orbital theory (MOT)

7. VALENCE BOND THEORY(i) It was presented by Heitler & London to explain

how a covalent bond is formed.It was extended by Pauling & Slater.(ii) The main points of theory are –(a) To form a covalent bond overlapping occurs

between half filled valence shell orbitals of thetwo atoms.

(b) Resulting bond acquires a pair of electrons withopposite spins to get stability.

(c) Orbitals come closer to each other from thedirection in which there is maximum overlapping

(d) So covalent bond has directional character.(e) Extent of overlapping strength of chemical

bond.(f) Extent of overlapping depends on two factors.(i) Nature of orbitals – p, d and f are directional

orbitals more overlapping s-orbitals non directional – less overlapping(ii) Nature of overlapping – Co-axial overlapping –

extent of overlapping more.Collateral overlapping – extent of overlapping lessOrder of strength of Co - axial overlapping –


p - p > s - p > s - s (When intramolecular distance remains same)

p - p p - s s - s

(g) As the value of n increases, bond strengthdecreases.

1 - 1 > 1 - 2 > 2 - 2 > 2 - 3 > 3 - 3(h) If n is same bond strength order will be

following2p - 2p > 2s - 2p > 2s - 2s1s - 2p > 2s - 2p > 3s - 3p(i) Electron which is already paired in valency shell

can enter into bond formation, it they can be unpairedfirst and shifted to vacant orbitals of slightly higherenergy of the same energy shell.

(j) This point can explain the trivalency of boron,tetravalency of carbon, penta-valency of phosphorusetc.

(k) Two types of bonds are formed on account ofoverlapping.

(A) Sigma (s) bond (B) Pi (p) bond

(A) Sigma () bond :

(i) Bond formed between two atoms by theoverlapping of half filled orbitals along their axis(end to end overlap) is called sigma bond.

(ii) bond is directional.

(iii) bond do not take part in resonance.

(iv) Free rotation is possible about a single bond.

(v) Maximum overlapping is possible betweenelectron clouds and hence it is strong bond.

(vi) There can be only s bond between two atoms.

Sigma bond are formed by three types of overlapping

(a) s - s overlapping (H2) - Two half filled s-orbitalsoverlap along the internuclear axis.

+

1s 1s

bond

1s 1s

(Formation of H molecule)2(1s orbital of hydrogen) (1s orbital of hydrogen)

(b) s - p overlapping (Formation of HF) – When half fill s-orbital of one atom overlap with half filled p-orbital ofother atom.

1s orbital of Hydrogen 2p of Fluorineorbital H–F

+

(c) p–p overlapping – (Coaxial) – It involves the coaxial overlapping between half filled p-orbitals of two differentatoms.

eg. Formation of Cl2, F2, Br2

– + + + – –

p orbital p orbital

(B) Pi () bond(i) The bond formed by sidewise (lateral) overlapping are known as bonds.(ii) Lateral overlapping is only partial, so bonds formed are weaker and hence more reactive than bonds (Repulsion

between nucleus is more as orbitals have to come much close to each other for bond formation)


Example: Formation of O2 molecule –

+

O molecule2

Pz

Py Py

Pz

Atomic orbitalof oxygen

Atomic orbitalof oxygen

(IV) In the different molecules if central atom have samenumber of lone pair of electron then bond angle will dependon electronegativities of A & B.

In ABx type of molecules if side atoms are same andEN of central atom increases the bond angle increases.

O.. ..

105°

H H

more repulsion

> 92°

H

S.. ..

H

less repulsion bp�bp

Electronegativity of 'O' > Electronegativity of 'S'Bond angle of – NH3 > PH3 > AsH3In ABx type molecules, if central atoms are same and

the EN of side atoms increases then bond angle decreases.

O.. ..

110°

Cl Cl

O.. ..

103°

F F

Electronegativity of Fluorine is greater than chlorinePF3 < PCl3 < PBr3 < PI3 (EN of side atom decrease)OF2 < Cl2O < Br2OSF2 < SCl2 < SBr2Bond angle depends on size of side atom, On increasing

size of side atom bond angle increases.Cl2O > H2O

O.. ..

105°H H

O.. ..

110°Cl Cl

9. RESONANCE(i) The concept of resonance was introduced by

Heisenberg (1920), and later developed byPauling and Ingold, to explain the properties ofcertain molecules,

Only py and pz of oxygen atom have unpairedelectron in each orbital for bonding.

Electronic configuration of oxygen is – 1s22s22px2

2py12pz

1

(iii) Free rotation about a bond is not possible.(iv) bond is weaker than bond (Bond energy

difference is 63.5 KJ or 15 K cal/mole)(v) bonds are non-directional, so do not determine

the shape of a molecule.(vi) bond takes part in resonance.(vii) bond formed by pure or unhybrid orbitals.

8. GILESSPIE AND NYHOM THEORY OR VSEPRTHEORY :(Valence shell electron pair repulsion theory)(I) If the cenral atom possess only bonded pairs ofelectrons along with identical atoms then shape of thecompound is symmetrical and according to Sidgwick &Powel.eg. BF3 — 120° — triangular

CH4 — 109° 28' — tetrahedralCO2 — 180° — linear

(II) If the central atom possess bonded pair of electronsas well as lone pair of electron, then shape of the moleculewill be unsymmetrical ie. the original bond angle willdisturbed due to repulsion between lone pair of electrons.

Similarly on having different type of side atoms,molecule becomes unsymmetrical due to unequal forceof repulsion between e–. Order of repulsion is -

l.p. – l.p. > l.p. – b.p. > b.p. – b.p.

Bond angle 1No. of lone pair of electron

(III) By increasing one lone pair of electron, bondangle is decreased approx by 2.5°.eg.:- CH4 NH3 H2O sp3

109° 107° 105° hybridisation


(ii) It has been found that the observed propertiesof certain compounds cannot be satisfactorilyexplained by writing a single lewis structure.The molecule is then supposed to have manystructures, each of which can explain most ofthe properties of the molecule but none canexplain all the properties of the molecules. Theactual structure is in between of all thesecontributing structures and is called resonancehybrid and the different individual structures arecalled resonating structures or canonical forms.This phenomenon is called resonance.

(iii) Let us discuss resonance in ozone, accordingto its resonating structure it should have onesingle bond (O – O = 1.48Å) but experimentsshow that both the bonds are same which canbe proved by its resonance hybrid as shownbelow.

OO O

OO O

OO O

Resonance hybridTo calculate bond order in the polyatomic molecule

or ion use following formula :

Bond order = Total number of bonds in a molecule

Resonating Structures

eg. O CO¯O¯

C—O Bond order 34

1.33

O O¯P

O¯

O¯P—O Bond order

45

1.25

O O¯Cl

O

OCl—O Bond order

47

1.75

SHAPES OF MOLECULES BASED ON VSEPR THEORY

Total no. No. of b.p. No. of General Type of Stereo Shape Exam.of hybrid (bond pairs) unshared formula hybridisation chemicalorbitals pair i.e. lp formula/str.

2 2 0 AB2 s p B—A—B

180° linear BeCl2

3 3 0 AB3 sp2

B B

BA120°

Trigonal

23

33

3

CO,GaF

,NO,BCl

3 2 1 AB2 sp2

B BA

<120°

V or Bent or23

2SO,O

,SnClangular

4 4 0 AB4 sp3

B

B

BB

A 109°28’ Tetrahedron 44

4

CH ,SiF ,NH

4 3 1 AB3 sp3

BB

BA

<109°

Trigonal NH3

pyramid CH3–

4 2 2 AB2 sp3

B B

A<109°

V or Bent or H2O

angular SF2


4 1 3 AB sp3BA

bond angleN.A. linear ClO–

5 5 0 AB5 sp3dB

B

B B

BA 120°

90° Trigonal

235

5

FXeO,SF,PF

bipyramidal

5 4 1 AB4 sp3d B

B B

BA 120°

Seesaw SF4

5 3 2 AB3 sp3dB

B

B

A<90° T-shaped ClF3, BrF3

5 2 3 AB2 sp3d

B

B

A Linear

3

22

I,XeF

,ICl

6 6 0 AB6 sp3d2

B

B B

B

B

BA 90°

All bond = 90°

Octahedral SF6

6FI

6 5 1 AB5 sp3d2B

B

B

B

BA <90°

All less than 90°

Square 4

5XeOF

,IF

pyramidal

10. HYBRIDISATIONConsider an example of Be compound

If it is formed without hybridisation then -

Cl Be Clp – s p – p

both the Be–Cl bonds should have different parametersand p-p bond strength is greater than s–p bond strength.

Practically bond strength and distance of both theBe-Cl bonds are same.

This problem may overcome if hybridisation of sand p-orbital occurs.

(i) It is introduced by pauling, to explain equivalentnature of covalent bonds in a molecule.

(ii) Definition : Mixing of different shapes andapproximate equal energy atomic orbitals, andredistribution of energy to form new orbitals, of sameshape & same energy. These new orbitals are calledhybrid orbitals. and the phenomenon is calledhybridisation.

Now after considering s-p hybridisation in BeCl2

Cl Be Clp – sp sp – p

bond strength of both the bonds will be equal.Characteristic of Hybridisation :

(i) Hybridisation is a mixing of orbitals and notelectrons. Therefore in hybridisation full filled, halffilled and empty orbitals may take part.

(ii) Number of the hybrid orbitals formed is alwaysbe equivalent to number of atomic orbital whichhave taken part in the process of hybridisation.

Structure ofhybrid orbital


(iii) Each hybrid orbital having two lobes, one is largerand other is smaller. Bond will be formed fromlarge lobe.

(iv) The number of hybrid orbitals on central atom ofa molecule or ion = number of s bonds + lone pairof electron.

(a) The 1st bond between two atoms will be sigma.(b) The other bond between same two atoms will

be bond.(c) The electron pair of an atom which do not take

part in bond formation called as lone pair ofelectron.

(v) One element can represent many hybridisation

state depending on experimental conditions forexample, Carbon showing sp, sp2 and sp3

hybridisation in its compounds.(vi) Hybrid orbitals are differentiated as sp, sp2, sp3

etc.(vii) The order of repulsion between lp & bp is : lp -

lp > lp - bp > bp - bp(viii) The directional properties in hybrid orbital is

more than atomic orbitals. Therefore hybridorbitals form stronger sigma bond. Thedirectional property of different hybrid orbitalswill be in following order.sp < sp2 < sp3 < sp3d< sp2d2 < sp3d3

DIFFERENCE BETWEEN HYBRIDISATION & OVERLAPPING

Overlapping Hybridisation

1. It occurs between orbitals of two atoms 1. It occurs among orbitals of the same atom2. Only half filled orbitals takes part in 2. Any type of orbital can participates

overlapping3. It occurs during bond formation bond 3. Process, just before overlapping.

formed after hybridisation4. Orbital of different energies may participates 4. It may takes place in ground or in excited

in excited states. state In ground state– NH3, NCl3, PH3, PCl3,

DETERMINATION OF HYBRIDISATION STATE :Method (I) : Count the following pair of electronarround the central atom :

(a) Count all pure bonded electron pairs (or bonds)

(b) Count all lone pair of electron(c) Count Co-ordinate bond

Method (II) : To predict hybridisation following formulamay be used :

Number of hybrid orbital = 12

[Total number of

valence electron in the central atom + total number ofmonovalent atoms – charge on cation + charge on anion]

eg. 4NH : H =

12

[5 + 4 – 1] = 4 : sp3 hybridisation.

SF4 : H = 12

[ 6 + 4 ] = 5 : sp3d hybridisation.

–2O4S : H = 12

[ 6 + 2 ] = 4 : sp3 hybridisation.

(‘O’ is divalent so add only charge on anion)

–O3N : H = 12

[5 + 1] = 3 : sp2 hybridisation.

Where, H is the number of hybrid orbitals.If such type of e– pairs are –

Hybridised Participation Geometryorbital orbital

two s p hybridisation s, pz linearthree sp2 hybridisation s, px, py trigonal planarfour sp3 hybridisation s, px , py, pz tetrahedralfive sp3d hybridisation s, px , py , trigonal

pz , dz2 bipyramidal

six sp3d2 hybridisation s, px , py , octahedralpz , dx2–y2

,dz2

seven sp3d3 hybridisation s, px , py , pentagonalpz , dx2–y2

, bipyramidaldz2, dxy

EXAMPLE OF HYBRIDISATION1. sp Hybridisation

BeCl2 : Be is 1s2 2s2. In the excited state one of the2s-electrons is promoted to vacant 2p orbital to accountfor its divalency.


Examples on sp hybridisation

Example s bond l.p.e. Hybridisation Bond angle Shape

BeH2 2 - sp 180° LinearBeF2 2 - sp 180° LinearBeCl2 2 - sp 180° LinearBeBr2 2 - sp 180° LinearBeI2 2 - sp 180° LinearCO2 2 - sp 180° LinearCO 1 1 sp 180° LinearNO2

+ 2 - sp 180° LinearC2H2 2 - sp 180° LinearHCN 2 - sp 180° LinearZnCl2 2 - sp 180° LinearHgCl2 2 - sp 180° LinearCdCl2 2 - sp 180° LinearR-Mg-X 2 - sp 180° Linear[Ag(CN)2]– 2 - sp 180° Linear[Cu(CN)2]– 2 - sp 180° LinearN2O 2 - sp 180° LinearN3

– 2 - sp 180° Linear2. sp2 Hybridisation :

BCl3 molecule , the ground state electronic configuration of central boron atom is 1s2 2s2 2p1. In the excited state,one of the 2s electrons is promoted to vacant 2p orbital as a result boron has three unpaired electrons.

Fig. Formation of sp2 hybrids and the BCl3 molecule

Fig. (A) Formation of sp hybrids from s and p orbitals ; (B) Formation of the linear BeCl2 molecule.


Examples on sp2 hybridisation

Example bond l.p.e. Hybridisation Bond angle Shape

BH3 3 - sp2 120° Trigonal planarBF3 3 - sp2 120° Trigonal planarBCl3 3 - sp2 120° Trigonal planarCH3 3 - sp2 120° Trigonal planarCH2=CH2 3,3 - sp2 120° Trigonal planarGraphite 3 - sp2 120° Trigonal planarHNO3 3 - sp2 120° Trigonal planarNO3

– 3 - sp2 120° Trigonal planarHNO2 2 1 sp2 <120° Angular (V)SO2 2 1 sp2 <120° Angular (V)SO3 3 - sp2 120° Trigonal planarHCO3

– 3 - sp2 120° Trigonal planarCO3

–2 3 - sp2 120° Trigonal planarSnCl2 2 1 sp2 <120° Angular (V)SnBr2 2 1 sp2 <120° Angular (V)SnI2 2 1 sp2 <120° Angular (V)AlCl3 3 - sp2 120° Trigonal planarGaCl3 3 - sp2 120° Trigonal planarPbCl2 2 1 sp2 <120° Angular (V)

3. sp3 hybridisation :CH4 molecule , the ground state electronic configuration of central carbon atom is 1s2 2s2 2p2. In the excited state,

one of the 2s electrons is promoted to vacant 2p orbital as a result carbon has four unpaired electrons

The structure of NH3 and H2O molecules can also be explained with the help of sp3 hybridisation.

Fig. Formation of NH3 molecule( Pyramidal) Fig . Formation of H2O molecule (Bent)


Other Example

(a) Four sigma bonds with zero lone pair electron :The following examples represent this condition.CH4, CF4, CCl4, CBr4, CI4, NH4

+, BF4–, AlF4

–, BeF4–2,

MgF4–2 [Zn(CN)4]–2, [Cd(CN)4]–2, [Hg(CN)4]–2

In above compounds, bond angle is 109° 28' &tetrahedron shape.(b) Three sigma bonds & one lone pair of electron :

(I) This condition is shown by following compounds& ions.

NH3, PH3, AsH3, SbH3, BiH3, NF3, PF3, NCl3, PCl3, :CH3, H3O+ ClO3

–

(II) sp3 hybridisation, pyramidal shape & bond anglewill be lreduced from 109° 28' .(c) Two sigma bonds & two lone pair of elctrons :

(I) This condition is shown by following compoundsand ions.

H2O , H2S, H2Se, H2Te, OCl2, OBr2, OF2, OI2 etc.(II) In all above examples, the central atom showing

sp3 hybridisation,angular shape and bond angle will be either less then

109° 28' or more than 109° 28'.

(d) BeCl2 (s) ,Cl

Cl

Be

Cl

Cl

Be

Cl

Cl

Be

Cl

Cl

steric

number = 4 (i.e. sp3 )

(e) BF4– ; B =

(f) , Diamond, CCl4 , ; SiCl4 ,

• SiO2 is a covalent network solid like diamond

• Structures of cyclic silicates : [Si3O9]6–

= Silicon ; O = Oxygen

Note : Oxygen atom bonded with two Si atoms cannot have negative charge. There is no oxygen–oxygenbond. All silicates contain only Si–O bond and there is noSi – Si Bond.

(g) NH2– (amide ion) NH2 – NH2

(hydrazine) Each N atom is tetrahedrally surrounded by one N,

two H and a lone pair. The two halves of the moleculesare rotated 95º about N – N bond and occupy a gauche(non-eclipsed) conformation. The bond length is 1.45 Å.

NH2OH (hydroxylamine) lp-lp repulsion

increases the N – O bond length.(h) P4 (White phosphorus)

All phosphorus atoms occupy all four vertexes oftetrahedron. There are six P–P bonds and L P–P–P is 60º.

Since bond angle is 60º (against normal tetrahedralbond angle, 109.5º) so, P4 molecule is a strained molecule.So it is chemically very reactive.

• P4O6


P – O bond length shows that the bridging bonds onthe edges are 1.65 Å and are normal single bonds. Thereis no P – P bonds.

• P4O10

The P – O bond lengths shows that the bridging bondson the edges are 1.60 Å but the P = O bonds on the cornersare 1.43 Å and this P = O is formed by pp – dp backbonding.

• H2O2 H2O2(g)

O – O bond length (148 pm) is larger than the expecteddue to the repulsions between the lone pairs on the twooxygen atoms.

It has book like structure (angle between the two pagesof the book 94º) and both the O atoms have two lonepairs each.

• SOCl2 (Thionyl chloride)

trigonal pyramidal

• XeO4

4. sp3 d hybridiation :

Important points regarding sp3d -(i) According to VSEPR theory lone pair will occupy

equatorial positions but not axial.

(ii) More electronegative atoms will prefer to occupyaxial positions.

(iii) Since, double bonds occupy more space.Therefore, they will also prefer equatorial positions.

PCl5 (g)

It is covalent in the gas but in solid state exists asionic solid consisting of [PCl4]+ (tetrahedral) and [PCl6]–

(octahedral). All P–Cl bonds are not of equal lengths. Hereaxial bonds are longer and weaker than equatorial bonds.Note : PF5 remains covalent and is trigonal bipyramidal inthe solid state. PBr5(s) exists as [PBr4]+ Br– in solution.

SF4 and

XeO2 F2

Cl F3

XeF2

5. sp3 d2 hybridization :

Important : Since, octahedral is a symmetrical figurehence


(a) positions of a lone pair can be any where(b) but if there are two lone pairs (max.) then these

must be in the trans position.

SF6 Bond angle = 90°

Due to over-crowding and maximum valency ofS, SF6 is much less active (almost inert) than SF4

[XeO6]4–

46

8]OXe[ is perxenate ion & H4XeO6 is called perxenic

acid. But H2 ]OXe[ 4

4 is called xenic acid.

XeOF4

XeF4

6. sp3 d3 Hybridization : Steric number = 7Geometry = Pentagonal bi-pyramidal

IF7 Bond angle = 72° & 90°

XeF6 (g)

or

Distorted octahedron with a nonbonding electron paireither at the center of a face or the midpoint of an edge.

• [XeF5] –

11. SOME SPECIAL BONDING SITUATIONS(a) Electron deficient bonding :

There are many compounds in which some electrondeficient bonds are present apart from normal covalentbonds or coordinate bonds which are 2c-2e bonds( twocentre two electron bonds). These electron deficient bondshave less number of electrons than the expected such asthree centre-two electron bonds(3c-2e) or banana bondpresent in diborane B2H6, Al2(CH3)6, BeH2(s) and bridgingmetal carbonyls.• B2H6 ( Diborane)

This molecule does not have any B – B bond like C –C bond in C2H6 (ethane). Both the B atoms are in sp3

hybridization state and each boron contains two types ofboron-hydrogen bond lengths.

This is an example of 3-centre 2-e– bond which isalso known as Banana bond. The bridging hydrogen atom


are not in the plane of the molecule, one is above the planeand the other is below.

• BeH2(s) the plane.

• Al

H C3

H C3

AlCH3

CH3C

C

HH H

HH H

Al2(CH3)6

But Al2Cl6 have covalent bond ( Halogen Bridge) onlyand there is no electron deficient bonding as depicted inthe given structure.

Cl

Cl

Al

Cl

Cl

Cl

Cl

Al

(b) Back Bonding :Back bonding generally takes place when out of two

bonded atoms one of the atom has vacant orbitals(generally this atom is from second or third period) andthe other bonded atom is having some non-bonded electronpair(generally this atom is from the second period).

Back bonding increases the bond strength anddecreases the bond length. For example, in BF3 the boronatom completes its octet by accepting two 2p-electronsof fluorine into 2p empty orbital.

• Decrease in B – F bond length is due to delocalisedp–p bonding between filled p-orbital of F atom and vacantp-orbital of B atom.

The extent of back bonding is much larger if the orbitalsinvolved in the back bonding are of same size, for examplethe extent of back bonding in boron trihalides is asfollows:

BF3 > BCl3 > BBr3

There is p-p back bonding in boron trihalide. Theextent of back bonding decreases from BF3 to BI3 becauseof increasing size of p-orbitals participating in back bondingthat is from 2p(in F) to 4p(in Br).

The extent of back bonding decreases if the atomhaving vacant orbitals is also having some non-bondedelectron pairs on it. So among the atoms of third periodthe extent of back bonding follows the order

Si > P > S > Cl• The extent of p-p overlapping

1Lewis acid character

?

• Trisilyl amine (SiH3)3N is planner but Trimethylamine (CH3)3N is pyramidal.

There is p–d delocalization of lone pair of electronon nitrogen atom and empty d-orbital of silicon.One ofthe 2s-orbital electrons jumps to the last Pz orbital anddoes not participate in sp2 hybridisation.

sp2 hybrid

sp3 hybrid (no vacant orbital in C)

• Silyl isocyanate (SiH3NCO) is linear but methylisocyanate (CH3NCO) is bent

SiH

HH

:

N

:

C O

vacantd-orbitals

H

HH

:

N

:

C O

No vacantorbitals

C

Lone pair on nitrogen is delocalised between N and Sithrough p–d back bonding. So silyl isocyanate is linear.

12. BOND PARAMETERS :(i) Bond Length (Bond distance)(ii) Bond Angle(iii) Bond Energy(i) Bond Length :- The average distance between

the nucleus of two atoms is known as bond lenght,


normally it is represented in Å. eg. A B It depends mainly on electronegativities of

constituent atomsCase–I: If electronegativity difference is zero, thenBond length = rA + rB

or dA –B = rA + rB

where, rA is covalent radius of ArB is covalent radius of BIf rA = rB , then, Bond length = 2rA or 2rB

Case II: If electronegative difference is not equalto zero then-

Bond length is given by shomaker & Stevensonformula, Bond length = rA + rB – 0.09 (XA – XB)

XA is electronegativity of AXB is electronegativity of Bwhere (XA – XB) Difference in electronegativities

of A and B

13. FACTORS AFFECTING BOND LENGTH :(a) Electronegativity difference :- Bond length

1ElectronegativityDifference

?

( While Bond Energy Electronegativity difference)H–F < H–Cl < H–Br < H–I

(b) Bond order or number of bonds :- Bond length

orderbondorbondsofNumber1

Bond energy µ Number of bondse.g. C–C, C = C, C C

Bond length 1.54 Å 1.34 Å 1.20 Å increasing? ? ? ? ?

Bond energy 80 140 180–200K.Cal. increasing? ? ? ??C—O C O C O1.43 Å 1.20 Å 1.13Å

C N C N C N

1.47 Å 1.28 Å 1.15 Å

(c) Resonace :- (due to resonance bond lengthaffected)eg. 1. BenzeneC—C bond length 1.54Åbut bond length is between

C=C bond length 1.34 Åsingle & double bond is = 1.39 Å

eg. 2. Bond length of C—O in CO2 is 1.15 Å, Resonanceoccurs in CO2 as follows-

O C O O — C O – + O C — O+ –

Bond length = 1.15 Å (Between double & triple bond)

(d) Hybridisation : - Bond length 1

% s character

3 3 sp sp

Ethane 1.54 Å3 2 sp sp

1.51 Å

3 sp sp

1.47 Å2 sp sp

2 1.46 Å2 sp sp

1.42 Å sp sp

1.37 Å

% s

-cha

ract

er in

crea

ses

(ii) Bond Angle :- The angle between any twoadjacent bond is known as bond angle. It is representedin degree (°), min (') and second (")

Factors affecting the bond angle-

(a) Number of bond : Bond angle Number ofbonds (Bond order)

109° 120° 180°

C C C C C C

(b) Hybridisation :Case I : When hybridisation is same, bonded atoms

are same but central atom and lone pair are different.

Then bond angle 1

Number of lone pair

Example :- CH4 3HN•• ••

••2OH

Hybridisation sp3 sp3 sp3

Bond angle 109° > 107° > 105° no lone one lone two

pair pair lone pairCase-II

When hybridisation is same, bonded atoms aresame, lone pair is same but central atom is different.Then bond angle µ electronegativity of central atom

Example:-3HN

••3HP

••3HAs

••

Bond angle 107° 93° 91° – Eletronegativity decreasing – Bond angle will decrease

Case-IIIWhen hybridisation is same, lone pair are same,

Central atom is same but bonded atoms are different.


sp3 OF2 103 – 105°sp3 Cl2O 109 – 111°sp3 Br2O 116 – 118°

Here, bond angle

1

electronegativity of bonded atom size of side

atom

EXAMPLE 4Which compound has the smallest bond angle in eachseries ?

(a) SbCl3 SbBr3 SbI3

(b) PI3 AsI3 SbI3

SOLUTION:

(a)

97.1º 98.2º 99ºCl, the most electronegative of the halogens in this

series, pulls shared electrons the most strongly away fromSb, reducing electron density near Sb. The consequenceis that the lone pair exerts the strongest influence on shapein SbCl3.

(b)

102º 100º 99ºPhosphorus is the most electronegative of the central

atoms. Consequently, it exerts the strongest pull on sharedelectrons, concentrating these electrons near P andincreasing bonding pair-bonding pair repulsions–hence,the largest angle in PI3. Sb, the least electronegative centralatoms, has the opposite effect.

(iii) Bond Energy (BE) : Bond energy may be difinedas-

(a) Bond formation energy : Energy releasedwhen any bond is formed is known as bond formationenergy or bond energy.

(b) Bond dissociation energy : Energy required todissociate any bond is known as bond dissociationenergy.

Calculation of released energy is more difficult thanthe dissociation energy, therefore dissociation energy of

bond is calculated and is assumed as bond energy orbond formation energy.

Case–I: In diatomic molecule : Bond energy = bond dissociation energyeg : N2 > O2 > H2 > F2

Case–II: For polyatomic molecule :-

eg: |

|

H

H

HCH Bond energy = 99.5 K. Cal/mole (per

C¾H bond) Bond dissociation energy is related to the state

of hybridisation.Factors affecting the bond energy : -

(a) Electronegativity difference(b) Bond order(c) Atomic size(d) Bond polarity(e) Resonance(f) Hybridisation(g) Lone pair electron(a) Electronegativity difference :- Bond energy µ

Electronegativity differenceeg. HF > HCl > HBr > HI(b) Bond order :- Bond energy µ Bond ordereg. C—C > C = C < – C C – 79 K. Cal, 143.3 K. Cal., 199.0 K. Cal.

(c) Atomic size :- Bond energy 1

Atomic size?

eg. CC < CN < NNException :- In case of halogen group, order of bond

energy is-Cl — Cl > Br — Br > F — F > I — IBecause of higher electron density and small size of

F atoms, repulsion between of two F atom,weakens the bond energy.

Other eg. S – S > O – OC – C > Si – Si > Ge – Ge

(d) Bond Polarity :- Bond energy polarityeg. H—F > H—Cl > H—Br > H—I(e) Resonance :- Bond energy increases due to

resonance eg. In benzene, bond energy of C–C increases due

to electrons of C = C.

Electronegativity ofbonded atom isdecreasing


(f) Hybridisation :-Bond energy % s-characterin hybrid orbitals.eg. sp¾sp > sp2¾sp2 > sp3¾sp3

% s-character 50% 33.3% 25%(g) Lone pair of electrons :- Bond energy

1lone pair of electrons

?

C C > N N > O O > F F××××××

××××××

××

××

××

××

× ×× ×

Size of F and O atoms small so their bond energyshould be high (small atomic radius) but it is actuallyless due to lone pair of electrons present on F and Oatoms, which repells each other in F–F and O–O type ofbonds.

Characteristic of Covalent Compound :

(i) Physical state :- Covalent compounds are found inall the three states - Gas, Solid & Liquid.

Separate molecules – In gaseous stateAssociate molecules – In liquid & solid state(Due to strong vander waal’s force and hydrogen

bonding among the molecules.)As the molecular weight increases, physical state

changes :eg. F2 and Cl2 Br2 I2, At2

gas liquid solid Top to bottom in a group, Vander waal’s force

increases between the molecules.(ii) Covalent solid : Those solids in which atoms arelinked together by covalent bonds, forms infinite threedimensional giant structure.

e.g. Diamond, Graphite, AlN, SiC, SiO2 etc.Molecular solid : Discrete (separate) molecules are

formed by covalent bonds and then the moleculesassociated due to intermolecular force of attraction.(Vander-waal’s force)

eg. Solid I2, dry ice (Solid CO2) etc.(iii) Conductivity : Mostly covalent compounds are badconductor of electricity. But few polar covalent

compounds due to self ionisation can conduct electricity.e.g. H2O, liq. NH3 etc.

H2O + H2O H3O+ + OH–

2NH3 NH4+ + NH2

–

Free ions are formed which can conduct electricity.Exceptions : Graphite, HCl in water.

(iv) Solubility : Non polar compound are soluble in nonpolar solvents. Non polar compounds forms Vander-waal’s bond with non polar solvent molecules.(v) Isomerism : Covalent bond is rigid anddirectional, so it shows isomerism.

eg. Organic compounds.(vi) Reaction : Reaction between covalent compoundsare slow. Because it involves breaking of old bonds andformation of new bonds.

14. MOLECULAR ORBITAL THEORY (MOT):MOT put forward by Hund & Mulliken, which can

be applied to explain the properties, which the old VBT(Valence bond theory) was unable to explain eg.Paramagnetic nature of O2 molecule, as per VBT (:O: :O:), it should be diamagnetic.Definition:

The atomic orbital lose their identity during moleculeformation (overlapping) and form new orbitals termedas molecular orbitals.

Characteristic of molecular orbitals:

(i) Molecular orbital formed by overlapping of atomicorbital of same energy

(ii) Number of molecular orbital formed = number ofatomic orbital involved in overlapping

(iii) Half of the molecular orbital have lower energyare called Bonding molecular orbital.

(iv) Half are of higher energy is termed as Antibondingmolecular orbital

(v) Electronic configuration in various molecularorbital are governed by same three rules.(a) Aufbau’s rule (b) Hund’s rule (c) Pauli’sexclusion principle

Table : Comparision of Bonding molecular orbital & Antibonding molecular orbital

Bonding molecular orbital (BMO) Antibonding Molecular orbital (ABMO)

Bonding MO is the result of the linear ABMO is result of linear combination of AOcombination of AO when their wave when their wave function are substractedfunction are added b = A + B a = A – B


BA2B

2A

2BA 2)( BA

2B

2A

2BA 2)(

It does not have node. It always have a node between two nuclei ofbonded atom.Charge density increase between two Charge density decrease in between two nuclei,nuclei resulting attraction between two leads to repulsion between two atoms.atomsEnergy of BMO is less, hense stable Energy of ABMO is high, hence unstable

Notation of molecular orbitals:

As atomic orbitals are known by letters s, p, d and fdepending on their shapes. Similarly for molecularorbital.

For bonding molecular orbital- etc.For antibonding molecular orbital- *, *, * etc.are used for different shapes of electron cloud.

ENERGY LEVEL DIAGRAM OF MOLECULAR ORBITALOn the basis of Aufbau’s rule - increasing order of

energies of various molecular orbitals is- (1s) < * (1s) < (2s) < * (2s) < (2pz) < (2px)= (2py) < *(2px) = * (2py) < * (2pz)

Energy level diagram for homonuclear diatomicmolecules like, O2, F2, Ne2

For O2 molecule-

2px)

( )2pz

2p

2p

2px)

2py)

2pz)

2p

Bond order : O = 2 2

O = 2.5 2+

O = 1.5 2

– O = 1.0 2

2–

Stability order –

O 2 > O O O2 2 2> > + – 2–

Bond length –

O >O2 2 > O O2 2 > 2– – +

2s)

2s

2s

2s)

1s

1s

s)

Incr

easi

ng e

nerg

y

Having two unpairedelectrons so paramagnetic

Bond order = 2

* 2py) *

*

*

s)*

Atomic orbital of oxygen atom

Atomic orbital of oxygen atom

Molecular orbital of oxygen Molecule


ENERGY LEVEL DIAGRAM FOR B2, C2 AND N2MOLECULES –

(1s) < * (1s) < (2s) < * (2s) < (2px) = (2py) < (2pz) < * (2px) = * (2py) < * (2pz)

For N2 molecule

*(2p )x

*(2p ) z

*(2p )y 2p

2p

(2p )x

2p

*(2s)

2s

2s

2s)

*(1s)

1s

1s

s)

2p )y

Incr

e asi

ng e

nerg

y in

N m

ole c

u le

2

Atomic orbitalof nitrogen atom

Atomic orbitalof nitrogen atom

Molecular orbitalof nitrogen molecule

Cause of exceptional behavior of molecular orbitalin B2, C2 and N2:

Energy of 2s and 2p atomic orbitals lie fairly close Due to small energy difference between 2s and

2pz orbitals, the interaction between them isquite large.

This results in loss of energy by (2s) and *(2s) and thus (2s) and * (2s) becomes morestable at the cost of (2px) and * (2px) whichgets unstablised (Higher energy).

Electronic configuration of molecules and theirrelated properties :-for writing electronic configuration of diatomicmolecules following two rules to be followed-

– Count the number of electrons present in twoatoms and then fill in the appropiate energy leveldiagram according to Aufbau rule.

– The pairing in (2px) and (2py) or * (2px) and* (2py) will take place only when each molecularorbital of identical energy has one electron.

– After writing the molecular orbital diagramfollowing parameter about molecules/ion may bepredicted.

(i) Bond order :- Bond order = ½ [Nb–Na]

Nb – Number of electron in bonding molecularorbital

Na – Number of electron in antibonding molecularorbital

(ii) Bond length :- (distance between two nuclei)

orderBond1lengthBond

If existmoleculethen:NN ab

exist not does MoleculeNNNN

ab

ab

(iii) Stability of molecules - stability of molecule Bond order of molecule

(iv) Dissociation energy - Bond dissociationenergy Bond order

(v) Magnetic property -(a) When electron in Molecular orbital are paired –

then the molecule is diamagnetic.

(b) When electron in Molecular orbital are unpaired –the the molecule is paramagnetic.

BONDING IN MOLECULES :(i) Hydrogen molecule-

Having two H atoms with one electron each (1s)1

1s)

1s(Atomic orbital of hydrogen atom)

1s)

Molecular orbital of H molecule2

1s(Atomic orbital of hydrogen atom)

E

Molecular orbital (M.O.) configuration of H2 = (1s)2 * (1s)°

Bond order = ½ [Nb – Na] = ½ [2 – 0] = 1 i.e. singlebond

Having paired electron, so diamagnetic.Stability quite stable (having single bond)


(ii) H2+ ion –

1s)

1s(Atomic orbital of H-atom)

1s)

Molecular orbital of H molecule2

+

1s(Atomic orbital of H -atom)

E

Molecular orbital (M.O.) configuration ofConfiguration of H2

+ = (1s)1 *(1s)°One electron in bonding molecular orbital.So paramagneticBond order = ½ [1– 0] = ½Less stable

(iii) H2– anion -

Molecular orbital (M.O.) configuration - (1s)2

*(1s)1

ParamagneticBond order = ½ [2 – 1] = ½

1s)

1s(Atomic orbital

of H-atom)

1s)Molecular orbital of H molecule2

–

1s(Atomic orbital

of H -atom)–

E

Stability is less than [H2+] because H2

– Containantibonding molecular orbital (ABMO) electron(iv) Helium molecule (He2) :

– Molecular orbital (M.O.) configuration (1s)2 * (1s)2

– Diamagnetic– Bond order = ½ [2 – 2] = 0 (zero)– Bond order zero indicates no linkage between

He atoms. Hence He2 molecule does not exist– Stability (He2) Highly unstable molecule

1s)

1s(Atomic orbital

of He-atom) 1s)

Molecular orbital of He molecule2

1s(Atomic orbital

of He-atom)

E

(v) Nitrogen molecule (N2) : (1s)2 (*1s)2 (2s)2

(*2s)2 (2p2x = 2p2

y ) (2pz)2

2p

*2pz

*2p = *2px y

2p = 2px y

2p

2s 2s

2s

s

2pz

N (MO)2

N(AO) N(AO)

M.O. Energy level diagram for N molecule2

N2 has a triple bond according to both the Lewis andthe molecular orbital models.

The bond order of N2 is 1/2(10 – 4) = 3. It containsone sigma and two p bonds.

• Anionic nitrogen species (N2– ) : Though 15

electrons but derived from N2, hence electronicconfiguration will be according to N2

Electronic configuration : (1s)2 (*1s)2 (2s)2

(*2s)2 (2p2x = 2p2

y ) (2pz)2 , (*2px)1

The bond order of N2 is 1/2(10 – 5) = 2.5. It isparamagnetic species.

• N2+ : Bond order = 2.5, out of N2

+ and N2¯, N2¯ isless stable though both have equal bond order but N2

– hasgreater number of antibonding electrons.(vi) Oxygen molecule (O2) : O2 : (1s)2 (*1s)2 (2s)2

(*2s)2 (2pz)2 (2p2x = 2p2

y ) (*2px1 = *2p1

y)

?*2pz

? ?2p = 2px y

?2s

2s

??2px

O (MO)

O(AO)

M.O. Energy level diagram for O molecule2

?2pz

O(AO)

??2py

??

2


The bond order of O2 is ? ?1 1N N [10 6] 22 2b a? ? ? ? .

So in oxygen molecule, atoms are held by a double bond.Moreover, it may be noted that it contains two unpairedelectrons in *2px and *2py molecular orbitals, therefore,O2 molecule should be paramagnetic, a prediction thatcorresponds to experimental observation. Several ionicforms of diatomic oxygen are known, including O2

+, andO2

2–. The internuclear O – O distance can be convenientlycorrelated with the bond order predicated by the molecularorbital model, as shown in the following table.

(vii) Fluorine molecule (F2) : (1s)2 (*1s)2 (2s)2

(*2s)2 (2pz)2 (2p2x = 2p2

y ) (*2px2 = *2p2

y)The molecular orbital picture of F2 shows a

diamagnetic molecule having a single fluorine-fluorinebond, in agreement with experimental data on this veryreactive molecule.

(viii) Neon molecule (Ne2) : (1s)2 (*1s)2 (2s)2

(*2s)2 (2pz)2 (2p2x = 2p2

y ) (*2px2 = *2p2

y)(*2pz)2

All the molecular orbitals are filled, there are equalnumbers of bonding and antibonding electrons and thebond order is therefore zero. The Ne2 molecule is a transientspecies, if it exists at all.Note : HOMO : Highest Occupied Molecular Orbital.

LUMO : Lowest Unoccupied Molecular Orbital

EXAMPLE 5Though O2 molecule is paramagnetic yet it is a colourlessgas. Why ?

SOLUTION:It is because the energy gap between HOMO and LUMOlevels in O2 molecule is so large that radiations of visiblelight cannot excite a e– from HOMO to LUMO. In fact O2gas shows absorption in UV zone. So it is colourless.

EXAMPLE 6Correct order of bond energy is:

(A) N2 > N2+ >N2

– >N22–

(B) N2+ >N2

– >N22– >N2

(C) N2 >N2¯ = N2

+ >N22–

(D) N2– >N2 = N2

+ >N22–

SOLUTION:(A) Bond order is directly proportional to the bond energy.

Bond order of N2 = 3 , N2+ , N2

– = 2.5 N22– = 2

But N2– has more electrons in antibonding MO’s and

thus N2+ is more stable than N2

–. So correct order ofbond energy will be N2 > N2

+ >N2– >N2

2–

EXAMPLE 7Which of the following species have a bond order of 3 ?

(A) CO (B) CN– (C) NO+ (D) O2+

SOLUTION:(A,B,C) Species CO, CN–, NO+ are isoelectronic with 14

electrons to N2 which has bond order of 3 (i.e. 10 4

3?

= 3),

so their bond order will be equal to three.

EXAMPLE 8

Which of the following are paramagnetic ?(A) C2 (B) O2

2– (C) Li2 (D) N2+

SOLUTION:(D) Species C2, O2

2–, Li2 have all the electrons paired butN2

+ has one unpaired electron in bonding molecular orbitalso it is paramagnetic.

15. DIPOLE MOMENT :(Ionic Nature in Covalent Bond)

(i) Polarity of any polar covalent bond or molecule ismeasured in terms of dipole moment.

(ii) For measurement of extent of polarity, Paulingintroduced the concept of dipole moment ().The product of positive or negative charge (q)and the distance (d) between two poles is calleddipole moment.Here : = q × d (magnitude of charge ×distance)

(iii) Dipole moment is a vector quantity i.e. it hasboth magnitude as well as direction.

(iv) Direction of dipole moment is represented byan arrow pointing from electropositive toelectronegative element and from central atomto lone pair of electrons.

central atom lone pair of electronor

(v) Unit of dipole moment is Debye


1 Debye = 1 × 10–18 esu cm.

= 1.6 × 10–29 coulomb metre

(vi) In the diatomic molecule dipole moment ()depends upon difference of Electronegativity i.e.dipole moment () Electonegativitydifference

order of dipole moment () : H–F > H–Cl > H–Br > H–I

dipole moment () = 0 for H–H, F–F, Cl–Cl,Br–Br, O=O

(vii) For polyatomic molecules dipole moments ()depends on the vector sum of dipole momentsof all the covalent bonds.

(viii) For PCl5 and SF6 , etc. dipole moment () = 0due to their regular geometry.

(ix) Benzene, naphthalene, biphenyl have dipolemoments () = 0 due to planar structure.

(x) If the vector sum is zero, than compound isnon-polar compound or symmetrical compound(and it is not essential that individual dipolemoments () of every bond should be zero).

Example : (A) BX3, CCl4, SiCl4, CH4, CO2, CS2,PCl5, SiH4 etc.

In these examples the bond B–F, C–Cl, C–H,C–O, P–Cl etc. are polar even thoughcompounds are non-polar.

(B) • •N H

H 3

1

2

=1.47D

4

NH3• •

H

• •P

H

H

H Electronegativity of P H

4

PH3• •

• •N F

F 3

1

2

=0.24 D

4

NF3• •

F

(xi) Dipole moment of H2O is 1.85 D which isresultant dipole moment () of two O–H bonds.

H

O

H

dipole moment () of H2O is more than dipolemoment () of H2S because electronegativityof oxygen is higher than sulphur.

(xii) Angular structure of molecule have greaterdipole moment.

APPLICATION OF DIPOLE MOMENT :(i) To determine polarity and geometry of molecule:

If dipole moment () = 0 compound is non polarand symmetrical

eg. CO2, BF3, CCl4, CH4. BeF2 etc.If dipole moment () 0 compound will be polar

and unsymmetrical.eg. H2O, SO2, NH3, Cl2O, CH3Cl, CHCl3 etc.

(ii) To calculate % ionic character :

Experimental value of dipole moment (μ)% Ionic character 100

Theoritical value of dipolemoment (μ)? ?

EXAMPLE 9A diatomic molecule has a dipole moment of 1.2 D. If itsbond distance is equal to 1.0Å then the fraction of anelectronic charge on each atom is :

SOLUTION:Assuming complete charge transfer then dipole moment= (4.8 × 10–10 esu) (10–8 cm) = 4.8 D

so % ionic character = 8.42.1

× 100 % = 25%

EXAMPLE 10The gaseous potassium chloride molecule has a measureddipole moment of 10.0 D, which indicates that it is a verypolar molecule. The separation between the nuclei in thismolecule is 2.67 × 10–8 cm. Calculate the percentage ioniccharacter in KCl molecule.

SOLUTION:Dipole moment of compound would have been completelyionic

= (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D

so % ionic character = 10.012.8 × 100%

= 78.125 % ~ 78% Ans.


(iii)To distinguish cis form or trans form:-(a) Dipole moment of cis isomers is normally higher

than trans isomers.eg. :

Cl—C—H

Cl—C—H

cis-form Polar ( )

Cl—C—H

H—C—Cl

Trans-formNon Polar ( )

(b) If two groups have opposite inductive effectthan trans-isomer will have greater dipole moment-

eg. :

H

C

C

CH3H

Cl

H

C

C

CH3H

Cl

(iv) To locate position of substituents in aromaticcompounds.

1Dipole moment (μ) Bond angle

(a) If same substituents are present in thesymmetrical position dipole moment () of benzene ringcompounds will be zero.

Cl

ClAngle 180°p-dichloro benzene

ClCl Cl

ClCl

Cl

Angle 120°m-dichloro benzene

Angle 60° o-dichloro benzene

(b) As angle between subtituents decrease value ofdipole moment () increase

EXAMPLE 11Why BeF2 has zero dipole moment whereas H2O has somedipole moment ?

SOLUTION:BeF2 has linear molecule and H2O has bent molecule.

= 0

0Some important orders of dipole moments ()HF > H2O > NH3 > NF3 H2O > H2SCH3Cl > CH3F > CH3Br > CH3I BF3 < NF3 < NH3

HF > H2O > SO2 > NH3 H2O < H2O2

16. HYDROGEN BONDDefinintion :

(i) It is an electrostatic attractive force betweencovalently bonded hydrogen atom of one moleculeand an electonegative atom (F, O, N)

(ii) It is not formed in ionic compounds(iii) Hydrogen bond forms in polar covalent

compounds, (not in non-polar)(iv) It is also known as dipole-dipole attraction

H+ — F– .... H+ — F–...... H+ — F–

Main condition for Hydrogen bonding :(i) Hydrogen should be covalently bonded with high

electronegative element like F, O, N(ii) Atomic size of electronegative element should

be small.Decreasing order of atomic size is–N > O > FDecreasing order of electronegativity is –F > O > N(4.0) (3.5) (3.0)

(iii) Strength of Hydrogen bond Electronegativity

of element 1

atomic size of element

(iv) Hydrogen bonding occurs in HCN, due to(–C N) triple bond (sp hybridisation), electro-negativities of carbon and nitrogen increases.


NC—H..........NC—H.........NC—H ––

Types of Hydrogen Bonding

Inter Molecular Intra Molecular

Homo Inter Molecular Hetero Inter Molecular

(A) Intermolecular Hydrogen bondHydrogen bond formation between two or more

molecules of either the same or different compoundsknown as Inter molecular Hydrogen bonding

These are two types.(i) Homointermolecular :- Hydrogen bond

between molecules of same compounds.

eg.

+H +H

+H

+H

+H +H +H+H

O– O– O– +H +H +H

F–

F–

F–

(ii) Hetro intermolecular :- Hydrogen bond between molecules of different compounds.

eg. alcohol, water

O — H + – + – + – + O — H O — H O — H

R H R H alcohol Water alcohol Water

(B) Intra molecular Hydrogen bond :- It takes place within the molecule.(i) Hydrogen bonded with electronegative elements of a functional group, form Hydrogen bond with another

electronegative element present on nearest position on the same molecule.(ii) This type of Hydrogen bond is mostly occured in organic compounds.(iii) It result in ring formation (Chelation).

eg.

O

N

H+

O–

Oo–nitrophenol

C

O

H

H–

Salicylaldehyde

O+

O

H+

o–fluorophenol

O

HO

2, 6-dihydroxyl benzoate

C

H O O

EFFECT OF HYDROGEN BOND ON PHYSICALPROPERTIES(i) Solubility :

(A) Inter molecular Hydrogen bonding(a) Few organic compounds (Non-polar) are

soluble in water (Polar solvent) due to Hydrogen bonding.eg. alcohol, acetic acid etc. are soluble in water.

–O — H O — H + – +

R H

Other examples–Glucose, Fructose etc, dissolve inwater.

(b) Ketone, ether, alkane etc. are insoluble (noHydrogen bond)

(c) Solubility order– CH3OCH3 < CH3OHPrimary amine > secondary amine > tertiary amine

(B) Intra molecular Hydrogen bonding:(a) It decreases solubility as it form chelate by

Hydrogen bonding, so Hydrogen is not free for othermolecule.


(b) It can not form H–bond with water molecule socan not dissolves.

C O–

O

H+

H

(Salicylaldehyde)(C) Inter molecular Hydrogen bond

O — H O C — H O — H +– +

p–hydroxy benzaldehydeO HH

C C

O — H O + –

It can form Hydrogen bond with water molecule soit can dissolved(ii) Viscosity:

Hydrogen bond associates molecules together, soviscosity increases

CH — OH2

CH — OH2CH OH3

CH — OH2

CH CH3 3 — O — CH3 — OHH O2

water

<

>

<

>

alcohol ether

CH — OHCH3 — OH

(iii)Melting point and boilling point(a) Due to intermolecular Hydrogen bond Melting

Point & Boiling Point of compounds increases.H2O > CH3OH > CH3 — O—CH3

(b) Trihydric alcohol > dihydric alcohol >monohydic alcohol

Monocarboxylic acid form stronger Hydrogen bondthan alcohol of comparable molecular weight. ThereforeBoiling Point of carboxylic acid is higher than alcohol.

(c) Decreasing order of Melting Point & BoilingPoint isomer amines-

1°–amine > 2°–amine > 3°–amine

R — NH2 > R — NH — R > R — N

R|

— R (nohydrogen with nitrogen atom)

(d) Boiling points of VA, VIA, VIIA hydridesdecreases on decreasing molecular weights.

VA VIA VIIANH3 H2O HF Boiling point HF > HI > HBr > HClPH3 H2S HCl H2O > TeH2 > SeH2 > H2SAsH3 SeH2 HBr SbH3 > NH3 > AsH3 > PH3

SbH3 TeH2 HI

(e) But sudden increase in boiling point of NH3, H2Oand HF is due to Hydrogen bonding

H2O > HF > NH3

Intramolecular Hydrogen bonding gives rise to ringformation, so the force of attraction among thesemolecules are vander waal’s force. So, Melting Pointand Boiling Point are low.(iv) Molecular weight :

Molecular weight of CH3COOH is double of itsmolecular formula, due to dimer formation occur byHydrogen bonding

R — C C — R O H —O+–

O — H O+ – (v) Physical state:

H2O is liquid while H2S is gas.

Water and Ice:- Both have Hydrogen bonding eventhen density of ice is less than water.

Volume of ice is more because of open cage likecrystal structure, from by association of water moleculeswith the help of Hydrogen bond.

H2O becomes solid (Ice) due to four hydrogen bondamong water molecule are formed in tetrahedral manner.

H H H H1 3

42

H H H H

O O O O

(vi) Base strength:

CH3NH2, (CH3)2 NH, (CH3)3 N, form Hydrogenbond with water. So, less hydrolysis i.e. it gives OH–

ions. While (CH3)4 N+ OH– (ammonium compound) will


give OH– ion in large amount due to no Hydrogenbonding.

H

— N

CH H — O

H H

3

H

— N

CH — H + OH

H

3–+

CH3

CH N CH

CH

3 3

3

(Ammonium compound.)NO Hydrogen atom bonded directly with Nitrogen atomso no hydrogen bonding occurs.

+

EFFECT OF INTRAMOLECULAR H–BONDING(i) Strength of acid

(a) The formation of intramolecular H–bonding inthe conjugate base of an acid gives extra stability toconjugate base and hence acid strength increases eg.Salicylic acid is stronger than benzoic acid and 2, 6 –dihydroxy benzoic acid > salicylic acid.

C C O O–

O

O O + H+

H

Conjugate base

2, 6-dihydroxyl benzoate ion

O

H H

H

+ H

C

O O

–1/2 –1/2O

O+

H

(b) C2H5SH is more acidic than C2H5OH. InC2H5OH, Hydrogen bond forms, so H+ is not free

(c) HF is weaker acid than HI, due to Hydrogenbond in H – F, H+ is not free(ii) Stability of chloral hydrate:-

If two or more OH group on the same atom arepresent it will be unstable, but chloral hydrate is stable(due to Hydrogen bonding).

Cl

Cl

Cl C C H Chloral hydrateO

OH

H

(iii) Maleic acid (cis) is stronger acid than fumaric acid(trans).

H H

H

O O

C C

C

H H

HOOC

C C

C C

C

C C

OH O–

OH O + H+

Stable conjugate base of maleic acidO

H

COOH

Fumaric acid (No-intramolecular Hydrogen bonding)

O(Maleic acid)

H

Note: The relative strength of various bonds is asfollows

Ionic bond > Covalent bond > Metallic bond >Hydrogen bond > Vander waal’s bond17. Vander Waal’s Forces

(a) This type of attractive forces occurs in case ofnon-polar molecules such as H2, O2, Cl2, CH4, CO2 etc.

(b) The existence of weak attractive forces amongthe non-polar molecule was first proposed by dutchscientist J.D. Vander Waal

(c) Vander waal’s force molecular weight Atomic weight Boiling point

Types of Vander Waal’s force : -(i) Dipole-Dipole attraction - It is again in betweentwo polar molecules such as HF and HCl

+

HF HCl(ii) Dipole - Induced dipole attraction : In this casea neutral molecule is induced as a dipole by anotherdipole as shown in fig.e.g. HCl Cl2

Before induction +

After induction + +

(iii) Induced dipole - induced dipole attraction orLondon dispersion forcebetween two non polar molecules as in Cl2, He etc.

+ +

Cl2 Cl2


1. Explain the significance of formal charge.2. Give one example having ionic and covalent bonding

both.3. If a carbon atom is sp2 hybridised how many

unhybrid p-orbitals will be there.4. Write formal charge of all the atoms in

(a) CO32– ion (b) NO2

– ion5. What change in energy is observed when a molecule

is formed from atom ?6. AlF3 is more ionic than AlCl3. Explain.

7. Enlist the drawbacks of octet rule.8. Define bond angle and bond length.

9. Considering X axis an internuclear axis, which ofthe following will not form a and why?

(a) 1s and 1s (b) 1s and 2px

(c) 2py and 2py (d) 1s and 2s

10. Differentiate between sigma and pi bonds.11. Give total number of and bonds in

(a) C6H6 (b) C6H5OH

(c) C7H8

12. What do you understand by a chemical bond. Namevarious types of bonds with one example.

13. Write various assumptions of VSEPR theory.

14. How does electronegativity of atoms affect thepolarity of a bond. Explain your answer with suitableexample.

15. Define hybridisation. Describe different types ofhybridisation taking one example of each.

16. Sodium metal vapourise on heating and vapourshave diatomic molecule of sodium (Na2). What typeof bonding is present in this molecule ?

17. H2S is a gas though its molecular weight is morethan H2O which is liquid. Explain.

18. Discuss the application of Fajan’s rule.

19. Boiling point of H2O is more than HF despite thefact that HF shows stronger hydrogen bondingexplain.

20. XeF2 is linear despite the fact it involves sp3dhybridisation. Explain.

21. Explain why water shows higher density at 277 Kand not at 273 K.

22. Do you expect P–Cl bond energy to be same inPCl3 and PCl5. Explain your answer.

23. What do you under stand by bond pair and lonepair? Explain with suitable example.

24. Comment on bond energies of N2– and N2

+.

25. The dipole moment of LiH is 1.964 × 10–29 Cm atinter atomic distance 1.596 Å. Calculate thepercentage ionic character.

26. The dipole moment of NH3 is more than that ofNF3 explain.

27. o-Hydroxybenzaldehyde is a liquid at roomtemperature while p-Hydroxybenzaldehyde is a highmelting solid. Explain.

28. I3– is known but F3

– is not why?

29. Define bond order in MOT.

30. Define resonance.

31. Give lewis dot structure for following.

XeF2, XeF4, XeF6, XeO2F2 XeO4

32. “Non-polar molecules necessarily have non polarbond”. Comment on this statement.

33. Why He2 molecule is not formed ?

34. Deduce the magnetic character of anion of KO2.

35. In which of the following molecule, is the Londonforce likely to be most important in determiningm.p. and b.p. and why?

ICl, Br2, HCl, H2S and CO.

36. CO2 is nonpolar while SO2 and H2O are polar.Explain.

37. Discribe the hybridisation in PCl5 molecule. Whyaxial bonds are longer than the equitorial bonds inPCl5?

38. Give SI unit for dipole moment.

39. Electronegativity of chlorine is same as that ofnitrogen yet it hardly forms hydrogen bond. Why?

40. Sugar is soluble in water though it is a covalentcompound. why?

41. Name the forces responsible for holding non polarmolecules together.

42. Define antibonding molecular orbitals.

43. Out of three isomeric forms of dichlorobenzenewhich isomer would have zero dipole moment.

EXERCISE–0 (RECALL YOUR UNDERSTANDING)


44. Among NH3, H2S and H2O, the compound withlowest B.P. is

45. Indicate the inter particle force operating in thefollowing cases.(a) H2O () (b) H2O(s)(c) Dry ice (d) HF()

(e) Diamond (f) Aluminium(g) KCl (h) H2

(i) Rock salt (j) CF4

46. Compare the relative stability of following speciesand indicate their magnetic behaviour O2, O2

+, O2–

and O22–

EXERCISE–1 (CHECK YOUR UNDERSTANDING)

Lewis Octet Rule1. If the atomic number of element X is 7 the lewis

diagram for the element is :

(a) X (b) X

(c) X (d) X

2. Which of the following Lewis diagram is incorrect?

(a) (b)

(c) (d)

3. What are the formal charges on central sulphur andeach terminal oxygen atoms in SO2?

(a) 0, 0, 0 (b) + 2, 0, – 1

(c) 0, – 1, + 1 (d) + 2, + 2, + 2

Ionic bond

4. An ionic bond A+B– is most likely to be formedwhen :(a) The ionization energy of A is high and the

electron affinity of B is low(b) The ionization energy of A is low and the electron

affinity of B is high(c) The ionization energy of A and the electron

affinity of B is high(d) The ionization energy of A and the electron

affinity of B is low5. Which forms a crystal of NaCl ?

(a) NaCl molecules (b) Na+ and Cl– ions(c) Na and Cl atoms (d) None of these

6. Which of the following pair of elements form acompound with maximum ionic character ?(a) Na and F (b) Cs and F(c) Na and Br (d) Cs and I

7. Two element have electronegativity of 1.2 and 3.0.Bond formed between them would be :(a) ionic (b) polar covalent(c) co-ordinate (d) metallic

8. Which one of the following pairs of elements ismost likely to form an ionic compound?(a) B and Cl2 (b) K and O2(c) O2 and Cl2 (d) Al and I2

9. Among Na+, Mg2+ and Al3+, the correct order ofease of formation of ionic compounds is :(a) Al3+ > Mg2+ > Na+ (b) Na+ > Mg2+ > Al3+

(c) Mg2+ > Al3+ > Na+ (d) Al3+ > Na+ > Mg2+

10. Which of the following shows the highest latticeenergy ?(a) RbF (b) CsF(c) NaF (d) KF

11. Which of the following have low lattice energy ?(a) Cs – F (b) Cs – Cl(c) Cs – Br (d) Cs – I

12. When two atoms combine to form a molecule :(a) energy is released(b) energy is obserbed(c) energy is neither released nor absorbed(d) energy may either released or absorbed

13. Which condition favours the bond formation ?(a) Maximum attraction and maximum potential

energy(b) Minimum attraction and minimum potential

energy(c) Minimum potential energy and maximum

attraction(d) None of the above


14. An electrovalent bond or ionic bond is formedbetween :(a) two electronegative atoms(b) two metals(c) electropositive and electronegative atoms(d) two electropositive atoms

15. Most favourable conditions for electrovalentbonding are :(a) low ionisation potential of one atom and high

electron affinity of the other atom(b) high electron affinity and high ionisation

potential of both the atoms(c) low electron affinity and low ionisation potential

of both the atoms(d) high ionisation potential of one atom and low

electron affinity of the outer atom.16. Ionic compounds in general possess both :

(a) high melting points and non-directional bonds(b) high melting points and low boiling points(c) directional bonds and low boiling points(d) high solubilities in polar and non-polar solvents

Fajan’s rule17. According to Fajan’s rule covalent character is

favoured by :(a) large cation and small anion(b) small cation and large anion(c) large cation and large anion(d) small cation and small anion

18. Which one of the following conbination of ion willhave highest polarisation ?(a) Pb2+, Br– (b) Pb4+, Br–

(c) Fe2+, Br– (d) Fe3+, Br–

19. Correct order of covalent character of alkaline earthmetal chloride in(a) BeCl2 < MgCl2 < CaCl2 < SrCl2(b) BeCl2 < CaCl2 < SrCl2 < MgCl2(c) BeCl2 > MgCl2 > CaCl2 > SrCl2(d) SrCl2 > BeCl2 > CaCl2 > MgCl2

20. Which of the following is in order of increasingcovalent character ?

(a) 4 2 3CCl BeCl BCl LiCl

(b) 4 2 3LiCl CCl BeCl BCl

(c) 2 3 4LiCl BeCl BCl CCl

(d) 2 4 3LiCl BeCl CCl BCl

21. Least melting point is shown by the compound :

(a) PbCl2 (b) SnCl4(c) NaCl (d) AlCl3

22. Which is most ionic according to Fajan’s rule?:

(a) AlF3 (b) Al2O3

(c) AlN (d) Al4C3

23. Which compound among the following has leastionic character?

(a) AlCl3 (b) AlI3

(c) MgI2 (d) CsI

24. In which of the following compound the cation haspseudo inert gas configuration?

(a) NaCl (b) AlCl3(c) CuCl (d) CaCl2

25. AgCl is colourless whereas AgI is yellow, becauseof :

(a) Ag+ have 18 electron shell to screen the nuclearcharge.

(b) Ag+ shows pseudo inert gas configuration.

(c) distortion of I– is more pronounced than Cl–

ion.

(d) existence of d – d transition.

Covalent Bond, Coordinate Bond

26. The maximum covalency of representative elementsis equal to (excluding 1st and 2nd period) :

(a) the number of unpaired p-electrons

(b) the number of paired d-electrons

(c) the number of unpaired s and p-electrons

(d) the actual number of s and p-electrons in theoutermost shell.

27. Which of the following contains both electrovalentand covalent bonds ?

(a) MgCl2 (b) H2O

(c) NH4Cl (d) none

28. The types of bond present in N2O5 are :

(a) only covalent (b) only ionic

(c) ionic and covalent (d) covalent & coordinate

29. Example of super octet molecule is :

(a) SF6 (b) PCl5(c) IF7 (d) All of these


30. The number of electrons involved in the bondformation in N2 molecule is :

(a) 2 (b) 4(c) 10 (d) 6

31. The octet rule is not obeyed in :(a) CO2 (b) BCl3(c) PCl5 (d) (b) and (c) both

32. For the formation of covalent bond the differencein the value of electronegativity should be :

(a) 1.7 (b) More than 1.7(c) 1.7 or more (d) equal to or less than 1.7

33. NH3 and BF3 combine readily because of theformation of :

(a) a covalent bond (b) a hydrogen bond(c) a coordinate bond (d) an ionic bond

34. The covalency of nitrogen in HNO3 is :(a) 0 (b) 3

(c) 4 (d) 535. Which one of the following molecules has a co-

ordinate as well as covalent bond ?(a) NH4Cl (b) AlCl3(c) NaCl (d) Cl2

36. Which of the following species are hypervalent ?

1. ClO4–, 2. BF3, 3. SO4

2– , 4. CO32–

(a) 1, 2, 3 (b) 1, 3

(c) 3, 4 (d) 1, 2

V.B.T., VSEPR Theory and Hybridisation

37. VBT is given by :(a) Hitler & london (b) Pauling & Slater(c) Hund & Muliken (d) Huckel & Hund

38. Number and type of bonds between two carbonatoms in CaC2 are :(a) one sigma () and one pi () bond(b) one and two bonds(c) one and one and a half bond(d) one bond

39. The total number of and bonds in C2(CN)4 are:(a) 9 and 9 (b) 9 and 18(c) 18 and 9 (d) 18 and 18

40. Acetylene consists of :(a) both sigma and pi bonds

(b) sigma bond only(c) pi bond only(d) none of these

41. Number of bonds in SO2 are :(a) two and two (b) two and one(c) two , two and one lone pair(d) none of these

42. The correct order towards bond angle is :(a) Bond angle does not depend on hybridisation.(b) sp < sp2 < sp3

(c) sp2 < sp < sp3

(d) sp3 < sp2 < sp43. Which of the following has been arranged in

increasing order of % p-character?(a) sp < sp2 < sp3 (b) sp3 < sp2 < sp(c) sp2 < sp3 < sp (d) sp2 < sp < sp3

44. Which is not true about CH4 molecule ?(a) Tetrahedral hybridisation(b) 109.5º bond angle(c) Four sigma bonds(d) One Ione pair of electrons on carbon

45. Choose the molecules in which hybridisation occursin the ground state ?(a) BCl3 (b) NH3

(c) PCl3 (d) BeF2

46. In C—C bond C2H6 undergoes heterolytic fission,the hybridisation of two resulting carbon atoms is :(a) sp2 both (b) sp3 both(c) sp2, sp3 (d) sp, sp2

47. The hybridization in PF3 is :(a) sp3 (b) sp2

(c) dsp3 (d) d2sp3

48. Which of the following compounds have bondangle as nearly 90º ?(a) CH4 (b) CO2

(c) H2O (d) SF6

49. sp2 – hybridisation is shown by :(a) BeCl2 (b) BF3(c) NH3 (d) XeF2

50. The hybridisation of carbon in diamond, graphiteand acetylene is –


(a) sp3, sp2, sp (b) sp3, sp, sp2

(c) sp2, sp3, sp (d) sp, sp3, sp2

51. The hybridization of the central atom in 2ICl is -

(a) dsp² (b) sp

(c) sp² (d) sp³

52. Each carbon in carbon suboxide is :

(a) sp² - hybridized

(b) sp³-hybridized

(c) sp-hybridized

(d) sp²-hybridized but linked with one co-ordinatebond

53. In which of the following pairs hybridisation of thecentral atom is different ?

(a) ClF3 , ClF3O

(b) ClF3O, ClF3O2

(c) [ClF2O]+, [ClF4O]–

(d) [ClF4O]–, [XeOF4]

54. Among the following pairs in which the two speciesare not isostructural is :

(a) SiF4 and SF4 (b) O3– and XeO3

(c) BH4– and NH4

+ (d) PF6– and SF6

55. Which among the following molecules have sp3dhybridisation with one lone pair of electrons on thecentral atom ?

(i) SF4 (ii) [PCl4]+

(iii) XeO2F2 (iv) ClOF3

(a) (i), (ii) and (iii) only

(b) (i), (iii) and (iv) only

(c) (i) and (iii) only

(d) (iii) and (iv) only

56. Match list l with List II and select the correctanswer using the codes given below the lists.

List I List II(Compound) (Shape)(a) CS2 1. Bent(b) SO2 2. Linear(c) BF3 3. Trigonal planer(d) NH3 4. Tetrahedral

5. Trigonal pyramidal

Code :(1) (2) (3) (4)

(a) 2 1 3 5(b) 1 2 3 5(c) 2 1 5 4(d) 1 2 5 4

57. Which is the right structure of XeF4 ?

(a)

:

Xe

FF

FF

(b)

:

XeF

FF

F :

(c):

XeF

F

F

F:

(d) Xe

FF

F

F:

58. In which of the following molecules number of lonepairs and bond pairs on central atom are not equal?(a) H2O (b) I3

–

(c) O2F2 (d) SCl259. Which has the smallest bond angle (X – S – X) in

the given molecules?(a) OSF2 (b) OSCl2(c) OSBr2 (d) OS2 .

60. Consider the following iodides :P3 As3 SbI3

102° 100.2° 99°


The bond angle is maximum in Pl3, which is :(a) due to small size of phosphorus(b) due to more bp–bp repulsion in P3(c) due to less electronegativity of P(d) none of these

61. Which one of the following species is not linear ?(a) CO2 (b) ClO2

(c) 3– (d) NO2+

62. The ion which is not tetrahedral in shape is :(a) BF4

– (b) NH4+

(c) XeO4 (d) ICl4–

63. Arrange the following in the increasing order ofdeviation from normal tetrahedral angle :(a) P4 < PH3 < H2O (b) PH3 < H2O < P4

(c) P4 <H2O < PH3 (d) H2O < PH3 < P4

64. The molecule/ion which has trigonal pyramidalshape is :(a) PCl3 (b) SO3

(c) CO32– (d) NO3

–

65. Which of the following pairs of compound has linearstructure ?(a) Cl2O, H2O (b) SO2, NO2

(c) OF2, H2O2 (d) BeCl2, CO2

66. CO2 is isostructure with : (I) HgCl2 (II) SnCl2(III) NO2 (IV) C2H2

The correct answer is :(a) I, IV (b) I, II(c) II, III (d) I, II, III

67. The pair having similar geometry is :(a) BF3, NH3 (b) BF3, AlF3

(c) BeF2, H2O (d) BCl3, PCl368. OF2 is :

(a) Linear molecule and sp hybridised(b) Tetrahedral molecule and sp3 hybridised(c) Bent molecule and sp3 hybridised(d) None of these

69. Structure of ICl–4 is :(a) trigonal(b) distorted trigonal bipyramid(c) octahedral(d) square planar

70. The bond angle H – N – H in NH3, 4NH and 2NH

decreases in the order as :

(a) 2NH > NH3 > 4NH

(b) 4NH > NH3 > 2NH

(c) NH3 > 2NH > 4NH

(d) NH3 > 4NH > 2NH

71. The angle between two covalent bond is maximumin :(a) H2O (b) CO2

(c) NH3 (d) CH4

72. Which is having highest bond angle :(a) PCl3 (b) PBr3

(c) PF3 (d) PI3

73. Which is linear PH4+ , H3S+ or NO2

+ ?(a) Phosphonium ion (PH4

+)(b) Sulphonium (H3S+)(c) Nitronium ion (NO2

+)(d) None of these

Polarity of Bond / Dipole Moment74. Which hydrogen is most polar ?

(a) LiH (b) CsH(c) HF (d) HI

75. The most polar bond is :(a) C – H (b) N – H(c) S – H (d) O – H

76. Which has maximum dipole moment ?

(a) (b)

(c) (d)

77. Of the following molecules, the one, which haspermanent dipole moment, is :(a) SiF4 (b) BF3

(c) PF3 (d) PF5

78. Which of the following has the least dipole moment?(a) NF3 (b) SO3

(c) XeO3 (d) NH3

79. Among the following compounds the one that ispolar and has central atom with sp3 hybridisationis:


(a) H2CO3 (b) SiF4

(c) BF3 (d) HClO2

80. The dipole moment of the given molecules are suchthat :

(a) BF3 > NF3 > NH3 (b) NF3 > BF3 > NH3

(c) NH3 > NF3 > BF3 (d) NH3 > BF3 > NF3

81. The correct order of dipole moment is :

(a) CH4 < NF3 < NH3 < H2O

(b) NF3 < CH4 < NH3 < H2O

(c) NH3 < NF3 < CH4 < H2O

(d) H2O < NH3 < NF3 < CH4

82. Carbon tetrachloride has no dipole moment becauseof :

(a) its planar structure

(b) its regular tetrahedral structure

(c) similar sizes of carbon and chlorine

(d) similar electron affinities of carbon and chlorine

83. The molecules which has zero dipole moment is :

(a) ClO2 (b) PCl3(c) XeF4 (d) CHCl3

84. The dipole moment of HCl is 1.03 D. If H–Cl bonddistance is 1.26 Å, what is the percentage of ioniccharacter in the H–Cl bond -

(a) 60% (b) 39%

(c) 29% (d) 17%

85. What should be the percentage ionic character inCsF when electronegativity difference is 3.3

(a) 90.9% (b) 0.09%

(c) 93.3% (d) 95.7%

86. What is the increasing order of ionic character inH2Se, H2S, H2O

(a) H2Se < H2S < H2O (b) H2Se > H2S > H2O

(c) H2Se < H2S > H2O (d) None of these

87. Which of the following molecules should not have = 0 :–

(a) H2 (b) CO2

(c) Cl2 (d) SO2

88. Which of the following compounds should havehigher dipole moment than the remaining three :–

(a) HF (b) H2O

(c) NH3 (d) NF3

Resonance

89. Resonating structures have different :(a) atomic arrangements(b) electronic arrangements(c) function groups(d) alkyl groups

90. Which among the following resonance structuresof N3

– satisfies the octet rule but is ruled out as aresonance structure?

(I) (II)

(III) (IV)

(a) IV only (b) I and IV only

(c) I only (d) II and III only91. Which one in the following is not the resonating

structure of CO2 :(a) O C = O (b) O = C = O

(c) –O – C O+ (d) +O C – O–

Special Bonding

92. Which of the following can not be completelyhydrolysed in cold water at room temperature ?(a) BCl3 (b) PCl3(c) BBr3 (d) CCl4

93. Which is the true statement about (SiH3)3N ?(a) It is trigonal planar.(b) It is trigonal pyramidal.(c) It is stronger lewis base than that of (CH3)3N.(d) It has a total of 9 sigma bonds.

Molecular Orbital Theory (MOT)

94. When two atomic orbitals combine they form :(a) two moleculear orbital(b) one molecular orbital(c) three molecular orbital

(d) four molecular orbital95. During the formation of a molecular orbital from

atomic orbitals of the same atom, probability ofelectron density is :(a) non zero in the nodal plane(b) maximum in the nodal plane(c) zero in the nodal plane(d) zero on the surface of the lobe


96. Which one of the following can not exist on thebasis of molecular orbital theory ?(a) H2

+ (b) He2+

(c) C2 (d) He2

97. Which of the following has fractional bond order ?(a) B2 (b) O2

2–

(c) F2 (d) H2–

98. In which of the following set, the values of bondorders will be 2.5 ?(a) O2

+, NO, NO2+, CN(b) CN, NO2+, CN–, F2

(c) O2+ , NO2+, O2

2+, CN–

(d) O22–, O2

–, O2+, O2

99. Among the following species, which has theminimum bond length ?(a) B2 (b) C2

(c) F2 (d) O2–

100. Correct order of bond energy is:(a) N2 > N2

+ >N2– >N2

2–

(b) N2+ >N2

– >N22– >N2

(c) N2 >N2¯ = N2

+ >N22–

(d) N2– >N2 N2

+ >N22–

101. Number of antibonding electrons in N2 is :(a) 4 (b) 10(c) 12 (d) 14

102. Pick out the incorrect statement.(a) N2 has greater dissociation energy than N2

+

(b) O2 has lower dissociation energy than O2+

(c) Bond length in N2+ is less than N2

(d) Bond length in NO+ is less than in NO.103. Which of the following pairs have identical values

of bond order ?(a) N2

+ and O2+ (b) F2 and Ne2

(c) O2 and B2 (d) C2 and N2

104. A simplified application of MO theory to thehypothetical ‘molecule’ OF would give its bondorder as :(a) 2 (b) 1.5(c) 1.0 (d) 0.5

105. Which of the following species donot have a bondorder of 3 ?(a) CO (b) CN–

(c) NO+ (d) O2+

106. Which of the following are paramagnetic ?(a) C2 (b) O2

2–

(c) Li2 (d) N2+

107. Which of the following species is paramagnetic ?

(a) NO– (b) O22–

(c) CN– (d) CO

108. The following molecules / species have been arrangedin the order of their increasing bond orders Identifythe correct order.

(I) O2 ; (II) O2– ;

(III) O22- ; (IV) O2

+

(a) I I I < I I < I < IV (b) IV < I I I < I I < I

(c) I I I < I I < IV < I (d) I I < I I I < I < IV

109. Which the following molecules / species haveidentical bond order and same magnetic properties?

(I) O2+ ; (II) NO ; (III) N2

+

(a) (I) , (II) only (b) (I) and I I I only

(c) (I) , (I I) and (I I I) (d) (I I) and (I I I) only110. Negative bond order means :

(a) molecule is unstable(b) molecule is stable

(c) molecule is neutral(d) none of them

Hydrogen bonding111. Pure phosphoric acid is very viscous because :

(a) it is a strong acid(b) it is tribasic acid

(c) it is hygroscopic(d) it has PO4

3– groups which are bonded by manyhydrogen bonds

112. Which of the following is least volatile ?(a) HF (b) HCl

(c) HBr (d) HI113. Which one of the following does not have

intermolecular H-bonding ?(a) H2O (b) o-nitro phenol

(c) HF (d) CH3COOH114. Which of the following exhibits H-bonding ?

(a) CH4 (b) H2Se(c) N2H4 (d) H2S


115. H – bonding is not present in :(a) NH3 (b) H2O(c) H2S (d) HF

116. Hydrogen bonding would not affect the boiling pointof :(a) HI (b) NH3(c) CH3OH (d) H2O

117. Which of the following compound has maximumnumber of H-bonds per mole ?(a) HF (b) PH3(c) H2O (d) OF2

118. Water (H2O) is liquid while hydrogen sulphide (H2S)is a gas because :(a) water has higher molecular weight(b) hydrogen sulphide is week acid(c) water molecular associate through hydrogen

bonding(d) sulphur has high electronegativity than oxygen

119. Which of the following compounds would havesignificant intermolecular hydrogen bonding ?HF, CH3OH, N2O4, CH4(a) HF, N2O4 (b) HF, CH4, CH3OH(c) HF, CH3OH (d) CH3OH, CH4

Intermolecular forces120. Which of the following has the highest boiling point?

(a) H2 (b) Ne(c) Xe (d) CH4

121. At ordinary temperature and pressure, amonghalogens chlorine is a gas, bromine is a Liquid andiodine is a solid. This is because :

(a) The specific heat is in the order Cl2 > Br2 > I2

(b) Intermolecular forces among molecule ofchlorine are the weakest and those in iodine thestrongest

(c) The order of density is I2 > Br2 > Cl2(d) The order of stability is I2 > Br2 > Cl2

122. Which of the following bonds/forces is weakest ?

(a) Covalent bond (b) Ionic bond

(c) Metallic bond (d) London force

123. In which molecule is the London dispersion forcelikely to be most important in determining boilingpoint?

(a) ICI (b) Br2

(c) H2S (d) CO

124. Iron is harder than sodium because :

(a) iron atoms are smaller.

(b) iron atoms are more closely packed.

(c) metallic bonds are stronger in sodium.

(d) metallic bonds are stronger in iron.

125. Which of the following is false ?

(a) Van der Waals forces are responsible for theformation of molecular crystals.

(b) Branching lowers the boiling points of isomericorganic compounds due to reduction in the vander Waals force of attraction.

(c) In graphite, van der Waals forces act betweenthe carbon layers.

(d) Boiling point of NH3 is greater than SbH3.

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)

1. Which of the following molecule is havingcomplete octet -

(a) BeCl2 (dimer) (b) BeH2 (dimer)

(c) BeH2 (s) (d) BeCl2(s)

2. In which of the following compounds octet iscomplete and incomplete for all atoms -

Al2Cl6 Al2(CH3)6 AlF3 BeCl2 BeH2

Dimer Dimer

(a) IC IC IC C C

(b) C IC IC C IC

(c) C IC C IC IC(d) IC C IC IC IC(Note : C for complete octet and IC for incompleteoctet.)

3. In the Born-Haber cycle for the formation of solidcommon salt (NaCl), the largest contributioncomes from -(a) the low ionization potential of Na(b) the high electron affinity of Cl(c) the low Hvap of Na(s)(d) the lattice energy


4. Compound with the highest melting point is -

(a) BaCl2 (b) CaCl2(c) BeCl2 (d) MgCl2

5. Which type of bond is not present in HNO2molecule -

(a) Covalent

(b) Co-ordinate

(c) Ionic

(d) Ionic as well as Co-ordinate

6. For two ionic solids, CaO and KI. Identify thewrong statement among the following -

(a) Lattice energy of CaO is much larger than thatof KI

(b) KI is soluble in benzene

(c) CaO has higher melting point

(d) KI has lower melting point

7. Which of the following pairs will form the moststable ionic bond ?

(a) Na and Cl (b) Mg and F

(c) Li and F (d) Na and F

8. The boiling point of ICl is nearly 40°C higher thanthat of Br2 although the two subtances have thesame relative molecular mass. This is bacause:

(a) ICl is ionic compound

(b) I-Cl bond is stronger than Br-Br bond

(c) ICl is polar covalent molecular while Br2 is nonpolar

(d) Ionization energy IP of Iodine is less than thatof Br

9. The M.P. of SnCl4 is less than of SnCl2, the suitablereason for the observed fact is:

(a) There is more charge on Sn+4

(b) The size of Sn+4 is small

(c) Ionic potential () of Sn+4 is high

(d) The shape of SnCl4 is tetrahedral

10. The number of & bond in the compoundrespectively are -

NC M(CO)3

EtC = C

NC

(a) 19, 11 (b) 19, 5(c) 13, 11 (d) 7, 3

11. The pair which have maximum value of ,would be -(a) Cs2O, NO2 (b) CO2, ZnO(c) BeO, Al2O3 (d) Cl2O, NO2

12. Ionic potential () of electropositive element willbe highest in which of the following compound -(a) CsCl (b) MgCl2

(c) AlF3 (d) SF6

13. In [Fe(CO)5], hybridisation state and number ofco-ordinate bonds are -(a) sp3d, 5 (b) dsp2,10(c) d2sp2,5 (d) dsp3, 10

14. Which is not correct:(a) Bond angle H–S–H < H–OH(b) Bond angle F–O–F < Cl–O–Cl(c) Bond angle H–P–H < H–N–H(d) Bond angle Cl–Sn–Cl > Cl–Hg–Cl

15. The AsF5 molecule is trigonal bipyramidal. Thehybrid orbitals used by the As atoms for bonding -

(a) yxzy–x p,p,s,d,d 222 (b) zyxxy p,p,p,s,d

(c) s,px, py,pz, 2zd (d) yx,y–x p,p,sd 22

16. Incorrect code regarding shape is -

(a) Linear : –3N , (CN)2, –

2ICl

(b) Pyramidal : –3CH , NH3, XeO3

(c) Trigonal planar : 333 CH,CH,CH

(d) Tetrahedral : SiH4, Ni(CO)4, [CuBr4]–2

17. Which of the following set is not correct -(a) SO3, O3,

4NH all have coordinate bonds

(b) H2O, NO2, –2ClO all are ‘V’ shape molecules

(c) 2

–2

–3 NO,Cl,II ; all are linear molecules

(d) SF4, SiF4, XeF4 are tetrahedral in shape18. Linear structure is assumed by -

(I) NCO– (II) CS2

(III)

2NO (IV) Solid BeH2

(a) all four (b) (II), (III) and (IV)(c) (I), (II) and (III) (d) (II) and (III)

19. The shapes of XeF4, XeF5– and SnCl2 are -

(a) octahedral, trigonal bipyramidal and bent


(b) Sq. pyramidal, pentagonal planar and linear(c) Sq. Planar, pentagonal planar and angular(d) See-saw, T-shaped and linear

20. Which of the following process are associated withchange of hybridization of the underlinedcompound?

(a) 3)OH(Al ppt. dissolved in NaOH

(b) 62HB is dissolved in THF

(c) 4SiF vapour is passed through liq. HF

(d) Solidification 5PCl vapour

Correct answer is -(a) 1, 2, 4 (b) 1, 3, 4(c) 2, 3, 4 (d) 1, 2, 3

21. Species having zero dipole moment:(a) XeF4 (b) SO2

(c) SF4 (d) CH2Cl222. Out of the two compounds shown below, the

vapour pressure of (b) at a particular temperatureis expected to be:

NO2

(1)

OH

and

OH

NO2

(2)

(a) Higher than that of (a)(b) Lower than that of (a)(c) Same as that of (a)(d) Can be higher or lower depending upon the size

of the vessel23. Which of the following molecules are expected to

exhibit intermolecular H-bonding?(I) Acetic acid (II) o-nitrophenol(III) m-nitrophenol (IV) o-boric acidSelect correct alternate :(a) I, II, III (b) I, II, IV(c) I, III, IV (d) II, III, IV

24. Which one is the most resonance stabilizedamongest the following -(a) NO3

– (b) NO2–

(c) SO2 (d) O3

25. In which of the following compounds resonancedoes not occurs -(a) H2O (b) SiO2

(c) SO3 (d) CO2

Correct answer is -(a) a and d (b) a and b(c) c and d (d) b, c and d

26. Resonance is not shown by -(a) C6H6 (b) CO2

(c) CO32– (d) SiO2

27. In PO43–, the formal charge on each oxygen atom

and the P–O bond order respectively are -(a) – 0.75, 0.6 (b) – 0.75, 1.0(c) – 0.75, 1.25 (d) – 3, 1.25

28. Incorrect order of melting point and boiling point -(a) NaCl < MgCl2 < AlCl3(b) HF > HBr > HCl(c) H2O > HF > NH3

(d) H2O > C2H5OH > CH3 – O–CH3

29. Pick out the wrong statement -(a) LiF has less solubility in water than LiI(b) Lattice energy of MgO is greater than Na2O(c) LiH is more stable than KH(d) KO2 is diamagnetic and colourless

30. A metal M readily forms its sulphate MSO4, whichis water soluble. It forms an insolublehydroxide M(OH)2 which is soluble in NaOH solu-tion, then M is -(a) Mg (b) Ca(c) Be (d) Ba

31. To a clear solution of a complex (X), a solution ofBaCl2 is added and a white ppt. is formed whichdoes not dissolves in dilute HCl. The compound Xis -(a) Nitrate (b) Bromide(c) Carbonate (d) Sulphate

32. Which of the following compound is not hydroly-sed with water -(a) SiCl4 (b) NCl3(c) CCl4 (d) BCl3

33. In which one of the following species has centralatom of different hybridization state than the otherthree?(a) SiH4 (b) BF4

–

(c) CH4 (d) [Ni (CN)4]–2


34. Which statement is correct(a) All the compounds having polar bond, have

dipole moment(b) SO2 is non-polar(c) H2O molecule is non polar, having polar bonds(d) PH3 is polar molecule having non polar bonds

35. Which of the following species are symmetrical(a) XeF4 (b) XeF6

(c) SO2 (d) NH3+2

Correct answer is -(a) a, b (b) b, c(c) c, d (d) a, d

36. Decreasing order of dipole moment of the follow-ing compounds is -

(a) A > B > C (b) C > A > B(c) C > B > A (d) A > C > B

37. Incorrect order of viscosity -(a) H2SO4 > HNO3

(b) H2O > CH3OH(c) o-nitro phenol > p-nitro phenol(d) Glycol > ether

38. Increasing strength of H–bonding (X --- H–X) in S,O,F,Cl,N is -(a) Cl, S, N,O, F (b) N, Cl, S, O, F(c) S, Cl, N, O, F (d) S, N, Cl, O, F

39. The intermolecular force in hydrogen fluoride is dueto -(a) Dipole-induced dipole interaction(b) Dipole -dipole interactions(c) Hydrogen bond(d) Disperson interaction

40. Which of the following compounds show intermo-lecular hydrogen bonding(A) o-nitrophenol (B) p-nitrophenol(C) phenol (D) salicylaldehyde(a) A and B (b) A and C(c) A and D (d) B, C and D

41. One of the resonating structure of SO4–2 is

S OO

O

OWhich set of formal charge on oxygen and bondorder is correct(a) 0.5 and 1.5 (b) 1.5 and 3(c) 2 and 3 (d) 1.5 and 1.5

42. A bonded molecule MX3 is T-shaped. Thenumber of non-bonding pairs of electron is -(a) 0(b) 2(c) 1(d) Can be predicted only if atomic number of M is

known43. The regular geometry of XeO2F2 is -

(a) Plane triangular (b) Trigonal bipyramidal(c) Square planar (d) Tetrahedral

44. Increasing order of bond length in NO, NO+ andNO– is -(a) NO > NO– > NO+ (b) NO+ < NO < NO–

(c) NO < NO+ < NO– (d) NO < NO+ = NO–

45. The correct order of increasing C–O bond lengthof CO, CO3

2–, CO2 is -(a) CO3

2– < CO2 < CO(b) CO2 < CO3

2– < CO(c) CO < CO3

2– < CO2

(d) CO < CO2 < CO32–

46. Pick out the incorrect statement -

(a) sp3d hybridisation involves 22 y–xd orbital

(b) Hybridised orbital form s-bond when overlapswith other orbitals

(c) SF2 molecule is more polar than CS2

(d) o-nitrophenol is more volatile than p-nitrophenol47. The nature of -bond in perchlorate (ClO4

–) ionis -(a) O(d) – C(p) (b) O(p) – C(p)

(c) O(p) – C(d) (d) O(d) – Cl(d)

48. The true statements fromt he following are -(1) PH5, NCl5 and BiCl5 do not exist(2) I3

+ has bent geometry(3) XeF4 is non-polar molecule(4) CO and C2

–2 has same bond order


(5) MgCl2 has more ionic nature than MgO(a) a, b, c, d (b) b, c, d(c) a, b, c (d) a, b, c, d, e

49. Which of the following have covalent, coordinateand ionic all three type of bonds -(1) NH4Cl (2) KNO3

(3) K3[Fe(CN)6] (4) H2CO3

(a) Only c (b) b and c only(c) a, b and c only (d) a, b, c and d

50. The group of substances in which at melting pointcovalent bond breaks up -(a) Nacl, KCl, CaCl2 (b) I2(S), CH4(S), dry ice(c) B4C, dimond, SiC (d) All of the above

51. The incorrect statement is -(a) have two nodal planes(b) Bond order of HeH+ is 0.5(c) In NCO–, C is sp hybridised(d) O3 is polar while O2 is non polar

52. In which of the following compounds ionic &covalent bonds are present -(a) KCl (b) SO2

(c) NaOH (d) CH4

53. Solid CH4 is -(a) Molecular solid (b) Ionic solid(c) Covalent solid (d) None of these

54. Which of the following compounds has the smallestbond angle in its molecule ?(a) SO2 (b) H2O(c) H2S (d) NH3

55. An ether is more volatile than an alcohol having thesame molecular formula. This is due to:(a) dipolar character of ethers.(b) alcohols having resonance structures.(c) inter-molecular hydrogen bonding in ethers.(d) inter-molecular hydrogen bonding in alcohols.

56. The pair of species having identical shapes formolecules of both species is :(a) CF4, SF4 (b) XeF2, CO2

(c) BF3, PCl3 (d) PF5, IF5.57. Which of the following pair of molecules will have

permanent dipole moments for both members?(a) SiF4 and NO2 (b) NO2 and CO2

(c) NO2 and O3 (d) SiF4 and CO2

58. The correct order of bond angles (smallest first) inH2S, NH3, BF3 and SiH4 is :(a) H2S < SiH4 < NH3 < BF3

(b) NH3 < H2S < SiH4 < BF3

(c) H2S < NH3 < SiH4 < BF3

(d) H2S < NH3 < BF3 < SiH4

59. The bond order in NO is 2.5 while that in NO+ is 3.Which of the following statements is true for thesetwo species?(a) bond length in NO+ is greater than in NO(b) bond length in NO is greater than in NO+

(c) bond length in NO+ is equal to that in NO(d) bond length is unpredictable

60. The states of hybridization of boron and oxygenatoms in boric acid (H3BO3) are respectively :(a) sp2 and sp2 (b) sp2 and sp3

(c) sp3 and sp2 (d) sp3 and sp3

61. Which one of the following has the regulartetrahedral structure ?(a) XeF4 (b) SF4

(c) BF4– (d) [Ni(CN)4]2–

(Atomic number : B = 5, S = 16, Ni = 28, Xe = 54)62. Which one of the following does not have sp2

hybridized carbon?(a) acetone (b) acetic acid(c) acetonitrile (d) acetamide

63. Lattice energy of an ionic compound depends upon:(a) charge on the ions only(b) size of the ions only(c) packing of ions only(d) charge on the ion and size of the ion

64. Based on lattice energy and other considerations,which one of the following alkali metal chlorides isexpected to have the highest melting point ?(a) LiCI (b) NaCl(c) KCl (d) RbCl

65. The molecular shapes of SF4, CF4 and XeF4 are :(a) The same with 2, 0 and 1 lone pairs of electrons

on the central atom, respectively.(b) The same with 1, 1 and 1 lone pair of electrons

on the central atoms, respectively.(c) Different with 0, 1 and 2 lone pairs of electrons

on the central atom, respectively.


(d) Different with 1, 0 and 2 lone pairs of electronson the central atom, respectively.

66. The number and type of bonds between two carbonatoms in calcium carbide are :(a) one sigma, one pi (b) one sigma, two pi(c) two sigma, one pi (d) two sigma, two pi

67. Which of the following molecules/ions does notcontain unpaired electrons?(a) O2

2– (b) B2

(c) N2+ (d) O2

68. A metal, M forms chlorides in +2 and +4 oxidationstates. Which of the following statements aboutthese chlorides is correct ?(a) MCl2 is more volatile than MCl4(b) MCl2 is more soluble in anhydrous ethanol than

MCl4(c) MCl2 is more ionic than MCl4(d) MCl2 is more easily hydrolysed than MCl4

69. In which of the following molecules/ions are all thebonds not equal?(a) SF4 (b) SiF4

(c) XeF4 (d) BF4–

70. Which of the following hydrogen bonds is thestrongest ?(a) O – H ... O (b) O – H ... F(c) F – H ... H (d) F – H ... F

71. The charge/ size ratio of a cation determines itspolarizing power. Which one of the followingsequences represents the increasing order of thepolarizing power of these cationic species, K+, Ca2+,Mg2+, Be2+?(a) K+ < Ca2+ < Mg2+ < Be2+

(b) Ca2 < Mg2+ < Be2+ < K+

(c) Mg2+ < Be2+ < K+ < Ca2+

(d) Be2+ < K+ < Ca2+ < Mg2+

72. In which of the following ionization processes, thebond order has increased and the magneticbehaviour has changed ?(a) O2 O2

+ (b) N2 N2+

(c) C2 C2+ (d) NO NO+

73. Which one of the following pairs of species havethe same bond order?(a) CN– and CN+ (b) O2

– and CN–

(c) NO+ and CN+ (d) CN– and NO+

74. The bond dissociation energy of B – F in BF3 is646 kJ mol–1 whereas that of C – F in CF4 is 515kJ mol–1. The correct reason for higher B – F bonddissociation energy as compared to that of C – F is:(a) stronger s bond between B and F in BF3 as

compared to that between C and F in CF4.(b) significant pp - pp interaction between B and F

in BF3 whereas there is no possibility of suchinteraction between C and F in CF4.

(c) lower degree of pp - pp interaction between Band F in BF3 than that between C and F in CF4.

(d) smaller size of B - atom as compared to that ofC - atom.

75. Using MO theory predict which of the followingspecies has the shortest bond length ?(a) O2

+ (b) O2–

(c) O22– (d) O2

2+

76. Among the following the maximum covalentcharacter is shown by the compound :(a) FeCl2 (b) SnCl2(c) AlCl3 (d) MgCl2

77. The hybridisation of orbitals of N atom in NO3–,

NO2+ and NH4

+ are respectively :(a) sp, sp2, sp3 (b) sp2, sp, sp3

(c) sp, sp3, sp2 (d) sp2, sp3, sp78. The structure of IF7 is :

(a) square pyramid (b) trigonal bipyramid(c) octahedral (d) pentagonal bipyramid

79. The number of types of bonds between two carbonatoms in calcium carbide is:(a) One sigma, one pi (b) Two sigma, one pi(c) Two sigma, two pi (d) One sigma, two pi

80. Which of the following has maximum number oflone pairs associated with Xe ?(a) XeF4 (b) XeF6

(c) XeF2 (d) XeO3

81. The molecule having smallest bond angle is :(a) NCl3 (b) AsCl3(c) SbCl3 (d) PCl3

82. In which of the following pairs the two species arenot isostructural ?(a) CO3

2– and NO3– (b) PCl4+ and SiCl4

(c) PF5 and BrF5 (d) AlF63– and SF6


83. Which one of the following molecules is expectedto exhibit diamagnetic behaviour?(a) C2 (b) N2

(c) O2 (d) S2

84. In which of the following pairs of molecules/ions,both the species are not likely to exist ?

(a) , He2-2 (b) H

-2, He2-

2

(c) H2+2 , He2 (d) H-

2, He2+2

85. Which of the following exists as covalent crystalsin the solid state ?(a) Iodine (b) Silicon(c) Sulphur (d) Phosphorus

86. Stability of the species Li2, Li –2 and Li +2 increasesin the order of :

(a) Li2 < Li –2 < Li +2 (b) Li –2 < Li +2 < Li2

(c) Li2 < Li –2 < Li +2 (d) Li –2 < Li2 < Li +287. Among the following oxoacids, the correct

decreasing order of acid strength is :(a) HOCl > HClO2 > HClO3 > HClO4

(b) HClO4 > HOCl > HClO2 > HClO3

(c) HClO4 > HClO3 > HClO2 > HOCl(d) HClO2 > HClO4 > HClO3 > HOCl

88. The correct statement for the molecule, CsI3, is :(a) it is a covalent molecule.(b) it contains Cs+ and(c) it contains Cs3+ and I¯ ions.

(d) it contains Cs+, I¯ and lattice I2 molecule.89. The hybridization of atomic orbitals of nitrogen in

NO2+, NO3

– and NH4+ are :

(a) sp, sp3 and sp2 respectively(b) sp, sp2 and sp3 respectively(c) sp2, sp and sp3 respectively(d) sp2, sp3 and sp respectively

90. The number of P—O—P bonds in tricyclicmetaphosphoric acid is :(a) zero (b) two

(c) three (d) four91. Molecular shapes of SF4, CF4 and XeF4 are

respectively :

(a) the same with 2, 0 and 1 lone pair of electronsrespectively.

(b) the same with 1, 1 and 1 lone pair of electronsrespectively.

(c) different with 0, 1 and 2 lone pair of electronsrespectively.

(d) different with 1, 0 and 2 lone pair of electronsrespectively.

92. Amongst H2O, H2S, H2Se and H2Te the one withhighest boiling point is :(a) H2O because of H-bonding.(b) H2Te because of higher molecular weight.(c) H2S because of H-bonding.(d) H2Se because of lower molecular weight.

93. The number of S–S bonds, in sulphur trioxide trimer(S3O9) is :(a) three (b) two(c) one (d) Zero

94. The correct order of hybridisation of the centralatom in the following species; NH3, [PtCl4]2–, PCl5and BCl3 is : [ Atomic number Pt = 78](a) dsp2, dsp3, sp2 and sp3

(b) sp3, dsp2, dsp3, sp2

(c) dsp2, sp2, sp3, dsp3

(d) dsp2, sp3, sp2, dsp3

95. The common feature of the species CN –, CO, NO+

are :(a) bond order three and isoelectronic.(b) bond order three and weak field ligand.(c) bond order two and acceptor.(d) isoelectronic and weak field ligands.

96. Specify the coordination geometry around andhybridisation of N and B atoms in a 1 : 1 complexof BCl3 & NH3.(a) N : tetrahedral sp3, B : tetrahedral sp3

(b) N : pyramidal sp3, B : pyarmidal sp3

(c) N : pyramidal sp3, B : planar sp2

(d) N : pyramidal sp3, B : tetrahedral sp3

97. Identify the correct order of boiling points of thefollowing compounds :

3 2 2 2CH CH CH CH OH1 3 2 2CH CH CH CHO

2 3 2 2CH CH CH COOH3

(a) 1 > 2 > 3 (b) 3 > 1 > 2(c) 1 > 3 > 2 (d) 3 > 2 > 1


98. Which of the following molecular species hasunpaired electron(s) ?(a) N2 (b) F2

(c) O2– (d) O2

2–

99. Which of the following are isoelectronic andisostructural ?NO3

– , CO32– , ClO3

– , SO3 .(a) NO3

– , CO32– (b) SO3 , NO3

–

(c) ClO3– , CO3

2– (d) CO32– , SO3 .

100. Among the following the molecule with the highestdipole moment is :(a) CH3CI (b) CH3Cl2(c) CHCI3 (d) CCl4

101. Which of the following represent the given modeof hybridisation sp2 – sp2 – sp – sp from left toright.(a) H2C = CH – C N(b) HC C – C CH(c) H2C = C = C = CH2

(d)

102. The number of lone pair(s) of electrons in XeOF4is :(a) 3 (b) 2(c) 1 (d) 4

103. Amongst the following the acid having –O–O– bondis :(a) H2 S2 O3 (b) H2 S2 O5

(c) H2 S2 O6 (d) H2 S2 O8

104. According to molecular orbital theory, which oneof the following statements about the molecularspecies O2

+ s correct ?(a) It is paramagnetic and has less bond order than

O2

(b) It is paramagnetic and more bond order thanO2

(c) It is diamagnetic and has less bond order thanO2

(d) It is diamagnetic and has more bond order thanO2

105. In which of the following the maximum number oflone pairs is present on the central atom ?(a) [ClO3]– (b) XeF4

(c) SF4 (d) I3–

106. Which of the following silicate is formed when threeoxygen atoms of [SiO4]4– tetrahedral units areshared?(a) Sheet silicate(b) Pyrosilicate(c) Three dimensional silicate(d) Linear chain silicate

107. The species having bond order different from thatin CO is :(a) NO¯ (b) NO+

(c) CN¯ (d) N2108. Among the following, the paramagnetic compound

is :(a) Na2O2 (b) O3(c) N2O (d) KO2

109. The percentage of p-character in the orbitalsforming P – P bonds in P4 is :(a) 25 (b) 33(c) 50 (d) 75

110. The nitrogen oxide(s) that donot contain(s) N—Nbond(s) is :(a) N2O (b) N2O3(c) N2O4 (d) N2O5

111. Assuming that Hund’s rule is violated, the bond orderand magnetic nature of the diatomic molecule B2is:(a) 1 and diamagnetic(b) 0 and diamagnetic(c) 1 and paramagnetic(d) 0 and paramagnetic

112. The shape of XeO2F2 molecule is

(a) trigonal bipyramidal

(b) square plannar

(c) tetrahedral

(d) see-saw

113. With respect to graphite and diamond, which ofthe statement(s) given below is (are) correct ?(a) Graphite is harder than diamond.(b) Graphite has higher electrical conductivity than

diamond(c) Graphite has higher thermal conductivity than

diamond(d) Graphite has lower C–C bond order than

diamond


114. The incorrect statement(s) about O3 is :(a) O-O bond lengths are equal.(b) Thermal decomposition of O3 is endothermic.(c) O3 is diamagnetic in nature.(d) O3 has a bent structure.

115. Assuming 2s-2p mixing is NOT operative, theparamagnetic species among the following is :(a) Be2 (b) B2

(c) C2 (d) N2

EXERCISE–3 (AIIMS SPECIAL)

These questions consist of two statements each,printed as Assertion and Reason. While answering theseQuestions you are required to choose any one of thefollowing four responses.

A. If both Assertion & Reason are True & theReason is a correct explanation of the Assertion.

B. If both Assertion & Reason are True but Reasonis not a correct explanation of the Assertion.

C. If Assertion is True but the Reason is False.

D. If both Assertion & Reason are False.

1. Assertion : Ionic compounds do not exhibit stereoisomerism

Reason : Ionic bonds are non directional.

2. Assertion : Ionic bonds are formed between metaland nonmetals

Reason : In ionic bonds electrons are shared

3. Assertion : D2O is better solvent for ioniccompounds, than H2O.

Reason : dielectric constant of D2O is higher thanH2O.

4. Assertion : Hydrated ionic radius of Li+ is maximumin IA group elements.

Reason : Lithium is a metal

5. Assertion : LiCl exhibits covalent character.

Reason : Lithium is lightest metal.

6. Assertion : K2CO3 do not gives CO2 gas on heating.

Reason : Value of for K+ is high.

7. Assertion : HCN is linear molecule

Reason : Carbon atom in HCN is sp hybridised

8. Assertion : CCl4 is not hydrolysed by water

Reason : Carbon in CCl4 is sp3 hybridised

9. Assertion : PCl3 has trigonal planar shape.

Reason : lp–lp repulsion is present in PCl3.

10. Assertion : CO2 molecule is non–polar while SO2is polar.

Reason : Carbon atom is smaller than sulpher.

11. Assertion : CO2 and SiO2 has same physical stateat room temp.

Reason : CO2 and SiO2 both contains C=O bondsand Si = O bond.

12. Assertion : CH3OH is soluble in water

Reason : CH3OH is ionic in nature

13. Assertion : Boiling point of H2O is greater thanC2H5OH.

Reason : Molecular wt. of H2O is higher thanC2H5OH.

14. Assertion : NO+ is more stable than NO–.

Reason : NO+ do not have electrones in antibondingorbitals.

15. Assertion : In MgO electrovalency of Mg is +2.

Reason : Mg shares two electrons with oxygen.

16. Assertion : Ionic reactions are faster than molecularreactions.

Reason : Ionic bonds are weaker than covalentbonds.

17. Assertion : LiI is more soluble in water than LiF.

Reason : LiI has more ionic character.

18. Assertion : Bond energy of H–H bond is greaterthan Cl–Cl bond.

Reason : H2 is more covalent than Cl2.

19. Assertion : Ionic compounds tend to be non–volalite.

Reason : Inter ionic forces in these compoundsare strong.

20. Assertion : NCl3 has pyramidal shape

Reason : In NCl3 central atom is sp3 hybridised.


21. Assertion : In NCl6– sp3d2 hybridisation is not

possible.

Reason : Nitrogen atom do not have vacant ‘d’orbitals.

22. Assertion : BF3 is planar while NF3 is non–planarmolecule.Reason : B–F bond is more polar than N–F bond.

23. Assertion : HF is more Ionic in nature than HIReason : HF is more polar than HI

24. Assertion : Density of ice is greater than waterReason : In ice H2O molecules are closely packed.

25. Assertion : The bond angle of PBr3 is greater thanthat of PH3 but the bond angle of NBr3 is less thanthat of NH3

Reason : Size of Br is less than H26. Assertion : Bond order of O2 and BN is same.

Reason : O2 and BN are isoelectronic27. Assertion : p-nitrophenol is more viscous than

o-nitrophenol.Reason : In p-nitrophenol, intermolecular H-bondingoccurs.

28. Assertion : Hydrogen do not follows octet rule inits compounds.Reason : Maximum covalency of H is one.

29. Assertion : NaCl is soluble in water LiClReason : In NaCl H > U

30. Assertion : H2O2 is not used as solvent for ioniccompoundsReason : Dielectric constant of H2O2 is low

31. Assertion : Ionic compounds exhibits electricalconductivity in solution state.Reason : In solution state electrons of ioniccompounds are free.

32. Assertion : –bonds participates in resonance.Reason : –bonds are covalent.

33. Assertion : SO2 is polar molecule.Reason : SO2 is covalent molecule.

34. Assertion : p-nitrophenol is more volatile than o–nitrophenol.Reason : Molecular wt. of p-nitrophenol is higherthan o-nitrophenol.

35. Assertion : N2+ is more stable than N2

–

Reason : N2+ has less electrons in antibonding

orbitals.

36. Assertion : SO4–2 is square planar in shape

Reason : SO4–2 has sp3d hybridisation

37. Assertion : NF3 molecule is more polar than NH3molecule.Reason : NF3 is pyramidal while NH3 is trigonalplanar.

38. Assertion : OF4 donot exists.Reason : Empty d–orbitals are absent in valenceshell of oxygen.

39. Assertion : NO2 is coloured gas.Reason : NO2 is inorganic compound

40. Assertion : NaCl in solid state is non–conductive.Reason : It is covalent compound.

41. Asertion : BeCl2 shows covalent character.

Reason : More polarization of Cl– by Be+2.

42. Assertion : MgO and NaF are isomorphousReason : Crystal structure of MgO and NaF isidentical.

43. Assertion : In hybridisation vacant orbitalparticipates.Reason : Hybridisation occurs only in orbitals.

44. Assertion : NF3 molecule is polar.Reason : N–F bonds are polar.

45. Assertion : Melting point of SiO2 is high.Reason : It is a covalent solid.

46. Assertion : In the halides of lithium LiF is insolublein waterReason : LiF is most ionic, among halides of lithium

47. Assertion : NaCl is soluble in non polar solvents.Reason : NaCl is a non-polar covalent compound.

48. Assertion : Ionic compounds are generallycolourless.Reason : They have no unpaired electron.

49. Assertion : Sigma bonds are stronger than bonds.Reason : Sigma bonds are covalent bonds.

50. Assertion : CO2 is non polar while SO2 is polarmolecule.Reason : S-O bonds are polar while C-O non polar.

51. Assertion : Between SiCl4 and CCl4 only SiCl4reacts with water.Reason : SiCl4 is ionic and CCl4 is covalent.

52. Assertion : Both H2O and SnCl2 are bent molecules.Reason : Both H2O and SnCl2 are sp2 hybridised.


53. Assertion : H2O has maximum density at 4°C.Reason : At 4°C more hydrogen bonds are formed.

54. Assertion : PbI4 is a stable compound.Reason : Iodine stabilizes higher oxidation state.

55. Assertion : NO is paramagnetic in nature.Reason : Bond order of NO is 2.5.

56. Assertion : Bond order in a molecule can assumeany value, positive or negative, integral or fractionalincluding zero.Reason : It depends upon the number of electronsin the bonding and antibonding orbitals.

57. Assertion : Ionic compounds tend to be non-volatile.Reason : Intermolecular forces in these are areweak.

58. Assertion : The atoms in a covalent molecule aresaid to share electons, yet some covalent moleculesare polar.Reason : In polar covalent molecules, the sharedelectrons spend more time on the average near oneof the atoms due to EN.

59. Assertion : Nitrogen is inreactive at roomtemperature but becomes reactive at elevatedtemperature (on heating or in the presence ofcatalysts).Reason : In nitrogen molecute, there is extensivedelocalization of electrons.

60. Assertion : Water is a good solvent for ioniccompounds but poor one for covalent compounds.Reason : Hydration energy of ions releases sufficientto overcome lattice energy and break hydrogenbonds in water while covalent bonded compoundsinteract so weakly that even van der Waals' forcesbetween molecules of covalent compounds cannotbe broken.

61. Assertion : Na2SO4 is soluble in water while BaSO4is insoluble.Reason : Lattice energy of BaSO4 exceeds itshydration energy.

62. Assertion : NO3– is planar.

Reason : N in NO3– is sp2 and no lone pair at central

atom.63. Assertion : N2 and NO+ both are diamagnetic

substances.

Reason : NO+ is isolectronic with N2.64. Assertion : The electronic structure of O3 is

O

O O

+

�.

Reason : O

O O

..

Structure is not allowed becauseoctet around 'O' can not be expanded.

65. Assertion : LiCl is predominantly a covalentcompound.Reason : Electonegativity difference between 'Li'and 'Cl' is too small.

66. Assertion : Bond order can assume any valueincluding zero.Reason : Higher the bond order, shorter is the bondlength and greater is the bond energy.

67. Assertion : The O–O bond length in H2O2 is shorterthan that of O2F2.Reason : H2O2 is an ionic compound.

68. Assertion : SiF62– is known but SiCl62– is not.

Reason : Size of fluorine is small and its lone pairof electrons interacts with d-orbitals of Si strongly

69. Assertion : Ozone is a powerful oxidizing agent incomparison to O2.Reason : Ozone is diamagnetic but O2 isparamagnetic.

70. Assertion : SeCl4 does not have a tetrahedralstructure.Reason : Se in SeCl4 has two lone pairs.

71. Assertion : B2 molecule is diamagnetic.Reason : The highest occupied molecular orbital isof type.

72. Assertion : BH4– is known while 3

6BH? is not.

Reason : B has very small atomic size.73. Assertion : Some molecules are polar.

Reason : The centre of negative charge and positivecharge do not coincide each other in some molecule.

74. Assertion : R3P=O exists but R3N=O does notexists.Reason : P is more electronegative than N.

75. Assertion : ClF3 has T-shape structure.Reason : It has two lone pair arrange at 180°(Angle).


EXERCIS-4 (ARCHIVES FROM VARIOUS MEDICAL ENTRANCE)

1. The boiling point of alcohol is higher than ether dueto : [AFMC 2001]

(a) H-bonding

(b) large size of alcohol

(c) presence of -OHgroup

(d) high molecular weight2. AlCl3 fumes in moist air because it is :

[AFMC 2001](a) covalent (b) volatile(c) hygroscopic (d) forms HCl in moist air

3. The highest dipole moment is of : [AFMC 2002](a) CF4 (b) CH3OH(c) CO2 (d) CH3F

4. The bonding present between the carbon atoms ofgraphite : [AFMC 2002](a) metallic (b) ionic(c) covalent (d) van der Waal’s forces

5. The shape of IF7 molecules is : [AFMC 2002](a) pentagonal bipyramidal(b) trigonal pyramidal(c) tetrahedral(d) square planar

6. Born-Haber cycle is used to determine :[AFMC 2002]

(a) electron affinity (b) lattice energy(c) crystal energy (d) all of these

7. The most polar bond is : [AFMC 2002](a) C – F (b) C – O(c) C – Br (d) C – S

8. Which of the following are iso-structural species?[AFMC 2003]

(a) NH4+ and NH2

–

(b) CH3– and CH3

+

(c) SO42–, PO4

3– and [BF4–]

(d) NH4+ and NH3

9. Which is planar molecule? [AFMC 2003](a) PCl3 (b) BCl3(c) NH3 (d) H3O+

10. The molecule of CO2 has angle 180º. It can beexplained on the basis of : [AFMC 2004](a) sp3 hydridisation (b) sp2 hydridisation(c) sp hydridisation (d) d2sp3 hydridisation

11. Oxygen molecule is : [AFMC 2004](a) diamagnetic (b) paramagnetic(c) ferromagnetic (d) ferrimagnetic

12. Which of the following molecules has trigonalplanar geometry ? [AIPMT 2005](a) IF3 (b) PCl3(c) NH3 (d) BF3

13. Which of the following would have a permanentdipole moment ? [AIPMT 2005](a) BF3 (b) SiF4(c) SF4 (d) XeF4

14. The correct sequence of increasing covalentcharacter is represented by [AIPMT 2005]

(a) LiCl < NaCl < BeCl2(b) BeCl2 < NaCl < LiCl(c) NaCl < LiCl < BeCl2(d) BeCl2 < LiCl < NaCl

15. The hybridisation present in IF3 is : [RPMT 2005](a) sp3d (b) sp3

(c) sp3d2 (d) sp3d3

16. The most polar bond is : [RPMT 2005](a) O—F (b) N—Cl

(c) N—F (d) N—N

17. The molecule having largest dipole moment amongthe following is : [RPMT 2005](a) CHI3 (b) CH4

(c) CHCl3 (d) CCI4

18. The correct sequence of hybridisation of methane,ethene and acetylene is : [RPMT 2005]

(a) sp, sp2, sp3 (b) sp2, sp3, sp

(c) sp3, sp2, sp (d) sp3, sp, sp2

19. The number of P–O–P bonds in the structure ofphosphorus pentaoxide and phosphorus trioxide arerespectively: [AIIMS 2005](a) 6, 6 (b) 5, 5(c) 5, 6 (d) 6, 5


20. In diborane, the two H—B—H angles are nearly:[AIIMS 2005]

(a) 60°, 120° (b) 95°, 120°(c) 95°, 150° (d) 120°, 180°

21. The high boiling point of water is due to whichreason : [RPMT 2005, 2006](a) co-ordinate bonding(b) covalent bond(c) electrostatic force of attraction(d) hydrogen bonding

22. In which of the following molecules are all the bondsnot equal ? [AIPMT 2006](a) CIF3 (b) BF3

(c) AIF3 (d) NF3

23. Which of the following species has a linear shape ?[AIPMT 2006]

(a) NO2– (b) SO2

(c) NO2+ (d) O3

24. The number of unpaired electrons in a paramagneticdiatomic molecule of an element with atomic number16 is [AIPMT 2006](a) 2 (b) 3(c) 4 (d) 1

25. Which of the following is not isostructural withSiCl4 ? [AIPMT 2006](a) SCl4 (b) SO4

2–

(c) PO43– (d) NH4

+

26. The electronegativity difference between N and Fis greater than that between N and H yet the dipolemoment of NH3 (1.5 D) is larger than that of NF3(0.2 D) [AIPMT 2006](a) in NH3 as well as in NF3 , the atomic dipole and

bond dipole are in the same direction.(b) in NH3 , the atomic dipole and bond dipole are

in the same direction whereas in NF3 these arein opposite directions.

(c) in NH3 as well as NF3, the atomic dipole andbond dipole are in opposite directions

(d) in NH3 the atomic dipole and bond dipole are inthe opposite directions whereas in NH3 theseare in the same directions.

27. Which of the following is not a correct statement?[AIPMT 2006]

(a) The electron-deficient molecules can act asLewis acids.

(b) The canonical structures have no real existence.(c) Every AB5 molecule does infact have square

pyramid structure.(d) Multiple bonds are always shorter than

corresponding single bonds.28. The hybridisation state of carbon in fullerene is:

[AFMC 2006](a) sp (b) sp2

(c) sp3 (d) sp3d29. The correct order of C–O bond length among CO,

CO32– and CO2 is : [AIPMT 2007]

(a) CO2 < CO32– < CO

(b) CO < CO32– < CO2

(c) CO32– < CO2 < CO

(d) CO < CO2 < CO32–

30. In which of the following pairs, the two speciesare isostructural ? [AIPMT 2007](a) SF4 and XeF4 (b) SO3

2– and NO3–

(c) BF3 and NF3 (d) BrO3– and XeO3

31. Which compound is highest covalent ?[AFMC 2007]

(a) LiCl (b) LiF(c) LiBr (d) LiI

32. Shape of XeF4 molecule is : [AFMC 2007](a) linear (b) pyramidal(c) tetrahedral (d) square planar

33. Hybridization present in ClF3 is : [RPMT 2007](a) sp2 (b) sp3

(c) dsp2 (d) sp3d34. Assertion : H–S–H bond angle in H2S is closer to

90° but H–O–H bond angle in H2O is 104.5°.Reason : lp - lp repulsion is stronger in H2S thanin H2O. [AIIMS 2007](a) If both assertion and reason are true and reason

is the correct explanation of assertion.(b) If both assertion and reason are true but reason

is not the correct explanation of assertion.(c) If Assertion is true but reason is false.

(d) If both assertion and reason are false.35. The angular shape of ozone molecule (O3) consists

of [AIPMT 2008](a) 1 sigma and 2 bonds(b) 2 sigma and 1 bonds


(c) 1 sigma and 1 bonds(d) 2 sigma and 2 bonds

36. The correct order of increasing bond angles in thefollowing triatomic species is [AIPMT 2008]

(a) NO2– < NO2

+ < NO2

(b) NO2– < NO2 < NO2

+

(c) NO2+ < NO2 < NO2

–

(d) NO2+ < NO2

– < NO2

37. Four diatomic species are listed below in differentsequences. Which of these presents the correctorder of their increasing bond order?

[AIPMT 2008]

(a) O2– < NO < C22– < He2

+

(b) NO < C22– < O2

– < He2+

(c) C22– < He2

+ < NO < O2–

(d) He2+ < O2

– < NO < C22–

38. Which of the following represents the Lewisstructure of N2 molecule ? [AFMC 2008]

(a) (b)

(c) N – Nxxx

xx

x x

x

x

xxx (d) N=N xx

xxx

x x

x

x

xxx

39. Hydrogen bond is strongest in : [AFMC 2008](a) S—H—O (b) O—H—S(c) F—H—F (d) O—H—N

40. The shape of PCl3 molecule is : [AFMC 2008](a) trigonal bipyramidal(b) tetrahedral

(c) pyramidal(d) square planar

41. H—O—H bond angle in H2O is 104.5º and not109º28’ because of : [RPMT 2008](a) lone pair-lone pair repulsion

(b) lone pair-bond pair repulsion(c) bond pair-bond pair repulsion

(d) high electronegativity of oxygen42. Assertion : The S–S–S bond angle in S8 molecule

is 105°.Reason : S8 has V-shape. [AIIMS 2008](a) If both assertion and reason are true and reason

is the correct explanation of assertion.

(b) If both assertion and reason are true but reasonis not the correct explanation of assertion.

(c) If Assertion is true but reason is false.

(d) If both assertion and reason are false.

43. Assertion : Fluorine molecule has bond order one.

Reason : The number of electrons in the antibondingmolecular orbitals is two less than that is bondingmolecular orbitals. [AIIMS 2008]

(a) If both assertion and reason are true and reasonis the correct explanation of assertion.




44. According to MO theory which of the followinglists rank the nitrogen species in terms of increasingbond order ? [AIPMT 2009]

(a) N2– < N2 < N2

2– (b) N22– < N2

– < N2

(c) N2 < N22– < N2

– (d) N2– < N2

2– < N2

45. In the case of alkali metals, the covalent characterdecreases in the order [AIPMT 2009]

(a) MCI > MI > MBr > MF

(b) MF > MCI > MBr > MI

(c) MF > MCI > MI > MBr

(d) MI > MBr > MCI > MF

46. In which of the following molecules/ions BF3,NO2

–, NH2– and H2O, the central atom is sp2

hybridised ? [AIPMT 2009]

(a) NO2– and NH2

– (b) NH2– and H2O

(c) NO2– and H2O (d) BF3 and NO2

–

47. What is the dominant intermolecular force on bondthat must be overcome in converting liquid CH3OHto a gas ? [AIPMT 2009](a) Hydorgen bonding

(b) Dipole-dipole interaction

(c) Covalent bonds

(d) London dispersion force

48. The type of hybridization present in each carbon ofbenzene is : [RPMT 2009]

(a) sp3 (b) sp2

(c) sp (d) none of these

49. Bond order of N2 molecule is : [RPMT 2009]

(a) 3 (b) 2

(c) 1 (d) 0


50. The number of unpaired electrons in Xe atom ofXeF2 is : [RPMT 2009](a) 3 (b) 5

(c) 4 (d) 251. Which of the following is an electron deficient

compound ? [RPMT 2009](a) B2H6 (b) NH3

(c) C2H6 (d) CCl452. Which of the following has zero dipole moment?

[RPMT 2009](a) ClF (b) PCl3(c) SiF4 (d) CFCl3

53. HCl molecule contains : [AFMC 2009](a) ionic bond (b) covalent bond(c) hydrogen bond (d) coordinate bond

54. The geometry of sulphate is : [AFMC 2009](a) square planar (b) tetrahedral

(c) square pyramidal (d) octahedral55. In which of the following pairs of molecules/ions,

the central atoms have sp2 hybridization ?

[AIPMT 2010]

(a) NO2– and NH3 (b) BF3 and NO2

–

(c) NH2– and H2O (d) BF3 and NH2

–

56. Which one of the following species does not existunder normal conditions ? [AIPMT 2010](a) Be2

+ (b) Be2

(c) B2 (d) Li257. The correct order of increasing bond angles in the

following species are : [AIPMT 2010](a) Cl2O < ClO2 < ClO2

–

(b) ClO2 < Cl2O < ClO2–

(c) Cl2O < ClO2– < ClO2

(d) ClO2– < Cl2O < ClO2

58. Which one of the following compounds is aperoxide ? [AIPMT 2010]

(a) KO2 (b) BaO2

(c) MnO2 (d) NO2

59. Which one of the following molecular hydrides actsas a Lewis acid ? [AIPMT 2010]

(a) NH3 (b) H2O

(c) B2H6 (d) CH4

60. The tendency of BF3, BCl3 and BBr3 to behave asLewis acid decreases in the sequence:

[AIPMT 2010]

(a) BCl3 > BF3 > BBr3 (b) BBr3 > BCl3 > BF3

(c) BBr3 > BF3 > BCl3 (d) BF3 > BCl3 > BBr3

61. Among the following which one has the highestcation to anion size ratio? [AIPMT 2010]

(a) CsI (b) CsF

(c) LiF (d) NaF

62. In which of the following molecules, the centralatom does not have sp3 hybridization?

[AIPMT 2010]

(a) CH4 (b) SF4

(c) BF–4 (d) NH+

4

63. How many bridging oxygen atoms are present inP4O10? [AIPMT 2010]

(a) 6 (b) 4

(c) 2 (d) 564. Some of the properties of the two species, NO–

3and H3O+ are described below. Which one of themis correct? [AIPMT 2010](a) Dissimilar in hybridization for the central atom

with different structures(b) Isostructural with same hybridization for the

central atom(c) Isostructural with different hybridization for the

central atom(d) Similar in hybridiation for the central atom with

different structures65. The type of hybrid orbitals used by iodine atom in

hypoiodous acid molecule are [AIIMS 2010](a) sp3 (b) sp2

(c) sp (d) sp3d

66. The correct order of increasing covalent characteris : [AIIMS 2010](a) LiCl, NaCl, BeCl2 (b) BeCl2, NaCl, LiCl

(c) NaCl, LiCl, BeCl2 (d) BeCl2, LiCl, NaCl67. Assertion : HOF bond angle is higher than HOCl

bond angle in HOX.Reason : Oxygen is more electronegative thanhalogens. [AIIMS 2010](a) If both assertion and reason are true and reason






68. Assertion : The HF2– ions exists in the solid state

and also in liquid state but not in aqueous solution.

Reason : The magnitude of hydrogen bonds inbetween HF–HF molecule is weaker than that inbetween HF and H2O molecules. [AIIMS 2010]





69. Which of the following has the minimum bondlength ? [AIPMT 2011]

(a) O2+ (b) O2

–

(c) O22– (d) O2

70. Which of the two ions from the list given belowthat have the geometry that is explained by the samehybridization of orbitals, NO2

–, NO3–, NH2

–, NH4+,

SCN– ? [AIPMT 2011]

(a) NO2– and NO3

– (b) NO4+ and NO3

–

(c) SCN– and NH2– (d) NO2

– and NH2–

71. Which of the following is least likely to behave asLewis base ? [AIPMT 2011]

(a) H2O (b) NH3

(c) BF3 (d) OH–

72. Name the type of the structure of silicate in whichone oxygen atom of [SiO4]4– is shared ?

[AIPMT 2011]

(a) Linear chain silicate

(b) Sheet silicate

(c) Pyrosilicate

(d) Three dimensional

73. The correct order of increasing bond length of C–H, C–O, C–C and C=C is : [AIPMT 2011]

(a) C–H < C=C < C–O < C–C

(b) C–C < C=C < C–O < C–H

(c) C–O < C–H < C–C < C=C

(d) C–H < C–O < C–C < C=C

74. Which of the following structures is the mostpreferred and hence of lowest energy for SO3 ?

[AIPMT 2011]

(a) (b)

(c) (d)

75. The pairs of species of oxygen and their magneticbehaviours are noted below. Which of the followingpresents the correct description ? [AIPMT 2011]

(a) –2O , 2

2O ? – Both diamagnetic

(b) O+, 22O ? – Both paramagnetic

(c) +2O , O2 – Both paramagnetic

(d) O, 22O ? – Both paramagnetic

76. Decreasing order of bond angle is: [AIIMS 2011]

(a) BeCl2 > NO2 > SO2

(b) BeCl2 > SO2 > NO2

(c) SO2 > BeCl2 > NO2

(d) SO2 > NO2 > BeCl277. The wrong statement about fullerene is :

[AIIMS 2011]

(a) it has 5-membered carbon ring

(b) it has 6-membered carbon ring

(c) it has sp2 hybridization

(d) it has 5-membered rings more than 6-memberedrings.

78. Assertion : R3P = O exists R3N = O does notexist. [AIIMS 2011]

Reason : P is more electronegative than N.






79. Which one of the following pairs is isostructural(i.e. having the same shape and hybridization)?

[AIPMT 2012](a) [BCl3 and BrCl3] (b) [NH3 and NO–

3](c) [NF3 and BF3] (d) [BF4

– and NH+4]

80. Bond order of 1.5 is shown by : [AIPMT 2012](a) O2

+ (b) O2–

(c) O22– (d) O2

81. Which of the following statements is not valid foroxoacids of phosphorus ? [AIPMT 2012](a) Orthophosphoric acid is used in the manufacture

of triple superphosphate(b) Hypophosphorous acid is a diprotic acid(c) All oxoacids contain tetrahedral four

coordinated phosphorus(d) All oxoacids contain atleast one P=O unit and

one P–OH group.82. Which of the following species contains three bond

pairs and one lone pair around the central atom ?[AIPMT 2012]

(a) H2O (b) BF3

(c) NH2– (d) PCl3

83. The pair of species with the same bond order is:[AIPMT 2012]

(a) O22– , B2 (b) O2

+ , NO+

(c) NO, CO (d) N2, O2

84. During change of O2 to O2– ion, the electron adds

on which one of the following orbitals?[AIPMT 2012]

(a) orbital (b) orbital(c) orbital (d) orbital

85. Four diatomic species are listed below. Identify thecorrect order in which the bond order is increasingin them: [AIPMT 2012]

(a) 22 2 2NO O C He

(b) 22 2 2O NO C He

(c) 22 2 2C He O NO

(d) 22 2 2He O NO C

86. N2 and O2 are converted into monocations, N2+

and O2+ respectively. Which of the following is

wrong? [AIIMS 2012](a) In N2

+, N—N bond weakens

(b) In O2+, the O–O bond order increases

(c) In O2+, paramagnetism decreases

(d) N2+ becomes diamagnetic

87. Which of the following is electron-deficient ?[NEET 2013]

(a) (SiH3)2 (b) (BH3)2

(c) PH3 (d) (CH3)2

88. Which one of the following molecules contains no bond ? [NEET 2013](a) H2O (b) SO2

(c) NO2 (d) CO2

89. Which of the following is a polar molecule ?[NEET 2013]

(a) SF4 (b) SiF4

(c) XeF4 (d) BF3

90. Which of the following is paramagnetic ?[NEET 2013]

(a) O2– (b) CN–

(c) NO+ (d) CO

91. XeF2 is isostructural with : [NEET 2013](a) Cl2– (b) SbCl3(c) BaCl2 (d) TeF2

92. Dipole-induced dipole interactions are present inwhich of the following pairs : [NEET 2013](a) Cl2 and CCl4 (b) HCl and He atoms

(c) SiF4 and He atoms (d) H2O and alcohol93. Which of the following molecules has the maximum

dipole moment ? [AIPMT 2014](a) CO2 (b) CH4

(c) NH3 (d) NF3

94. Which one of the following species has planetriangular shape ? [AIPMT 2014](a) N3 (b) NO3

–

(c) NO2 (d) CO2

95. Which of the following species contains equalnumber of and bonds? [AIPMT-2015](a) HCO3

– (b) XeO4

(c) (CN)2 (d) CH2(CN)2

96. Which of the following pairs of ions are isoelectronicand isostructural? [AIPMT-2015](a) CO3

2–, SO32– (b) ClO3

–, CO32–

(c) SO32–, NO3

– (d) ClO3–, SO3

2–


97. Which of the following options represents thecorrect bond order? [AIPMT-2015](a) O2

– > O2 > O2+ (b) O2

– < O2 < O2+

(c) O2– > O2 < O2

+ (d) O2– < O2 > O2

+

98. The correct bond order in the following species is[AIPMT-2015]

(a) O22+ > O2

+ > O2– (b) O2

2+ < O2– < O2

+

(c) O2+ > O2

– < O22+ (d) O2

– < O2+ > O2

2+

99. The shapes of SF4 and XeF2 respectively are :[AIIMS-2015]

(a) trigonal bipyramidal and trigonal bipyramidal(b) see-saw and linear(c) T-shape and linear(d) square planar and trigonal biypramidal

100. Consider the statement I. Bond length in N2

+ is 0.02 Å greater than in N2. II. Bond length of NO+ is 0.09 Å less than in NO.III. O2

2– has shorter bond length than O2.Which of the following statements are true?

[AIIMS-2015](a) I and II (b) II and III(c) I, II and III (d) I and III

101. Predict the correct order among the following :

[NEET-2016]

(a) Ione pair - bond pair > bond pair - bond pair >lone pair - lone pair

(b) lone pair - lone pair > lone pair - bond pair >bond pair - bond pair

(c) lone pair - lone pair > bond pair - bond pair >lone pair - bond pair

(d) bond pair - bond pair > lone pair - bond pair >lone pair - lone pair

102. Consider the molecules CH4, NH3 and H2O. Whichof the given statement is false ? [NEET-2016]

(a) The H–C–H bond angle is CH4 is larger than theH–N–H bond angle is NH3

(b) The H–C–H bond angle is CH4 the H–N–H bondangle in NH3 and the H–O–H bond angle in H2Oare all greater than 90º.

(c) Then H–O–H bond angle in H2O is larger thanthe H–C–H bond angle in CH4

(d) The H–O–H bond angle in H2O is smaller thanthe H–N–H bond angle in NH3

103. Which of the following molecules has more thanone lone pair? [AIIMS-2016]

(a) SO2 (b) XeF2

(c) SiF4 (d) CH4

104. Four diatomic species are listed below in differentsequences. Which of these represents the correctorder of their increasing bond order?

[AIIMS-2016]

(a) 2222 ONOHeC

(b) 2222 CNOOHe

(c) 2222 HeCNOO

(d) 2222 HeOCNO

105. Assertion : OHC

bond angle is less than the

normal tetrahedral bond angle.Reason : Lone pair-lone pair repulsion decreasesbond angle. [AIIMS-2016](a) If both assertion and reason are true and reason

is the correct explanantion of assertion.(b) If both assertion and reason are true but reason

is not the correct explanation of assertion.(c) If assertion is true but reason is false.(d) If both assertion and reason are false.

106. The species, having bond angles of 120º is :[NEET-2017]

(a) PH3 (b) CIF3

(c) NCl3 (d) BCl3107. Match the interhalogen compounds of Column I

with the geometry in column II and Assign thecorrect code. [NEET-2017]Column I Column II(a) XX’ (i) T-shape(b) XX3' (ii) Pentagonal bipyramidal(c) XX5' (iii) Linear(d) XX7' (iv) Square-pyramidal

(v) TetrahedralCode :

(a) (b) (c) (d)(a) (iii) (iv) (i) (ii)(b) (iii) (i) (iv) (ii)(c) (v) (iv) (iii) (ii)(d) (iv) (iii) (ii) (i)


108. Which of the following pairs of compounds isisoelectronic and isostructural? [NEET-2017](a) BeCl2, XeF2 (b) TeI2, XeF2

(c) IBr2–, XeF2 (d) IF3, XeF2

109. Which one of the following pairs of species havethe same bond order ? [NEET-2017]

(a) CO, NO (b) O2, NO+

(c) CN–, CO (d) N2, O2–

110. Hybridistion states of C in CH3+ and CH4 are :

[AIIMS-2017](a) sp2 & sp3 (b) sp3 & sp2

(c) sp2 & sp2 (d) sp3 & sp3

111. Which of the following substances has the leastcovalent character ? [AIIMS-2017](a) Cl2O (b) NCl3(c) PbCl2 (d) BaCl2

112. Assertion : Benzene exhibit two different bondlengths, due to C–C single and C=C double bonds.Reason : Actual structure of benzene is a hybrid offollowing two structures. [AIIMS-2017]

(a) Assertion is correct, reason is correct; reasonis a correct explanation for assertion.

(b) Assertion is correct, reason is correct; reasonis not a correct explanation for assertion.

(c) Assertion is correct, reason is incorrect.(d) Assertion is incorrect, reason is correct.

113. In the structure of ClF3, the number of lone pairsof electrons on cetral atom ‘Cl’ is: [NEET-2018](a) one (b) Three(c) four (d) two

114. Consider the following species :CN+, CN–, NO and CNWhich one of these will have the highest bond order?

[NEET-2018](a) NO (b) CN(c) CN+

(d) CN.

115. Which of the following cantain at least one lone pairin all of its halide. [AIIMS-2018(M)](a) Xe (b) Se(c) Cl (d) N

116. Which is incorrect regarding S and P mixing (alongZ-axis.) [AIIMS-2018 (M)]

(a) Nodal plane(s) present in ABMO(b) Nodal plane is absent in BMO(c) MO formed may have higher energy than parent

AO(d) MO formed are asymmetric

117. Assertion :The surface tension of water is morethan other liquid. [AIIMS-2018 (M)]Reason : Water molecules have strong intermolecular H-bonding as attractive force.(a) If both assertion and reason are true and reason



118. Which molecule pair do not have identical structure[AIIMS-2018 (E)]

(a) I3–, BeF2 (b) HClO, SO2

(c) BF3, ICl3 (d) BrF4–, XeF4

119. Which contain at least one e– in 2p bonding MO

[AIIMS-2018 (E)](a) O2 (b) B2

(c) C2 (d) Li2120. What is impact on benzene in magnetic field :

[AIIMS-2018 (E)](a) Strong attract (b) Weakly attract(c) Strongly repel (d) weak repel

121. For geometric isomers of 3–hexene :[AIIMS-2018 (E)]

C H2 5

C H2 5

C=CH

H

& H

C H2 5

C=CC H2 5

H

(a) M.P. is high and dipole moment high for trans(b) M.P. is low and dipole moment low for trans(c) M.P. is high and dipole moment low for trans(d) M.P. is low and dipole moment high for trans

122. For N3– which statement is wrong.

[AIIMS-2018 (M)](a) Iso electronic with CO2

(b) NH2OH and N3– have same O.N. on nitrogen

atom(c) N–N bond length are same(d) HN3 have linear shape


123. Which pair of diatomic species do not have samebond order ? [AIIMS-2018 (M)](a) B2

–, C2 (b) O22–, F2

–

(c) N2+, O2

– (d) B22–, C2

124. ClF2–, ClF4

–find out number of lone pair and geom-etry. [AIIMS-2018 (M)](a) 3 – Linear, 2 – Square planar(b) 3 – Square planar, 2 – Linear(c) 0 – Linear, 3 – Square planar(d) 2 – Linear, 2 – Square planar

125. Which have correct order of dipole moment :[AIIMS-2018 (M)]

(a) SO2 > H2O (b) NF3 > NH3

(c) BF3 < NH3 (d) SO2 < SO3

126. Assertion : Aldehyde have lower boling point thanether.Reason : Aldehydes are less polar than ether.

[AIIMS-2018 (M)](a) If both assertion and reason are true and reason


is not the correct explanation of assertion.(c) If assertion is true but reason is false.


127. In which of the following shape is same but hybrid-ization is different:

[AIIMS-2018 (E)]

(a) ICl2–, XeF2 (b) SO2, NO2+

(c) SO2, NH2– (d) CO2, SO2

128. Correct order of bond angle is :

[AIIMS-2018 (E)]

(a) SO2 < H2S (b) SO2 < H2O

(c) NH3 < H2O (d) NH3 < SO2

129. % s-character of N–H bond is maximum in :[AIIMS-2018 (E)]

(a) N2H2 (b) N2H4

(c) NH3 (d) NH4+

130. Assertion : SO2 is more covalent than SeO2

Reason : Covalent radius of Se is more than S[AIIMS-2018 (E)]



(c) If assertion is true but reason is false.(d) If both assertion and reason are false.


ANSWER KEY


1. Formal charge helps in selection of lowest energystructure from given various Lewis structures.

2. NaOH3. One

4. (a) (b)

5. A decrease in energy is observed.6. F– is smaller than Cl– and is less polarisable by

Al3+. Hence it is more ionic.7. Octet rule is applied only upto second period

elements of the periodic table. It is not applicablefor(i) incomplete octet of central atom.(ii) odd-electron molecules(iii) explanded octet of central atoms. e.g., XeO4

8. Bond length is defined as equilibrium distancebetween the nuclei of two bonded atoms in amolecule. Bond angle is the angle between the twobonds.

9. (a) (b)

(c) (d)

(c) is correct answer because in this case, theoverlapping will be lateral thereby formed a p bond.

10. 1. Sigma bond is formed by the overlapping alonginternuclear axis , whereas pi bond is formedby the lateral or sideways overlapping.

2. Pi bond can’t be formed by s orbitals whilesigma bond can be formed by s orbitals.

3. Sigma bond is stronger than pi bond.4. Sigma bonds define the hybridisation and

geometry of a molecule while pi bonds have nodirect impact on these.

11. (a) ;

(b) C6H5OH; ;

(c)

12. Chemical bond is defined as electrostatic force ofattraction which holds two or more atoms togetherin a molecule.Chemical bond is of three types(a) Ionic / electrovalent e.g. NaCl(b) Covalent e.g. CH4(c) Co-ordinate e.g. H3N ® BF3

13. The main postulates of VSEPR theory are asfollows :(i) The shape of a molecule depends upon the

number of valence shell electron pairs [bondedor nonbonded) around the central atom.

(ii) Pairs of electrons in the valence shell repelone another since their electron clouds arenegatively charged.

(iii) These pairs of electrons tend to occupy suchpositions in space that minimise repulsion andthus maximise distance between them.

(iv) The valence shell is taken as a sphere with theelectron pairs localising on the sphericalsurface at maximum distance from oneanother.

(v) A multiple bond is treated as if it is a singleelectron pair and the two or three electron pairsof a multiple bond are treated as a single superpair.

(vi) Where two or more resonance structures canrepresent a molecule, the VSEPR model isapplicable to any such structure.

The repulsive interaction of electron pairsdecreases in the order :lone pair (lp) - lone pair (lp) > lone pair (lp) - bondpair (bp) > bond pair (bp) -bond pair (bp)


14. Due electronegativity difference between twobonded atoms, the shared pair of electron is shiftedtowards more electronegative atom. As a result, adipole is created within the bond. This type ofcovalent bond is called polar covalent bond.

15. The intermixing of atomic orbitals of same or nearlysame energy so as to give same number of neworbitals of equivalent energy is known ashybridisation.Salient features of hybridisationThe main features of hybridisation are as under

1. The number of hybrid orbitals is equal to the numberof the atomic orbitals that get hybridised.

2. The hybridised orbitals are always equivalent inenergy and shape.

3. The hybrid orbitals are more effective in formingstable bonds than the pure atomic orbitals.

4. These hybrid orbitals are directed in space in somepreferred direction to have minimum repulsionbetween electron pairs and thus a stable arrangementis obtained. Therefore, the type of hybridisationindicates the geometry of the molecules.

Type of Hybridisation Example

sp CO2, CHCHsp2 BF3, CH2=CH2sp3 CH4, NH3

sp3d PCl5, SF4sp3d2 SF6, ICl6Åsp3d3 IF7, XeF6

16. Covalent.17. This is due to inter molecular H-bonding in H2O

which is missing in H2S.18. Application & Exceptions of Fajan’s Rules :

The properties which are governed by nature ofbond (predominantly covalent or ionic) can beexplained on the basis of Fajan’s rule. Followingare few illustrations(i) Ag2S is less soluble than Ag2O in H2O because

Ag2 S is more covalent due to bigger S2– ion.(ii) Fe(OH)3 is less soluble than Fe(OH)2 in water

because Fe3+ is smaller than Fe2+ and thuscharge is more.Therefore, Fe(OH)3 is more covalent thanFe(OH)2.

(iii) The colour of some compounds can beexplained on the basis of polarisation of theirbigger negative ions.

For example : AgCl is white AgBr, AgI,Ag2CO3 are yellow. Similarly, SnCl2 is whitebut SnI2 is black. PbCl2 is white but PbI2 isyellow.The bigger anions are more polarised andhence their electrons get excited by partialabsorption of visible light.

(iv) Variation of melting point [melting pointof covalent compound < melting point ofionic compound]:BeCl2 , MgCl2 , CaCl2, SrCl2, BaCl2? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??

Ionic character increases, melting pointincreases; since size of cation increases & sizeof anions is constant.CaF2, CaCl2, CaBr2, CaI2? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??

Covalent character increase, melting pointdecrease; since size of anions increase & sizeof cations is constant.

19. Since electronegativity of F is more than O thestrength of H-bonding in HF is more than that inH2O yet the b.p. of water is more. This is due toH2 fact that there are two H-bond donor and twoH-bond acceptor sites in water unhide in HF, onlyone H-bond donor and acceptor (Active) sites exists.

20.

21. In water molecule four H-bonds exists per watermolecular. This one water molecule is linked withfour other molecules with H-bonding, tetrahedrally.Thus giving rise to a cage like structure collapsesin ice. As ice melts, this cage like structure collapsesand the molecules come closer there by makingdensity of water more than ice.At 273 K, sufficient H-bonding is present in water.On heating the H-bonding collapses and moleculescome closer. This happens. upto 277 K. After thistemp. the increase of volume due to expansion ofliquid becomes significant against a decreases involume due to breaking of H-bonding. Thus after277 K, there is a net increase in volume and adecrease in density. Hence density of water is max.at 277 K.


22. PCl3 – sp3 hybridisationPCl5 – sp3dP–Cl bond in PCl3 involves sp3 – p overlap whilethat in PCl5 involves sp3d–p overlap. Hence thebond energy will be different.

23. The electron pair involved in formation of covalentbond is called bond pair while the electron pair whichis not involved in bond formation is called a lonepair. For example in NH3 three electron of N areinvolved in bond formation while one electron pairis not involved in bonding.

24. M.O. configuration of N2Q : s1s2 s*1s2 s2s2

s*2s2 2

2

2

2y

z

p

p

? ??? ?? ??? ?

s2s2 1

0

*2

* 2y

z

p

p

? ??? ?? ??? ?

M.O. configuration of N2Å : s1s2 s*1s2 s2s2

s*2s2 2

2

2

2y

z

p

p

? ??? ?? ??? ?

s2px1

Bond order of N2– =

12 .[10 – 5] = 2.5

Bond order of N2+ =

12 .[9 – 4] = 2.5

Though bond order is same for N2– and N2

+ yetbond energy of N2

+ will be more due to less no. ofelectrons in antibonding molecular orbitalsBond energy N2

+ > N2–

25. µ = q × d1.964 × 10–29 = q × 1.596 × 10–10

or q = 1.2306 × 10–19 columbcharge on one e– = 1.6 × 10–19 columbcharge on one e– = 1.6 × 10–19 columbs

% ionic character = 19

191.2306 10

1.6 10

?

??

? × 100

= 76.82%

26.

The dipole moment in NH3 acts in the direction H®Nand the dipole moment due to L.P. increases thedipole moment of NH3.

In NF3 the direction of dipole moment is N®F andthe dipole moment due to LP partially neutralisesthe bond dipole moments, hence the lower dipolemoment for NF3

27.

o-Hydroxybenzaldehyde shows intramolecular H-bonding while p-Hydroxybenzaldehyde showsintermolecular H-bonding. During melting process.more energy is required to break intermolecular H-bonds additionally, thereby increasing its m.p.

28. I3– is known but F3

– is not , because I has vacantd-orbital and can expand its octet whereas F doesn'thave this facility.

29. Bond order is defined as one half the differencebetween the number of electrons present in bondingand antibonding orbitals.

B.O. = 12 (Nb – Na)

30. When all the properties of any compound can’t beexplained by a single structure and it requires twoor more structures to explain all its properties, thisphenomenon is called. resonance.

31. XeF2 XeF4

XeF6 XeO2F2

XeO4

32. The statement is incorrect. A non polar molecularmay contain polar bonds in it. BF3 is a non polarmolecular but it has three polar B–F bonds at 120º.Dipole moment being and vector quantity is zerofor the molecule.


33. The configuration for He2 is s1s2 s*1s2

Bond order = 12 [2– 2] = 0

Hence He2 molecular is not formed.34. Anion O2

–

M.O. configuration s1s2 s*1s2 s2s2 s*1s2 s2Ox2

2 * 1

2 * 0

2 2

2 2y y

z z

p p

p p

? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?

since it has one unpaired electron,

it is paramagnetic.35. Br2.. It is the only non polar molecule.36. CO2 having sp hybridisation leads to zero dipole

moment. Linear

These two have bent structure, hence non zerodipole moment.

37.

Thus five orbitals (one’s, 3p and one d) are availablefor hybridisation giving trigonal bipyramidalgeometry shown below.

Here three P–Cl bonds are in a plane and have 120ºangle. These bonds are called equitorial bonds. Theother two bonds above and below the equitorialplane and these bonds are called axial bonds. Therebonds are at 90º from equitorial plane.Since the repulsion between axial bond pairs is morethan that in equitorial ones the axial bonds are littlelonger than the equitorial bonds.

38. Coulomb meter.39. Due to bigger size of chlorine atom.40. Sugar is soluble in water as it is capable of forming

hydrogen bond with water.

41. London or dispersion forces are responsible forholding non polar molecule together.

42. The molecule orbital formed by destructiveinterference of electron waves of combining atomicorbitals are called antibonding molecular orbitals.

43. p-Dichlorobenzene44. H2S due to no H-bonding.45. (a) H-bond

(b) H-bond(c) vander Waal’s force(d) H-bond(e) Covalent(f) Metallic(g) Ionic(h) Covalent(i) ionic(j) covalent

46. O2 : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2 (p2p2x= p2p2y ) (p*2px1 = p*2p1y)Bond order = 1/2(10 – 6) = 2.0,ParamagneticO2+ : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2 (p2p2x= p2p2y ) (p*2px1 = p*2p0y)Bond order = 1/2(10 – 5) = 2.5.ParamagneticO2

– : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2 (p2p2x= p2p2y ) (p*2px2 = p*2p1y)Bond order = 1/2(10 – 7) = 1.5.ParamagneticO2

2– : (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2 (p2p2x= p2p2y ) (p*2px2 = p*2p2y)Bond order = 1/2(10 – 8) = 1.Diamagnetic.

47. K(s) + 12 F2(g) KF(s) ; –562.6 kJ mol–1

K(s) K(g) ; 89.6 kJ mol–1

K(g) K+(g) + e– ; 419.0 kJ mol–1

21

F2(g) F (g) ; 12 158.2 = 79.1 kJ mol–1

F(g) + e– F (g) ; – 332.6 kJ mol–1

Let DH lattice (KF) be U , then according to Hess’slaw of constant heat summation.– 562.6 = 89.6 + 419.0 + 79.1 – 332.6 – USo, U = 817.7 kJ mol–1

Ans. Lattice energy of KF = 817.7 kJ mol–1


EXERCISE–1 (CHECK YOUR UNDERSTANDING)1. (c) 2. (c) 3. (a) 4. (b) 5. (b) 6. (b) 7. (a) 8. (b) 9. (b) 10. (c)

11. (d) 12. (a) 13. (c) 14. (c) 15. (a) 16. (a) 17. (b) 18. (b) 19. (c) 20. (c)21. (b) 22. (a) 23. (b) 24. (c) 25. (c) 26. (d) 27. (c) 28. (d) 29. (d) 30. (d)31. (d) 32. (d) 33. (c) 34. (c) 35. (a) 36. (b) 37. (a) 38. (b) 39. (a) 40. (a)41. (c) 42. (d) 43. (a) 44. (d) 45. (c) 46. (c) 47. (a) 48. (d) 49. (b) 50. (a)51. (d) 52. (c) 53. (c) 54. (a) 55. (b) 56. (a) 57. (c) 58. (b) 59. (a) 60. (b)61. (b) 62. (d) 63. (d) 64. (a) 65. (d) 66. (a) 67. (b) 68. (c) 69. (d) 70. (b)71. (b) 72. (d) 73. (c) 74. (c) 75. (d) 76. (a) 77. (c) 78. (b) 79. (d) 80. (c)81. (a) 82. (b) 83. (c) 84. (d) 85. (a) 86. (a) 87. (d) 88. (b) 89. (b) 90. (a)

91. (a) 92. (d) 93. (a) 94. (a) 95. (c) 96. (d) 97. (d) 98. (a) 99. (b) 100. (a)101. (a) 102. (c) 103. (a) 104. (b) 105. (d) 106. (d) 107. (a) 108. (a) 109. (c) 110. (a)111. (d) 112. (a) 113. (b) 114. (c) 115. (c) 116. (a) 117. (c) 118. (c) 119. (c) 120. (c)121. (b) 122. (d) 123. (b) 124. (d) 125. (d)

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)1. (d) 2. (c) 3. (d) 4. (a) 5. (d) 6. (b) 7. (b) 8. (c) 9. (c) 10. (a)

11. (d) 12. (d) 13. (d) 14. (d) 15. (c) 16. (c) 17. (d) 18. (c) 19. (c) 20. (b)

21. (a) 22. (a) 23. (c) 24. (a) 25. (b) 26. (d) 27. (c) 28. (a) 29. (d) 30. (c)

31. (d) 32. (c) 33. (d) 34. (d) 35. (d) 36. (c) 37. (c) 38. (c) 39. (c) 40. (d)

41. (a) 42. (b) 43. (b) 44. (b) 45. (d) 46. (a) 47. (c) 48. (a) 49. (c) 50. (c)

51. (b) 52. (c) 53. (a) 54. (c) 55. (d) 56. (b) 57. (c) 58. (c) 59. (b) 60. (a)

61. (c) 62. (c) 63. (d) 64. (b) 65. (d) 66. (b) 67. (a) 68. (c) 69. (a) 70. (d)

71. (a) 72. (d) 73. (d) 74. (b) 75. (d) 76. (c) 77. (b) 78. (d) 79. (d) 80. (c)

81. (c) 82. (c) 83. (a) 84. (c) 85. (b) 86. (b) 87. (c) 88. (b) 89. (b) 90. (c)

91. (d) 92. (a) 93. (d) 94. (b) 95. (a) 96. (a) 97. (b) 98. (c) 99. (a) 100. (a)

101. (a) 102. (c) 103. (d) 104. (b) 105. (d) 106. (a) 107. (a) 108. (d) 109. (d) 110. (d)

111. (a) 112. (d) 113. (b) 114. (b) 115. (c)

EXERCISE–3 (AIIMS SPECIAL)1. (a) 2. (c) 3. (d) 4. (b) 5. (b) 6. (c) 7. (a) 8. (b) 9. (d) 10. (b)

11. (d) 12. (c) 13. (c) 14. (c) 15. (c) 16. (c) 17. (c) 18. (c) 19. (a) 20. (b)

21. (a) 22. (b) 23. (a) 24. (d) 25. (d) 26. (c) 27. (a) 28. (b) 29. (a) 30. (c)

31. (c) 32. (b) 33. (b) 34. (d) 35. (a) 36. (d) 37. (d) 38. (a) 39. (b) 40. (c)

41. (a) 42. (a) 43. (a) 44. (b) 45. (a) 46. (b) 47. (d) 48. (a) 49. (b) 50. (c)

51. (c) 52. (c) 53. (c) 54. (d) 55. (b) 56. (a) 57. (c) 58. (a) 59. (c) 60. (a)

61. (a) 62. (a) 63. (b) 64. (b) 65. (c) 66. (b) 67. (d) 68. (a) 69. (b) 70. (c)

71. (d) 72. (b) 73. (a) 74. (c) 75. (c)


EXERCIS-4 (ARCHIVES FROM VARIOUS MEDICAL ENTRANCE)1. (a) 2. (d) 3. (d) 4. (c) 5. (a) 6. (d) 7. (a) 8. (c) 9. (d) 10. (c)

11. (b) 12. (d) 13. (c) 14. (c) 15. (a) 16. (c) 17. (c) 18. (c) 19. (a) 20. (b)

21. (d) 22. (a) 23. (c) 24. (a) 25. (a) 26. (b) 27. (c) 28. (d) 29. (d) 30. (d)31. (d) 32. (d) 33. (d) 34. (b) 35. (b) 36. (b) 37. (d) 38. (a) 39. (c) 40. (c)41. (a) 42. (c) 43. (a) 44. (d) 45. (d) 46. (d) 47. (a) 48. (d) 49. (a) 50. (a)51. (a) 52. (c) 53. (b) 54. (b) 55. (b) 56. (b) 57. (d) 58. (d) 59. (c) 60. (b)61. (b) 62. (b) 63. (a) 64. (a) 65. (a) 66. (c) 67. (d) 68. (a) 69. (a) 70. (a)71. (c) 72. (c) 73. (a) 74. (d) 75. (c) 76. (a) 77. (d) 78. (c) 79. (d) 80. (d)81. (b) 82. (d) 83. (a) 84. (a) 85. (d) 86. (d) 87. (d) 88. (a) 89. (a) 90. (a)91. (a) 92. (b) 93. (c) 94. (b) 95. (b) 96. (d) 97. (b) 98. (d) 99. (b) 100. (a)

101. (b) 102. (c) 103. (b) 104. (b) 105. (a) 106. (d) 107. (b) 108. (c) 109. (c) 110. (a)111. (d) 112. (c) 113. (d) 114. (d) 115. (a) 116. (d) 117. (a) 118. (c) 119. (a) 120. (d)121. (c) 122. (b) 123. (a) 124. (a) 125. (c) 126. (d) 127. (c) 128. (d) 129. (a) 130. (b)

ISOMERISM (CHEMISTRY) 2.1

ISOMERISM

Chapter

2INTRODUCTION

Two or more than two organic compounds havingthe same molecular formula and molecular weight but

different physical and chemical properties are calledisomers and the phenomenon is called isomerism.

ISOMERISM

Structural Isomerism Stereoisomerism

Geometrical

ConfigurationalChain Position Ring Chain Functional Metamerism Tautomerism

Conformational

Optical

1. STRUCTRAL ISOMERISM1.1 CHAIN ISOMERISM :

The compounds which have same molecular formula, samefunctional group, same position of functional group ormultiple bond or substituent but different arrangement ofcarbon chain (different parent name of compound) showschain isomerism.

EXAMPLE 1 :

CH3— CH2— CH2— CH3CH3 CH CH3

CH3

1–Butane (4C) 2–Methyl propane (3C)

EXAMPLE 2 :CH3— CH2— CH == CH2 CH2 C CH3

CH 3 Butene (4C) 2–Methyl-1-propene (3C)

EXAMPLE 3 :CH3— CH2— CH2— OH CH 3 CH CH2 OH

CH 21–Butanol (4C) 2–Methyl–1–propanol (3C)

EXAMPLE 4 :

2-Methylbutane 2-Methylbutane

CH—CH2—CH , CH —CH—CH CH3 3 2 3|CH2

||CH3

|

CH3|

Identical compounds(not isomers)

1.2 POSITION ISOMERISM :The compounds which have same molecular formula, samefunctional group, same parent carbon chain but differentposition of functional group or multiple bond orsubstituents, shows position isomerism.

EXAMPLE 5 :CH2 == CH — CH2 CH3 CH3 CH == CH CH3

But–1–ene But–2–ene

EXAMPLE 6:CH3 CH2 CH2 CH2 OH

1–Butanol

|OH

CH — CH — CH—CH3 2 3

2–Butanol

2.2 ISOMERISM (CHEMISTRY)

EXAMPLE 7:CH3 CH2 CH2 CH2 Cl

1–Chlorobutane

|Cl

CH — CH — CH—CH3 2 3

2–Chlorobutane

Example of Chain and Position Isomers :(i) C4H10 have two isomers : Both butane and isobutaneare chain isomers.CH3 CH2 CH2 CH3 CH — CH— CH3 3

|CH3

Butane Isobutane(ii) C5H12 have three isomers : All of three structuresare chain isomers because only carbon chain (parent) isdifferent.

CH3 CH2 CH2 CH2 CH3 Pentane

|Cl3

CH — CH — CH— CH3 2 3

2–Methyl butane

CH — C— CH3 3 |CH3

|CH3

CH — C— CH 3 3|CH3

2,2–Dimethylpropane

(iii) C6H14 has 5 isomers

(a) CH3CH2CH2CH2CH2CH3 Hexane

(b) CH — CH — 3 2 2 3 CH — CH— CH |CH3

2–Methyl pentane

(c) CH — CH — CH — 3 2 2 3 CH— CH |CH3

3–Methyl pentane

(d) CH — CH — — 3 3 CH CH |CH3

|CH3

2,3–Dimethyl butane

(e)|CH3

|CH3

H — C— CH 3 2 3C — CH|CH3

2,2–Dimethyl butane

• a – b, b – d, a – c, c – d Chain Isomers• b – c, d – e Position Isomers

(iv) C7H16 has 9 isomersCH — CH — CH — CH — CH — CH — CH 3 2 2 2 2 2 3

CH — CH — CH — CH — CH— CH3 2 2 2 3

CH — CH — CH — C— CH3 2 2 3

CH — CH — C— CH — CH3 2 2 3

CH — CH — CH — CH— CH — CH3 2 2 2 3

CH — CH— CH— CH — CH3 2 3

CH — CH— CH — CH— CH3 2 3

CH — CH — CH— CH — CH3 2 2 3

CH — CH— C— CH3 3

1.

2.

3.

4.

5.

6.

7.

8.

9.

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3

|CH3 |

CH3

|CH2 — CH3

|CH3

Heptane

2–Methylhexane

3–Methylhexane

2,2–Dimethyl pentane

3,3-Dimethylpentane

2,3–Dimethylpentane

2,4–Dimethylpentane

3–Ethylpentane

2,2,3–Trimethylbutane


(v) C3H6Cl2 has 4 isomers : Position of chlorine atomis different in all the structure, so these are positionIsomers.

1. H C— CH — CH — Cl3 2

|Cl

1.1-Dichloropropane

2. H C— CH — CH — Cl3 2 2

|Cl

1.3-Dichloropropane

3.|

|Cl

H — C 3 3C — CH|Cl

2.2-Dichloropropane

4.|

|Cl

H — C 3 3C H— CH|Cl

1.2-Dichloropropane

(vi) C8H10 has 4 aromotic isomers

CH3

CH3 (o,m,p)

CH — CH2 3

1.3 RING CHAIN ISOMERISM (RCI)

Same molecular formula but different mode of linking(open chain or closed chain) of carbon atoms.

C H3 4CH2

CH2 CH2

— [open chain]

[closed chain or ring]

They have same molecular formula so they are Ringchain isomers.

EXAMPLE 8:

Relate a, b and c:– C H3 4

CH — C CH3

CH

CH2 CH

(a)

CH 2(b)

(c)

—— C —— CH 2

SOLUTION:a – b Functional Isomersa – c, b – c Ring-chain Isomers, FunctionalIsomers

Special points :(a) Alkenes with cycloalkane and alkynes(Alkadienes) with cycloalkenes show Ring-chainIsomerism.(b) Ring-chain Isomers are also Functional Isomers butpriority is given to Ring-chain Isomers.

EXAMPLE 9:CH3—CH CH—CH3 and CH2 CH2

CH2 CH2

are :–

(1) Ring-chain Isomers(2) Chain Isomers(3) Functional Isomers(4) Position Isomers

Ans. (1)OR

(1) Functional Isomers(2) Position Isomers(3) Chain Isomers(4) Metamerism

Ans. (1)OR

(1) Functional Isomers(2) Position Isomers(3) Ring-chain Isomers(4) 1 and 3 Both

Ans. (4)

EXAMPLE 10:Relate structures a, b, c and d.(a) CH3—CH2—C CH (b) CH2 C CH—CH3

(c) —

CH3

(d)

a, b. Functional Isomersa, c. Ring-chain Isomers and Functional Isomersb, c. Ring-chain Isomers and Functional Isomersa, d. Ring-chain Isomers and Functional Isomersc, d. Chain Isomersb, d. Ring-chain Isomers and Functional Isomers

EXAMPLE 11:

CH — CH — CH — CH3 2 2 2

5 4 3 2

|1CN

and

CH — CH — CH — CH3 2 3

4 3 2

|1CN

are ?


SOLUTION:Molecular formula same, Functional group same, positionof Functional group same but different parent carbon atomchain so both are Chain isomers

EXAMPLE 12 :How many minimum carbons required for Chain isomerismand Position isomerism in alkanes ?

SOLUTION:4, 6 ( Do it yourself)

EXAMPLE 13:How many minimum carbons required for Chain isomerismand Position isomerism in alkenes ?


EXAMPLE 14:How many minimum carbons required for Chain isomerismand Position isomerism in alkynes ?


1.4 FUNCTIONAL ISOMERISM

Same molecular formula but different functional groups.Following compounds show Functional isomerism, as theyhave same molecular formula and different functionalgroup.(i) Alcohol and ether

Ex. CH3—CH2—OH and CH3—O—CH3

(ii) Aldehydes and ketones

Ex. CH3 CH2 C H

O

and CH3 C CH3

O(iii) Acids and ester

Ex. CCH3 OH

O

and CH O

O

CH3

(iv) Cyanide and isocyanideEx. CH3—CH2—CH2—CN andCH3—CH2—CH2—NC

(v) Nitro and Nitrite

Ex. CH2 CH3 NO

O and CH3—CH2—O—N O

(vi) Keto and enol

Ex. CCH3 CH3

O

and CCH2

OH

CH3

(vii) Amide and Oxime

Ex. CCH3 NH2

O

and CH3—CH NOH

(viii) 1°, 2°, 3° aminesEx. (i) CH2—CH2—CH2—NH2

(ii) CH3—NH—CH2—CH3

(iii)NCH3 CH3

CH3

(ix) Alcoholic and phenolic compounds :

Ex. CH OH2

and

OHCH3

(x) Alkyl halides do not show Functional isomerism.

EXAMPLE 15:C4H10O Isomers [4 alcohol and 3 ethers] total 7 isomersare possible.Alcohol : CH3—CH2—CH2—CH2—OH and

CHCH3 CH2 CH3

OH

CHCH3 CH2 OH

CH3

and

CCH3 OH

CH3

CH3

Ethers : CH3—O—CH2—CH2—CH3,C2H5—O—C2H5,

OCH3 CH CH3

CH3

EXAMPLE 16:C5H12O Isomers [8 alcohols and 6 ethers] total 14 isomersare possible.

(i) CH3—O—CH2—CH2—CH2—CH3

(ii) OCH3 CH CH2 CH3

CH3


(iii) OCH3 CHCH2 CH3

CH3

(iv) C2H5—O—C3H7

(v) OCH3 C CH3

CH3

CH3

(vi) CH3 CH2 O CH CH3

CH3

EXAMPLE 17:C4H10O (2 aldehyde and 1 ketone) total 3 isomers arepossible.

(i) CH3 CH2 CH2 HC

O

(ii) CH3 CH

CH3

C H

O

(iii) CH3 C

O

CH2 CH3

EXAMPLE 18:C4H8O2 (2 acids and 4 esters) total 6 isomers are possible.

(i) CH3 CH2 CH2 C

O

OH

(ii) CH3 C

O

OC H2 5

(iii) CH3 CH

CH3

C OH

O

(iv) C H2 5 C OCH3

O

(v) H C OC H3 7

O

(vi) C O

O

H CH CH3

CH3

1.5 METAMERISMSame molecular formula, same polyvalent Functional groupbut different alkyl groups attached to polyvalent Functionalgroup.

Polyvalent Functional group [More than one valency]are :—O— , —S—, C

O O

O C

O

NH, , ,

—NH—,

C

O

OC

O

C

O

N,N

Both are metamers.

EXAMPLE 19:CH3—O—CH2—CH2—CH3 ; OCH3 CH CH3

CH3

EXAMPLE 20:

CH3—O—CH2—CH2—CH3 ; OCH3 CH CH3

CH3

1–Methoxypropane 2–Methoxypropane

Important: (i) Both are metamers and Position isomerbut priority is given to metamerism.(ii) Metamers may be Position isomer or Chain isomerbut priority is given to metamerism.

EXAMPLE 21:CH3—CH2—NH—CH2—CH3

N–Ethyl ethanamine

CH3—NH—CH2—CH2—CH3 N–Methyl propanamine

They are only metamers not Chain Isomers

EXAMPLE 22:CH3—NH—CH2—CH2—CH2—CH3

CH —NH—CH —CH—CH3 2 3 |CH3

Both are metamers and Chain isomer

EXAMPLE 23:CH —C— CH —CH3 2 3 CH —CH —2 2

|O|

2–Hexanone


CH —CH —C —CH —CH3 2 2 2 —CH3|O|

3–Hexanone

Both are metamers and Position isomer

EXAMPLE 24:CH —C—CH —CH —CH —CH3 2 2 2 3

|O|

2–Hexanone

CH —CH —C—CH—CH3 2 3 |CH3

|O|

2–Methyl–3–pentanone

Both are Metamers, Chain and Position isomers

EXAMPLE 25:

Structures CH — C— C H3 6 5 O—|O|

and

C H CH6 5 3 — C—O—|O|

are

Ans. Both are metamers.

1.6 TAUTOMERISM OR DESMOTROPISM :Tautomerism was introduced by “Laar”. It's also calleddesmotropism.• Desmotroism means bond turning. [Desmos = Bond;Tropos = Turn]• Tautomers have same molecular formula but differentstructural formula due to wandering nature of activehydrogen between two atoms.

CH — C— H2

-Hydrogen or active H

-Hydrogen or active H|H

|O|

EXAMPLE 26:

H— C— C —H|H

|O|

|H

H— C— C—H |O—H

|H

—

EXAMPLE 27:CH— C— CH3 2|

O| |

H

CH— C CH3 2|O—H

keto ene + ol = enol

Note : (1) Tautomers are also F. I. and exist in dynamicequilibrium is used to show tautomerism

(2) By shifting of H–atom bond also change itsposition.

Condition for Tautomerism :

(a) For carbonyl compounds :- Carbonyl compoundshaving atleast one –H show tautomerism

(i) CH — C— H3 |O|

3 H, show tautomerism.

(ii) CH — C— CH3 3 |O|

6 H, show tautomerism

(iii) CH — CH— C—H3 |CH3

|O|

1 H, show tautomerism

(iv) H— C — H |O|

No H, No tautomerism

(v) C—H|O|

No H, No Tautomerism

(vi)C—CH3

|O|

3 H, shows tautomerism (Aceto

phenone)

(vii) Ph C

O

Ph No H, No tautomerism

(Benzophenone)

(viii) Ph C

O

CH2 C

O

Ph 2 H, shows tautomerism

(ix) H

O

HHH 4H, shows tautomerism

(x)

HO

H

H

HO

-H, attached sp2 carbon does not take

part in tautomerism


EXAMPLE 28:Which of the following show keto enol tautomerism.

(1) O

O

(2)

OHH

O

(3)

O

O

(4) CH OHCH

Ans. (1), (2) and (4)(b) For nitro compounds : Nitro compounds having

atleast one a – H show tautomerism

CH2

O

OH

CH2 NO

OHN

Nitro form Aci nitro form(acidic form so soluble in base)

Note : Nitro compounds with at least one a–H aresoluble in NaOH.

EXAMPLE 29:Which of the following show tautomerism.

(1) CH3CH2—NO2 (2) (CH3)2CH—NO2(3) (CH3)3C—NO2 (4) Ph—CH2—NO2

Ans. 1, 2 and 4

EXAMPLE 30:Which of the following do not soluble in NaOH.

(1) CH3CH2—NO2 (2) (CH3)2CH—NO2(3) (CH3)3C—NO2 (4) Ph—CH2—NO2

Ans. Only 3(c) H—CN and H—N C are tautomers [alsoFunctional isomers] while R—CN and R—N C areonly Functional isomers.

H—C N C N—HActive H

(d) H NO

O and H—O—N O are tautomers.

Important : Enol Content

1. CH2 C H

OH

CH2 C H

OH “keto” ( 99%) “enol” (1%)

2. CH3 C CH2

O H

CH3 C CH2

OH 99% 1%

Less stable or unstable

3. HO

OHSP2

“keto” ( 1%) “enol” (stable by resonance and aromatic nature) ( 99%)

4. CCH3 CH2 C OC H2 5

OOCCH3 CH C

OOH

OC H2 5

Aceto acetic esteraqueous sloution

liquid state

keto : enol = 93 : 7

keto : enol = 25 : 75AAE

(a) In aqueous solution

CCH3 CH2 C O C H2 5

OO

OH H OH H

• Keto from is stablised by intermolecular H–Bonding.(b) In liquid state

CCH3 CH C OC H2 5

OO H• Enol form is stabilised by intramolecular H–Bonding.

(1) Enol content number of >C O group.(2) If number of >C O groups are equal then

proportional to number of –H.(3) Group —OH attached to sp2 carbon or double bond

is less stable or unstable.(4) More active H, more take part in tautomerism.(5) Stability of enol form depnds on (i) Resonance and

(ii) H – Bond.

EXAMPLE 31:Arrange the following in correct order of enol content.

1. CH3 C H

O

2. CH3 C

O

CH3


3. CCH3 CH2 C H

OO

4. CCH3 CH2 C

OO

CH3

Ans. 1 < 2 < 3 < 4

EXAMPLE 32:Which have maximum stable enol form

(1)H

O OH

(2)

OHH

OH

(3)

O

O

OH

OH

(4)O

OH

Ans. 4. Stable by Resonance and is Aromatic

OTHER EXAMPLES

EXAMPLE 33:Aromatic isomers of C7H8O.

(a)

CH2 OH

(b)

CH3

OH

(c)

CH3

OH(d)

OH

CH3

(e)

O CH3

• a, b – Functional isomers• b, c – Position isomers• c, d – Position isomers

• a, d – Functional isomers• a, e – Functional isomersNote : Alcoholic and phenolic groups are Functional

isomers.

EXAMPLE 34:

CH3 CH2 CH2 C

O

OH and

CH3 CH2 CH HC

O

OH

are

(a) Position isomers(b) Functional isomers(c) Geometrical isomers(d) Chain isomers

Ans. (b) Functional group different so Functional isomers.Note : If Functional group is changed then they will benever Chain, Position isomer and metamers.

EXAMPLE 35:CH3—S—CH2—CH3 and CH3—CH2—S—CH3 are –Ans. Identical

EXAMPLE 36:

CH3 OC

O

C C H6 5

O and

C H6 5 OC

O

C

O

CH3 are –

Ans. Identical

EXAMPLE 37:Relationship between the given compounds-

(a) (i) CH3 — CH2—CH2 — CH3 Butane

Size of main chain = 4Size of side chain = 0

(ii) CH CH CH3— — 3

CH3

2–MethylpropaneSize of main chain = 3Size of side chain = 1

Structure (i) & (ii) are chain isomers.


(b) (i) 1–Ethylcyclohexane

Size of main chain = 6 Size of side chain = 2

(ii) 1,4–Dimethylcyclohexane

Size of side chain = 6 Size of side chain 1 = 1 Size of side chain 2 = 1Structure (i) & (ii) are chain isomers.

(c) (i) Cyclohexane

Size of main chain = 6Size of side chain = 0

(ii) 1,2,3–Trimethylcyclopropane1,2,3–Trimethylcyclopropane

Size of main chain = 3Size of side chain 1 = 1Size of side chain 2 = 1Size of side chain 3 = 1

Structure (i) & (ii) are chain isomers.

(d) 3 2 2

3 2

(but 1 ene)

(but 2 ene)

H C CH CH CHposition isomers

H C CH CH CH

(e) 2 2 3

3 2 3

(pent 1 yne)

(pent 2 yne)

HC C CH CH CHposition isomers

H C C C CH CH

(f) (i) CH3—CH2OH (ii) CH3—O—CH3Ethanol MethoxymethaneFunctional groups –OH Function groups – O –

Structure (i) and (ii) are functional isomers.

(g) (i) CH —C—OH3

O

Ethanoic acid

Functional groups — COOH

(ii) H—C—OCH3

O

Methyl methanoate

Functional groups O

—C—O—

Structure (i) and (ii) are functional isomers.

(h) (i) C2H5—O—C2H5

Diethyl ether

Hydrocarbon groups —C2H5, —C2H5

(ii) C3H7—O—CH3

Methyl propyl ether

Hydrocarbon groups —C3H7, —CH3

Structure (i) & (ii) are metamers.

(i) (i)O

CH —C—CH3 3 (ketone)

(ii)OH

CH C—CH2 3—— (enol)

Structure (i) and (ii) are tautomers.

2. STEREO ISOMERISMTwo or more than two compounds having same molecularformula, same structural formula but differentarrangements of atoms or groups in space.

2.1 CONFIGURATIONAL ISOMERISM

Stereo isomers which have following characteristics.(i) Stereo isomer which cannot interconvert in each

other at room temperature due to restricted rotationknown as Geometrical isomerism.

(ii) Stereo isomer which cannot super impose on eachother due to chirality know as optical isomerism.

(i) Geometrical isomerism (G. I) : Alkenes (>C C<),oximes (>C N—) and azo compounds[—N N—], show G. I. due to restricted rotation aboutdouble bond while cycloalkanes show G. I. due to restrictedrotation about single bond.

G.I. in Alkenes :Reason :- Restricted rotation about double bond : It isdue to overlapping of p–orbital.

CH3

H

CH3

H


EXAMPLE 38:

CH3

HC C

CH3

H

CH3

HC C

CH3

H

cis–2–butene trans–2–butene

Condition for Geometrical isomerism :

1. Only those alkenes show G. I. in which "Each sp2

carbon have different atoms or groups"

a

bC C

a

bC C

a

b

x

y

Geometrical isomers Geometrical isomers

C Ca

a

y

y C

C

p p

q q

Not Geometrical isomers Not Geometrical isomers

EXAMPLE 39:Which of the following show Geometrical isomerism–

(1) 1,1–diphenyl–1–butene(2) 1,1–diphenyl–2–butene(3) 2,3–dimethyl–2–butene(4) 3-phenyl–1–butene

Ans. (2)

NOMENCLATURE SYSTEMS OF GEOMETRICALISOMERS :(a) Cis–Trans System : If same groups at same sidethen cis and if same groups at different side then trans.

C Ca

b

a

bC C

a b

ab

[Same groups, same side] cis [Same groups different side] trans

EXAMPLE 40:

C Ca

y

a

xC C

z

y

z

x

cis cis

EXAMPLE 41:

C CCH3

CH2

H

H CH3

trans-2–pentene

C

C

Cl

Br

Cl

Br

It does not show Geometrical isomersSo no cis–trans

EXAMPLE 42:Isomers of C2H2Cl2

(i)

(ii)

(iii)

(ii) and (iii) are geometrical isomersProperties of Cis–Trans Isomers :(i) Stability — cis < trans

[less repulsion betweensimilar groups in trans]

(ii) Dipole moment — cis > trans(iii) Polarity — cis > trans

[more , more polarity)(iv) Solubility — cis > trans

[because cis > trans ](v) Boiling point — cis > trans

[ionic character](vi) Melting point — cis < trans

[more packing capacityof trans.]

Dipole moment [] :

EXAMPLE 43:

CH3 C H

CH3 C H

H C CH3

CH3 C H

c is 0 trans = Zero

EXAMPLE 44:

CH3 C H

CH2C

H CH3

CH3 C H

CCl H

trans 0 cis 0


EXAMPLE 45:

CH3 C H

C ClH

trans 0

Question : If dipole moment of chlorobenzene is m, then

dipole moment of

Cl

ClCl is –

Ans. Zero(b) Syn/Anti naming : This is applicable to C=N, N=N.If H-atom of carbon and H/OH/NH2 of –N are at sameside of double bond then known as syn isomer and if atopposite side, known as anti isomer.

C = NCH3

H

OH

..anti

C = NCH3

H OH

..

syn

(c) E – Z System:E (Entgegen) : When high priority groups are

opposite side.Z (Zussaman) : When high priority groups are same

side.

C CLP

HP

HP

LPC C

HP

LP

HP

LP

'E' 'Z'

HP – High priority and LP – Low priority

Priority Rules :

Rule I : Priority is proportional to atomic number of atomwhich is directly attached to sp2 carbon.

C C

F

Cl [HP]

Br

[HP] I C C

[HP] Cl CH3

N H2 [HP]C

ClClCl

'Z' 'E'Rule II : If rule-I is failed then consider next atom

C C[HP] F

CH3

[LP] CH H C3 2C CH3

H

CCH3

CH3

CH3

[LP]

[C, C, H]

[HP]

[C, C, C]

'Z'

decreasing order of atomic numberRule III :– If multiple bond is present then consider themas :-

C C C C

C C

C C C C

C C

C C

C OCO

OC N C

NN

N]

N CC C

C

C

NH

H

H

OH

O

O [HP]

[O, O, O]

[HP]

[O, O, H]

'Z'

2

2 6 6

3 3 2 3 2

Priority order for some group is :Br > Cl > OH > NH COOH CHOCH OH CN C H C CHC(CH ) CH CH CH(CH )

Rule IV : If isotopes are present then consider atomicweight.

C CD

H CH3

CH2 CH3[HP] [HP]

'Z'Ex. To decide the seniority among –CCH, –CH = CH2, –CH2–CH3l their hypothetical equivalents are compared.

C C| |

HC C| |C C

(for C C H)

>

)CHCHfor( 2

HH||

HCC||CC

>

)3CH–2CHfor(HH||

HCC||HH

EXAMPLE 46:

C CCl

HCH3

CH3[HP] [HP]

'Z'

EXAMPLE 47:

C CH

CH3

Cl

CH3

'E'


EXAMPLE 48:

C CCH Cl2CH2

COOH

CH3

CH C

[Cl, H, H]

[O, O, O]

[HP]

[LP]

'E'

EXAMPLE 49:

C CCH I2

N C

CCH3

CH3 CH3

CBr3 [Br, ]Br, Br

[I, H, H]

'E'

GEOMETRICAL ISOMERS IN OXIMES [>C = N] :• Oximes show G. I. due to restricted rotation about

double bond.• Only those oximes show Geometrical isomerism in

which sp2 carbon have two different groups.[CH3—CH O + H2N—OH]

CH3—CH N—OH (oxime)

EXAMPLE 50:Acetaldoximes has two Geometrical isomers –

CH3 C H

N OH syn

When H and OH are on the same side

CH3 C H

NHO

anti

When H and OH are on the opposite side

EXAMPLE 51:Ph—CH N—OH Benzaldoxime

Ph C H

N OH [syn.]

Ph C H

NHO [Anti]


(1) CH3—CH2—CH N—OH(2) H2C N—OH

(3) CH3 C

N OH

CH3

(4) CH3 C

N OH

CH CH2 3

Ans. (1), (4)GEOMETRICAL ISOMERS IN AZO COMPOUNDS :( N

..N..

)

N

N

Ph

Ph

(syn)

N

N

Ph

(Anti)

Ph

Ph—N N—Ph (Azo benzene)GEOMETRICAL ISOMERS IN CYCLOALKANES :Cycloalkanes show Geometrical isomers due to restrictedrotation about single bond. Only those cyclo alkanes showGeometrical isomers in which atleast two different carbonshave two different groups.

Me

H

Me

H

Me

H

Me

H

Me

H Me

H

Two Geometrical isomers cis

trans


(1) HClH

Cl(2) Cl

HClH

H Cl

(3) Cl

Cl Br

Br

HH

(4)

Me

Me

Me

Me

Ans. (1), (2) and (4)

NUMBER OF GEOMETRICAL ISOMERS INPOLYENES :R1—CH CH—CH CH .................... CH CH—R2

(a) If R1 R2 then number of Geometrical isomers = 2n

[n = number of double bonds.]


EXAMPLE 54:CH3—CH CH—CH CH—CH CH—CH2CH3

As n = 3r number of Geometrical isomers = 23 = 8(b) If R1 = R2 then number of Geometrical isomers =2n – 1 + 2p – 1

where 2np ? (when n is even)

and 1

2

np (n is odd)

EXAMPLE 55:CH3—CH CH—CH CH—CH CH—CH3 [n = 3]Number of Geometrical isomers = 22 + 21

[p = 3 1

2

] = 4 + 2 = 6

Special Point : CH2 CH—CH CH2

[13–butadiene] show Geometrical isomerism due toresonance.CH2 CH—CH CH2

CHCH 2 CH CH 2

C C

CH 2

HCH 2

HC C

HCH 2

CH 2

Hcis trans.

–

–

–

(ii) Optical isomerism :Optical isomers : Two or more than two compound havesame molecular formula, same structural formula butdifferent direction of rotation of PPL (plane polarised light).

All the symmetrical molecules can not changed thevibrational plane of PPL, hence these compounds areoptically inactive. All dissymmetrical compound willchange the vibrational plane of PPL, hence thesecompounds are optically active.

Non polarised Nicol prism Polarised light

Optical activity : Tendency to rotate plane of PPL in aparticular direction.If a compound rotates plane of PPL in clockwise directionthen it will be dexterorotatory or d or (+) and if acompound rotates plane of PPL in anticlockwise directionthen it will be Laevorotatory or l or (–).

PPL Polarimeter Dextro rotatory

Laevo rotatory Optically inactive

Specific Rotation []:Specific rotation is the number of degrees of rotation

observed if a 1 dm tube is used and the compound beingexamined is present to the extent of 1g/ml.'

[]t =

c here t – temperature (20 – 25°C),

= wavelength light used (D line of sodium = 5893 A°)Note : Specific rotation does not depend upon l & C butchanges with respect to wavelength of light (l) &temperature (t).

EXAMPLE 56:Observed angle of rotation +70° for 2g/ml solution ofcompound X in a sample tube of 1 dm length. Find thespecific rotation of compound X.

[] = 70

2 1

= + 35°

Condition for optical activity : "Molecule should bechiral".

Chiral object : Chiral = unsymmetrical whichcannot devided into two equal parts.

Symmetrical or achiral Mirror image of an achiralPlane of symmetry object can superimpose

on that object

Chiral carbon atom :When a carbon atom is linked to 4 diffrent groups it

is said to be chiral carbon atom.


EXAMPLE 57:Which of the following doesnot have chiral carbon atom.

(A*) (B*)

(C) (D*)

Important points :(1) If a compound is having only 1 chiral C–atom

then it is always optically active.(2) If a compound have more than one chiral carbon

then it may or may not be optically active.(3) Optically active compounds are non-

superimposable.(4) Non-superimposable mirror images are said to be

enantiomers of each other.

mirror

(5) Presence of chiral centres is not the criteria foroptically active compound but chirality is responsible foroptical activity. That is absence of both element ofsymmetry (POS & COS).

Plane of symmetry (POS) :It is an imaginary plane which bisects the molecule in

two equal parts in such a way that one half of the moleculeis mirror image of other half.

1. 2.

3. 4.

Centre of symmetry (COS) :

It is a point of rotary where if we draw the lines inopposite direction & these lines meet to the same group atsame distance.

1.

2.

EXAMPLE 58:Which of the following is/are optically active or opticallyinactive ?

(i) (ii)

Neither plane of symmetry Optically active Nor, centre of symmetry

Optically active

(iii) (iv)

Optically active

(v)

(vi)

Note (1) : Any compound having 1 chiral centre willbe optically active except alkylated ammonia oralkylated carbanion derivative.

(i)

(A)Enantiomers (cancel the optical rotation)

(ii)

(B)Enantiomers (cancel the optical rotation)


Reason : Compound A & B are not optically activeinspite of having 1 chiral centre because of rapidinterchanging into (conversion) its enantiomers with thefrequency of 2.3 × 1010 hertz and cancel out the opticalrotation.

Projection formula in optical isomerism :(I) Wedge-dash projection formula :

It is a convenient way of depicting three dimensionalstructure in two dimension. In this projection four bondsof a tetrahedral molecule is shown by two lines (in theplane) one wedge (up the plane) and one dash (down theplane).

(II) Fischer projection formula :It is also a convenient way of depicting three

dimensional structure in two dimension.Rules for writing Fischer projection formula :

(i) The structure is drawn in the form of cross (+) withthe chiral carbon at the intersection of horizontal &vertical lines.

(ii) On vertical line, main chain is taken with first carbonat the top.

(iii) The horizontal lines represent the bonds directedtowards the viewer and vertical lines represents awayfrom the viewer.

(a)

OHOH||

CHOHCCH*

2 (glyceraldehyde can be

represented in two different Fischer projection as)

(b)

2

3

NH|

COOHHCCH * (Alanine can be represented

in two different Fisher projections as)

EXAMPLE 59:Draw the wedge–dash projection formula of :

(A) Cl H

CF3

CH3

(B) Me OH

NHMe

Ph

SOLUTION:

(A) = C

CH3Cl

CF3

H (B) = C

PhMe

NHMe

OH

Configurational nomenclature in optical isomers :(I) D - L System (Relative configuration) :

This method is used to relate the configuration ofsugars and amino acids by the help of enantiomers ofglyceraldehyde. The configuration of (+)-glyceraldehydehas been assigned as D and the compounds with the samerelative configuration are also assigned as D, & those with(-) glyceraldehyde are assigned as L.

EXAMPLE 60:

Sugars have several asymmetric carbons. A sugarwhose highest numbered chiral centre (the penultimatecarbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designatedas D-sugar, one whose highest numbered chiral centrehas the same configuration as L-glyceraldehyde isdesignated as L-sugar.


(II) R and S configurations in Fischer projection :(absolute configuration)

Rule I : The priorities of groups which are attachedwith the asymmetric C-atom are assigned by sequencerule (CIP rule).

Rule II : The lowest priority group is brought to thebottom of Fischer projection by two or even simultaneousexchanges.

Rule III : Then an arrow is drawn from first prioritygroup to second priority group & then to third prioritygroup. If the arrow is clockwise the configuration assignedto the projection is R & If it is anticlockwise theconfiguration assigned is S.

(i)

(ii)

(iii)

(iv) º

EXAMPLE 61:Assign R or S configuration to each of the followingcompounds.

(A) H OH

CHO

CH OH2

C (B) Br COOH

Cl

H

C

(C) H C5 2 H

OH

CH3

C (D)

Ans. (A) R (B) R (C) R (D) 2 R, 3S

R and S configuration in wedge-dash formula :Step 1 : Decide the priority of groups by sequence

rule.Step 2 : Shift the lowest prior group to dash by even

simultaneous exchanges.Step 3 : Draw an arrow from first prior group to

third prior group via second prior group.Step 4 : If the direction of arrow is clockwise the

configuration is R and if anticlockwise it is S.Step 5 : Draw the Fischer projection formula having

equivalent configuration to the wedge-dash formula.

EXAMPLE 62:

(a)

Here the lowest priority group is already on dash, sothere is no need for exchanges.


(b)

A. Enantiomers :

Stereoisomers which are non-superimposable mirrorimages of each other are called enantiomers.

EXAMPLE 63:In the following molecule there is no centre of symmetryor plane of symmetry. Therefore its two mirror imageisomers are non-superimpossable and hence enantiomers.

COOH

OH

CH3

H

COOH

CH3

HHO

object (I) mirror image (II)

EXAMPLE 64:In the following molecule there is no centre of symmetryor plane of symmetry. Therefore its two mirror imageisomers are non-superimpossable and hence enantiomers.

Enantiomers are of two types : (a) Dextrorotatory(d or +) (b) Laevorotatory (l or – )

Properties of enantiomers :-

(1) Enantiomers are always mirror image isomers.(2) They have same molecular formula, structural

formula but have different orientation in space.(3) They have dissymmetry/chirality.(4) Every enantiomer is optically active.(5) The two enantiomers can rotate the plane-polarised

light with equal magnitude and opposite signs.(6) Enantiomers can be distinguished only by polarimeter.(7) They have similar physical properties except the sign

of optical rotation.(8) Enantomers cannot be separated by physical metthod

of separation

B. Racemic mixture :

A homogeneous mixture of equal amounts ofenantiomers is called a racemic mixture or racemicmodification. A racemic modification is always opticallyinactive. When enantiomers are mixed together, the rotationcaused by the molecules of one enantiomer is exactlycancelled by an equal and opposite rotation caused by themolecules of its another enantiomer.

d + equimolar

Net rotation zero

[External compensation]

The prefix (±) is used to specify the racemic natureof the particular sample. e.g. (±) Lactic acid, (±) Alanine.

C. Meso compound :

A meso compound is one whose molecules aresuperimposable on their mirror images even though theycontain chiral centres.Properties of meso compound :-

(1) The optical stereoisomers which have more than oneasymmetric carbon atoms but have a plane ofsymmetry are called meso compound.

(2) They are achiral (optical rotation = 0).(3) They have [] = 0 due to internal compensation

of optical rotation.

(4) They are diastereomer of d – pair. So, it has differentphysical properties than d – -pair..

(5) They are non resolvable.

COOH

OH H

COOH

OH H

COOH

OH H

COOH

OH H

Plane of symmetry Achiral molecule

Let one chiral C rotates PPL from OH to COOH,

then net rotation zero. (due to internal compensation)

optically inactive

EXAMPLE 65:Consider the stereoisomers of 2, 3-Butanediol


In all the possible isomers I & II are enantiomers. ButIII & IV are not enantiomers since they have plane ofsymmetry they are superimposable to each other (allsymmetrical compounds are superimposable to their mirrorimages). Thus III & IV are identical & meso compounds.

So total stereoisomers of 2, 3-butanediol is 3 (twoenantiomers and one meso isomer).

D. Optical diastereomers :The optical isomers which are neither mirror image

nor superimposable to each other are called diastereomers.Diastereomers have different physical and chemicalproperties and they can be easily separated by physicalmethods.Properties of diastereomers :-

(1) Diastereomers are not mirror image isomers.(2) They have same molecular formula, structural

formula but have different orientation in space.(3) Diastereomers may or may not be optically active.(4) They have different physical properties.(5) Diastereomers can be separated by any suitable

physical metthod of separation

EXAMPLE 66:Let us consider the stereoisomers of 3-chlorobutan-2-ol

There are 4 stereoisomers of 3-chlorobutan-2-ol. Inwhich (I & II) & (III & IV) are enantiomeric pairs. (I &III) or (I & IV) or (II & III) or (II & IV) all the isomersin each pair are neither mirror image nor superimposableto each other. Therefore these pairs are opticaldiastereomers.

EXAMPLE 67:

(1) COOH

OH H

OH H

COOH

(2) COOH

OH H

H HO

COOH

(1) and (2) are diastereomers.

EXAMPLE 68:

(I & II are diastereomers)

E. OPTICAL RESOLUTION :

(Chemical method of separation by using opticallyactive reagent)

Separation of Racemic Mixture :- The enantiomersin a d – pair have identical physical properties, so theycannot be separated by normal physical methods ofseparation. The special method is used for separation ofd – pair known as optical resolution.

F. NUMBER OF OPTICAL ISOMERS (NUMBER OFSTEREOISOMERS) :

EXAMPLE 69:Let us draw the total stereoisomers of

SOLUTION: n = 3 (odd chiral centres with similar ends.) Total isomers= 23–1 = 22 = 4


EXAMPLE 70:

CHO CH CH CH CH CH OH2

OH OH OH OH

, find the

number of optical isomers.

SOLUTION:n = 4

Optically Active = 2n = 24 = 16 and Meso = 0(dissimilar ends)

EXAMPLE 71:

CH3 CH CH CH3

OH OH

, find the number of optical

isomers.

SOLUTION:n = 2 (even)Enantiomers = 22–1 = 2, Optically active = 2 + 1 = 3,

Meso = 2 122

? = 2° = 1 (Similar ends)

CH 3

H

HO

OH

H

CH 3

CH 3

HO

H

H

OH

CH 3

CH 3

CH 3

H

H OH

OH

EXAMPLE 72:

CH3 CH CH CH CH3

OH OH OH

* find the number of optical

isomers.

SOLUTION:n = 3 (odd); Optically active = 23–1 = 4

Meso = 1

22n?

= 2

EXAMPLE 73:

CH3 CH CH CH

OH Br CH3

CH CH CH3

Br OH

SOLUTION:n = 5Optically active = 25–1 = 16,

Meso = 1

22n?

= 4,

2.2 CONFORMATIONAL ISOMERISM : In case of alkane structural isomer start from four

carbon atoms which is chain isomer. Butconformational isomer starts from two carbon atomswhich arises due to free rotation about C–C s bond.

Infinite number of conformers are possible. Amongthe infinite number of comformers the least stable issaid to be eclipsed form and the most stable form issaid to be staggered.

So these two forms can not be isolated/seperatedfrom each other.

(a) Saw horse Projection formula : CH3–CH3 (Ethane)

rotation H(a)

H(b)

H(c)

H(a)

H(c) H(b)

Eclipsed Staggered

(b) Newmann Projection formula : CH3–CH3 (Ethane)

1mol/kJ5.12

Strains :

1. Angle strain : Any atoms tends to have bond anglethat match to those of its bonding orbitals and any deviationfrom normal bond angles are accompanied by angle strain.

Let us illustrate angle strain in cyclo alkane as follows:Angle strain = 1/2 (Tetrahedral value – Given value)


A.S. = 12

(109°28’ – 60°) = 12

(49° 28’)

= 12

(48° 88’) = 24° 44’

A.S. = 12

(109°28’ – 90) = 12

(19°28’) = 12

(1888’) = 9°44’

A.S. = 00 44’

2. Torsional strain : Any pair of tetrahedral carbonattached to each other tends to have staggered formbetween their bonds and any deviation from staggeredform are accompanied by torsional strain.

3. Vander Waal Strain : If the two non bondedgroups/atoms are brought closer than the sum of theirvan der Waals radii, the interaction between them becomesrepulsive then it is known as van der Waals strain. Generallyvan der Waals strain increases with the size of groups anddecreases with the distance between groups.

(I) Conformational Analysis of ethane : CH3–CH3

º60 º120

Eclipsed Staggered

Eclipsed Staggered

Energy - Level Diagram of ethane


(II) Conformational Analysis of Propane : CH3–CH2–CH3(a) Saw horse Projection formula

Eclipsed Staggered

(b) Newman Projection Formula

Potential energy diagram of propane is exactly same as ethane but energy barrier is slightly more than ethane.(III) Conformational Analysis of Butane :

Ethyl group -hydrogen atom repulsion is less than methyl-methyl repulsion. So draw the newman between C2 - C3for detailed analysis.


I/VII = Fully eclipsed;II/VI = Gauche form;III/V = Partially eclipsed;IV = Anti formStability Order : IV > II > III > I (Anti > Gauche > Partially eclipsed > Fully eclipsed)P.E. Order : IV < II < III < I


Potential Energy Diagram of Butane :

Ex. Draw conformational isomers of the compound CH – CH –CH – CH3 3

CH3 CH3

with respect to C2 and C3 carbon atoms.

Ans.CH3

H

Eclipsed

H

CH3CH3

H C3

º60

CH3

H

H

CH3 CH3

CH3

º120

CH3

H

HCH3

H C3

H C3

º180

CH3

H

CH3CH3

HH C3

Staggered

(IV) Conformational analysis in case of intramolecular hydrogen bonding :In case of G – CH2 – CH2 – OH, where G = – OH, – NH2 , – F, – NR2, – NO2, – COOH, – CHO the gauche form

is more stable than the anti form due to intramolecular hydrogen bondingEx. F–CH2–CH2–OH (2-Fluoroethanol)

Stability : Gauche form > anti form.

Ex. (Glycol).


Gauche form is most stable due to hydrogen bonding.Stability Order : II > IV > III > I

EXAMPLE 74:Draw the most stable conformation of O2N – CH2 – CH2 – OH

SOLUTION:

(gauche form)

(V) Interconversion of projection formulae :Tartaric Acid : COOH – CHOH – CHOH – COOH no. of Stereoisomers : 3

1. Meso form

EXAMPLE 75:Conformational isomers of meso tartaric acid

OH

COOH

HOH

H

COOH

H OH

COOH

OH

COOH

H

OH

COOH

HCOOH

OH

H

OH

COOH

H HOH

COOH

H OH

COOH

OH

H

COOH

2. d/l form º


EXAMPLE 76:Conformational isomers of optical active (d/l ) tartaric acid.

H OH

COOHCOOH

OHH

OH

COOH

HH

COOHOH

H OH

COOHOH

HCOOH

OH

COOH

HCOOH

OH H

OH

COOH

H

H

OH

COOH

H OH

COOH H

COOH

OH

CONFORMATIONAL ANALYSIS OF CYCLOHEXANE :

(I) Chair form :The most stable conformation of cyclohexane ring is the chair form. In this non-planar structure the C–C bonds

angles are all close to 109.5º. This conformation is almost free from all the strains (like angle strain and torsional strain.)

Axial and equatorial bonds in chair form of cyclohexane :

The 12 hydrogen atoms of chair conformation of cyclohexane can be divided into two groups. Six hydrogenscalled axial hydrogens, hence their bonds parallel to a vertical axis that passes through the rings centre. These axialbonds are directed up & down on adjacent carbons. The second set of six hydrogens called equatorial hydrogens andare located approximately along the equator of the molecule.

(II) Boat form :Another conformation which is known as boat form has exactly eclipsed conformations.


In boat form of cyclohexane 6 hydrogens areequatorial, 4 hydrogens are axial and two hydrogensare flagpoles. It is an unstable conformation ofcyclohexane due to torsional strain among axialhydrogens and due to van der waals strain caused bycrowding between the "flagpole" hydrogens.

Conformational inversion (Ring flipping) incyclohexane :

Like alkanes cyclohexane too is conformationallymobile. Through a process known as ring inversion, chair-chair interconversion, or more simply ring flipping onechair conformation is converted to another chair throughthe half chair, twist boat and boat form.

By ring flipping all axial bonds convert to equatorialand vice-versa. The activation energy for cyclohexanering inversion is 45 kJ/mol. It is a very rapid process witha half-life of about 10–5 sec. at 25°C.

The relative energy profile of various conformations ofcyclohexane

because of the greater stability of the chair form, morethan 99% of the molecules are estimated to be in a chairconformation of any given moment. Twist boat form ofcyclohexane is chiral.

Conformational analysis of monosubstitutedcyclohexanes :

In ring inversion in methylcyclohexane the two chairconformations are not equivalent. In one chair the methyl

group is at axial ; in the other it is at equatorial. At roomtemperature 95% of the methylcyclohexane exist inequatorial methyl group whereas only 5% of the moleculehave an axial methyl group.

1, 3-diaxial repulsion :A methyl group is less crowded when it is at equatorial

than when it is at axial. The distance between the axialmethyl groups at C-1 and two hydrogens at C-3 & C-5 isless than the sum of their vander waal radii which causesvander waal strain in the axial conformation this type ofcrowding is called 1, 3-diaxial repulsions. When the methylgroup is equatorial, it experience no significant crowding.

EXAMPLE 77:Draw the most stable conformation of

(a) 1, 2-dimethylcyclohexane.(b) cyclohexane-1, 3-diol

SOLUTION:

(a)

(b)


Structural Isomerism

1. What is isomerism? Give an example.2. What is positional isomerism? Give an example.3. What is Metamerism? Give one example.4. Write all cyclic isomers of C3H6O.5. Identify the molecular weight of the compound ‘X’

containing carbon and hydrogen atoms only with 3and 2 bonds in one molecule.

6. Write the ring chain isomers of But-2-ene ?7. Identify the relationship between the given

compounds.

(a) and

(b) and

(c) and (d) CH3

COOCH3 & CH3COCH2OH8. Identify the relationship between the given

compounds.

(a) and (b)

Cl

Br

and Cl

Br

(c) and

(d) and

Stereoisomerism9. Write the definition of geometrical isomers & give

one example.10. Write the definition of enantiomer with example.11. Write the definition of Diastereomer with one example.12. Write the definition of meso compound with one

example.

EXERCISE–0 (RECALL YOUR UNDERSTANDING)13. Draw the structure of chair & boat form of

cyclohexane and also draw the axial & equitorial Hatom in chair form of cyclohexane.

14. Write the stablity order of conformations of ethyleneglycol ?

15. Define the racemic mixture ?16. Find the total number of geometrical isomers of

following compounds.

(a) (b) 17. Indicate whether each of the following compound is

'E' or 'Z'.

(a)

(b)

(c)

(d)

18. Find total number of chiral carbon atoms in thefollowing compounds :

(a) (b)

(c) (d)

19. Find R/S configuration of following compounds.

(a) (b)

(c) (d)


20. Find relationship between the given pairs.

(a) and

(b) and

21. Write the newman projection formula along C1–C2bonds in staggered form of following compounds?

(a) (b)


Structural Isomerism

1. Isomers have essentially identical :(a) Structural formula (b) Chemical properties(c) Molecular formula (d) Physical properties

2. Compound with same molecular formula but differentstructural formula are called :(a) Isomers (b) Isotopes(c) Isobars (d) Isoelectric

3. Which compound is not the isomer of 3-Ethyl-2-methylpentane ?(a)

(b)

(c)

(d)

4. What is the correct relationship between the followingcompounds ?

3

3223

CH|

CHCHCHCHCH ,

3

22223

CH|CHCHCHCHCH

(a) Chain isomers (b) Position isomers(c) Functional isomers (d) Identical

5. CH3 – CH2 – NH – CHO ;

2

3

NH|

CHOCHCH

I II

Which type of isomerism is observed between Iand II.(a) Chain isomers (b) Position isomers(c) Functional isomers (d) Metamers

6. The correct relationship among the following pairsof given compounds is

(I,II) (II,III)

(1) Functional Isomers Metamers

(2) Metamers Functional Isomers

(3) Metamers Metamers

(4) Functional Isomers Functional Isomers

7. Which of the given set of molecules have similarmolecular formula :(a) Nonane,2-methylheptane(b) 3-Isopropylcyclopent-1-ene, 3-methylhexane(c) 3-Methylcyclopent-1-ene,penta-1,3-diene(d) Ethylcyclohexane, oct-2-ene


8. Which of the following pairs of compounds are chainisomers.(a) Isobutyl alcohol and s-pentyl alcohol(b) Isobutyl alcohol and t-butyl alcohol(c) Secpentyl alcohol and neopentyl alcohol(d) Ethyl alcohol and ethylene glycol

9. A position isomer of 2-pentanone is :(a) 3-Pentanone (b) 3-Methyl-2-butanone(c) 1-Pentanal (d) 2,2-Dimethylpropanal

10. Monocarboxylic acids are functional isomers of :(a) Alcohols (b) Ethers(c) Esters (d) Aldehydes and ketones.

11. CH3CONH2 & HCONHCH3 are called :(a) Position isomers (b) Chain isomers(c) Tautomers (d) Functional isomers

12. Which of the following pair of compounds are notisomers ?(a) Propyne and cyclopropene(b) Propyne and propadiene(c) Propene and cyclopropene(d) 1–Propanol and methyoxyethane

13. Saturated acyclic ethers are isomeric with :(a) Aldehydes(b) Ketones(c) Both aldehydes and ketones(d) Alcohols

14. How many positional isomers are possible fordimethylcyclohexane ?(a) 3 (b) 4(c) 5 (d) 6

15. How many aromatic isomers are possible fortrichlorobenzene (C6H3Cl3) ?(a) 2 (b) 3(c) 4 (d) 5

16. The number of ether isomers represented by formulaC4H10O is (only structural) :(a) 4 (b) 3(c) 2 (d) 1

17. Total number of 2° amine isomers of C4H11N wouldbe (only structural) :(a) 4 (b) 3(c) 5 (d) 2

18. How many structural isomers of all the tertiaryalcohols with molecular formula C6H14O ?(a) 2 (b) 3(c) 4 (d) 5

19. The number of structural isomers for C5H10 is :(a) 8 (b) 6(c) 9 (d) 10

20. Molecular formula C4H10O represent :(a) Two primary alcohol(b) One secondary alcohol(c) One tertiary alcohol(d) All of these

Geometrical isomerism

21. Stereoisomers have different :(a) Molecular formula (b) Structural formula(c) Configuration (d) Molecular mass

22. What should be the minimum conditions to showgeometrical isomerism ?(a) Restricted rotation about double bond or ring(b) Groups which are responsible to show

geometrical isomerism differ in their relativedistance

(c) Two different groups at both restricted atoms.(d) All of these

23. Which can show the cis-trans isomerism ?(a) ClCH2CH2Cl (b) Cl2C==CH2

(c) Cl2C==CCl2 (d) ClCH ==CHCl24. Which of the following compounds will not show

geometrical isomerism ?(a) Azomethane (b) 1-Bromo-2-chloroethene(c) 1-Phenylpropene (d) 2-Methyl-2-butene

25. Which of the following will not show geometricalisomerism ?(a) CH3 – N = N – CH = CH2

(b) (c) CH3 – CH = N – OH(d) Cl–CH=C=CH–Cl

26. Among the given compounds identify the pair ofgeometrical isomers :-


(a) I & II (b) I & III(c) II & IV (d) III & IV

27. Geometrical isomerism is shown by :

(a) HC = CCH3H

H (b) HC = CCH3l

CH3

(c) HC = CCH3

CH3

CH3 (d) HC = C CH3BrBr

28. Which of the following compound does not havegeometrical isomers ?(a) 2-Pentenoic acid (b) 2-Butenoic acid(c) 3-Pentenoic acid (d) 3-Butenoic acid

29. How many geometrical isomers are possible for thegiven compound ?Ph – CH = CH – CH = CH – COOH(a) 3 (b) 4(c) 2 (d) 1

30. How many geometrical isomers are possible for thegiven compound ?CH3 – CH = CH – CH = CH – CH = CH2

(a) 2 (b) 4(c) 6 (d) 8

31. How many geometrical isomers are possible for thegiven compound ?

(a) 2 (b) 4(c) 6 (d) 8

32. No. of Geometrical isomers for following compoundis :

(a) 8 (b) 16(c) 32 (d) 10

33. Which carbonyl compounds can give one oxime onlyon reaction with hydroxyl amine ?(a) HCHO (b) CH3CHO(c) PhCHO (d) CH3COPh

34. Select the correct options for molecular formulaC2H2Cl2 :(a) The total number of isomers is 4.(b) All the structures show geometrical isomerism.(c) All isomers have 5s bonds and one p bond.(d) Its has linear shape.

CIP Rules (E/Z Naming) & Physical Properties of G.I

35. Identify (Z) - 2 - pentene :-

(a) (b)

(c) (d)

36. The 'E'-isomer is/are :

(a)

(b)

(c)

(d)

(a) only (a) (b) only (b)(c) (a) and (b) (d) All

37. The correct stereochemical formula of Trans-3-chloro-1-phenylbut-1-ene is :

(a)

(b)

(c)

(d)

38. Which of the following is a syn isomer :

(a) (b)

(c) (d)

39. Which of the following is not true for maleic acidand fumaric acid ?(a) Configurational isomers


(b) Stereo isomers(c) Z and E isomers(d) Constitutional isomers

40. Which of the following is incorrect statement :(a) Geometrical isomers are not mirror image isomer.(b) A compound having double bond (restricted

bond) always show geometrical isomerism.(c) Acyclic compoubd having single bond does not

show geometrical isomerism.(d) Cyclodecene can show cis & trans form.

Optical Isomerism

41. The necessary and sufficient condition for a moleculeto be optically active :(a) It must contain asymmetric carbon atoms(b) It must be chiral atom(c) It must be identical with its mirror image(d) It must be non-superimpossable with its mirror

image42. Chiral molecules are those which are :

(a) Superimpossable on their mirror image(b) Not superimpossable on their mirror image(c) unstable molecules(d) capable of showing geometrical isomerism

43. Number of chiral carbon atoms in the compound w,x, y and z respectively would be :

(a) 3, 0, 2, 1 (b) 2, 1, 0, 1(c) 3, 1, 2, 1 (d) 1, 1, 2, 0

44. Which of the following compounds possesses achiral centre ?

(a) (b)

(c) (d)

45. will shows :

(a) Geometrical isomerism only(b) Optical isomerism only(c) Geometrical and optical isomerism(d) Neither geometrical nor optical isomerism

46. The molecule 3–penten–2–ol can exhibit :(a) Optical isomerism(b) Geometrical isomerism(c) Metamerism(d) TautomerismThe correct answer is :(a) a and b (b) a and c(c) b and c (d) a and d

47. Which of the following isomerism is not shown bythe compounds having molecular formula C4H10O ?(a) Metamerism(b) Position isomerism(c) Geometrical isomerism(d) Optical isomerism

48. The structure shows

CH3C = C

H

CH3C — COOH

H

CH3

(a) Geometrical isomerism(b) Optical isomerism(c) Geometrical & optical isomerism(d) none

49. Which of the following compounds exhibitsstereoisomerism ?(a) 2-methylbutene-1(b) 3-methylbutyne-1(c) 3-methylbutanoic acid(d) 2-methylbutanoic acid

50. Which of the following is an optically activecompound ?(a) 1–Butanol (b) 1–Propanol(c) 2–Chlorobutane (d) 4–Hydroxyheptane

51. Meso form of tartaric acid is :(a) Dextorotatory(b) laevorotatory(c) neither Laevo not dextro rotatory due to internal

compensation


(d) A mixture of equal quantities of dextro andleavorotatory forms

52. The following two compounds are :

and

(a) Enantiomers (b) Diastereomers(c) Identical (d) Epimers

53. Stereo isomers which are not mirror image of eachother, are called : -(a) Enantiomers (b) Tautomers(c) Metamers (d) Diastereomers

54. Which of the following amine is optically active ?(a) CH3NH2

(b) CH3NHCH3

(c) CH3CH2CH2 –

(d) sec-Butylamine55. Which of the following compound has ‘S’

configuration ?

(a) (b)

(c) (d)

56. The correct configuration assigned for givencompound :

(a) 2R, 3R (b) 2S, 3S(c) 2R, 3S (d) 2S, 3R

57. Which of the following compound has 'D'configuration ?

(a) (b)

(c) (d)

58. represents the Fischer projection

formula :(a) D (b) L(c) d (d) l

59. The correct IUPAC name of D-Glucose is :

(D-Glucose)

(a) (2D, 3D, 4L, 5D) 2, 3, 4, 5, 6-pentahydroxyhexanal

(b) D-2, 3, 4, 5, 6-pentahydroxyhexanal(c) 6-oxo-(2D, 3L, 4D, 5D) - 2, 3, 4, 5, 6-

pentahydroxohexane(d) (2D, 3L, 4D, 5D) - 2, 3, 4, 5, 6-

pentahydroxyhexanal60. A recemic mixture contains dextrorotatory and

laevorotatory isomers in the proportion –(a) 2 : 1 (c) 1 : 1(c) 1 : 5 (d) 3 : 1

61. The priority of groups OH, COOH, CHO, OCH3attached to a chiral carbon is in order :(a) OH > COOH > CHO > OCH3(b) OCH3 > OH > CHO > COOH(c) OCH3 > OH > COOH > CHO(d) OCH3 > COOH > CHO > OH

62. Which of the following is not true for maleic acidand fumaric acid ?(a) Configurational isomers(b) Stereo isomers(c) Z and E isomers(d) Optical isomers

Conformational Isomerism63. The eclipsed and staggered conformation of ethane

is due to –


(a) Free rotation about C–C single bond(b) Restricted rotation about C–C single bond(c) Absence of rotation about C–C bond(d) None of the above

64. The two structures I & II represents :

(I) (II)(a) Conformational isomers(b) Stereoisomers(c) Constitutional isomers(d) Identical

65. In the complete rotation of butane from 0º to 360ºthe gauche (straggered) conformation appears(a) once (b) Twice(c) Thrice (d) Four times

66. Which of the following is associated with Torsionalstrain ?(a) Repulsion between bond pair of electrons(b) Inductive effect(c) Bond angle strain(d) Attraction of opposite charges

67. In the given conformation C2 is rotated about C2–C3bond anticlockwise by an angle of 120° then theconformation obtained is :

CH3

H

HH

H

CH3

(a) Fully eclipsed conformation

(b) Partially eclipsed conformation(c) Gauche conformation

(d) Staggered conformation68. The pair of structures given below represent

(a) Enantiomers (b) Diastereomers(c) Structural isomers (d) conformational isomers

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)Isomerism

1. CH3CHOHCH2CHO and CH3CH2CH2COOHconstitute a pair of :-(a) Position isomers (b) Metamers(c) Optical isomers (d) Functional isomers

2. The minimum number of carbon atoms present in anorganic compound to be able to show positionisomerism is :-(a) 3 (b) 4(c) 2 (d) 5

3. Which of the following compound is isomeric withpropanoic acid :-

(a) CH 3 C OC H2 5

O

(b) CH 2 CH 2

O

C H

OH

(c) CH3—CH(OH)—CH3

(d) CH3O—CH2—CH2OH

4. The pair of structures represents :-

MeH

H H

Me

MeH

H H

H

Cl CH Cl2

(a) Enantiomers (b) Position isomers(c) Conformers (d) None

5. The simplest alkanol exhibiting optical activity is(a) n-butyl alcohol (b) Isobutyl alcohol(c) s-butyl alcohol (d) t-butyl alcohol

6. C C H

C

H C3

H C3

H

COOHH C3

Exhibits :-

(a) Tautomerism(b) Optical isomerism(c) Geometrical isomerism(d) Geometrical and optical isomerism


7. In which of the following cyclic chain is the mainchain ?

(a)

(b)

(c)

(d)

8. How many structure isomers could be obtained fromthe alkane C6H14 ?(a) Four (b) Five(c) Six (d) Seven

9. The compound which is not isomeric with diethylether is(a) n-Propyl methyl ether(b) Butanol -1(c) 2-Methyl propan-1-ol(d) Butanone

10. Dimethylcarbinol is isomeric with :(a) Isopropyl alcohol (b) Isobutyl alcohol(c) sec-butyl alcohol (d) Ethylcarbinol

11. The number of dihydric phenols possible with themolecular formula C6H6O2 is(a) 2 (b) 3(c) 4 (d) 5

12. Structural isomers possible for C4H8Br2 are(a) 9 (b) 8(c) 7 (d) 6

13. Which of the following is not an isomer of allylalcohol ?(a) Acetone(b) 1–Propanol(c) 1,2 - Epoxypropane(d) Cyclopropanol

14. Total number of structure isomers of C4H10O is :(a) 7 (b) 4(c) 3 (d) 8

15. Which of the following shows cis-trans isomerism ?(a) 2-Butyne (b) 2-Butene(c) 2-Butanol (d) 2-Chloro-1-butene

16. What is the relation between 3-Ethylpentane and3-Methylhexane ?(a) Chain isomers (b) Position isomers(c) Functional isomers (d) No relation

17. Which can not show geometrical isomerism ?(a) H2N2

(b) 1-Chloro-2-phenylethene

(c) C = N–OHH C3

H C3

(d) HO–N = N–OH18. Which of the following is a pair of metamers ?

(a)

(b)

(c)

(d)

19. The lowest molecular weight alkane, which isoptically active ?(a) 3-methylhexane(b) 2, 3-dimethylbutane(c) 2, 3, 3-trimethylbutane(d) 2-methylhexane

20. o–Cresol & benzyl alcohol are :(a) Functional isomers (b) Position isomers(c) Chain isomers (d) All the above

21. The two compounds given below are :

(a) Enantiomer (b) Identical(c) Meso compound (d) Diastereomers

22. Degree of unsaturation in is :


(a) 3 (b) 4(c) 7 (d) 6

23. Total number of stereoisomers of compound is :

BrOH||

CHCHCHCH 33

(a) 2 (b) 4(c) 6 (d) 8

24. Find the number of total structurally isomeric 3°amides with molecular formula C5H11NO :(a) 4 (b) 3(c) 2 (d) 5

25.BrBrBr|||

CH–CH–CH–CH–CH 33

Total number of stereoisomers in above compoundis :(a) 6 (b) 4(c) 8 (d) 16

26. Which of the following compound can not showgeometrical isomerism ?

(a) (b)

(c) (d)

27.

OHBrCl|||

CH–CH–CH–CH–CH 33

Total number of stereoisomers in above compoundis :(a) 6 (b) 4(c) 8 (d) 16

28. The absolute configuration of the following compoundis :

(a) 2 S, 3 R (b) 2 S, 3 S(c) 2 R, 3 S (d) 2 R, 3 R

29. Which of the following compound can showgeometrical isomerism ?

(a)

(b)

(c)

(d)

30. and are

(a) enantiomers (b) diastereomers(c) identical (d) none of these

31. Total number of geometrical isomers in the givencompound is :

C = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = CC = CH – DC = CD – HC = C

(a) 2 (b) 4(c) 6 (d) 8

32. The total number of ketones (including opticalisomers) with the molecular formula C6H12O is :(a) 4 (b) 5(c) 6 (d) 7

33. Total number of geometrical isomers in the givencompound is :

(a) 3 (b) 6(c) 8 (d) 16

34. Which of the following statements is not correct :(a) Enantiomers are Eessentially chiral and optically

active


(b) Diastereomers are not neccesarily chiral andoptically active

(c) All geometrical isomers are diastereomers(d) All diastereomers are chiral and optically active

35. Which of the following will form only one oxime onreaction with NH2OH solution ?

(a) I, II (b) II, III(c) I, IV (d) II, III, IV

36. Which is incorrect statement about geometricalisomers ?(a) Geometrical isomers can be separated by

fractional distillation.(b) In two geometrical isomers the distance between

two particular groups at the ends of the restrictedbond must be changed.

(c) In cycloalkenes, geometrical isomerism existacross C=C with ring size equal to or greaterthan 8 carbon atom.

(d) does't show geometrical isomerism

because it has only 7 C atoms in ring.37. The incorrect order with respect to the properties

mentioned for the following pair of compounds is :

(a) < (Stability)

(b) < (Stability)

(c) > (Boiling point)

(d) > dipole moment ()

(a) (b), (c) and (d) (b) (b) and (c)(c) (a) and (d) (d) (c) and (d)

38. How many meso steroisomers are possible for 2, 3,4–pentanetriol :(a) 1 (b) 2(c) 3 (d) None

39. The total number of isomers for C4H8 is(a) 5 (b) 6(c) 7 (d) 8

40. The correct order/s for the given pair of isomers is

(a) > (Melting point)

(b) <

(Dipole moment)

(c) > (Boiling point)

(d) > (Water solubility)

41. The total number of isomeric alcohols with themolecular formula C4H9OH is(a) 2 (b) 3(c) 4 (d) 5

42. Consider the following organic compound :

To make it a chiral compound, the attack should beon carbon atom no.(a) 1 (b) 3(c) 4 (d) 7

43. The number of possible open chain (acyclic) isomericcompounds for molecular formula C5H10 would be(a) 8 (b) 7(c) 6 (d) 5

44. Which of the following pairs of structures representenantiomers ?

(a) and

(b) rFkk


(c) and

(d) and

45. If optical rotation produced by is +

36º then that produced by is

(a) –36º (b) 0º(c) +36º (d) unpredictable

46. Which one among the following is not diastereomericpair ?

(a) I and III (b) I and II(c) II and III (d) I and IV

47. The number of optically active stereoisomers possiblefor butane – 2, 3 – diol :(a) 1 (b) 2(c) 3 (d) 4

48. How many stereoisomers are there for tartaric acid ?(a) 2 (b) 3(c) 4 (d) 8

49. Which of the following pair of isomers can not beseparated by fractional crystallisation or distillation -(a) Maleic acid and Fumaric acid(b) (+)-Tartaric acid and meso - tartaric acid

(c)

2

3

NH|

COOHCHCH and COOHCHCHNH 222

(d) (+)-lactic acid and (–)-lactic acid50. How many stereoisomers (geometrical and optical)

are possible for the following compound ?CH3CH = CHCH2CH(OH)COOH(a) 1 (b) 2(c) 3 (d) 4

51. The Baeyer’s angle strain is expected to be maximumin(a) Cyclodecane (b) Cyclopentane(c) Cyclobutane (d) Cyclopropane

52. In the Newmann projection formula of the least stablestaggered form of n–butane, Which of the followingreason is the cause of its unstability ?(a) Vander Waal’s strain(b) Torsional strain(c) Combination of both.(d) None of these

53. Find relation between these two compounds A andB :

OH

HHO

HCH3

COOH

OH

OHCH3

HCOOH

H

(A) (B)

(a) Enantiomer (b) Diasteromer(c) Meso (d) Identical

54. Which of the following statement regarding thepopulations of different conformations of opticallyactive butane-2, 3- diol is true(a) The most populated conformer will have the

hydroxy group at anti-position(b) All staggered conformations will be equally

populated and are major(c) The most populated conformer will have hydroxyl

groups at guache position(d) Relative populations of different conformers are

not predicted


55. In the following structures, which two forms arestaggered conformation of ethane ?

(a) 1 and 4 (b) 2 and 3(c) 1 and 2 (d) 1 and 3

56. The most stable conformation of ethylene glycol is.(a) Anti (b) Gauche(c) Partially eclipsed (d) Fully eclipsed

57. The most stable conformation of 3-fluorobutan-2-olis :(a) Fully eclipsed form(b) Partially eclipsed form(c) Gauche form(d) Anti form

58. In the energy diagram-dihedral angle of theconformations of ethane, the staggered forms arefound at(a) The crests(b) The troughs(c) Any point between crests and troughs(d) On alternate crests

59. The correct order of stability of various conformersof 2-fluoroethanol is –(a) Anti > Gauche > Fully eclipsed > Partially

eclipsed(b) Gauche > Anti > Partially eclipsed > Fully eclipsed(c) Anti > Partially eclipsed > Fully eclipsed >

Gauche(d) Fully eclipsed > Partially eclipsed > Gauche >

Anti60. Which of the following is an achiral molecule?

(a) (b)

(c) (d)

61. Incorrect about the compounds I, II, III is :

, ,

(a) I & II are diastereomers(b) I & III are identical(c) II & III are diastereomers(d) I & II are optically active

62. Select the correct statement/s :(a) Eclipsed and staggered ethanes give different

products on reaction with chlorine in presenceof light.

(b) The conformational isomers can be isolated atroom temperature.

(c) Torsional strain in ethane is minimum at dihedralangles 60°, 180° and 300°.

(d) Steric strain is minimum in staggered gaucheform of n-butane.

63. Meso–tartaric acid H OHH OH

COOH

COOH

is optically inactive

due to the presence of :-(a) Molecular symmetry(b) Molecular asymmetry(c) External compensation(d) Two asymmetric carbon atoms

64. Which is optically active molecule :-

(a) C H 6 5

O

C OH (b) CH 3

OH

CH C H2 5


(c) C H 6 5

H

CH OH (d) CH 3

CH3

CHC H 6 5

65. The number of stereo isomers of glucose (a six carbonsugar) is :-(a) 8 (b) 12(c) 16 (d) 24

66. Which of the following compoundis optically inactive:

(a) CH 3 CH2 CH CH 3

OH

(b) H CH3

(c) CH

H C3

C CH

CH3

(d)

H C3 CH3

Cl H

67. Which of the following has Z-configuration :

(a) CH

H C3 CC H2 5

H

(b) CHOCH2

BrC

CH(CH )3 2

CH2 CH3

(c) CBr

ClC

H

D

(d) All the above68. Which of the following does not contain any

asymmetric carbon but can show enantiomerism:-(a) Lactic acid (b) 1,3-pentadiene(c) Tartaric acid (d) 2,3-pentadiene

69. Which of the following represents the structurehaving cis arrangement around each double bond :-

(a)

(b)

(c)

(d)

70. Which is a pair of geometrical isomers :-

, ,

,

(a) I and II (b) I and III(c) II and IV (d) III and IV

71. Which one of the following is a meso–compound.

(a) (b)

(c) (d)

72. The minimum number of carbon atoms in ketone toshow metamerism :–(a) 3 (b) 4(c) 5 (d) 6

73. Which of the following is optically active :–

(a) (b)

(c) (d)

74. BrCH2–CH2–CH=O and CH3–CH2–

Br

C=O are

(a) Functional isomers (b) Position isomers(c) Chain isomers (d) Metamers

75. Which are metamers :-

(a) CH3–O–CH2CH2CH3, CH3–O–CH CH3

CH3

(b) C2H5–O–C2H5, CH3CH2CH2CH2OH(c) CH3–O–C2H5, CH3–CH2–O–CH3


(d) CH3–C

O

–CH3, CH3–CH2–C

O

–H

76. Which is incorrect statement :-

(a)

HMeMe

HHH

and

HHH

MeHMe

are

conformations

(b)Cl

H

Cl

H is a meso-compound

(c) and are Geometrical

isomers

(d) C=C=CH

Me

H

Me and C=C=C

H

Me

H

Meare Enantiomers

77. The total number of configurational isomers of thegiven compound are :-

CH3–CH=CHCHOHCHOHCH3

(a) 2 (b) 4

(c) 6 (d) 8

78. Which is a pair of geometrical isomers :-

(a) Cl

BrC=C

Me

H and

Br

ClC=C

H

Me

(b) Ph

MeC=N

OH and

Me

PhC=N

OH

(c) Br Br and BrBr

(d) Me

HC=C

Ph

Me and

H

MeC=C

Ph

Me79. Which compound is optical active –

(a) CH 3 C COOH

H

H

(b) CH 3 C COOH

OH

H

(c) CH 3 C COOH

CH3

OH

(d) CH 3 C COOH

CH3

Cl

80. Which of the following is not a metamer of C4H10O(a) Diethyl ether(b) Methyl n-propyl ether(c) 2–Methoxy propane(d) Isobutyl alcohol


1. Assertion : Alkanes containing more than 3-carbonatoms can exhibit chain isomerism.Reason : Because all the carbon atoms in alkanesare sp3 hybridized.

2. Assertion : Butane and 2-methyl butane cannot bechain isomers.Reason : Butane is a straight chain alkane while 2-methyl butane is a branched chain alkane.

3. Assertion : The melting point of fumaric acid ishigher than that of maleic acid.Reason : The molecules of fumaric acid are moresymmetric than those of maleic acid and hence itgets closely arranged in the crystal lattice.

4. Assertion : trans–1–chloro propene has higher dipolemoment than cis–1–chloro propene.Reason : The resultant of two vectors in trans–1–

chloropropene is more than in cis–1–chloro propene.5. Assertion : Meso tartaric acid is optically inactive.

Reason : Its optically inactivity is due to externalcompensation.

6. Assertion : The boiling point of Cis–1,2–dichloroethene is higher than corresponding trans-isomer.Reason : Because the cis isomer of 1,2–dichloroethene is more polar than trans.

7. Assertion : Enol form of cyclohexane–1,3,5–trioneis more stable than its ketoform.Reason : It contains –hydrogen atoms.

8. Assertion : Trihydroxyglutaric acid (HCOO—CHOH—CHOH—CHOH—COOH) exists in fourstereoisomeric forms, two of which are opticallyactive while the other two are meso–forms.


Reason : It contains two asymmetric and onepseudo–asymmetric carbon atom.

9. Assertion : 1,2–propadiene exhibits opticalisomerism.Reason : Its mirror image is non–superimposable.

10. Assertion : lactic acid shows gemetricalisomerism.Reason : Because it does not have chiral carbon.

11. Assertion : Organic compounds which do notcontain chiral carbon atoms can not be opticallyactive.Reason : An organic compound is optically activeonly when its mirror image is non–superimposableirrespective of the fact whether it contains a chiralcarbon atom or not.

12. Assertion : Ethanol cannot show position isomerism.Reason : Ethanol cannot show isomerism.

13. Assertion : Meso tartaric acid is optically inactiveReason : Because it has plane of symmetry.

14. Assertion : Benzaldehyde forms two oximes onreacting with NH2OH.Reason : The two oximes arise due to geometricalisomerism around C N bond.

15. Assertion : The boiling point of cis 1, 2–dichloroethene is higher than that of corresponding trans–isomer.Reason : Cis– 1,2–dichloro ethene has higher momentas compared to that of the trans–isomer.

16. Assertion : Neopentyl bromide and 2–bromo–2–methyl butane are enantiomers.Reason : Both have same molecular formula as wellas structure formula.

17. Assertion : CH3–CH2–Br and CH3–CH2–I areFunctional isomers.Reason : They have different Functional groups.

18. Assertion : Organic compounds which do notcontain chiral carbon atoms can be optically active.Reason : Presence of chiral carbon is an essentialcondition for optical activity.


1. The configuration of the given compound is :[AIPMT-2005]

(a) E (b) R(c) S (d) Z

2. Which one of the following pair represents stereoisomerism :- [AIPMT-2005](a) Linkage isomerism and Geometrical isomerism(b) Chain isomerism and Rotational isomerism(c) Optical isomerism and Geometrical isomerism(d) Structural isomerism and Geometrical isomerism.

3. Correct configuration of the following is : -[AIIMS-2005]

CH3

H OH

OHCH3

H

(a) 2S, 3S (b) 2S, 3R(c) 2R, 3S (d) 2R, 3R

4. Alcohols are isomeric with : [AFMC 2005](a) acids (b) ethers(c) esters (d) aldehydes

5. Which compound is optical active - [RPMT-2005]

(a) CH – C – COOH3

H

H

(b) CH – C – COOH3

OH

H

(c) CH – C – COOH3

OH

CH3

(d) CH – C – COOH3

Cl

CH3

6. Which of the following is not chiral ?[AIPMT-2006]

(a) 2-Butanol (b) 2, 3-Dibromo pentane(c) 3-Bromo pentane (d) 2-Hydroxy propanoic acid

7. Among the following which one can have a mesoform ? [AIPMT-2006](a) CH3CH(OH)CH(CI)C2H5

(b) CH3CH(OH)CH(OH)CH3

(c) C2H5CH(OH)CH(OH)CH3

(d) HOCH2CH(CI)CH3

8. Among the following which one can have a mesoform – [AIIMS-2006]


(a) CH3CH(OH)CH(Cl)C2H5

(b) CH3CH(OH)CH(OH)CH3

(c) C2H5CH(OH)CH(OH)CH3 (d) HOCH2CH(Cl)CH3

9. Among the following L–serine is : - [AIIMS-2006]

(a)

CH OH2

NH2

H CO H2

(b)

NH2

H

CO H2

HOH C2

(c) CH OH2

H

CO H2

N H2

(d)

CH OH2

H

CO H2

N H2

10. CH3–CHCI–CH2–CH3 has a chiral centre which oneof the following represents its R configuration :

[AIPMT-2007]

(a) H — C — CH3

C H2 5

CI

|

|(b) CI — C — CH3

C H2 5

H

|

|

(c) H — C — CI

CH3|

|C H2 5

(d) H C — C — CI3

H

|

|

C H2 5

11. The total number of possible isomeric trimethylbenzene is : [RPMT 2007](a) 2 (b) 3(c) 4 (d) 6

12. How many stereoisomers does this molecule haveCH3CH = CHCH2CHBrCH3 [AIPMT-2008]

(a) 8 (b) 2(c) 4 (d) 6

13. Which of the following compound is expected to beoptically active? [AFMC-2008](a) (CH3)2 CHCHO (b) CH3CH2CH2CHO

(c) CH3CH2CHBrCHO (d) CH3CH2CBr2CHO14. The number of metamers of the compound with

molecular formula C5H10O is : [AFMC 2009](a) 1 (b) 3

(c) 8 (d) 615. In the following the most stable conformation of

n-butane is : [AIPMT-2010]

(a) (b)

(c) (d)

16. Which of the following is optically active :-

[AIIMS-2010]

(a)

OH

OH

(b)

OHCH3

(c)

OH

(d)

CH3

17. Find the number of stereo isomers of 1,2-dihydroxycyclopentane: [AIIMS-2011](a) 1 (b) 2

(c) 3 (d) 418. Which of the following acids does not exhibit optical

isomerism ? [AIPMT (Pre) 2012](a) Maleic acid (b) -amino acids

(c) Lactic acid (d) Tartaric acid


19. Given,

Which of the given compounds can exhibittautomerism? [NEET-2015](a) l and ll (b) l and lll(c) ll and lll (d) l, ll and lll

20. Point out incorrect sawhorse drawing(s) for thefollowing compound. [AIIMS-2015]

Br

CH3

CH3

H

BrH

(a) H

HCH3

CH3

Br

Br

(b) H

BrCH3

CH3

Br

H

(c) H

BrCH3

CH3

Br

H (d) H

CH3

CH3

H

Br

Br

21. The correct statement regarding the comparison ofstaggered and eclipsed conformations of ethane, is:

[NEET-2016](a) The staggered conformation of ethane is more

stable than eclipsed conformation, becausestaggered conformation has no torsional strain.

(b) The staggered conformation of ethane is less stablethan eclipsed conformation, because staggeredconformation has torsional strain.

(c) The eclipsed conformation of ethane is morestable than staggered conformation, becauseeclipsed conformation has no torsional strain.

(d) The eclipsed conformation of ethane is morestable than staggered conformation even thoughthe eclipsed conformation has torsional strain.

22. Which of the following biphenyls is optically active[NEET-2016]

(a) (b)

(c) (d)

23. Among the following, the achiral amion acids is[AIIMS-2016]

(a) 2-ethylalanine(b) 2-methylglycine(c) 2-hydroxymethylserine(d) tryptophyan.

24. Assertion : Staggered conformation of ethane is 12.5kJ mol–1 more stable than the eclipsed conformation.Reason : The tow conformation of ethane cannotbe separated at room temperature [AIIMS-2016](a) If both assertion and reason are true and reason

is the correct explanantion of assertion.(b) If both assertion and reason are true but reason


25. Assertion : 2,4-Dimethyl hex-2-ene has 4stereoisomerReason : It show geometrical isomerism

[AIIMS-2018 (E)](a) If both assertion and reason are true and reason



26. Which of the following are not enantiomer pair.[AIIMS-2018 (E)]

C2H5

H

CH3

Cl

C2H5

H

Cl

CH l3

A B

C2H5

Cl

H

CH3

C2H5

Cl

CH3

H

C D(a) A & B (b) A & B(c) B & D (d) C & D


1. The phenomenon of existence of two or morecompounds possessing the same molecular formulaebut different properties is known as isomerism. Ex.C2H5OH & CH3–O–CH3.

2. When two or more compounds differ in the positionof substituent atoms or functional groups on the samecarbon skeleton, then they are called position isomersand the phenomenon is termed as position isomerism.Eg. Cl–CH2–CH2–CH3 (1-chloropropane) & CH3–CH(Cl)–CH3 (2-chloropropane).

3. Metamerism arises due to different alkyl chains oneither side of the polyvalent functional group in themolecule.Eg. CH3 –CH2–O–CH2–CH3 and CH3 –O–CH2–CH2–CH3

4. , ,

5. H–CºC–H, Mol wt. = 26

6.

7. (a) Chain Isomers (b) Chain Isomers(c) metamers (d) Functional isomers

8. (a) Homologes (b) Position isomers(c) Functional isomers (d) Metamers

9. Two compounds having same structural formula butdifferent orientation of atoms or groups in space dueto restricted rotation of bonds are known asgeometrical isomers.

Eg.

10. Stereoisomers which are non-superimposable mirrorimages of each other are called enantiomers.Eg.

11. The stereoisomers which are neither mirror imagenor superimposable to each other are calleddiastereomers.

Eg. (a) &

(b) &

12. A meso compound is one whose molecules aresuperimposable on their mirror images even thoughthey contain chiral centres. A meso compound isoptically inactive since it has a plane of symmetry.

Eg. plane

13.

ANSWER KEY



EXERCISE–1 (CHECK YOUR UNDERSTANDING)1. (c) 2. (a) 3. (b) 4. (a) 5. (c) 6. (c) 7. (d) 8. (c) 9. (a) 10. (c)

11. (d) 12. (c) 13. (d) 14. (b) 15. (b) 16. (b) 17. (b) 18. (b) 19. (d) 20. (d)21. (c) 22. (d) 23. (d) 24. (d) 25. (d) 26. (c) 27. (b) 28. (d) 29. (b) 30. (b)31. (b) 32. (b) 33. (a) 34. (c) 35. (a) 36. (c) 37. (d) 38. (c) 39. (d) 40. (b)41. (d) 42. (b) 43. (a) 44. (a) 45. (c) 46. (a) 47. (c) 48. (b) 49. (d) 50. (c)51. (c) 52. (a) 53. (d) 54. (d) 55. (c) 56. (d) 57. (c) 58. (a) 59. (d) 60. (b)61. (c) 62. (d) 63. (a) 64. (c) 65. (b) 66. (a) 67. (c) 68. (c)

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)1. (d) 2. (c) 3. (b) 4. (b) 5. (c) 6. (b) 7. (b) 8. (b) 9. (d) 10. (d)

11. (b) 12. (a) 13. (b) 14. (a) 15. (b) 16. (a) 17. (c) 18. (b) 19. (a) 20. (a)21. (a) 22. (c) 23. (b) 24. (d) 25. (b) 26. (a) 27. (c) 28. (b) 29. (d) 30. (b)31. (b) 32. (d) 33. (b) 34. (d) 35. (b) 36. (d) 37. (d) 38. (b) 39. (b) 40. (d)41. (d) 42. (b) 43. (c) 44. (b) 45. (b) 46. (a) 47. (b) 48. (b) 49. (d) 50. (d)51. (d) 52. (a) 53. (d) 54. (c) 55. (c) 56. (b) 57. (c) 58. (b) 59. (b) 60. (a)61. (d) 62. (c) 63. (a) 64. (b) 65. (c) 66. (d) 67. (d) 68. (d) 69. (b) 70. (c)71. (d) 72. (c) 73. (c) 74. (a) 75. (a) 76. (a) 77. (d) 78. (d) 79. (b) 80. (d)

EXERCISE–3 (AIIMS SPECIAL)1. (b) 2. (b) 3. (a) 4. (a) 5. (c) 6. (a) 7. (b) 8. (a) 9. (d) 10. (d)

11. (d) 12. (c) 13. (a) 14. (a) 15. (a) 16. (d) 17. (d) 18. (c)

EXERCIS-4 (ARCHIVES FROM VARIOUS MEDICAL ENTRANCE)1. (b) 2. (c) 3. (a) 4. (b) 5. (b) 6. (c) 7. (b) 8. (b) 9. (a) 10. (a)

11. (b) 12. (c) 13. (c) 14. (b) 15. (a) 16. (b) 17. (c) 18. (a) 19. (d) 20. (a)21. (a) 22. (c) 23. (c) 24. (b) 25. (d) 26. (a)

14. Gauche > Anti > Partially eclipsed > Fully eclipsed15. A homogeneous equimolar mixture of enantiomers is

called a racemic mixture or racemic modification. Aracemic mixture is always optically inactive.

16. (a) 4 (b) 217. (a) = E (b) = E

(c) = E (d) = Z18. (a) 1 (b) 2

(c) 3 (d) 5

19. (a) R (b) S(c) R (d) (R, R)

20. (a) Enantiomers (b) Positional isomers

21. (a) (b)

CHEMICAL EQUILIBRIUM (CHEMISTRY) 3.1

INTRODUCTION

1. CHEMICAL REACTIONSymbolic representation of any chemical change in terms of reactants and products is called chemical reaction.

Types of chemical reaction:

(a) On the basis of physical state:

Homogeneous reaction Heterogeneous reactionAll reactants and products are in same phase. Reactants and products are in two or more phases.

N2(g) + 3H2(g) 2NH3(g) Zn(s) + CO2(g) ZnO(s) + CO(g)

(b) On the basis of direction:Reversible reaction Irreversible reaction(i) Chemical reaction in which products Chemical reaction in which products cannot be

can be converted back into reactants. converted back into reactants.

N2 + 3H2 2NH3 AgNO3 + NaCl AgCl + NaNO3

3Fe + 4H2O Fe3O4 + 4H2 NaCl + H2SO4 NaHSO4 + HCl

H2 + I2 2HI Zn + H2SO4 ZnSO4 + H2

(ii) Proceed in forward as well as backward Proceed only in one direction (forward direction)direction.

(iii) These attain equilibrium These do not attain equilibrium.(iv) Reactants are never completely converted Reactants are nearly completely converted into

into products products.(v) Generally thermal dissociations are held Generally thermal decompositions are held

in closed vessel. in open vessel.

(1) PCl5(g) PCl3(g)+Cl2(g) 2KClO3(s) 2KCl(s) + 3O2(g)

(2) 2HI(g) H2(g) + I2(g)

(vi) Neutralisation reaction except strong Neutralisation reaction of strong acid andacid and strong base. strong base.

HCl + NH4OHNH4Cl + H2O HCl + NaOH NaCl + H2O

CHEMICAL EQUILIBRIUM

Chapter

3

3.2 CHEMICAL EQUILIBRIUM (CHEMISTRY)

2. RATE OF REACTIONThe change in concentration of reactants or products

in unit time is known as rate of the reaction.Rate of reaction

= change in concentration of the reactants( )

time taken for the change

– dcdt

reactants.

Here negative sign indicate that concentration of thereactants decrease with increase in time.

Rate of reaction

= (+)change in concentration of the products

time taken for the change

dcdt

products

Here positive sign indicate that concentration of theproducts increase with increase in time.

Note: The concentration change may be positive ornegative but the rate of reaction is always positive.

Unit of rate of reaction =moles/litre

second =moles

litre ×second = mol L–1 s–1

3. LAW OF MASS ACTION[By Guldberg and Waage]

Rate at which a substance reacts [Active Mass ofthe substance]

Active Mass= Molar concentration i.e. Moles/Litres

= Wt of substance (gram)Molar wt.×Vol.(Litre)

It is represented in square brackets i.e. [ ] e.g. [A],[N2] etc.

The rate of a chemical reaction at a particulartemperature is proportional to the product of active massesof reactants raised to the powers of their stoichiometriccoefficients.

Ex. aA + bB productsRate of reaction [A]a [B]b

Rate = k [A]a [B]b, where k is the rate constant of the reaction.

EXAMPLE 1:Four vessel each of volume V = 10 Lilres contains

(A) 16 g CH4 (B) 18 g H2O(C) 35.5 g Cl2 (D) 44 g CO2Which container will contain same molar concentration

and same active mass as that in (A)?

SOLUTION:

(A) [ CH4] =16

16 10 = 0.1 M

(B) [H2O] =18

18 10 = 0.1 M

(C) [Cl2] =35.5

71 10 = 0.05 M

(D) [CO2] =44

44 10 = 0.1 M

Hence, (B) and (D) has same molar concentration asthat in (A)

EXAMPLE 2:8.5 g ammonia is present in a vessel of 0.5 litre capacitythen find out the active mass of ammonia?

SOLUTION:

? ?38.5NH 1 mole/litre

17× 0.5? ?

a. State of Chemical EquilibriumState of equilibrium means the balance of driving forcesi.e. the factors taking the reaction in forward directionand the backword direction.

The equilibrium state represents a compromisebetween two opposing tendencies.

(c) On the basis of heat: Exothermic reaction Endothermic reaction(i) Heat is evolved in these type of chemical Heat is absorbed in these type of chemical reactions.

reactions.R P + x kcal R P – x kcal

(ii) Change in heat energy (with product) Q = (–) veQ = (+) ve

(iii) Change in enthalpy or internal energy H = (+) veH = (–) ve

Note: (i) Whenever questioner doesn't ask about direction then we take forward direction only.(ii) In a reversible reaction if forward reaction is exothermic then the backward reaction will be endothermic and vice-

versa.


(i) System try to minimise energy.(ii) System try to maximise entropy.In a reversible reaction like–

1 2Reactants

R R 1 2Products

P P

Initially only reactants are present. R1 and R2 combineto form P1 and P2. As soon as P1 and P2 are formed, theystart the backward reaction. As concentrations of R1 andR2 decrease rate of forward reaction decreases and rateof backward reaction increases. Ultimately a stage isreached when both the rates become equal. Such a stateis known as “ Chemical Equilibrium” or “state ofEquilibrium”.

At equilibrium:(i) Rate of forward reaction (rf) = rate of backward

reaction (rb)(ii) Concentration (mole/litre) of reactant and product

remains constant with respect to time.

Forward reaction (r )f

Backward reaction (r )b

rate

TimeEquilibrium r = r f b

(R + R P + P )1 2 1 2

(P + P R + R )1 2 1 2

Reactant

Product

Concentration

TimeEquilibrium

Product

Concentration

TimeEquilibrium

Reactant [R ] and [R ]1 2

[P ] and [P ]1 2

Physical Equilibrium

ChemicalEquilibrium

Equilibrium in chemical process is called chemical

equilibrium.

For examplephase changes like H O(l) 2 H O(g) ;

Solvation like

2

Equilibrium in physical process is called

equilibrium.physical

For exampleH (g) + Cl (g) 2HCl(g)2 2

H O2

excess NaCl(aq)NaCl(s)

Types of equilibria on the basis of process

Types of equilibria on basis of physical state

Homogeneous equilibrium Heterogeneous equilibriumWhen all reactants and products When more than one phase areare in same phase present

H2(g) + Cl2(g) 3Fe(s) + 4H2O(g) 2HCl(g) Fe3O4(s) + 4H2(g)

SO2(g) + NO2(g) 2Na2O2(s) + 2H2O() SO3(g) + NO(g) 4NaOH + O2(g)

b. Characteristics of Chemical Equilibrium:

1. The nature and the properties of the equilibriumstate are the same regardless of the direction fromwhich it is achieved. It can be achieved from bothdirection.

2. Equilibrium is dynamic in nature.It means that reaction has not stopped. It appearsthat no change is occuring but But both theopposing reactions are proceeding at the same rate.So there is no net change.Thus equilibrium is notstatic in nature.

3. A catalyst can change the rate of approach ofequilibrium but does not change the state ofequilibrium. By using catalyst, the equilibrium canbe achieved in different (more/less) time, but therelative concentrations of reactants and productsare same irrespective of the presence or absenceof a catalyst.


4. Equilibrium can be observed by constancy of someobservable properties like colour, pressure,concentration, density, temperature, refractiveindex etc.which may be suitable in a givenreaction.

5. At equilibrium, free energy changeG = 06. Equilibrium state can be affected by changing

factors like pressure, volume, concentration andtemperature etc.(Le Chateliers Principle).

7. System moves toward an equilibrium statespontaneously even if it is disturbed. It will returnto original state.

EXAMPLE 3:(1) Consider the following cases–

() 2HI

H +2 2I

Equilibrium state has been attained

()

T >T >T >T1 3 42

Temperatures becomes constant

Furnace Metal Rod

)

H2O( )l

Level of waterbecomes constant

The flow of energy in case () is same as that in–(A) (B) (C) and (D) None

SOLUTION: (D)None, Because in and , the flow of energy or matter istaking place only in one direction. While in equilibriumstate, the flow of energy takes place in both directionsequally. Thus () is a dynamic equilibrium while states in and are called steady state (static equilibrium).

4. EQUILIBRIUM CONSTANT (K)For a general reaction

aA + bB cC + dD,Forward reaction rate rf = kf [A]a [B]b ,Backward reaction rate rb = kb [C]c[D]d ,

At equilibrium rf = rbkf [A]a [B]b = kb [C]c [D]d

The concentrations of reactants & products atequilibrium are related by

f

b

KK = KC =

[ ] [ ][ ] [ ]

c d

a bC DA B

Kc is a constant and is called the equilibriumconstant in terms of concentration.

where all the concentrations are at equilibrium andare expressed in moles/litre.e.g. PCl5 (g) PCl3 (g) + Cl2(g)

KC = 3 2

5

[PCl ] [Cl ][PCl ]

e.g. N2(g) + 3H2(g) 2NH3(g)

KC = 2

33

2 2

[NH ][N ] [H ]

e.g.12 H2(g) +

12 I2(g) HI(g)

Kc = 1/ 2 1/ 22 2

[HΙ][H ] [ ]

EXAMPLE 4:In a reaction A (g) + B (g) C (g)+ D(g) , A, B, aremixed in a vessel at temperature T. The initial concentrationof A was twice the initial concentration of B. After theequilibrium is reached, concentration of C was thrice theconcentration of B Calculate KC.

SOLUTION:Let concentration of B initially is ‘a’ mole/litre

A + B C + D n = 0at t = 0 2a a 0 0at t = teq 2a – x a – x x xGiven that

x = 3 (a – x) x = 34

a KC = [ ] [ ][ ] [ ]C DA B

Kc = .

(2 )( )x x

a x a x Kc =

234

3 324 4

a

a aa a

Kc =95

= 1.8

KP Equilibrium constant in terms of partialpressure. It is defined for the equilibrium reaction whichcontains at least one gaseous component.


e.g. aA(g) + bB(g) cC(g) + dD(g)

KP =[ ] [ ][ ] [ ]

c dC D

a bA B

P PP P

where various pressures are the partial pressures of variousgases substancs.

EXAMPLE 5:Calculate the expression for Kc and Kp if initially a molesof N2 and b moles of H2 is taken for the following reaction.

N2 (g) + 3H2 (g) 2NH3 (g) (n < 0)(P, T, V given)

SOLUTION:

N2 (g) + 3H2 (g) 2NH3 (g) (n < 0)(P, T, V given)

At t = 0 a b 0 t = teq (a – x) (b – 3x) 2x

[N2] = a x

V

, [H2] = 3b x

V

, [NH3] = 2xV

KC =

2

3

2

3

xV

a x b xV V

= 2 2

34

( )( 3 )x V

a x b x

Total no. of moles at equilibrium = a + b – 2x

2[ ]NP =

( )2

a xa b x

. P,

2[ ]HP =

( 3 )2

b xa b x

. P,,

3[ ]NHP =

(2 ).2

x Pa b x

KP =3

2 2

2NH

3N H

[P ]

[P ][P ]

=

2

3

2 .2

( 3 ).2 2

x Pa b x

a x b x PPa b x a b x

KP =

2 2

2

34

4

4 .( 2 )( )( 3 ).

( 2 )

x Pa b xa x b xP

a b x

= 2 2

2 3( 2 ) .4

( )( 3 )a b x x

P a x b x

Relation between Kp & KC

PV = nRT or, P = nV RTT

P =CRT where C =nv = (moles per litre)

PC = [C] RT ; PD = [D] RT ; PA = [A] RT ;PB = [B] RT

KP = [ ] ( ) [ ] ( )[ ] ( ) [ ] ( )

c c d d

a a bC RT D RTA RT B RTb

= [ ] [ ][ ] [ ]

c d

a bC DA B (RT)(c+d) – (a+b)

Kp = Kc(RT)n

Where n = (c + d) – (a + b), calculation of ninvolves only gaseous components.

n = sum of the number of moles of gaseousproducts – sum of the number of moles of gaseousreactants. n can be positive, negative, zero or evenfraction.

CaCO3 (s) CaO (s) + CO2(g)

n = 1 (because there is only one gas component inthe products and no gas component in the reaction)

Kp = Kc.(RT)

Unit of Equilibrium contants:

Unit of Kp is (atm)n .Unit of Kc is (mole/Lit)n = (conc.)n

Note: In fact, equilibrium constant does not carryany unit because it is based upon the activities of reactantsand products and activities are unitless quantities. Underordinary circumstances, where activities are not known,above types of equilibrium constant and their units areemployed.

For pure solids and pure liquids, although theyhave their own active masses but they remainconstant during a chemical change (reaction).Therefore, these are taken to be unity for the sakeof convenience.

e.g. CaCO3(s) CaO(s) + CO2(g)

KC = [ CO2], KP = 2COP

EXAMPLE 6:

Calculate Kp and Kc if initially a moles of PCl5 is takenPCl5 (g) PCl3(g) + Cl2(g)


Sol.PCl5 (g) PCl3(g) + Cl2(g)At t = 0 a 0 0At t = teq (a – x) x x

[PCl5] = a x

V

, [PCl3] =xV , [Cl2 ] =

xV

KC =

2xv

a xv

Total no. of moles = a – x + x + x = a + x

[5PClP ] =

( )a x Pa x , [

3PClP ] = .x P

a x ,

[2ClP ] =

.x Pa x

KP = .xP xP

a x a xa x P

a x

= 2

2 2x P

a x

5. APPLICATIONS OF EQUILIBRIUM CONSTANT Predicting the direction of the reaction

Reaction Quotient (Q)At each point in a reaction, we can write a ratio of

concentration terms having the same form as theequilibrium constant expression. This ratio is called thereaction quotient denoted by symbol Q.

It helps in predicting the direction of a reaction.

The expression Q =[ ] [ ][ ] [ ]

c d

a bC DA B at any time during

reaction is called reaction quotient.

The reaction quotient is a variable quantity withtime.

It helps in predicting the direction of a reaction. if Q > Kc reaction will proceed in backward

direction until equilibrium in reached. if Q < Kc reaction will proceed in forward

direction until equilibrium is established.

if Q = Kc Reaction is at equilibrium.

eg. 2A(g) + B(g) C(g) + D(g)QC = Reaction quotient in terms ofconcentration

QC = 2[ ][ ][ ] [ ]C DA B

KC = 2

[ ] [ ][ ] [ ]

eq eq

eq eq

C DA B

[Here all the conc. are at equilibrium]

EXAMPLE 7:In the following reaction which one oxide is more stable.

2XO (g) X2(g) + O2(g); K1 = 1 × 1024

2XO2 (g) X2(g) + 2O2(g) ; K2 = 2.5 × 1010

SOLUTION:

K1 > K2 So the stability of XO2 > XO

EXAMPLE 8:

For the reaction NOBr (g) NO(g) +12 Br2 (g)

KP = 0.15 atm at 90°C. If NOBr, NO and Br2 aremixed at this temperature having partial pressures 0.5 atm,0.4 atm and 0.2 atm respectively, will Br2 be consumedor formed ?

SOLUTION:

QP = 2

1/ 2[ ]

[ ]Br NO

NOBr

P PP

=1/ 2[0.2] [0.4]

[0.50] = 0.36

KP = 0.15

QP > KPHence, reaction will shift in backward direction Br2 will be consumed

Predicting the extent of the reaction

K = [Product][Reactant]

K or Kc pNegligible Very large

reaction hardly proceeds

Both reactant and products are significant

reaction proceeds almost to completion

10-3 10+31

Case-I: If K is large (K > 103) then productconcentration is very very larger than the reactant([Product] >>[Reactant]) Hence concentration of reactantcan be neglected with respect to the product. In this case,the reaction is product favourable and equilibrium will bemore in forward direction than in backward direction.


Case-II: If K is very small (K < 10–3)[Product] << [Reactant]

Hence concentration of Product can be neglected ascompared to the reactant.

In this case, the reaction is reactant favourable.

EXAMPLE 9:The KP values for three reactions are 10–5, 20 and 300then what will be the correct order of the percentagecomposition of the products.

SOLUTION:Since Kp order is 10–5 < 20 < 300 so the percentagecomposition of products will be greatest for Kp = 300. Calculating equilibrium concentrations

The concentration of various reactants and productscan be calculated using the equilibrium constant and theinitial concentrations.

EXAMPLE 10:1 mole of N2 and 3 moles of H2 are placed in 1L vessel.Find the concentration of NH3 at equilibrium, if equilibriumpressure is 1 atm and the equilibrium constant at 400K

is 4

27

SOLUTION:N2 (g) + 3H2(g) 2NH3(g) (n < 0)1mol 3 mol 0(1 – x) (3 – 3x) 2xPeq = 1 atm, T = 400 K

KC =2

33

2 2

[NH ][N ][H ] =

2

3(2 )

(3 3 ) (1 )x

x x = 4

272

4(1 )x

x = 1 x = (1 – x)2 x2 – 3x + 1 = 0

x = 3 9 4

2

x =3 5

2

x =3 2.24

2

or x = 3 2.24

2

x = 5.24

2 = 2.62 or, x = 0.76

2 x = 0.38 (since × cannot be greater than 1) [NH3] = 0.38 × 2 = 0.76

EXAMPLE 11:1.1 mole of A mixed with 2.2 mole of B and the mixture iskept in a 1 litre at the equilibrium A + 2B 2C + D is

reached. At equilibrium 0.2 mole of C is formed then thevalue of KC will be:

SOLUTION:A + 2B 2C + D

Initial mole 1.1 2.2 0 0At Eq. 1.1 – x 2.2 – 2x 2x x

1.1 – 0.1 2.2 – 0.2 0.2 0.11 2 0.2 0.1

Active mass11

21 0.2 0.1

KC = 2

2[ ] [ ][ ][ ]C DA B =

2 2 110 10 10

1 2 2

= 1

1000 = 0.001

EXAMPLE 12:The Kp for the reaction, N2O4 2NO2 is 640 mm at775K. Calculate the percentage dissociation of N2O4 atequilibrium pressure of 160 mm. At what pressure thedissociation will be 50% ?

SOLUTION:Suppose number of moles of N2O4 taken = 1 mol

Degree of dissociation = N2O4 2NO2

Initial moles 1 mol 0At Eq. (1–) mol 2 molTotal number of moles at equilibrium= 1 – + 2 = 1 + Let P be the total pressure at equilibrium, then

= 2 4N OP =

(1 )(1 ) P

2NOP =

2(1 )

P

Kp = 2

2 4

2NO

N O

PP =

2(1 )11

P

P

= 24

(1 )(1 )

P

Substituting Kp = 640 mm and P = 160 mm

640 =2

24

1

160

= 0.707 ot 70%For 50% dissociation, a = 0.5, Substituting the value

of Kp and a in equation (i)


640 =2

24(0.5)

1 (0.5) P

P = 480 mm

EXAMPLE 13:At 540 K, 0.10 mol of PCl5 are heated in a 8.0 L flask.The pressure of the equilibrium mixture is found to be 1.0bar. Calcualted Kp and Kc for the reaction.

SOLUTION:PCl5 PCl3 + Cl2

Initial moles 0.1 0 0At. Eq. 0.1 – x x xTotal number of moles at equilibrium= (0.1 – x) + x + x = 0.1 + xBut total number of moles,

n = PVRT = 1 1

1bar 80.083 bar mol 540

LL K K? ?

??

= 0.18

0.1 + x = 0.18x = 0.08

[PCl5] = 0.1 0.08

8

M = 2.5 10–3 M

[PCl3] = 0.08

8 M = 0.01 M

[Cl2] = 0.08

8 M = 0.01 M

Kc = 3 2

5

[PCl ] [Cl ][PCl ] = 3

0.01 0.012.5 10

= 4 10–2

Kp = Kc (RT)n = 4 10–2 (0.083 540) = 1.79 bar

6. CHARACTERISTICS OF EQUILIBRIUMCONSTANT & FACTORS AFFECTING IT:

a. Equilibrium constant does not depend uponconcentration of various reactants, presence ofcatalyst, direction from which equilibrium isreached

b. The equilibrium constant does not give any ideaabout time taken to attain equilibrium.

c. K depends on the stoichiometry of thereaction.

(a) If two chemical reactions at equilibrium havingequilibrium constants K1 and K2 are added thenthe resulting equation has equilibrium constantK = K1 . K2

Equation constant

A (g) B(g) K1

B (g) C(g) K2

On adding A(g) C(g) K = K1 . K2

(b) If the reaction having eq. constant K1 is reversed

then resulting equation has eq. constant 1

1K

A(g) + B(g) C(g) + D(g)

On reversing, C(g) + D(g) A(g) + B(g) K = 1

1K

(c) If a chemical reaction having equilibriumconstant K1 is multiplied by a factor n thenthe resulting equation has equilibrium constantK = (K1)n , n can be fraction

eg. D2(g) 2A(g) K1

Multiplying by 12

,12 D2(g) A(g) K = (K1)1/2 = 1K

EXAMPLE 14:The value of Kc for the reaction, N2(g) + 2O2(g) 2NO2(g) at a certain temperature is 400. calculate the valueof equilibrium constant for.

(i) 2NO2(g) N2(g) + 2 O2(g) ;

(ii) 1/2 N2(g) + O2(g) NO2(g)

SOLUTION:Equilibrium constant (Kc) for the reaction N2 (g)+2O2(g) 2NO2(g) is

Kc =2

22

2 2

[NO ][N ][O ] = 400

(i) For the reaction 2NO2(g) N2(g)+ 2O2(g),

K’C =2

2 22

2

[N ][O ][NO ] =

1

cK

K’c = 4001

= 0.0025 mole litre–1

(ii) For the reaction 1/2 N2(g) + O2(g) NO2(g)

K’’c =2

1/22 2

[NO ][N ] [O ] = cK

K’’c = 400 = 20 litre–1/2 mole–1/2


EXAMPLE 15:What should be the relationship between K2 and K1, if theequilibrium constant of the reaction given below are K1and K2 respectively ?

2SO2(g) + O2(g) 2SO3(g)

SO2(g) + 12 O2(g) SO3(g)

(1) K2 = K1 (2) K2 = 1K

(3) K2 = K1 (4) 2K2 = K1

SOLUTION:

2SO2(g) + O2(g) 2SO3(g) +12 O2(g) SO3(g)

K1 = 2

32

2 2

[SO ][SO ] [O ] K2 =

3

2 2

[SO ][SO ][O ]

Therefore, K2 = 1K

d. Equilibrium constant is dependent only on thetemperature.It means Kp and Kc will remain constant at constant

temperature no matter how much changes are made inpressure, concentration, volume or catalyst. However if temperature is changed,

log 2

1

kk = 2.303

HR

1 2

1 1T T

;

H = Enthalpy of reaction

If T2 > T1 then K2 > K1 provided H = +ve(endothermic reaction)

K2 < K1 if H = –ve (exothermic reaction)

In the above equation, the unit of R and H/T shouldbe same.

EXAMPLE 16:From the following data:

(i) H2(g) + CO2(g) H2O(g) + CO(g)

K2000K = 4.4

(ii) 2H2O(g) 2H2(g) + O2(g)

K2000K = 5.31 ×10–10

(iii) 2CO(g) + O2(g) 2CO2(g)

K1000K = 2.24 ×1022

State whether the reaction (iii) is exothermic orendothermic?

SOLUTION:

Equation (iii) = – [2 × (i) + (ii)]

K2000 (iii) = 2 2 101 2

1 1(4.4) 5.31 10K K

= 9.7 × 107

T K reaction is exothermic.

7. HOMOGENEOUS LIQUID SYSTEM:FORMATION OF ETHYL ACETATE

The reaction between alcohol and acid to form ester is anexample of homogeneous equilibrium in liquid system.

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) +H2O(l)

KC = 3 2 5 2

3 2 5

[CH COOC H ][H O][CH COOH][C H OH]

EXAMPLE 17:In an experiment starting with 1 mole of ethyl alcohol, 1mole of acetic acid and 1 mole of water at T°C, theequilibrium mixture on analysis shows that 54.3% of theacid is esterfied. Calculate the equilibrium constant of thisreaction.

SOLUTION:

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) +

H2O(l)Initial 1 1 0 1At equilibrium 1 – x 1 – x x 1 + x

1 – 0.543 1 – 0.543 0.543 1 + 0.543

(54.3% of 1 mole =1 54.3

100

= 0.543 mole)

Hence given x = 0.543 moleApplying law of mass action:

KC =[ester][water][acid][alcohol] =

0.543 1.5430.457 0.457

= 4.0

EQUATION INVOLVING IONS:Equilibrium involving ions always take place in aquous

medium . In case of expression of KC concentration ofion is taken.

Ex. Ag+(aq.) + Cl–(aq.) AgCl(s) Kc = + –1

[Ag ][Cl ]


8. HETEROGENOUS EQUILIBRIUM:For pure solid and pure liquid, active mass is takento be unity i.e. 1 as they remain constant throughoutthe reaction

CaCO3 (s) CaO (s) + CO2 (g)

KP = 2COP , KC = [CO2 (g)]

[CaCO3(s)] =moles

volume =

3

3

CaCO

CaCO

WM

V

= 3

3

CaCO

density CaCOM = constant

K = 2

3

[CaO(s)][CO (g)][CaCO (s)]

3K.[CaCO (s)]

[CaO(s)] = [CO2(g)]

KC = [CO2(g)]

H2O(l) H2O(g)

KP = 2H O(g)P , KC = [H2O (g)]

[For pure solid and pure liquid active mass is takenas unity i.e. = 1]

EXAMPLE 18:In a reaction C(s) + CO2 (g) 2CO(g), the equilibriumpressure is 12 atm. If 50% of CO2 reacts. Calculate KP.

SOLUTION:

C(s) + CO2 (g) 2CO(g)t = 0 a 0

t = teq a – 2a

22a

Peq = 12 atm

2COX = 13 , COX =

23

2COP =13 × 12 = 4

COP =23 × 12 = 8

KP =8 8

4

= 16

9. DEGREE OF DISSOCIATION (A):It is the fraction of one mole dissociated into the products.

(Defined for one mole of substance)So, = no. of moles dissociated / initial no. of moles

taken= fraction of moles dissociated out of 1 mole.Note: % dissociation = × 100Suppose 5 moles of PCl5 is taken and if 2 moles of

PCl5 dissociated then = 25 = 0.4

Let a gas An dissociates to give n moles of A as follows-

An (g) n A (g)

t = 0 a 0

t = teq a – x n.x = xa x = a.

a – a = a(1-) n a

Total no. of moles = a – a + n a

= [1 + (n – 1)] a

Significance of n

n = sum of stoichiometric coefficient of product

sum of coefficient of reactants

(i) for PCl5(g) PCl3(g) + Cl2(g) (n = 2)

(ii) for 2NH3(g) N2(g) + 3H2(g) (n = 32 +

12 = 2)

(iii) for 2HI(g) H2(g) + I2(g) (n = 1)

EXAMPLE 19:Calculate the degree of dissociation and Kp for the followingreaction.

PCl5(g) PCl3(g) + Cl2(g)t = 0 a 0 0t = t a –x x xSince for a mole, × moles are dissociated

SOLUTION:

For 1 mole, xa moles = are dissociated

x = a PCl5(g) PCl3(g) + Cl2(g)

At t = teq a – a a a

Total no. of moles at equilibrium = a + a = a (1 + )


5PClP =(1 )

(1 )a P

a ,

3PClP = .

1a Pa

,

2ClP = (1 )a

a .P

KP =

2

111

P

P

KP = 2

2.

1P

(Remember)

Observed molecular weight and Observed VapourDensity of the mixture

Observed molecular weight of An(g)

= nmolecular weight of A (g)total no.of moles at equilibrium

= .

(1 ( 1) )tha M

a n

Mobs = [1 ( 1) ]thM

n

where Mth = theoritical molecular weight (n = atomicity)

Mmixture = [1 ( 1) ]nAM

n ,

nAM = Molar mass of gas An

Vapour density (V.D).: Density of the gas dividedby density of hydrogen under same temp & pressure iscalled vapour density.

D = vapour density without dissociation = 2

nAM

d = vapour density of mixture = observed v.d. = mix2

M

Dd = 1 + (n – 1)

0( 1) ( 1)

T oM MD dn d n M

where MT = Theoritical molecular wt. M0 = observedmolecular wt. or molecular wt. of the mixture atequilibrium.

Note: It is not applicable for n = 1 [eg. Dissociationof HI & NO].

EXAMPLE 20:The vapour density of a mixture containing NO2 and N2O4is 38.3 at 33°C calculate the no. of moles of NO2 if 100gof N2O4 were taken initially.

SOLUTION:

N2O4(g) 2NO2(g)Mmix = 2 × 38.3 = 76.6

Mmix = 1thM

=

921 = 0.2

N2O4 2NO2

t = 0 a 0t = t a – a 2a

no. of moles of NO2 = 2a = 2 100 0.2

92

= 0.435

EXAMPLE 21:When sulphur in the form of S8 is heated at 900 K, theinitial pressure of 1 atm falls by 29% at equilibrium. Thisis because of conversion of some S8 to S2. Find the valueof equilibrium constant for this reaction.

SOLUTION:S8(g) 4S2(g)

Initial pressure 1 atm 0Eq. Pressure (1 – x) atm 4x atmBut x = 0.29 atm

8SP = 1 – x = 1 – 0.29 = 0.71 atm

2SP = 4x = 4 029 = 1.16 atm

Kp = 4

2

8

( )pSpS =

4(1.16)0.71

= 2.55 atm3

10. EXTERNAL FACTORS AFFECTING EQUILIBRIUM

Le Chatelier's Principle:If a change is applied to the system at equilibrium, thenequilibrium will be shifted in that direction in which it canminimise the effect of change applied and the equilibriumis established again under new conditions.

a. Effect of concentration: If the concentration of acomponent is increased, reaction shifts in a directionwhich tends to decrease its concentration. e.g. In thefollowing example.

N2 (g) + 3H2(g) 2NH3(g)

[reactant] Forward shift

[Product] Backward shift If concentration of reactant is increased at

equilibrium then reaction shifts in the forward direction . If concentration of product is increased then

reaction shifts in the backward direction


Note: The addition of any solid component does notaffect the equilibrium.

b. Effect of volume: If volume is increased, pressure decreases hence

reaction will shift in the direction in which pressureincreases that is in the direction in which numberof moles of gases increases and vice versa.

If volume is increased then, forng > 0 reaction will shift in the forward directionng < 0 reaction will shift in the backwarddirectionng = 0 reaction will not shift. eg. H2(g) + I2(g)

2HI(g) (No effect)

Explanation:(i) ng > 0, eg. PCl5(g) PCl3(g) + Cl2 (g)

QC =

32

5

( )( )

( )

PClCl

PCl

nnV V

nV

?

QC1V

for ng > 0

On increasing V, QC, decreases.Now, for QC < KC reaction will shift in forwarddirection.

Thus, if, Volume QC (Forward shift)

Volume QC (Backward shift)

(ii) ng < 0, eg. N2(g) + 3H2(g) 2NH3(g)

QC =

3

2 2

2

3

( )

( ) ( )

NH

N H

nV

n nV V

QC V2 for ng < 0

V QC (Backward shift) ;

V QC (Forward shift)

c. Effect of pressure: On increasing pressure,equilibrium will shift in the direction in which pressuredecreases i.e. no. of moles in the reaction decreases andvice versa.

P no. of moles(i) For ng = 0 No. effects

(ii) For ng > 0, PCl5(g) PCl3(g) + Cl2(g)

Qp = 3 2

5

( ). ( )( . )

PCl Cl

PCl

X P X PX P

QP P [X = mole fraction]

P ; QP ; (Forward shift) [P = Total pressureat equilibrium]

P ; QP ; (Backward shift)

(iii) For n < 0, eg. N2(g) + 3H2(g) 2NH3(g)

QP = 3

2 2

2

3

( ) )

[( ). ][( ) ]NH

N H

X P

X P X P

QP 21

P

P ; QP ; (Forward shift) ;

P ; QP ; (Backward shift)

d. Effect of catalyst: Due to catalyst, the state ofequilibrium is not affected i.e. no shift will occur ascatalyst lowers the activation energy of both the forward& reverse reaction by same amount, thus altering theforward & reverse rate equally and hence, the equilibriumwill be attained faster i.e time taken to reach the equilibriumis less.

e. Effect of inert gas addition:(i) At constant volume: Inert gas addition has no

effect at constant volume.(ii) At constant pressure: If inert gas is added then

to maintain the pressure constant, volume isincreased. Hence equilibrium will shift in thedirection in which larger no. of moles of gas isformed(i) ng > 0, reaction will shift in the forward

direction(ii) ng < 0, reaction will shift in the backward

direction(iii) ng = 0, no effect

6. Effect of temperature:(i) Exothermic reaction: The reaction in which heat

is evolvedA(g) + B(g) C(g) + D(g) + Heat H = – ve

eg. N2(g) + 3H2(g) 2NH3(g) + Heat

T K' will decrease (from vant’ hoffequation)

log 1

2

KK =

2 1

º 1 12.303

HR T T


log 1

2

KK < 0

log K1 – log K2 > 0 log K1 > log K2 K1 > K2Reaction will shift in backward direction.T K will increases.Reaction will shift in forward direction.

(ii) Endothermic reaction: energy consumed.A(g) + B(g) C(g) + D(g) – Heat H = + veT K Forward ;

T K Backward

EXAMPLE 22:What will be the amount of dissociation, if the volume isincreased 16 times of initial volume in the reactionPCl5 PCl3 + Cl2 ?

(1) 4 times (2) 14 times

(3) 2 times (4) 15 times

Ans. (1)

SOLUTION:x V? = 16x ? . Thus, 4 times.

EXAMPLE 23:At certain temperature, the equilibrium constant (Kc) is16 for the reaction

SO2(g) + NO2(g) SO3(g) + NO(g)If we take one molw each each of all the four gases

in 1 L container, what be comcentration of NO and NO2at equilibrium ?

SOLUTION:SO2(g) + NO2(g) SO3(g) = NO(g)Suppose x moles of SO2 react with x moles of NO2 to

form x moles of SO3 and x moles of No to attain equilibrium.The equilibrium concentration, therefore woud be

[SO2] = (1 – x) mol L–1 ; [NO2] = (1–x) mol L–1 ;[SO3] = (1 + x) mole L–1 ; [NO] = (1 + x) miole L–1

Kc = 3

2 2

[SO ][NO][SO ][NO ] =

(1 )(1 )(1 )(1 )

x xx x

= 16

2

2(1 )(1 )

xx

= 16 or

(1 )(1 )

xx

= 4

1 + x = 4 – 4x or 5x = 3

or x =35 = 0.6 mole

[NO2] = (1 – x) = ( 1 – 0.6) = 0.4 mol L–1

[NO] = (1 + x) = (1 + 0.6 ) = 1.6 mol L–1

EXAMPLE 24:Nitrogen and hydrogen react to form ammonia as per thereaction

1/2 N2(g) + 3/2H2(g) NH3(g)When the mixture of the three gases is in equilibrium

predict whether the amount of ammonia increases ordecreases if

(i) The pressure on the system is increased,(ii) The temperature of the system is raised,(iii) The concentration of hydrogen is increased.

SOLUTION:(i) When pressure is increased, equilibrium shifts to

that direction in which pressure decreases i.e. inthe direction in which the number i of moles ofgases decreases. hence, the reaction shifts in theforward direction and thus the amount of NH3increases.

(ii) As the forward reaction is exthormic, increasesof temperature will shifts the equilibrium in theback directiuon (endothermic direction) and thusthe amout of NH3 decreases.

(iii) On increasing the concentration of H2 theequilibrium will shifts in the forward direction andthus the amount of NH3 increases.

EXAMPLE 25:An endothermic reaction

A(g) + 2B(g) 2C(g)is in equilibrium at a certain temperature can we increasethe amount of C by

(i) Adding catalyst(ii) Increasing pressure(iii) Increasing temperature

SOLUTION:(i) No, because catalyst does not distrub the

equilibrium state(ii) Yes, because increase in pressure will shift the

equilibrium in forward direction as the number ofmoles of products is less than that of reactants.

(iii) Yes, because increase in temperature would shiftthe reaction in the forward direction 9endothermicdirection)

EXAMPLE 26:The volume of a closed reaction vessel in which theequilibrium.

2SO2(g) + O2(g) 2SO3(g) is halved, Now(A) The rates of forward and backward reactions will

remains the same.


(B) The equilibrium will not shift.(C) The equilibrium will shift to the right.(D) The rate of forward reaction will become double

that of reverse reaction and the equilibrium willshift to the right.

SOLUTION:In the reaction

2SO2(g) + O2(g) 2SO3(g)In this reaction three moles (or volumes) of reactants

are converted into two moles (or volumes) of productsi.e. there is a decreases in volume and so if the volume ofthe reaction vessel is halved the equilibrium will be shiftedto the right i.e. more product will be formed and the rateof forward reaction will incerases i.e. double that ofreverse reaction.

11. APPLICATION OF LE CHATELIER’S PRINCIPLE:PRACTICAL EQUILIBRIUM SITUATIONS

a. Vapour Pressure of Liquid:It is the pressure exerted by the vapours over it’sliquid when it is in equilibrium with the liquid. Vapourpressure of water is also called aqueous tension.

H2O () H2O (g) ; KP = 2H OP = constant at

fixed temperatureHence V.P. of liquid is independent of pressure, volumeand concentration change.e.g. at 25°C, vapour pressure of water 24 mm ofHgRelative Humidity

= 2

2

Partial pressure of H O vapoursVapour pressure of H O at that temp.

b. Formation of diamond:C (graphite) C (diamond) – Heat; H = + ve

Density Low Density HighVolume High Volume LowFormation of diamond is favourable at high pressureand high temperature

c. Melting of ice:H2O (s) H2O (); H = + ve

Density Low Density HighVolume High Volume Low

Melting of ice is favourable at high temperature andhigh pressure.

d. Boiling of water:H2O() H2O(g)

Density High Density LowVolume Low Volume High

On incerasing pressure, equilibrium will shift in thedirection in which volume is decreasing i.e. backward.Hence, on incerasing pressure, the boiling pointincreases.

e. Formation of ammonia by Haber’s process:N2 (g) + 3 H2 (g) 2NH3(g)H = – 22.4 Kcal/mol.(i) The reaction will shift in the forward direction at

low temperature, but at very low temperature therate of reaction becomes very low; thus moderatetemperature is used for this reaction.

(ii) At high pressure, reaction will shift in forwarddirection to form more product.

f. Manufacturing of SO3 by contact process2SO2(g) + O2(g) 2SO3(g) + 45.2 kcalHigh pressure (1.5 to 1.7 atm), Low temperature(500°C), Higher qunatity of SO2 and O2 arefavourable conditions for the formation of SO3.

g. Manufacturing of NO by Birkeland–Eyde processN2(g) + O2(g) 2 NO(g) – 43.2 kcal(i) No effect on change of pressure(ii) High temperature (1200 °C to 2000 °C), High

concentration of N2 and O2 are favourablecondition for the formation of NO.

EXAMPLE 27:Which of the following conditions should be morefavourable for increasing the rate of forward reaction inthe equilibrium H2 H + H (H = +ve) ?

(1) 2000 C temperature and 760 mm pressure.(2) 3500 C temperature and 100 cm pressure.(3) 3500 C temperature and 1 mm pressure.(4) All are wrong.

Ans. (3)

SOLUTION:In H2 H + H, heat has to be provided to dissociate H2into H. Therefore, the reaction is endothermic(H will positive). So, temperature should be high. Since,one mole of H2 forms two atoms of H, so volume isincreasing (n is positive) so pressure should be low forincreasing the rate of forward reaction.

EXAMPLE 28:The equilibrium constant of the reaction at 25°C

CuSO4.5H2O (s) CuSO4.3H2O(s) + 2H2O(g)is 1.084 × 10–4 atm2. Find out under what conditions ofrelative humidity, CuSO4.5H2O will start loosing its water ofcrystallization according to above reaction. (Vapour pressureof water at 25°C is 24 mm of Hg).


SOLUTION:

KP = 2

2H OP so

2H OP = 41.084 10

= 1.041 ×10–2 atm 8 mm of Hg• If in a room, pressure of water is greater than 8

mm of Hg then CuSO4.3H2O will absorb waterfrom air and will form CuSO4.5H2O & will keepabsorbing until partial pressure of H2O becomes8 mm of Hg.

• If 2H OP < 8 mm of Hg then CuSO4.5H2O will

loose water of crystallization and reaction willmove in forward direction.

i.e. If relative humidity <824 < 33.33%

then CuSO4.5H2O will loose water ofcrystallization.

12. THERMODYNAMICS OF EQUILIBRIUMFor a general reaction, mA + nB pC + qD, G

is given by-G = Gº + 2.303 RT log10Qwhere G = Gibb's Free energy change

Gº = Standard Gibb's Free energy changeQ = reaction quotient

Since, at equlibrium, Q = KHere K is thermodynamic equilibrium constant

replacing Kc or Kp ;'o\p\

K = ( ) ( )( ) ( )

p qC D

m nA B

a aa a ; Here aX denotes the activity of X.

In fact, ‘ax’ is the ratio of the activity of substance atequilibrium and its activity in standard condition. That iswhy it is unitless and K is also unitless.Note: (i) Themodynamic equilibrium constant is unitless

since activity is unitless.(ii) For pure solids & pure liquids, activity is unity.(iii) For gases (ideal behaviour), the activity is its

partial pressure (in atm).(iv) For components in solution, activity is molar

concentration. At equilibrium, G = 0 Gº = – 2.303 RT log10KNow since, Gº = Hº – TSºwhere Hº = Standard enthalpy change of the reactionS°Standard entropy change – 2.303 RT log10K = Hº – TSº

log10 K = –º

2.303H

.1

RT + º

2.303S

R

If plot of ln k vs 1T is plotted then it is a straight line

with slope = – HR

, and intercept =

SR

0 T – 1

log

K

or 1/T

0 T – 1

log

K

or 1/T

Endothermic reaction Exothermic reaction

Slope =º

2.303H

R

= tan y intercept = º

2.303S

R

If at temperature T1, equilibrium constant is K1 and atT2, it is K2 then ;

log10K1 = º

2.303H

R

.1

1T +

º2.303

SR

...(i)

log10K2 = º

2.303H

R

. 2

1T

+ º

2.303S

R

...(ii)

[Assuming Hº and Sº remains constant in thistemperature range.]

Subtract eq. (ii) from (i) we get Vant Hoff equation-

log1

2

KK

=0

2.303H

R

2 1

1 1T T

Note: (i) H should be substituted with sign.

(ii) Unit of H/T and gas constant R should besame.

(iii) For endothermic (H > 0) reaction value ofthe equilibrium constant increases with the risein temperature

(iv) For exothermic (H < 0) reaction, value ofthe equilibrium constant decreases withincrease in temperature

Condition for Spontaneity: G < 0 for spontaneousprocess or reaction.

Since, G = H – TS H – TS < 0 T > H/S• G > 0 for non-spontaneous process or reaction.• G = 0 for equilibrium.


EXAMPLE 29:Variation of equilibrium constant K with temperature T isgiven by van’t Hoff equation,

log K = log A – 2.303H

RT

A graph between log K and T–1 was a straight line asshown in the figure and having = tan–1 (0.5) and OP =10. Calculate:

(a) H° (standard heat of reaction) when T = 300 K,

(b) A (pre-exponential factor),(c) Equilibrium constant K, at 300 K,(d) K at 900 K if H° is independent of temperature.

SOLUTION:

(a) log10 K = log10 A – 2.303H

RT

It is an equation of a straight line of the type y = c+ mx

Slope ‘m’ = tanq = 2.303H

R

0.5 = 2.303 8.314H

H° =9.574 J mol–1

(b) Intercept ‘c’ = log10 A = 10 A = 1010

(c) log K = 10 –9.574

2.303 8.314 298

K = 9.96 × 109

(d) log2

1

KK

= 2.303H

R

1 2

1 1T T

log 299.96 10

K

=9.574

2.303 8.314

1 1298 798

On solving K2 = 9.98 × 109

Ans. (a) 9.574 J mol–1 ; (b) A = 1010 ; (c) 9.96 ×109;(d) 9.98 × 109

Simultaneous Equilibrium:If in any container there are two or more equilibria

existing simultaneously involving one or more than onecommon species. Then in both/all equilibrium theconcentration of common species is the total concentrationof that species due to all the equilibria under consideration.e.g. A(s) X(g) + Y(g)

t = 0 a 0 0t = teq a – t t t + u

B(s) Z(g) + Y(g)b 0 0b – u u u + t

KC1 = t (u + t)

KC2 = (u + t) u

EXAMPLE 30:102 g of solid NH4HS is taken in the 2L evacuated flask at57°C. Following two equilibrium exist simultaneously

NH4HS (s) NH3 (g) + H2S (g)

NH3 (g) 12 N2 (g) +

32 H2 (g)

one mole of the solid decomposes to maintain boththe equilibrium and 0.75 mole of H2 was found at theequilibrium then find the equilibrium concentration of allthe species and KC for both the reaction.

SOLUTION:

Moles of NH4HS = 10251 = 2

NH4HS (s) NH3 (g) + H2S (g)1CK

2 0 0 1 1 – x 1

NH3 (g) 12 N2 (g) +

32 H2 (g)

2CK

1 – x 2x 3

2x

Given that moles of H2 =32x

= 0.75 x = 12

KC1 =

1 (1 ) 12 2 8

x [Since V = 2 L]

KC2 =

3 / 2 ½34 4

12

x x

x

=

3 / 2 ½3 18 8

14

= 3 / 2

3 / 2 1 4 (3)(3)64 1 16



1. The equilibrium constant expression for a gas

reaction is Kc = 4 5

3 24 6

2

[NH ] [O ][NO] [H O] write the balanced

chemical equation corresponding to this expression.2. The equilibrium constant for the reaction N2 + 3H2

2NH3(g) is K. What will be the equilibrium

constant for the reaction 12 N2 +

32 H2 NH3 ?

3. Under what condition, a reversible process becomesirreversible ?

4. The value of equilbrium constant depends on what?5. What does the equilibrium constant K < 1 indicate?6. For an exothermic reaction, what happens to the

equilibrium constant if temperature is increased?7. What are the conditions for getting maximum yield

of NH3 by Haber’s process?8. Which measurable property becomes constant in

water water vapour equilibrium at constanttemperature.

9. Write expression for Kp and Kc for the reactionCaCO3(s) CaO(s) + CO2(g).

10. State and explain the Law of Mass Action.11. What are Kc and Kp ? Find out a relationship between

them.12. Define ‘Homogeneous Equilibria and Heterogeneous

Equilibria’. Give two examples of each of them.13. Write the expressions for the equilbrium constant,

Kc for each of the following reactions :(i) 2NOCl(g) 2NO(g) + Cl2(g)(ii) 2 Cu (NO3)2 (s) 2CuO(s) + 4 NO2(g) +

O2(g)(iii) CH3COOC2H5(aq) + H2O()

CH3COOH(aq) + C2H5OH(aq)(iv) Fe3+ (aq) + 3OH– (aq) Fe (OH)3(s)(v) I2(s) + 5F2(g) 2IF5()

14. Explain why pure liquid and solids can be ignoredwhile writing the equilbrium constant expression.

15. What qualitative information can you obtain fromthe magnitude of equilbrium constant?

16. A reaction A (g) + B(g) 2C(g) is an equilibriumat a certain temperature. Can we increases theamount of products by ( i) adding catalyst(ii) increasing pressure?

17. Why does ice melt showly at higher altitudes?18. The value of Kc for the reaction, N2(g) + 2O2(g)

2NO2(g) at a certain temperature is 400.calculate the value of equilibrium constant for.(i) 2NO2(g) N2(g) + 2 O2(g) ;(ii) 1/2 N2(g) + O2(g) NO2(g)

19. If concentration are expressed in moles L–1 andpressure in atmospheres, what is the ratio of Kp toKC for the reaction2 SO2 (g) + O2(g) 2SO3(g) at 25ºC?

20. Describe the effect of :(a) addition of CH3OH(b) removed of CO(c) removal of CH3OH, on the equilibrium of the

reaction :2H2(g) + CO(g) CH3OH(g)

21. Discuss the effect of temperature on the equilibriumconstant. How does it change for (a) exothermicreaction (b) endothermic reaction (c) reaction havingzero heat of reaction?

22. A sample of HI (g) is placed in a flask at a pressureof 0.2 atm. At equilbrium, the partial pressure ofHI(g) is 0.04 atm. What is Kp for the givenequilibrium?

23. The following reaction has attaiend equilibriumCO(g) + 2H2(g) CH3OH(g) ,Hº = – 92.0kJ mol–1

What will happen if(i) volume of the reaction vessel is suddenly

reduced to half?(ii) the partial pressure of hydrogen is suddenly

doubled?(iii) an inert gas is added to the system at constant

volume.24. What is Kc for the following equilibrium when the

equilibrium concentration of each substance is:[SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M?2SO2(g) + O2(g) 2SO3(g)

25. Nitric oxide reacts with Br2 and gives nitrosylbromide as per reaction given below:2NO (g) + Br2 (g) 2NOBr (g)When 0.087 mol of NO and 0.0437 mol of Br2 aremixed in a closed container at constant temperature,


0.0518 mol of NOBr is obtained at equilibrium.Calculate equilibrium amount of NO and Br2.

26. At 1400 K, Kc = 2.5×10–3 for the reaction CH4(g) +2H2S(g) CS2(g) + 4H2(g) . A 10.0L reactionvessel at 1400 K contains 2.00 mole of CH4, 3.0 molof CS2, 3.0 mole of H2 and 4.0 mole of H2S. Thenin which direction reaction will proceed to reachequilibrium.

27. At 473 K, equilibrium constant, Kc, for thedecomposition of phosphorus pentachloride, PCl5is 8.3 × 10–3. If decomposition is depicted as :PCl5(g) PCl3(g) + Cl2(g),rHº = 124.0 kJ mol–1

(a) Write an expression for Kc for the reaction ?(b) What is the value of Kc for the reverse reaction

at the same temperature ?(c) What would be the effect on Kc if (i) more PCl5

is added (ii) pressure is increased (iii) temperatureis increased?

28. At a certain temperaute and a total pressure of 105

Pa, iodine vapour contain 40% by volume of iodineatoms[2 (g) 2I(g)]. Calculate Kp for the equilbrium.

29. Consider the following reactionN2O4 (g) 2NO2 (g) H = 58.6 kJWhat will be the effect of the following changes onthe concentration of N2O4 at equilibrium ?(i) Increasing the pressure(ii) Increasing the temperature(iii) Increasing the volume

(iv) Adding more NO2 (g) to the system withoutchanging temperature and pressure

(v) Adding catalyst.30. At 523 K, 1 litre of partially dissociated PCl5 at 1

atm weighs 2.695 g. Calculate the percentagedissociation of PCl5 at 523 K.

31. A sample of pure PCl5 was introduced into anevacuated vessel at 473 K. After equilibrium wasattained, concentration of PCl5 was found to be0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3,what are the concentrations of PCl3 and Cl2 atequilibrium? (Given : 4.15 = 2.04)

PCl5 (g) PCl3 (g) + Cl2(g)32. At 1127 K and 1 atm pressure, a gaseous mixture of

CO and CO2 in equilibrium with soild carbon has90.55% CO by massC (s) + CO2 (g) 2CO (g)Calculate Kc for this reaction at the abovetemperature.

33. Dihydrogen gas used is produced by reactingmethane from natural gas with high temperaturesteam. The first stage of two stage reaction involvesthe formation of CO and H2. In second stage, COformed in first stage is reacted with more steam inwater gas shift reaction,CO (g) + H2O (g) CO2 (g) + H2 (g) ; Kp = 10.1If a reaction vessel at 400°C is charged with anequimolar mixture of CO and steam such thatPCO = PH2O = 4.0 bar, what will be the partial

pressure of H2 at equilibrium? (Given : 0.1 = 0.316)


Equilibrium, Equilibrium constant & law of mass action

1. In a reaction A + B C + D the rate constant offorward reaction & backward reaction is k1 and k2then the equilibrium constant (k) for reaction isexpressed as –

(a) KC = 2

1

KK (b) KC = 1

2

KK

(c) KC = K1 × K2 (d) KC = K1 + K2

2. The equilibrium constant of the reaction SO2(g) +½O2(g) SO3(g) is 4 × 10–3 atm–1/2. Theequilibrium constant of the reaction 2SO3(g) 2SO2(g) + O2(g) would be :

(a) 250 atm (b) 4 × 103 atm(c) 0.25 × 104 atm (d) 6.25 × 104 atm

3. Molar concentration of 96 g of O2 contained in a 2litre vessel is -(a) 16 mol/litre (b) 1.5 mol/litre(c) 4 mol/litre (d) 24 mol/litre

4. log p

c

KK + log RT = 0 is a relationship for the reaction:

(a) PCl5 PCl3 + Cl2(b) 2SO2 + O2 2SO3(c) H2 + 2 2H(d) N2 + 3H2 2NH3


5. For the following gases equilibrium, N2O4 (g) 2NO2 (g), Kp is found to be equal to Kc. This isattained when :(a) 0ºC (b) 273 K(c) 1 K (d) 12.19 K

6. In the chemical reaction N2 + 3H2 2NH3 atequilibrium, state whether :(a) Equal volumes of N2 & H2 are reacting(b) Equal masses of N2 & H2 are reacting(c) The reaction has stopped(d) The same amount of ammonia is formed as is

decomposed into N2 and H2

7. Ratio of active masses of 22 g CO2, 3g H2 and 7gN2 in a gaseous mixture :(a) 22 : 3 : 7 (b) 0.5 : 3 : 7(c) 1 : 3 : 1 (d) 1 : 3 : 0.5

8. The equilibrium concentration of [B]eq for thereversible reaction A B can be evaluated by theexpression:

(a) KC[A]e–1 (b)

f

b

kk [A]e

–1

(c) kf kb–1[A]e (d) kf kb[A]–1

9. In a chemical equilibrium, the rate constant for thebackward reaction is 7.5 10–4 and the equilibriumconstant is 1.5 the rate constant for the forwardreaction is :(a) 2 10–3 (b) 5 10–4

(c) 1.12 10–3 (d) 9.0 10–4

10. For which reaction is Kp = KC :(a) 2 NOCl(g) 2NO(g) + Cl2(g)

(b) N2(g) + 3H2(g) 2NH3(g)

(c) H2(g) + Cl2(g) 2HCl(g)

(d) 2SO2(g) + O2(g) 2SO3(g)11. For the reaction

CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g)Which one is correct representation :

(a) KP = ? ?2

2H OP (b) KC = [H2O]2

(c) KP = KC(RT)2 (d) All12. For a reaction N2 + 3H2 2NH3, the value of KC

does not depends upon :(1) Initial concentration of the reactants

(2) Pressure(3) Temperature(4) catalyst(a) Only c (b) a,b,c(c) a,b,d (d) a,b,c,d

13. At 1000 K, the value of Kp for the reaction :

A(g) + 2B(g) 3C(g) + D(g) is 0.05 atmosphere.The value of KC in terms of R would be :(a) 20000 R (b) 0.02 R(c) 5 10–5 R (d) 5 10–5 R–1

14. When alcohol (C2H5OH) and acetic acid are mixed

together in equimolar ratio at 27ºC, 100

3 % is

converted into ester. Then the KC for the equilibriumC2H5OH() + CH3COOH () CH3COOC2H5() + H2O().(a) 4 (b) 1/4(c) 9 (d) 1/9

15. Rate of reaction curve for equilibrium can be like :[rf = forward rate , rb = backward rate]

(a) (b)

(c) (d)

16. The reaction A(g) + B(g) C(g) + D(g) isstudied in a one litre vessel at 250°C. The initialconcentration of A was 3n and that of B was n. Whenequilibrium was attained, equilibrium concentrationof C was found to the equal to the equilibriumconcentration of B. What is the concentration of Dat equilibrium?(a) n/2 (b) (3n – 1/2)(c) (n – n/3) (d) n

17. In the reaction C(s) + CO2(g) 2CO(g), theequilibrium pressure is 12 atm. If 50% of CO2 reactsthen Kp will be :(a) 12 atm (b) 16 atm(c) 20 atm (d) 24 atm


18. What is the minimum mass of CaCO3 (s), belowwhich it decomposes completely, required to establishequilibrium in a 6.50 litre container for the reaction :CaCO3(s) CaO(s) + CO2(g) Kc = 0.05 mole/litre(a) 32.5 g (b) 24.6 g(c) 40.9 g (d) 8.0 gm

19. For the reaction 3 A (g) + B (g) 2 C (g) at agiven temperature, Kc = 9.0. What must be thevolume of the flask, if a mixture of 2.0 mol each ofA, B and C exist in equilibrium?(a) 6 L (b) 9 L(c) 36 L (d) None of these

20. The role of catalyst in a chemical reaction is :(a) To help attain equilibrium in a shorter time.(b) To lower the activation energy.(c) To shift the equilibrium in such a way as to

increase the concentration of the product(d) Both 1 & 2

21. The equilibrium constant (Kp) for the reactionPCl5(g) PCl3(g) + Cl2(g) is 16. If the volumeof the container is reduced to one half its originalvolume, the value of Kp for the reaction at the sametemperature will be :(a) 32 (b) 64(c) 16 (d) 4

22. The equilibrium constant for the reaction : N2(g) +O2(g) 2NO(g) at 2000 K is 4 104 In presenceof a catalyst the equilibrium is established ten timesfaster at the same temperature. What is the value ofequilibrium constant in presence of catalyst : -(a) 40 104 (b) 4 10–4

(c) 4 104 (d) None23. The equilibrium constant for the reaction : H2(g) +

I2(g) 2HI(g) is 64. If the volume of the containeris reduced to one fourth of its original volume, thevalue of the equilibrium constant will be :(a) 16 (b) 32(c) 64 (d) 128

24. A3(g) 3A (g)

In the above reaction, the initial concentration of A3is “a” moles/lit. If x is degree of dissociation of A3.The total number of moles at equilibrium will be :

(a) a – 3ax

(b) 3a

– ax

(c) 2a ax?? ?

? ?? ? (d) a + 2ax

25. ‘a’ moles of PCl5, undergoes, thermal dissociationas : PCl5 PCl3 + Cl2, the mole fraction of PCl3at equilibrium is 0.25 and the total pressure is 2.0atmosphere. The partial pressure of Cl2 at equilibriumis :(a) 2.5 (b) 1.0(c) 0.5 (d) None

26. 4.5 moles each of hydrogen and iodine heated in asealed ten litre vessel. At equilibrium 3 moles of HIwere found. The equilibrium constant for H2(g) +I2(g) 2HI(g) is :(a) 1 (b) 10(c) 5 (d) 0.33

27. 1.50 moles each of hydrogen and iodine were placedin a sealed 10 litre container maintained at 717 K. Atequilibrium 1.25 moles each of hydrogen and iodinewere left behind. The equilibrium constant, Kc forthe reaction. H2(g) + I2(g) 2HI(g) at 717 K is(a) 0.4 (b) 0.16(c) 25 (d) 50

28. For the reaction equilibrium :

N2O4 (g) 2NO2(g) ; the concentration of N2O4and NO2 at equilibrium are 4.8 × 10–2 and 1.2 × 10–2

mol/L respectively. The value of Kc for the reaction is :(a) 3 × 10–3 M (b) 3 × 103 M(c) 3.3 × 102 M (d) 3 × 10–1 M

29. The equilibrium constant for the reaction :

N2(g) + O2(g) 2NO(g) at temperatureT is 4 × 10–4. The value of Kc for the reaction.

NO(g) 12 N2(g) +

12 O2(g) at the same

temperature is :(a) 0.02 (b) 50(c) 4 × 10–4 (d) 2.5 × 10–2

30. For the reaction 3A(g) + B(g) 2C(g) at a giventemperature, Kc = 9.0. What must be the volume ofthe flask, if a mixture of 2.0 mol each of A, B and Cexist in equilibrium ?(a) 6 L (b) 9 L(c) 36 L (d) None of these

31. Sulfide ion in alkaline solution reacts with solid sulfurto form polysulfide ions having formulas S2

2–, S32–,


S42– and so on. The equilibrium constant for the

formation of S22– is 12 (K1) & for the formation of

S32– is 132 (K2), both from S and S2–. What is the

equilibrium constant for the formation of S32– from

S22– and S

(a) 11 (b) 12(c) 132 (d) None of these

32. 1 mole N2 and 3 mol H2 are placed in a closedcontainer at a pressure of 4 atm. The pressure fallsto 3 atm at the same temperature when the followingequilibrim is attained.

N2(g) + 3H2(g) 2NH3(g). The equilibrium constantKp for dissociation of NH3 is :

(a) 31 × (1.5)0.5 atm–2 (b) 0.5 × (1.5)3 atm2

(c) 3

20.5 × (1.5) atm3 × 3

(d) –2

33 × 3 atm

0.5 × (1.5)

33. Chemical equilibrium is dynamic in nature because–(a) The equilibrium in maintained quickly

(b) Conc. of reactants and products become sameat equilibrium

(c) Conc. of reactants and products are constant butdifferent

(d) Both forward and backward reactions occur atall times with same speed

34. Which of the following statements is false in caseof equilibrium state –

(a) There is no apparent change in properties withtime

(b) It is dynamic in nature(c) It can be attained from either side of the reaction

(d) It can be attained from the side of the reactantsonly

35. At any moment before a reversible reaction attainsequilibrium it is found that –

(a) The rate of the forward reaction is increasingand that of backward reaction is decreasing

(b) The rate of the forward reaction is decreasingand that of backward reaction is increasing

(c) The rate of both forward and backward reactionsis increasing

(d) The rate of both forward and backward reactionsis decreasing

36. A chemical reaction A B is said to be inequilibrium when -(a) Complete conversion of A to B hastaken place(b) Conversion of A to B is only 50% complete(c) Only 10% conversion of A to B has taken place(d) The rate of transformation of A to B is just equal

to rate of transformation of B to A in the system37. According to Law of Mass action, the rate of

reaction is directly proportional to -(a) molarities of the reactants(b) normalities of the reactants(c) molalities of the reactants(d) mole fractions of the reactants

38. In a chemical equilibrium, the equilibrium constant isfound to be 2.5. If the rate constant of backwardreaction is 3.2 × 10–2, the rate constant of forwardreaction is -(a) 8.0 × 10–2 (b) 4.0 × 10–2

(c) 3.5 × 10–2 (d) 7.6 × 10–3

39. K1 and K2 are the rate constants of forward andbackward reactions. The equilibrium constant K ofthe reaction is -(a) K1 × K2 (b) K1 – K2

(c) 1

2

KK (d)

1 2

1 2

+–

K KK K

40. The value of KP for the reaction H2(g) + I2(g) 2HI(g) is 50. What is the value of KC

(a) 30 (b) 40(c) 50 (d) 70

41. In which of the following reaction, the value of KPwill be equal to KC –

(a) N2(g) + O2(g) 2NO (g)

(b) PCl5 (g) PCl3 (g) + Cl2(g)

(c) 2NH3 (g) N2(g) + 3H2(g)

(d) 2SO2 (g) + O2(g) 2SO3 (g)42. Select the correct expression regarding the relation

between KP and KC for the reaction

aX(g) + bY(g) bZ(g) + aW(g) -

(a) KP = KC(RT)a+b (b) KP = 2( )CK

a b+

(c) KP = KC RT (d) KP = KC


43. The equilibrium constant KC for the decompositionof PCl5 is 0.625 mole / lit at 300ºC. Then the valueof KP is -(a) 2.936 atm (b) 0.0625 atm(c) 6.25 atm (d) 0.00625 atm

44. The reaction A(g) + B(g) C(g) + D(g) proceeds toright hand side upto 99.9%. The equilibrium constantK for the reaction will be -(a) 104 (b) 105

(c) 106 (d) 108

45. For the reaction, 2NO2 (g) 2NO (g) + O2(g),KC = 1.8 × 10–6 at 185ºC. At 185ºC, the value of KC

for the reaction - NO(g) + 12 O2 (g) NO2(g)

is–(a) 0.9 × 106 (b) 7.5 × 102

(c) 1.95 × 10–3 (d) 1.95 × 103

46. The equilibrium concentration of X, Y and YX2 are4, 2 and 2 moles respectively for the equilibrium2X(g) + Y(g) YX2(g).The value of KC is -(a) 0.625 (b) 0.0625(c) 6.25 (d) 0.00625

47. For N2(g) + 3H2(g) 2NH3(g) + Heat –(a) KP = KC (b) KP = KC RT(c) KP = KC (R T)–2 (d) KP = KC (R T)–1

48. In the reaction

C2H4(g) + H2(g) C2H6(g), the equilibrium constantcan be expressed in units of -(a) litre–1 mol–1 (b) mol2 litre–2

(c) litre mol–1 (d) mol litre–1

49. Equilibrium concentration of HI, I2 and H2 is 0.7,0.1 and 0.1 moles/litre. Calculate the equilibriumconstant for the reaction :

I2(g) + H2(g) 2HI(g) –

(a) 0.36 (b) 36(c) 49 (d) 0.49

50. The equilibrium constant for the reaction Zn (s) +CO2(g) ZnO (s) + CO (g) is

(a) 2

co

co

PP (b)

[ZnO][Zn]

(c) 2

ZnO CO

Zn CO

P PP P (d)

2

2

Zn CO

ZnO CO

P PP P

51. For the reaction

C (s) + CO2 (g) 2CO (g)the partial pressure of CO2 and CO are 2.0 and 4.0atm respectively at equilibrium. The Kp for thereaction is –(a) 0.5 (b) 4.0(c) 8.0 (d) 32.0

52. The unit of equilibrium constant for the reactionH2 + I2 2HI is –(a) Mole–1 litre (b) Mole–2 litre(c) Mole litre (d) None

53. P

C

KK for the gaseous reaction –

(1) 2 A + 3 B 2C

(2) 2 A 4B

(3) A + B + 2C 4Dwould be respectively -(a) (RT)–3 , (RT)2, (RT)º(b) (RT)–3 , (RT)–2, (RT)–1

(c) (RT)–3 , (RT)2, (RT)(d) None of the above

54. In Which of the following equilibria, the value of KPis less than KC -

(a) H2 + I2 2Hl

(b) N2 + 3H2 2NH3

(c) N2 + O2 2NO

(d) CO + H2O CO2 + H2

Reaction Quotient

55. In a 20 litre vessel initially 1 – 1 mole CO, H2O CO2is present, then for the equilibrium of

CO + H2O CO2 + H2 following is true :(a) H2, more then 1 mole(b) CO, H2O, H2 less then 1 mole(c) CO2 & H2O both more than 1 mole(d) All of these

56. Vapour density of PCl5 is 104.16 but when heatedto 230°C its vapour density is reduced to 62. Thedegree of dissociation of PCl5 at this temperaturewill be :(a) 6.8 % (b) 68%(c) 46% (d) 64%


57. A reaction mixture containing H2, N2 and NH3 haspartial pressure 2 atm, 1 atm and 3 atm respectivelyat 725 K. If the value of KP for the reaction, N2 +3H2 2NH3 is 4.28 10–5 atm–2 at 725 K, inwhich direction the net reaction will go :(a) Forward(b) Backward(c) No net reaction(d) Direction of reaction cannot be predicted

58. Consider following reactions in equilbrium withequilibrium concentration 0.01M of every species

(I) PCl5(g) PCl3(g) + Cl2(g)

(II) 2HI(g) H2(g) + I2(g)

(III) N2(g) + 3H2 (g) 2NH3(g)Extent of the reactions taking place is :(a) I > II > III (b) I < II < III(c) II < III < I (d) III < I < II

59. The role of catalyst in reversible reaction is-(a) To increase the rate of forward reaction(b) Decrease the rate after equilibrium(c) Allow equilibrium to be achieved quickly(d) None

Degree of dissociation

60. In a container equilibrium N2O4 (g) 2NO2 (g)is attained at 25°C. The total equilibrium pressure incontainer is 380 torr. If equilibrium constant of aboveequilibrium is 0.667 atm, then degree of dissociationof N2O4 at this temperature will be

(a) 13 (b)

12

(c) 23 (d)

14

61. The extent of dissociation of PCl5 at a certaintemperature is 20 % at one atm pressure. Calculatethe pressure at which this substance is halfdissociated at the same temperature.(a) 0.123 (b) 0.246(c) 0.826 (d) 0.111

62. Consider the reactions

(i) PCl5(g) PCl3(g) + Cl2(g)

(ii) N2O4(g) 2NO2(g)The addition of an inert gas at constant volume

(a) will increase the dissociation of PCl5 as well asN2O4

(b) will reduce the dissociation of PCl5 as well asN2O4

(c) will increase the dissociation of PCl5 and step upthe formation of NO2

(d) will not disturb the equilibrium of the reactions

63. At 248º C , the kP for the reaction , SbCl5 (g) SbCl3 (g) + Cl2 (g) is 1.07 atm at a total pressure of1 atm . Calculate the degree of dissociation of SbCl5.(a) 0.516 (b) 0.718(c) 0.321 (d) None of these

64. 4 moles of A are mixed with 4 moles of B, when2 moles of C are formed at equilibrium, accordingto the reaction, A + B C + D. The equilibriumconstant is :(a) 4 (b) 1

(c) 2 (d) 4

65. 4 moles of PCl5 are heated at constant temperaturein closed container. If degree of dissociation for PCl5is 0.5 calculate total number of moles at equilibrium:(a) 4.5 (b) 6(c) 3 (d) 4

66. The dissociation of CO2 can be expressed as 2CO2 2CO + O2. If the 2 moles of CO2 is taken

initially and 40% of the CO2 is dissociated equilibriumthen total number of moles at equilirbrium :(a) 2.4 (b) 2.0(c) 1.2 (d) 5

67. In the reaction 2P(g) + Q(g) 3R(g) + S(g). If2 moles each of P and Q taken initially in a 1 litreflask. At equilibrium which is true :(a) [P] < [Q] (b) [P] = [Q](c) [Q] = [R] (d) None of these

68. In a 0.25 litre tube dissociation of 4 moles of NO istake place. If its degree of dissociation is 10%. Thevalue of Kp for reaction 2NO N2 + O2 is :

(a) ? ?21

18 (b) ? ?21

8

(c) 1

16 (d) 132

69. The vapour density of N2O4 at a certain temperatureis 30. What is the percentage dissociation of N2O4at this temperature ?


(a) 53.3 % (b) 106.6%(c) 26.7% (d) none

70. A sample of mixture of A(g), B(g) and C(g) underequilibrium has a mean molecular weight (observed)is 80.

The equilibrium is A(g) B(g) + C(g)(mol. (mol. (mol.wt. wt. wt.= 100) = 60) = 40)

Find the degree of dissociation for A(g).(a) 0.25 (b) 0.5(c) 0.75 (d) 0.8

71. If in the reaction N2O4(g) 2NO2(g), is the partof N2O4 which dissociates then the number of molesat equilibrium will be-(a) 3 (b) 1(c) (1–)2 (d) (1 + )

72. At 444º C, the equilibrium constant K for the reaction

2AB(g) A2(g) + B2(g) is 1

64 . The degree of dissociation

of AB will be -(a) 10% (b) 20%(c) 30% (d) 50%

73. For the reaction A(g) + B(g) C(g) + D(g), thedegree of dissociation would be

(a) + 1

KK

(b) K + 1

(c) ± 1K (d) K – 1

74. For the reaction : N2O3(g) NO(g) + NO2(g); totalpressure = P, degree of dissociation = 50%. ThenKp would be –(a) 3P (b) 2P

(c) 3P

(d) 2P

75. An unknown compound A dissociates at 500ºC togive products as follows -

A(g) B(g) + C(g) + D(g)Vapour density of the equilibrium mixture is 50 whenit dissociates to the extent to 10%. What will be themolecular weight of Compound A –(a) 120 (b) 130(c) 134 (d) 140

76. N2O4 dissociates as N2O4(g) 2NO2(g) at 273 Kand 2 atm pressure. The equilibrium mixture has adensity of 41. What will be the degree of dissociation-(a) 14.2% (b) 16.2%(c) 12.2% (d) None

77. At 250ºC and 1 atmospheric pressure, the vapourdensity of PCl5 is 57.9 . What will be the dissociationof PCl5 –(a) 1.00 (b) 0.90(c) 0.80 (d) 0.65

Thermodynamics of Equilibrium

78. The correct relationship between free energy changein a reaction and the corresponding equilibriumconstant K is : [RPMT 2008](a) – Go = RT ln K (b) G = RT ln K(c) – G = RT ln K (d) Go = RT ln K

79. Which of the following is not favourable for SO3formation

2SO2 (g) + O2 (g) 2SO3 (g); H = – 45.0kcal(a) High pressure(b) High temperature(c) Decreasing SO3 concentration(d) Increasing reactant concentration

80. In an equilibrium reaction for which G° = 0, thevalue of equilibrium constant K =(a) 0 (b) 1(c) 2 (d) 10

81. The effect of temperature on equilibrium constantis expressed as (T2 > T1)

log K2/log K1 = 2 1

1 12.303

HR T T

. For endo-

thermic reaction false statement is

(a) 2 1

1 1T T

= positive

(b) H = positive(c) log K2 > log K1

(d) K2 > K1

Le-chatelier’s principle

82. A reaction in equilibrium is represented by thefollowing equation –


2A(s) + 3B(g) 3C(g) + D(g) + O2 if the pressureon the system is reduced to half of its original value(a) The amounts of C and D decreases(b) The amounts of C and D increases(c) The amount of B and D decreases(d) All the amounts remain constant

83. In which of the following equilibrium reactions, theequilibrium would shift to right side, if total pressureis decreased :

(a) N2 + 3H2 2NH3

(b) H2 + I2 2HI

(c) N2O4 2NO2

(d) H2 + Cl2 2HCl84. In a vessel containing SO3, SO2 and O2 at

equilibrium, some helium gas is introduced so thatthe total pressure increases while temperature andvolume remain constant. According to Le-Chatelierprinciple, the dissociation of SO3,(a) Increases (b) Decreases(c) Remains unaltered (d) None of these

85. On cooling of following system at equilibrium CO2(s) CO2(g)

(a) There is no effect on the equilibrium state(b) More gas is formed(c) More gas solidifies(d) None of above

86. For an equilibrium H2O(s) H2O() which ofthe following statements is true.(a) The pressure changes do not affect the

equilibrium(b) More of ice melts if pressure on the system is

increased(c) More of liquid freezes if pressure on the system

is increased(d) The degree of advancement of the reaction do

not depend on pressure.87. In the Haber process for the industrial manufacture

of ammonia involving the reaction,

N2(g) + 3H2(g) 2NH3(g) at 200 atm pressurein the presence of a catalyst, a temperature of about500ºC is used. This is considered as optimumtemperature for the process because(a) yield is maximum at this temperature(b) catalyst is active only at this temperature

(c) energy needed for the reaction is easily obtainedat this temperature

(d) rate of the catalytic reaction is fast enough whilethe yield is also appreciable for this exothermicreaction at this temperature.

88. The effect of increasing the pressure on the followingequilibrium 2A + 3B 3A + 2B is -(a) Forward reaction is favoured

(b) Backward reaction is favoured(c) No effect

(d) None of these89. When NaNO3(s) is heated in a closed vessel, O2 is

liberated and NaNO2(s) is left behind. At equilibrium–

(a) Addition of NaNO3(s) favours forward reaction(b) Addition of NaNO2(s) favours reverse reaction

(c) Increasing pressure favours reverse reaction(d) Decreasing temperature favours forward reaction

90. aA + bB cC + dDIn above reaction low pressure and high temperature,conditions are shift equilibrium in back direction socorrect set :(a) (a + b) > (c + d), H > 0

(b) (a + b) < (c + d), H > 0(c) (a + b) < (c + d), H < 0

(d) (a + b) > (c + d), H < 091. In which of the following reactions, increase in the

pressure at constant temperature does not affect themoles at equilibrium.

(a) 2NH3(g) N2(g) + 3H2(g)

(b) C(g) + 12 O2(g) CO(g)

(c) H2(g) + 12 O2(g) H2O(g)

(d) H2(g) + I2(g) 2 HI(g)92. Change in volume of the system does not alter the

number of moles in which of the followingequilibrium

(a) N2(g) + O2(g) 2NO(g)

(b) PCl5(g) PCl3(g) + Cl2(g)

(c) N2(g) + 3H2(g) 2NH3(g)

(d) SO2Cl2(g) SO2(g) + Cl2(g)


93. The conditions favourable for the reaction :

2SO2(g)+O2(g) 2SO3(g) ; H° = – 198 kJ are:

(a) low temperature, high pressure(b) any value of T and P

(c) low temperature and low pressure(d) high temperature and high pressure

94. The yield of product in the reaction

2A(g) + B(g) 2C(g) + Q kJwould be lower at :

(a) low temperature and low pressure(b) high temperature & high pressure

(c) low temperature and high pressure(d) high temperature & low pressure

95. Consider the reaction, CaCO3(s) CaO(s) +CO2(g) ; in closed container at equilibrium. Whatwould be the effect of addition of CaCO3 on theequilibrium concentration of CO2 -

(a) Increases(b) Decreases

(c) Remains unaffected(d) Data is not sufficient to predict it

96. In the melting of ice, which one of the conditionswill be more favourable –

(a) High temperature and high pressure(b) Low temperature and low pressure

(c) Low temperature and high pressure(d) High temperature and Low pressure

97. In the reaction, 2SO2 (g) + O2 (g) 2SO3 (g) +X cals, most favourable condition of temperatureand pressure for greater yield of SO3 are -(a) Low temperature and low pressure

(b) High temperature and low pressure(c) High temperature and high pressure

(d) Low temperature and high pressure98. On adding inert gas to the equilibrium

PCI5(g) PCI3(g) + CI2(g) at constant pressure.The degree of dissociation will remain –

(a) Unchanged (b) Decreased(c) Increased (d) None of these

99. Dissociation of phosphorus pentachloride is favouredby –(a) High temperature and high pressure(b) High temperature and low pressure(c) Low temperature and low pressure(d) Low temperature and high pressure

100. Adding inert gas to system N2(g) + 3H2(g) 2NH3(g)at equilibrium at constant volume will be –(a) N2 and H2 are formed in abundance(b) N2, H2 and NH3 will have the same molar

concentration(c) The production of ammonia increases(d) No change in the equilibrium

101. In the reaction N2(g) + 3H2(g) 2NH3(g), theforward reaction is exothermic and the backwardreaction is endothemic. In order to produce moreheat it is necessary –(a) To add ammonia(b) To add N2 and H2

(c) Increasing the concentration of N2,H2 and NH3equally

(d) None of the above102. The reaction in which the yield of the products can

not be increased by the application of high pressureis –

(a) PCl3 (g) + Cl2 (g) PCl5 (g)

(b) N2 (g) + 3H2 (g) 2NH3 (g)

(c) N2 (g) + O2 (g) 2NO (g)

(d) 2SO2 (g) + O2 (g) 2SO3 (g)103. Factors affecting KC is/are -

(a) Increasing concentration of the reactant(b) Presence of catalyst(c) Method of writing balanced equation

(or stoichiometry of reaction)(d) Time taken by the chemical reaction

104. In the reaction A (g) + B (g) C (g), thebackward reaction is favoured by(a) Increase in pressure(b) Decrease in pressure(c) Neither increase nor decrease in pressure(d) Data unpredictable


1. Ammonia dissociates into N2 and H2 such that degreeof dissociation is very less than 1 and equilibriumpressure is P0 then the value of is [if Kp for2NH3(g) N2 (g) + 3H2(g) is 27 × 10–8 P0

2:(a) 10–4 (b) 4 × 10–4

(c) 0.02 (d) can’t be calculated.2. At a temp, T, a compound AB4(g) dissociates as

2AB4(g) AA2(g) + 4B2(g) with a degree ofdissociation x, which is small compared with unity.The expression of Kp in terms of x and total pressureP is ;(a) 8P3x5 (b) 256P3x5

(c) 4Px2 (d) None of these3. One mole of ammonium carbamate dissociate as

shown below at 500 K.

NH2COONH4(s) 2NH3(g) + CO2(g)If the pressure exerted by the released gases is3.0 atm, the value of KP is -(a) 7 atm (b) 3 atm(c) 4 atm (d) 8 atm

4. The equilibrium constant, Kp for the reaction

2SO2(g) + O2(g) 2SO3(g)is 4.0 atm–1 at 1000 K. What would be the partialpressure of O2 if at equilibrium the amount of SO2and SO3 is the same ?(a) 16.0 atm (b) 0.25 atm(c) 1 atm (d) 0.75 atm

5. If for 2A2B(g) 2A2(g) + B2(g), Kp = TOTALPRESSURE (at equilibrium) and starting thedissociation from 4 mol of A2B then :(a) degree of dissociation of A2B will be (2/3)(b) total no. of moles at equilibrium will be (14/3)(c) at equilibrium the no. of moles of A2B are not

equal to the no. of moles of B2 .(d) at equilibrium the no. of moles of A2B are equal

to the no. of moles of A2 .

6. KP for the reaction A(g) + 2B(g) 3C(g) + D(g);is 0.05 atm. What will be its KC at 1000 K in termsof R –

(a) R105 5–

(b) 5–105R

(c) 5 × 10–5 R (d) None of these

7. If K1, K2, K3 are equilibrium constant for formationof AD, AD2, AD3 respectively as followsA + D AD,

AD + D AD2, AD2 + D ADAD3 . Thenequilibrium constant ‘K’ for A + 3D AD3 isrelated as(a) K1 + K2 + K3 = K(b) logK1 + logK2 + logK3 = log K(c) K1 + K2 = K3 + K(d) log K1 + logK2 = logK3 + log K

8. If PCl5 is 80% dissociated at 523 K. Calculate thevapour density of the equilibrium mixture at 523 K –(a) 75.9 (b) 57.9(c) 97.5 (d) 95.7

9. For the reaction A2(g) + 3B2 2C2(g) the partialpressure of A2, B2 at equilibrium are 0.80 atm and0.40 atm respectively. The pressure of the system is2.80 atm. The equilibrium constant Kp will be(a) 50 (b) 5.0(c) 0.02 (d) 0.2

10. On decomposition of NH4HS, the followingequilibrium is established :

NH4HS(s) NH3(g) + H2S (g)If the total pressure is P atm, then the equilibriumconstant KP is equal to(a) P atm (b) P2 atm2

(c) P2 / 4 atm2 (d) 2P atm11. A liquid is in equilibrium with its vapour at its boiling

point. On the average the molecules in the two phaseshave equal :(a) inter molecular forces(b) potential energy(c) kinetic energy(d) none of these.

12. The volume of a closed reaction vessel in whichthe equilibrium :

2SO2(g) + O2(g) 2SO3(g)sets is halved, Now -(a) The rates of forward and backward reactions

will remain the same.(b) The equilibrium will not shift.(c) The equilibrium will shift to the right.



(d) The rate of forward reaction will become doublethat of reverse reaction and the equilibrium willshift to the right.

13. Ammonia gas at 15 atm is introduced in a rigid vesselat 300ºK. At equilibrium the total pressure of thevessel is found to be 40.11 atm at 300ºC. The degreeof dissociation of NH3 will be:(a) 0.6 (b) 0.4(c) Unpredictable (d) None of these

14. At room temperature, the equilibrium constant forthe reaction P + Q R + S was calculated to be4.32. At 425°C the equilibrium constant became 1.24× 10–2. This indicates that the reaction(a) is exothermic(b) is endothermic(c) is difficult to predict(d) no relation between H and K

15. In a 2 litre flask, the reaction takes place as :COCl2(g) CO(g) +Cl2(g)The equilibrium conc. of [COCl2] was found to be0.4. If the excess of COCl2 is added to the system,the equilibrium reestablishes and [COCl2] becomes1.6. What is the equilibrium conc. of [CO] ?(a) Half of the former value(b) Thrice of the former value(c) Remains unaltered(d) Twice of the former value

16. Equilibrium constant for the following equilibriumis given at 0ºC.Na2HPO4 . 12H2O (s) Na2HPO4 . 7H2O (s) +5H2O(g) KP = 31.25 × 10–13

the vapour pressure of water is

(a) 15 ×10–3 atm (b) 0.5 × 10–3 atm

(c) 5 × 10–2 atm (d) 5 × 10–3 atm.17. 4.0 gms of hydrogen react with 9.023×1023

molecules of chlorine to form HCl gas. The totalpressure after the reaction was found to be 700mm. The partial pressure of HCl will be –(a) 3900 mm (b) 600 mm(c) 700 mm (d) 350 mm

18. Consider the general hypothetical reaction,A(s) 2B(g) + 3C(g)If the concentration of C at equilibrium is doubled,then after the equilibrium is re-established, theconcentration of B will be

(a) two times the original value(b) one half of its original value

(c) 1/2 2 times the original value

(d) 2 2 times the original value

19. In a reaction vessel of 2 litre capacity 3 moles ofN2 reacts with 2 moles of O2 to produce 1 mole ofNO. What is the molar concentration of N2 atequilibrium ?(a) 1.25 (b) 1.50(c) 0.75 (d) 2.0

20. For the reaction (a) and (b) : A B + C ....(a)D 2E ....(b)

Given 1PK :

2PK : : 9 : 1

If the degree dissociation of A and D be same thenthe total pressure at equilibria (a) and (b) are in theratio (Assume reaction are started with equal numberof moles of A and D).(a) 3 : 1 (b) 36 : 1(c) 1 : 1 (d) 0.5 : 1

21. In the following reaction, 16 g hydrogen hasundergone change at equilibrium.

CO(g) + 2H2(g) CH3OH(g) + x K calWhich of the following statement is correct ?(a) 4x Kcal of heat will be produced(b) 1 mol of CH3OH will be formed(c) 28 g CO has undergone reaction(d) None of the above

22. What should be the active mass in g mol L–1 when0.585 g NaCl is present in 100 mL(a) 0.1 (b) 0.5(c) 1.0 (d) 2.0

23. Pure ammonia is placed in a vessel at a temperaturewhere its dissociation constant () is appreciable.At equilibrium :(a) Kp does not change significantly with pressure.(b) a does not change with pressure(c) concentration of NH3 does not change with

pressure.(d) concentration of hydrogen is less than that of

nitrogen24. What should be the respective active masses of

methyl alcohol and carbon tetrachloride, if theirdensities are 0.5 and 1.2 g/mL ?


(a) 15.62 and 7.79 (b) 16.65 and 7.40(c) 15.46 and 7.80 (d) 15.40 and 6.50

25. For I2(g) 2I(g), Kc at 1000 K is 10–6. Onemole I2 is added in a one-litre container. Which ofthe following expressions is correct for the abovesystem at equilibrium ?

(a) [I2] + [I] = 1 + x (b) [I2] = 12 [I]

(c) [I2(g)] >> [I(g)] (d) Both (a) and (c)26. If equilibrium constant for the reaction

2HI(g) H2(g) + I2(g) is Kc then the equilibriumconcentration of HI will be -

(a) 21/2

c

[H ][K ] (b)

1/2

c

22

][K]][I[H

(c) [I2][Kc]–1/2 (d) All of the above.

27. If 0.5 mol of H2 is reacted with 0.5 mol of I2 in a10 L container at 444°C and at same temperaturevalue of equilibrium constant Kc is 49, the ratio of[HI] and [I2] will be -

(a) 7 (b) 71

(c) 71

(d) 49

28. What should be the value of KC for the reaction2SO2(g) + O2(g) 2SO3(g). If the amount are SO3= 48g, SO2 = 12.8 and O2 = 9.6 at equilibrium andthe volume of the container is one litre ?(a) 64 (b) 0.30(c) 42 (d) 8.5

29. If 0.5 mole H2 is reacted with 0.5 mole I2 in a ten-litre container at 444°C and at same temperature valueof equilibrium constant KC is 49, the ratio of [H] and[I2] will be :

(a) 7 (b) 17

(c) 17 (d) 49

30. Three moles of N2 react with 2 mol of O2 in a two-litre container to form one mole of NO. What willbe the number of moles of O2 per litre at equilibrium?(a) 0.75 (b) 1.50(c) 1.25 (d) None of these

31. If 2/9 of 1 mol of HI is dissociates, the equlibriumconstant of disintegration of acid at sametemperature will be -

(a) 64 (b) 641

(c) 49 (d) 491

32. For the reaction A + 2B 2C + D, initialconcentration of A is a and that of B is 1.5 times thatof A. Concentration of A and D are same atequilibrium. What should be the concentration of Bat equilibrium ?

(a) 4a

(b) 2a

(c) 34a

(d) All of the above

33. For N2O3 NO + NO2, if total pressure is Patm and amount of dissociation is 50%, the valueof Kp will be -(a) 3 P (b) 2 P

(c) 3P

(d) 2P

34. One mole of PCl5 is heated in a closed container ofone litre capacity. At equilibrium, 20% PCl5 is notdissociated. What should be the value of Kc ?(a) (3.2)–1 (b) 3.2(c) 2.4 (d) 42

35. In the reaction, N2 + O2 2NO, the moles/litreof N2, O2 and NO respectively 0.25, 0.05 and 1.0 atequilibrium, the initial concentration of N2 and O2will respectively be :(a) 0.75 mol/litre, 0.55 mole/litre(b) 0.50 mole/litre, 0.75 mole/litre(c) 0.25 mole/litre, 0.50 mole/ litre(d) 0.25 mole/litre, 1.0 mole/litre

36. For the reaction, N2O3 NO + NO2, the valueof equilibrium constant Kp at fixed temperature is4. Wat will be the amount of dissocidation at sametemperature and 5 atmospheric pressure ?

(a) 31

(b) 32

(c) 97

(d) 42


37. In the reaction PCl5 PCl3 + Cl2, a mol ofPCl5 are initially taken. If the amount x getsdissociated and total pressure is P then the value of

3PClp × P–1 will be -

(a) 1 x

x? (b) 1 ax

x?

(c) x

a x? (d) – x

a x

38. Value of KC at 300°C for N2 + O2 2NO is 9 10–4 and equilibrium amounts of N2 and O2 are used.The concentration of NO at equilibrium will be(a) 0.0148 a (b) 0.296 a(c) 0.148 a (d) 0.0296 a

39. One mole PCl5 is heated in a closed container of onelitre capacity. At equilibrium 20% PCl5 is notdissociated. What should be the value of KC ?(a) (3 – 2)–1 (b) 3.2(c) 2.4 (d) 42

40. In the reaction, PCl5 PCl3 + Cl2, one mol eachof PCl5 and PCl3 are initially present and atequilibrium x mol of PCl5 remain. What will be thetotal number of moles in the reaction atequilibrium?(a) 1 – x (b) 2 – x(c) 3 – x (d) 4 – x

41. In the reaction 2A(g) + B(g) 3 C(g) + D(g) twomoles each of A and B are initially taken in a one-liter flask. Which of the follwoing expressions iscorrect for the above sysytem at equlibrium ?(a) [A] – [D] = 2 – 3x (b) [A] – [D] = 1 – x(c) [A] + [D] = 2 + x (d) None of the above

42. In the reaction, PCl5 Cl2, the amount of eachPCl5, PCl3 and Cl2 is 2 mole at equilibrium and totalpressure is 3 atmosphere. The value of KP will be(a) 1.0 atm (b) 3.0 atm(c) 2.9 atm (d) 6.0 atm

43. How many moles per litre of PCl5 has to be taken toobtain 0.1 mole Cl2, if the value of equilibrium contantKC is 0.04 ?(a) 1.15 (b) 0.25(c) 0.35 (d) 0.05

44. If the amount of dissociation is 0.5 , the value of

Kp for the reaction N2O3 NO + NO2 will be -

(a) equal to the pressure of the system

(b) 82

of the pressure of the system

(c) 83 of the pressure of the system

(d) 5 times of the pressure of the sysytem45. When helium gas is added in the reaction,

COCl2 CO + Cl2, at equilibrium the amountof CO -(a) increases (b) decreases(c) remains unchanged (d) All of the above

46. At 227°C, 60% of 2 gram moles of PCl5 getsdissociated in a two litre container. The value of Kpwill be(a) 450 R (b) 400 R(c) 50 R (d) 100 R

47. 1 M solution of glucose reaches dissociationequilibrium, according to the equation given below-C6H12O2 6HCHOWhat is the concentration of HCHO at equilibrium,if equilibrium constant for reaction6HCHO C6H12O6 is 6 × 1022

(a) 9.6 × 10–4 m (b) 1.6 × 10–4 m(c) 1.6 × 10–3 m (d) 9.6 × 10–3 m

48. One mole of N2O4(g) at 300 K is kept in a closedvessel at 1 atm pressure. It is heated to 600 K when20% by mass of N2O4(g) decomposes to NO2(g)then Kp will be -(a) 1.2 atm (b) 2.4 atm(c) 2.0 atm (d) 0.4 atm

49. The reaction, PCl5 PCl3 + Cl2 is statrted in afive litre container by taking one mole of PCl5. If 0.3mole of PCl5 is there at equilibrium, the total moleand concentration of PCl5 and KC will respectivelybe

(a) 0.70, 0.14, 49

150 (b) 0.30, 0.12, 23

100

(c) 0.10, 0.07, 23

100 (d) 0.05, 20, 49

15050. 20% of N2O4 molecules are dissociated in a sample

of gas at 27°C and 760 torr. Calculate the densityof mixture-(a) 1.27 gL–1 (b) 9.24 gL–1

(c) 4.98 gL–1 (d) 3.12 gL–1

51. At 70°C and 1 atm, N2O4 is 66% dissociated intoNO2. What volume will 10 g N2O4 occupy under


these conditions -(a) 4.08 L (b) 6.08 L(c) 3.08 L (d) 4.78 L

52. One mole of N2O4(g) at 300 K is left in a closedcontainer under one atm. It is heated to 600 K when20% by mass of N2O4 (g) decomposes to NO2(g).The resultant pressure is :(a) 1.2 atm (b) 2.4 atm(c) 2.0 atm (d) 1.0 atm

53. For the following gases equilibrium.N2O4(g) 2NO2(g)Kp is found to be equal to Kc. This is attained whentemperature is(a) 0°C (b) 273 K(c) 1 K (d) 12.19 K

54. For the reaction : CO(g) + 12 O2(g) CO2(g),

Kp/Kc is :(a) RT (b) (RT)–1

(c) (RT)–1/2 (d) (RT)1/2

55. For the reaction : 2NO2(g) 2NO(g) + O2(g)Kc = 1.8 × 10–6 at 184° C and R = 0.083 JK–1 mol–1.When Kp and Kc are compared at 184°C, it is foundthat :(a) Kp > Kc

(b) Kp < Kc

(c) Kp = Kc

(d) Kp Kc depends upon pressure of gases56. PCl5 dissociation a closed container as :

PCl5(g) PCl3(g) + Cl2(g)If total pressure at equilibrium of the reaction mixtureis P and degree of dissociation of PCl5 is , the partialpressure of PCl3 will be :

(a) .1

P

(b) 2.

1 –P

(c) .– 1

P

(d) .1 –

P

57. For the reaction : 2HI (g) H2(g) + I2(g), thedegree of dissociated () of HI(g) is related toequilibrium constant Kp by the expression :

(a) 1 + 2

2pK

(b) 1 + 2

2pK

(c) 2

1 + 2p

p

KK (d)

2

1 + 2p

p

K

K

58. The equilibrium constant for the reaction

A(g) + 2B(g) C(g)is 0.25 dm6 mol–2. In a volume of 5 dm3, whatamount of A must be mixed with 4 mol of B to yield1 mol of C at equilibrium.(a) 3 moles (b) 24 moles(c) 26 moles (d) None of these

59. For the reaction H2(g) + I2(g) 2HI(g), thestandard free energy is Gº > 0. The equilibriumconstant (K) would be ________?(a) K = 0 (b) K < 1(c) K > 1 (d) K = 1

60. The equilibrium constant Kc for the reaction,

A(g) + 2B(g) 3C(g) is 2 × 10–3

What would be the equilibrium partial pressure ofgas C if initial pressure of gas A & B are 1 & 2 atmrespectively.(a) 0.0625 atm (b) 0.1875 atm(c) 0.21 atm (d) None of these

61. A 20.0 litre vessel initially contains 0.50 mole eachof H2 and I2 gases. These substances react and finallyreach an equilibrium condition. Calculate theequilibrium concentration of HI if Keq = 49 for thereaction H2 + I2 2HI.(a) 0.78 M (b) 0.039 M(c) 0.033 M (d) 0.021 M

62. At a total pressure of 2 atmospheres, the number ofmoles of N2, H2 and NH3 in a gaseous mixture are1,3 and 2. The partial pressure of NH3 in atmospheresis –(a) 1.0 (b) 0.5(c) 0.33 (d) 0.67

63. The KC for the reaction A + B C + D is 9. Ifone mole of each of A and B are mixed and there isno change in volume, the number of moles of Cformed is–(a) 0.50 (b) 0.75(c) 0.90 (d) 1.5

64. ‘a’ moles of PCl5 undergo thermal dissociation as –

PCl5 PCl3 +Cl2, the mole fraction of PCl3 atequilibrium is 0.5. The total pressure is 2.0


atmosphere. The partial pressure of Cl2 at equilibriumis –(a) 2.5 (b) 1.0(c) 0.5 (d) None

65. The equilibrium constant of the reactionA + B C + D is 10. If rate constant of forwardreaction is 203, the rate constant of backwardreaction is –(a) 20.3 (b) 10.3(c) 2.03 (d) 203

66. The equilibrium constants for the reaction X2 2X at300K and 600K are 10–8 and 10–3 respectively. Thereaction is –(a) Exothermic (b) Endothermic(c) Thermo neutral (d) Slow

67. The reversible reaction

[Cu(NH3)4]2+ + SO32– [Cu (NH3)3SO3] + NH3

is at equilibrium. What would not happen if ammoniais added –(a) [SO3

2– ] would increase(b) [Cu (NH3)3SO3] would increase(c) The value of equilibrium constant would not

change(d) [Cu(NH3)4]2+ would increase

68. The equilibrium constant for the reaction2 X (g) + Y (g) 2Z (g) is 2.25. What would bethe concentration of Y at equilibrium with 2.0 molesof X and 3.0 moles of Z in one litre vessel atequilibrium–(a) 1.0 moles (b) 2.25 moles(c) 2.0 moles (d) 4.0 moles

69. The relation between KP and KC for the reactionA + B C + 2D is –(a) KP = KC [RT]–1 (b) KP.KC –1 = RT(c) KC KP

–1 = RT (d) KP = KC [RT]3

70. In the following reactions –

(1) 2N2O5 (g) 4NO2 (g) + O2 (g) and

(2) 4NO2 (g) + O2 (g) 2N2O5 (g)Choose the correct fact –(a) (a) KP > KC (b) KP = KC

(b) (a) KP > KC (b) KP < KC

(c) (a) KP = KC (b) KP < KC

(d) (a) KP < KC (b) KP > KC

71. In a 10 litre box 2.5 mole hydroiodic acid is taken.After equilibrium

2HI H2 + I2

the concentration of HI is found to be 0.1 mol –1

The concentration of [H2] at equilibrium in mol –1 ,is –(a) 2.4 (b) 0.15(c) 1.5 (d) 7.5 × 10–2

72. The statement not applicable to an irreversiblereaction is –(a) It goes to completion in the forward direction(b) On increasing the concentration of the reactants,

the rate of the reaction increases(c) The removal of the reaction products makes the

reaction faster(d) The addition of the reaction products does not

influence the rate of reaction73. KP will be equal to KC under which of the following

conditions for the reaction–

aA + bB cC + dD(a) (a + b) > ( c + d )(b) (a + b) – ( c + d ) = 0(c) (c + d) > ( a + b )(d) (a + c) = ( b + d )

74. For the equilibrium reaction :

2HCl (g) H2 (g) + Cl2 (g) the equilibriumconstant is 1.0 × 10–5 what is the concentration ofHCl if the equilibrium concentrations of H2 and Cl2are 1.2 ×10–3 M and 1.2 × 10–4 M respectively –(a) 12 × 10–2 M (b) 12 × 10–4 M(c) 12 × 10–3 M (d) 12 × 104 M

75. The value of ng for the reaction :

2Hg(l) + Cl2 (g) Hg2Cl2 (s) is –(a) – 1 (b) + 2(c) – 2 (d) 0

76. For the reaction H2 (g) + I2 (g) 2HI (g) at 721K, the value of equilibrium constant KC is 50. Whenthe equilibrium concentration of both is 0.5 M, valueof KP under the same conditions will be –(a) 0.02 (b) 0.2(c) 50.0 (d) 50/RT

77. The value of equilibrium constant for a reactiondepends on-(a) Temperature (b) Pressure(c) Volume (d) Catalytic agent


78. The correct expression for equilibrium constant, KC,for the reaction :

H2 (g) + O2 (g) H2O (g) is –(a) [H2O]/[H2] [O2] (b) / .(c) [H2O]/[H2] [O2]1/2 (d) /.

79. For the reaction :

PCl3 (g) + Cl2 (g) PCl5(g) at 250º C the Value ofKC is 26, then the value of KP at the same temp. willbe –(a) 0.61 (b) 0.57(c) 0.83 (d) 0.46

80. For an exothermic reaction in which the number ofmoles of reactants are more than the number of molesof products. In order to displace the reaction in thereverse direction, what are the favourable conditions.(a) High pressure, low temp. and high conc. of the

product(b) Low pressure, high temp. and high conc. of the

products(c) Low pressure, low temp. and high conc. of the

products(d) High pressure, high temp. and low conc. of the

products


The following questions 1 to 18 consists of twostatements each, printed as Assertion and Reason.While answering these questions you are to chooseany one of the following four responses.

(A) If both Assertion and Reason are true and theReason is correct explanation of the Assertion.

(B) If both Assertion and Reason are true but theReason is not correct explanation of theAssertion.

(C) If Assertion is true but the Reason is false.(D) If Assertion & Reason are false.

1. Assertion (A) : For an exothermic reaction, extentof reaction increases with increase in temperature.

Reason (R) : With increase in temperature, therate of forward reaction increases more.

2. Assertion (A) : The active mass of pure solid andpure liquids is taken unity.

Reason (R) : The active mass of pure solid andpure liquids depends on density and moleular massThe density and molecular mass of pure solid andpure liquids are constant.

3. Assertion : For the reaction,

H2(g) + I2(g) 2HI(g), Kp = Kc

Reason : Kp of all gases reactions is equal to Kc.

4. Assertion (A) : In the presence of catalyst, thevalue of equilbrium constant K increases

Reason (R) : Catalysts increases the rate offorward and bacward reaction to same extent.

5. Assertion : A net reaction can occur only if a systemis not at equilibrium.

Reason : All reversible reactions occur to reach astate of equilibrium.

6. For the reaction, N2 + O2 2NO, increase inpressure at equilibrium has no effect on thereaction.

Reason : moles of gaseous product – molesof gaseous reactant = 0.

7. Assertion (A) : SO2(g) + 21

O2(g) SO3(g) +

heat. Forward reaction is favoured at lowtemperature and high pressure.Reason (R) : Reaction is endothermic.

8. Assertion : The reaction quotient , Q has the sameform as the equilibrium constant Keq, and isevaluated using any given concentrations of thespecies involved in the reaction, and not necessarilyequilibrium concentrations.Reason : If the numerical value of Q is not the sameas the value of equilibrium constant, a reaction willoccur.Here both Assertion and Reason are correct butReason is not a correct explanation or Assertion.

9. Assertion (A) : Solubility of a gas in liquidsincreases with increase in pressure of the gas inequilbrium with solution.Reason (R) : The dissolution of a gas in liquids isan exothermic process.

10. Assertion (A) : Solubility of a gas in waterdecreases with increase in temperature.Reason (R) : Dissolution of a gas in water is anexothermic process.


11. Assertion (A) : The value of K increases whenconcetration of the reactants are increased.

Reason (R) : With increases of concetration ofreactants the equilbrium shifts in forward direction.

12. Assertion (A) : For the reaction,

H2 (g) + I2 (g) 2HI(g), Kp =Kc.

Reason (R) : Kp of all gaseous reactions is equalto Kc.

13. Assertion : If the equation for a reaction is reversed,the equilibrium constant is inverted and if the equationis multiplied by 2, the equilibrium constant is squared.

Reason : The numerical value of an equilibriumconstant depends on the way the equation for thereaction is written.

14. Assertion : Kp = Kc for all reactions.

Reason : At constant temperature, the pressure ofthe gas is not proportional to the concentration.

15. Assertion (A) : Effect of temperature of KC or KP

depends on enthalpy change.

Reason (R) : Increases in temperature shifts theequilbrium in exothermic direction and decrease intemperature shifts the eqilbrium position inendothermic direction.

16. Assertion (A) : On opening a sealed soda bottledissolved carbon dioxide gas escapes.

Reason (R) : Gas escapes to reach the newequilbrium condition of lower pressure.

17. Assertion : A catalyst does not influences the valuesof equilibrium constant.

Reason : Catalysts influence the rate of both forwardand backward reactions equally.

18. Assertion (A) : H and E are almost the samefor the reaction, N2(g) + O2(g) 2NO(g).

Reason (R) : All reactants and products are gases.

19. Assertion (A) : Equilbrium constant has meaningonly when the corresponding balanced chemicalequation is given at equilbrium.

Reason (R) : Its value changes for the newequation obtained by multiplying or dividing theoriginal equation by a number

20. Assertion (A) : Catalyst affects the final state ofthe equilibrium.

Reason (R) : It enables the system to attain a newequilbrium state by complexing with reagents.

21. Assertion (A): Kp is related to Kc by the relation,Kp = Kc (RT)n

Reason (R) : Kp has same units as Kc.

22. Assertion (A) : Reaction quotient Q is equal toKeq when the reaction is in equilibrium.

Reason (R) : If a catalyst is added to the reactionat equilibrium, the value of Q remains no longerequal to Keq.

23. Assertion : For PCl5(g) PCl3(g) + Cl2(g). Ifmore Cl2 is added the equilibrium will shift inbackward direction hence equilibrium constant willdecrease.

Reason : Addition of inert gas to the equilibriummixture at constant volume, alter the equilibrium.


1. A + 2B 2C + D initial concentration of B was1.5 times that of A, but the equilibrium concentrationof A and B are found to be equal. The equilibriumconstant for the reaction is : [AFMC 2001](a) 4 (b) 8(c) 12 (d) 16

2. For the manufacture of ammonia by the reaction N2+ 3H2 2NH3 + 21.9 K cal, the favourableconditions are : [AFMC 2001](a) Low temperature, low pressure & catalyst(b) Low temperature, high pressure & catalyst(c) High temperature, low pressure & catalyst(d) High temperature, high pressure & catalyst

3. If the concentration of the reactant are doubled in areversible chemical equation having two reactant (inequilibrium) then the equilibrium constant will :

[AFMC 2003]

(a) become one fourth (b) doubled

(c) halved (d) remains same

4. Of the following which change will shift the reactiontowards the product ? [AIIMS 2004]

I2(g) 2I(g), H°r (298 K) = + 150 kJ

(a) Increase in concentration of I2

(b) Decrease in concentration of I2


(c) Increase in temperature(d) Increase in total pressure

5. CaCO3(s) CaO(s) + CO2(g) reaction in a linegoes to completion because : [AFMC 2005](a) CaO does not react to CO2 to give CaCO3.(b) backward reaction is very low.

(c) CO2 formed escapes out.(d) None of the above.

6. For the chemical equilibrium,

CaCO3(s) CaO(s) + CO2(g),H° can be determined from which of the followingplots ? [AIIMS 2005]

(a)

(b)

(c)

(d)

7. In the two gaseous reactions (i) and (ii) at 250ºC [AIPMT 1994, 2005]

(i) NO (g) + 12 O2 (g) NO2 (g), K1

(ii) 2NO2 (g) NO (g) + O2 (g), K2

The equilibrium constants K1 and K2 are related as:

(a) K2 = 1

1K (b) K2 = K1

1/2

(c*) K2 = 1

21

K(d) K2 = K1

2

8. For the reaction, CH4 + 2O2 (g) CO2 (g) + 2H2O (), rH = – 170.8 kJ mol–1

Which of the following statements is not true ?[AIPMT 2006]

(a) At equilibrium, the concentrations of CO2 (g) andH2O (l) are not equal.

(b*) The equilibrium constant for the reaction is

given by Kp = 2

4 2

[CO ][CH ] [O ]

(c) Addition of CH4 (g) or O2 (g) at equilibrium willcause a shift to the right.

(d) The reaction is exothermic.

9. In the reaction : H2 + I2 2HIIn a 1 L flask 0.4 moles of each H2 and I2 are taken.At equilibrium 0.5 moles of HI are formed. Whatwill be the value of equilibrium constant Kc ?

[RPMT 2006](a) 20.2 (b) 25.4(c) 0.284 (d) 11.1

10. When hydrogen molecules decomposed into its atomswhich conditions give maximum yield of hydrogenatoms ? [UP CPMT 2006, RPMT 2007](a) High temperature and low pressure(b) Low temperature and high pressure(c) High temperature and high pressure(d) Low temperature and low pressure

11. 1 mole of H2 and 2 moles of I2 are taken initially ina 2 litre vessel. The number of moles of H2 atequilibrium is 0.2. Then the number of moles of I2and HI at equilibrium are :

[UP CPMT 2006, RPMT 2007](a) 1.2, 1.6 (b) 1.8, 1.0(c) 0.4, 2.4 (d) 0.8, 2.0

12. If a mixture of CO and N2 in equal amount havetotal 1 atm pressure,find out partial pressure of N2in mixture. [RPMT 2007](a) 1 atm (b) 0.50 atm(c) 2 atm (d) 3 atm


13. For reaction, 2NOCl (g) 2NO (g) + Cl2 (g),KC at 427ºC is 3 × 10–6 L mol–1. The value of Kp isnearly: [AFMC 2007](a) 7.5 × 10–5 (b) 2..5 × 10–5

(c) 2..5 × 10–4 (d) 1.72 × 10–4

14. The value of the equilibrium constant of the reaction

HI (g) 12 H2(g) +

12 I2 (g) is 8.0.

The equilibrium constant of the reaction H2(g) + I2(g) 2HI (g) will be : [AIPMT 2008](a) 1/16 (b*) 1/64(c) 16 (d) 1/8

15. If concentration of OH– ions in the reaction

Fe(OH)3(s) Fe3+(aq) + 3OH–(aq)

is decreased by 14 times, then equilibrium

concentration of Fe3+ will increase by[AIPMT 2008]

(a) 8 times (b) 16 times(c*) 64 times (d) 4 times

16. The value of 1PK and 2PK for the reactions

X Y + Z ... (a) and A 2B ... (b)are in the ratio of 9 : 1. If the degree of dissociationof X and A be equal, then total pressure at equilibrium(a) and (b) are in the ratio [AIPMT 2008](a) 3 : 1 (b) 1 : 9(c*) 36 : 1 (d) 1 : 1

17. The dissociation equilibrium of a gas AB2 can berepresented as2AB2 (g) 2 AB(g) + B2 (g)The degree of dissociation is ‘x’ and it is smallcompared to 1. The expression relating the degreeof dissociation (x) with equilibrium constant Kp andtotal pressure P is. [AIPMT 2008](a) (2 Kp/P) (b*) 2Kp/P)1/3

(c) (2 Kp/P)1/2 (d) (Kp/P)18. In which of the following reactions, the

concentration of the product is higher than theconcentration of reactant at equilibrium ?(K = equilibrium constant) [AIIMS 2008]

(a) A B ; K = 0.001

(b) M N ; K = 10

(c) X Y ; K = 0.005

(d) R P ; K = 0.01


N2 (g) + 3H2 (g) 2NH3 (g)Unit of Kc = L2 mol–2.Reason : For the reaction,N2 (g) + 3H2 (g) 2NH3 (g)

Equilibrium constant, Kc = 2

33

2 2

[NH ][N ][H ] .

[AIIMS 2008](a) If both assertion and reason are true and reason



(c) If Assertion is true but reason is false.(d) If both assertion and reason are false.

20. Two moles of each reactant A and B are taken in areaction flask. They react in the following manner,

A (g) + B (g) C (g) + D (g)At equilibrium, it was found that the concentrationof C is triple to that B. The equilibrium constant forthe reaction is : [AIIMS 2009](a) 4.5 (b) 6(c) 9 (d)


2NO(g) + O2(g) 2NO2 (g)increase in pressure favours the formation of NO2.Reason : The reaction is exothermic.

[AIIMS 2009](a) If both assertion and reason are true and reason


is not the correct explanation of assertion.(c) If Assertion is true but reason is false.(d) If both assertion and reason are false.

22. The dissociation constant for acetic acid and HCNat 25ºC are 1.5 × 10–5 and 4.5 × 10–10, respectively.The equilibrium constant for the equilibrium,CN– + CH3COOH HCN + CH2COO–

[AIPMT 2009](a) 3.0 × 105 (b) 3.0 × 10–5

(c) 3.0 × 10–4 (d) 3.0 × 104


23. Which reaction is not affected by change in pressure?[AFMC 2009]

(a) H2 + I2 2HI

(b) 2C + O2 2CO

(c) N2 + 3H2 2NH3

(d) PCl5 PCl3+ Cl224. N2O4(g) 2NO2 – Q

The unit of Kp for the give reaction is :[RPMT 2009]

(a) atmosphere (b) atmosphere2

(c) atmosphere–1 (d) None of these

25. In the equilibrium reaction 2HI (g) H2 + I2which of the following expressions is true?

[RPMT 2009](a) Kp = Kc (b) Kc = 2Kp

(c) Kp > Kc (d) Kc = Kp (RT)2

26. For an equilibrium reaction, the rate constants forthe forward and the backward reaction are 2.38 ×10–4 and 8.15 × 10–5 respectively. What will be theequilibrium constant for the reaction ?

[RPMT 2010](a) 92.2 (b) 29.2(c) 20.2 (d) 2.92

27. What will be the partial pressure of He and O2respectively, if 200 ml of He at 0.66 atm and 400 mlof O2 at 0.52 atm pressure are mixed in 400 ml vesselat 20ºC ? [AFMC 2010](a) 0.33 and 0.56 (b) 0.33 and 0.52(c) 0.38 and 0.52 (d) 0.25 and 0.45

28. The reaction, 2A(g) + B(g) 3C(g) + D(g)is begun with the concentrations of A and B both atan initial value of 1.00 M. When equilibrium isreached, the concentration of D is measured andfound to be 0.25 M. The value for the equilibriumconstant for this reaction is given by the expression:

[AIPMT 2010](a) [(0.75)3 (0.25)] [(1.00)2 (1.00)](b*) [(0.75)3 (0.25)] [(0.50)2 (0.75)](c) [(0.75)3 (0.25)] [(0.50)2 (0.25)](d) [(0.75)3 (0.25)] [(0.75)2 (0.25)]

29. Match List I (Equations) with List II (Types ofprocesses) and select the correct option.

[AIPMT 2010]

List I List IIEquations Types of process

(a) Kp > Q (i) Non-spontaneous(b) G° < – RT In Q (ii) Equilibrium(c) Kp = Q (iii) Spontaneous and

endothermic

(d) T > HS

(iv) Spontaneous

(a) a - (i), b - (ii), c - (iii), d - (iv)(b) a - (iii), b - (iv), c - (ii), d - (i)(c) a - (iv), (b -(i), c - (ii), d - (iii)(d) a - (ii), b - (i), c - (iv), d - (iii)

30. In which of the following equilibrium Kc and Kp arenot equal? [AIPMT 2010]

(a) 2NO(g) N2(g) + O2(g)

(b) SO2(g) + NO2(g) SO3(g) + NO(g)

(c) H2(g) + 2(g) 2H(g)

(d*) 2C(s) + O2(g) 2CO2(g)31. The following equilibria are given

(I) N2 + 3H2 2NH3; K1

(II) N2 + O2 2NO; K2

(III) H2 + 12

O2 H2O; K3

The equilibrium constant for the reaction,

2NH3 + 52

O2 2NO + 3H2O

in terms of K1, K2 and K3 will be : [AIIMS 2010]

(a) K1K2K3 (b) 1 2

3

K KK

(c) 2

1 3

2

K KK (d)

32 3

1

K KK

32. The following equilibria are given :

N2 + 3H2 NH3 K1

N2 + O2 2NO K2

H2+ 12

O2 H2O K3

The equilibrium constant of the reaction :

2NH3 + 52

O2 2NO + 3H2O in terms of K1,K2 and K3 is : [AIIMS 2011]


(a) 1 2

3

K KK (b)

21 3

2

K KK

(c) 3

2 3

1

K KK (d) K1 K2 K3

33. For the reaction N2(g) + O2(g) 2NO(g), theequilibrium constant is K1. The equilibrium constantis K2 for the reaction 2NO(g) + O2(g) 2NO2(g).What is K for the reaction NO2(g) ½N2(g) +O2(g) ? [AIPMT 2011](a) 1 / (2K1K2) (b) 1 / (4K1K2)(c) [1 / K1K2]½ (d) 1 / (K1K2)

34. In the reaction, H2(g) + Cl2(g) 2HCl(g)[AFMC 2011]

(a) Kp Kc (b) Kp = Kc(c) Kp > Kc (d) Kp < Kc

35. Given the reaction between 2 gases represented byA2 and B2 to give the compound AB(g).A2(g) + B2(g) 2 AB(g).At equilibrium, the concentrationof A2 = 3.0 × 10–3 Mof B2= 4.2 × 10–3 Mof AB=2.8 × 10–3 M. [AIPMT 2012]lf the reaction takes place in a sealed vessel at 527°C,then the value of KC will be :(a) 2.0 (b) 1.9(c) 0.62 (d) 4.5

36. For the reversible reaction :N2(g) + 3H2(g) 2NH3(g) + heatThe equilibrium shifts in forward direction :

[AIPMT 2014](a) by increasing the concentration of NH3(g)(b) by decreasing the pressure(c) by decreasing the concentrations of N2(g) and

H2(g)(d) by increasing pressure and decreasing

temperature37. For a given exothermic reaction, Kp and Kp

’ are theequilibrium constants at temperatures T1 and T2respectively. Assuming that heat of reaction isconstant in temperatures range between T1 and T2,it is readily observation that: [AIPMT 2014](a) Kp > Kp

’ (b) Kp < Kp’

(c) Kp = Kp’ (d) Kp =

1

pK?

38. If the value of an equilibrium constant for a particularreaction is 1.6 × 1012, then at equilibrium the systemwill contain [AIPMT-2015](a) all reactants(b) mostly reactants(c) mostly products(d) similar amounts of reactants and products

39. Which of the following statements is correct for areversible process in a state of equilibrium?

[AIPMT-2015](a) G = – 2.30RT log K(b) G = 2.30RT log K(c) Gº = – 2.30RT log K(d) Gº = 2.30RT log K

40. Consider the reaction equilibrium

Ice Water – xkcal(Greater volume) (Lesser volume)The favourable conditions for forward reaction are:

[AIIMS-2015](a) low temperature, high pressure and excess of

ice.(b) low temperature, low pressure and excess of

ice.(c) high temperature, low pressure and excess of

ice.(d) high temperature, high pressure and excess of

ice.41. Kp for the reaction A B is 4. If initially only A is

present then what will be the partial pressure of Bafter equilibrium? [AIIMS-2016](a) 1.2 (b) 0.8(c) 0.6 (d) 1

42. For the following reaction in gaseous phase

CO(g) + 12

O2(g) CO2(g), Kp / Kc is:

[AIIMS-2016](a) (RT)1/2 (b) (RT)–1/2

(c) (RT) (d) (RT)–1

43. A 20 litre container at 400 K contains CO2(g) atpressure 0.4 atm and an excess of SrO neglect thevolume of solid SrO). The volume of the containeris now decreased by moving the movable pistonfitted in the container. The maximum volumeof the container, when pressure of CO2attains its maximum value, will be : (Given that :SrCO3(s) SrO(s) + CO2(g), Kp = 1.6 atm)

[NEET-2017]


(a) 5 litre (b) 10 litre(c) 4 litre (d) 2 litre

44. The equilibrium constants of the following are :

N2 + 3H2 2NH3 K1

N2 + O2 2NO K2

H2 + ½ O2 H2O K3

The equilibrium constant (K) of the reaction

2NH3 + 25

O2 K 2NO + 3H2O, will be :

[NEET-2017](a) K1K3

3 / K2 (b) K2K33 / K1

(c) K2 K3 /K1 (d) K23K3 / K1

45. Which of the following conditions will favourmaximum formation of the product in the reaction

A2(g) + B2(g) X2(g) rH = –XkJ ?[NEET-2018]

(a) Low temperature and high pressure(b) High temperature and low pressure(c) High temperature and high pressure(d) Low temperature and low pressure

46. 2ICl I2 + Cl2 KC = 0.14

Initial concentration of ICl is 0.6 M

then equilibrium concentration of I2 is :

[AIIMS-2018 (M)]

(a) 0.37M (b) 0.128 M

(c) 0.224 M (d) 0.748 M

47. A + 2B 2C K = ?

2 mole each A and B present in 10 lt so that C formis 1 mole, Calculate KC.

[AIIMS-2018 (E)]

(a) 1.5 (b) 6.67

(c) 0.15 (d) 2.3


ANSWER KEY

EXERCISE–0 (RECALL YOUR UNDERSTANDING)1. 4 NO(g) + 6H2O(g) 4NH3(g) + 5 O2(g)

2. K3. If one of the products (gaseous) is allowed to escape

out (i.e., in open vessel).4. The value of equilbrium constant depends upon

(i) nature of the reaction (ii) temperature.5. The reaction does not proceed much in the forward

direction.6. K = kf / kb. In exothermic reaction, with increases

of temperature, kb increases much more than kf.Hence K decreases.

7. High concentrations of N2 and H2 at lowtemperature, high pressure.

8. Vapour pressure.

9. Kc = 2[CO (g)]1 . Taking active masses of solids as

unity, Kc = [CO2(g)]. Similarly, Kp = 2COp .

10. Statement : At a given temperature, the product ofconcentrations of the reaction products raised to therespective stoichiometric coefficient in the balancedchemical equation divided by the product ofconcentrations of the reactants raised to theirindividual stoichiometric coefficients has a constantvalue. This is known as the Equilibrium Law or Lawof Chemical Equilibrium.

11. KC = c d

a b[C] [D][A] [B] KP =

c dC D

a bA B

[P ] [P ][P ] [P ]

Kp = Kc(RT)n

12. (i) Homogeneous equilibrium When all reactantsand products are in same phase present

H2(g) + Cl2(g) 2HCl(g);

2

c2 2

[HCl]K

[H ][Cl ]?

SO2(g) + NO2(g) SO3(g) + NO(g) ;

3c

2 2

[SO ][NO]K

[SO ][NO ]?

(ii) Heterogeneous equilibrium When more than onephase are present

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g);

42

c 42

[H ]K

[H O]?

2Na2O2(s) + 2H2O() 4NaOH(l) +O2(g) ; KC = [O2]

13. (i) KC = 2

22

[NO(g)] [Cl (g)][NOCl (g)]

(ii) Kc = [NO2(g)]4 [O2(g)]

(iii) Kc = 3 2 5

3 2 5 2

[CH COOH(aq)[C H OH(aq)][CH COOC H (aq)][H O( )]

(iv) Kc = 3 31

[Fe (aq)[OH (aq)]? ?

(v) KC = 52

1[F (g)]

14. [Pure liquid] or [Pure solid]

= No. of molesvolume of L =

Mass/mol. massVolume

= MassVolume ×

1Mol. mass =

DensityMol. mass (Density &

Molecular weight are remain constant).15. (i) Large value of equilabium constant (> 103)

shows that forward reaction is favoured, ie..concentration of products is much large thanthat of the reactants at equibrium.

(ii) Intermediate value of K (10–3 to 103) showsthat the concentration of the reactant andproducts are comparable.

(iii) Low value of K (< 10–3) shows at backwardreaction is favoured, i.e. concenctration ofreactants is much large than that of the products.

16. (i) No change because catalyst does not disturb thestate of equilbrium.

(ii) No change because np = nr.17. Melting of ice in endothermic process accomanied

by decreases of volume. At altitudes, pressure aswell as temperature is low. In the equilbrium, Ice

Water, both the factors favour backwardprocess. Hence the melting is slow.

18. Given : Equilibrium constant (Kc) for the reactionN2 (g)+ 2O2(g) 2NO2(g) is KC = 400


Formulae : Kc = 2

22

2 2

[NO ][N ][O ] = 400

Asked : (i) For the reaction 2NO2(g) N2(g)+

2O2(g) , K’C = 2

2 22

2

[N ][O ][NO ] =

c

1K

Substitution & Calculation : K’c = 1

400 = 0.0025

Asked : For the reaction 1/2 N2(g) + O2(g)NO2(g)

Substitution & Calculation :

K’’c = 2

1/22 2

[NO ][N ] [O ] = cK

K’’c = 400 = 20

19. Given :ng = np – nr = – 1.Asked : Kp / KC = ?Formulae : Kp = Kc (RT)–1


Kp/Kc = (RT)–1 = 1

0.821 298? = 0.04

20. (a) Equilibrium will shift in the backward direction(b) Equilibrium will shift in the backward direction

(c) Equilibrium will shift in the forward direction21. (a) Increase the temperature equilibrium constant

decreases.(b) Increase the temperature equilibrium constant

increases.(c) Equilibrium constant remain same.

22. Given : Pressure of HI = 0.2 atm (initially),Pressure of HI = 0.04 atm (equilibrium)

2HI(g) H2(g) + I2(g)

Initial pressure 0.2 atm 0 + 0

At. eqm. 0.04 atm 0.16

2 atm 0.16

2 atm.

= 0.08atm = 0.08 atm(Decrease in the pressure of HI = 0.2–0.04 = 0.16atm)

Formulae Kp = 2 2H I

2HI

p p

p

?


20.08 atm × 0.08 atm

(0.04 atm) = 4.0

23. KC = 3

22

[CH OH][CO][H ] Kp =

3

2

CH OH

CO H

pp p?

(i) When volume of the vessel is reduced to half,the concentration of each reactant of productbecomes double. Thus.

QC = 3

22

2[CH OH]2[CO] {2 | H |}? =

14 Kc

As Qc < Kc, equilibriums will shift in the forwarddirection producing more of CH3OH to makeQc = Kc.

(ii) Qp = 3CH OH

CO

pp ×

2

2H

1(2p ) =

14 Kp.

Again Qp < Kp, equilibrium will shift in theforward direction to make Qp = Kp.

(iii) As volume remains constant molarconcentrations will not change. Hence there isnot effect on the state of equilibrium.

24. Given : [SO2] = 0.60M, [O2] = 0.82M and [SO3] =1.90MAsked : KC = ?

Formulae : Kc = 32

2 2

[SO ][SO ] [O ]

?

Substitution & Calculation :2

2(1.90M)

(0.60M) (0.82M) = 12.22

25. Given : 0.087 mol of NO and 0.0437 mol of Br2.Substitution & Calculation :0.0518 mol of NOBr is formed from 0.0518 mol ofNO and 0.0518/2 = 0.0259 mol of Br2. At equilibrium Amount of NO = 0.087 – 0.0518 =0.0352 molAmount of Br2 = 0.0437 – 0.0259 = 0.0178 mol

26. Given : 2.00 mole of CH4, 3.0 mol of CS2, 3.0mole of H2 and 4.0 mole of H2S.Substitution & Calculation :

QC =

4

2

3 310 102 4

10 10

? ? ? ?? ? ? ?? ? ? ?

? ?? ?? ?? ?? ?? ?

= –2243 1032

?

= 7.59 × 10–2 > Kc

so, reaction will proceed in backward direction.


27. Given : PCl5(g) PCl3(g) + Cl2(g),

(a) Kc = 3 2

5

[PCl (g)] [Cl (g)][PCl (g)]


(b) k’ = c

1K = 3

18.3 10??

= 120.48

(c) (i) No effect.(ii) No effect.(iii) As given reaction is endothermic, on increasing

the temperature, kf will increase. As Kc = f

b

kk ,

Kc will increase with increase of temperature.28. Given : Total pressure of 105 Pa, iodine vapour

contain 40% by volume of iodine atoms.Asked : Kp = ?Substitution & Calculation :Partial pressure of atoms

(pI) = 40

100 × 105 Pa = 0.4 × 105 Pa

Partial pressure of

(pI2) = 60

100 × 105 Pa = 0.60× 105 Pa

Kp = 2

2I

I

pp =

5 2

5(0.4 10 )0.60 10

??

= 2.67 × 107 Pa29. (i) Backward

(ii) Forward(iii) Forward(iv) Backward(v) No change

30. Given : WPCl5 = 2.695 g, Volume = 1 litre, Pressure= 1 atm, Temp. = 523 KAsked : = ?.

Formulae : PV = wM RTT

M0 = wRTPV


2.695 0.0821 5231 1

? ?? = 115.7

For PCl5, Mt = 31 + 5 × 35.5 = 208.5

= t 0

0

M – MM =

208.5 – 115.7115.7 = 0.80.

= 80%.31. Given : Concentration of PCl5 = 0.5 × 10–1 mol

L–1, Kc = 8.3 × 10–3.Asked : Concentrations of PCl3 and Cl2 atequilibrium = ? PCl5(g) PCl3(g) + Cl2(g)At eqm. 0.5×10–1 molL–1 x mol L–1 x mol L–1

(suppose)

Formulae : KC = 3 2

5

[PCl ] [Cl ][PCl ]


Kc = 2

10.5 10x

?? = 8.3 × 10–3 (Given)

or x2 = (8.3 × 10–3) (0.5 × 10–1) = 4.15 × 10–4

or x = 44.15 10?? = 2.04 × 10–2 M = 0.02 M

32. Given : If total mass of the mixture of CO and CO2is 100 g, thenCO = 90.55 and CO2 = 100 – 90.55 = 9.45 gAsked : KC = ?

Formulae : Kp = 2

2CO

CO

pp

Substitution & Calculation :Number of moles of CO = 90.55/28 = 3.234Number of moles of CO2 = 9.45/44 = 0.215

pCO = 3.234

3.234 0.215? × 1 atm = 0.938 atm

2COp = 0.215

3.234 0.215? × 1 atm = = 0.062 atm

Kp = 2

2CO

CO

pp =

2(0.938)0.062

= 14.19

ng = 2 – 1 = 1 Kp = Kc (RT) or

Kc = pKRT

= 14.19

0.0821 1127? = 0.153

33. Given : PCO = PH2O = 4.0 bar, Kp = 10.1Asked : Partial pressure of H2 at equilibrium


Formulae : Kp = 2 2

2

CO H

CO H O

(p ) (p )(p ) (p )

Substitution & Calculation :Suppose the partial pressure of H2 at equilbrium =p barCO(g) + H2O (g) CO2(g) + H2(g)Initial pressure 4.0 bar 4.0 bar

At eqm. (4 – p) (4 – p) p p

Kp = 2

2(4 )p

p? = 0.1 (Given)

4p

p? = 0.1 = 0.316

p = 1.264 – 0.316 p or 1.316 p = 1.264 or p = 0.96bar

EXERCISE–1 (CHECK YOUR UNDERSTANDING)1. (b) 2. (d) 3. (b) 4. (b) 5. (d) 6. (d) 7. (d) 8. (c) 9. (c) 10. (c)

11. (d) 12. (c) 13. (d) 14. (b) 15. (a) 16. (a) 17. (b) 18. (a) 19. (a) 20. (d)21. (c) 22. (c) 23. (c) 24. (d) 25. (c) 26. (a) 27. (b) 28. (a) 29. (b) 30. (a)31. (a) 32. (b) 33. (d) 34. (d) 35. (b) 36. (d) 37. (a) 38. (a) 39. (c) 40. (c)41. (a) 42. (d) 43. (a) 44. (c) 45. (b) 46. (b) 47. (c) 48. (c) 49. (c) 50. (a)51. (c) 52. (d) 53. (a) 54. (b) 55. (b) 56. (b) 57. (b) 58. (b) 59. (c) 60. (b)61. (a) 62. (d) 63. (b) 64. (b) 65. (b) 66. (a) 67. (a) 68. (a) 69. (a) 70. (a)71. (d) 72. (b) 73. (a) 74. (c) 75. (a) 76. (c) 77. (c) 78. (a) 79. (b) 80. (b)81. (a) 82. (b) 83. (c) 84. (c) 85. (c) 86. (b) 87. (d) 88. (c) 89. (c) 90. (d)91. (d) 92. (a) 93. (a) 94. (d) 95. (c) 96. (a) 97. (d) 98. (c) 99. (b) 100. (d)

101. (b) 102. (c) 103. (c) 104. (b)

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)1. (c) 2. (a) 3. (c) 4. (b) 5. (a) 6. (a) 7. (b) 8. (b) 9. (a) 10. (c)

11. (c) 12. (d) 13. (b) 14. (a) 15. (d) 16. (d) 17. (b) 18. (c) 19. (a) 20. (b)21. (a) 22. (a) 23. (a) 24. (a) 25. (d) 26. (d) 27. (a) 28. (b) 29. (a) 30. (a)31. (d) 32. (b) 33. (c) 34. (b) 35. (a) 36. (b) 37. (c) 38. (d) 39. (b) 40. (d)41. (a) 42. (a) 43. (c) 44. (a) 45. (c) 46. (a) 47. (b) 48. (d) 49. (a) 50. (d)51. (d) 52. (b) 53. (d) 54. (c) 55. (a) 56. (a) 57. (d) 58. (c) 59. (b) 60. (b)61. (b) 62. (d) 63. (b) 64. (b) 65. (a) 66. (b) 67. (b) 68. (a) 69. (b) 70. (b)71. (d) 72. (c) 73. (d) 74. (a) 75. (a) 76. (c) 77. (a) 78. (c) 79. (a) 80. (b)

EXERCISE–3 (AIIMS SPECIAL)1. (d) 2. (a) 3. (c) 4. (d) 5. (a) 6. (a) 7. (c) 8. (b) 9. (b) 10. (a)

11. (d) 12. (c) 13. (a) 14. (d) 15. (c) 16. (a) 17. (a) 18. (b) 19. (a) 20. (d)21. (c) 22. (c) 23. (d)

EXERCIS-4 (ARCHIVES FROM VARIOUS MEDICAL ENTRANCE)1. (a) 2. (b) 3. (a) 4. (c) 5. (c) 6. (a) 7. (c) 8. (b) 9. (d) 10. (a)

11. (a) 12. (b) 13. (d) 14. (b) 15. (c) 16. (c) 17. (b) 18. (b) 19. (a) 20. (c)21. (b) 22. (d) 23. (a) 24. (a) 25. (a) 26. (d) 27. (b) 28. (b) 29. (c) 30. (d)31. (d) 32. (c) 33. (c) 34. (b) 35. (c) 36. (d) 37. (a) 38. (c) 39. (c) 40. (d)41. (b) 42. (b) 43. (a) 44. (b) 45. (a) 46. (b) 47. (b)

INTRODUCTIONSolids are characterised by the state of matter in which

particles are closely packed and held together by stronginter molecular attractive force.

1. PROPERTIES OF SOLIDS(a) In solid state the particles are not able to move

randomly.(b) They have definite shape and volume.(c) Solids have high density.(d) Solids have high and sharp melting point which

is depend on the strength or value of bindingenergy.

(e) They are very low compressible.(f) They show very slow diffusion.

2. TYPES OF SOLIDS :(a) Crystalline solids :

(a) In this type of solids the atoms or molecule arearranged in a regular pattern in the threedimensional network.

SOLID STATE

Chapter

4(b) They have well defined geometrical pattern, sharp

melting point, definite heat of fusion andanisotropic nature. Anisotropic means they exhibitdifferent physical properties in all directions.e.g. The electrical and thermal conductivities aredifferent directions.

(c) They are generally incompressible.(d) The general examples of crystalline solids are as

Quartz, diamond etc.(b) Amorphous Solids :

(a) In this type of solids, the arrangement of buildingconstituents is not regular.

(b) They are regarded as super cooled liquids withhigh viscosity in which the force of attractionholding the molecules together are so great, thatthe material becomes rigid but there is noregularity in structure.

(c) They do not have sharp melting points.(d) They are isotropic as they exhibit same physical

properties in all the directions.(e) The general examples of this solids are as glass,

Rubber, plastics etc.

Difference between crystalline and amorphous solids

Property Crystalline solids Amorphous solids

1. Shape They have definite and regular They do not have definite and regulargeometrical form. geometrical form.

2. Melting point They have definite melting point They do not have definite melting point.3. Heat of fusion They have a definite heat of fusion They do not have definite heat of fusion.4. Compressibility They are rigid and incompressible. These may be compressed to any

appreciable extent.5. Cutting with a They are given cleavage i.e. they They are given irregular cleavage

Sharp edged tool break into two pieces with plane i.e.they break into two pieces withsurfaces. irregular surface.

4.2 SOLID STATE (CHEMISTRY)

6. Isotropy and They are anisotropic. They are isotropic.Anisotropy

7. Cooling curve

Note: Amorphous solids show smooth cooling curve while crystalline solids show two breaks in cooling curve. In the case of crystallinesolids two breaks points ‘a’ and ‘b’ are appear. These points indicate the beginning and the end of the process of crystallization. In this timeinterval temperature remains constant.

Some Important Characteristics of Various types of Crystals

S. Characteristics Ionic Crystals Covalent Molecular Metallic CrystalsNo. Crystals Crystals

1 Units that Cations and Atoms Molecules Positive ions in a "sea oroccupy lattice anions pond" of electrons.points

2. Binding Electrostatic Shared vander Waals Electrostatic attractionforces attraction electrons or Dipole- between positively

between ions dipole charged ions andnegatively chargedelectrons.

3. Hardness Hard Very hard Soft Hard or softGraphiteis soft

4. Brittleness Brittle Interme- Low Lowdiate

5. Melting point High Very high Low Varying from moderateto high

6. Electrical Semi cond- Non-con- Bad condu- Good conductorsConduction uctor due to ductor ctor

crystal impe- Graphiterfections,con- is goodductor in fused conductorstate

7. Solubility in Soluble Insoluble Soluble as insolublePolar solvents well as

insoluble8. Heat of NaCl(s) Graphite NH3(s) Cu(s)

Vaporisation 170.75 718.43 23.55 304.59(kJ mol–1)

9. Heat of NaCl – NH3 (s) Cu(s)fusion (kJ mol–1) 28.45 – 5.65 13.016

SOLID STATE (CHEMISTRY) 4.3

S. Characteristics Ionic Crystals Covalent Molecular Metallic CrystalsNo. Crystals Crystals

10. Example NaCl, KNO3 Diamond, H2O(s), Na, Cu, Ag, Fe,CsCl, Na2SO4 graphite, CO2(s), Pt, alloysZnS Quartz Sulphur,

(SiO2), Sugar, Iodine,SiC noble gases

Note: Isomorphism: The occurence of a given substance in more than one solid crystalline forms have differentphysical properties is known as polymorphism.(1) This property when occurs in elements is known as allotropy.(2) Sometimes we come across examples of chemically different solids which crystallise in the same crystallineshape. Such substances are said to be Isomorphous (same shape). viz.(1) Na2HPO4.12H2O and Na3AsO4.12H2O (2) K2SO4 , K2CrO4(3) ZnSO4.7H2O , MgSO4.7H2O , FeSO4.7H2O (4) KMnO4 , KClO4(5) K2SO4.Al2(SO4)3.24H2O, K2SO4. Cr2(SO4)3.24H2O.

EXAMPLEA solid X melts slightly above 273K and is a poorconductor of heat and electricity. To which of thefollowing categories does it belong :

(1) Ionic solid (2) Covalent solid(3) Metallic (4) Molecular Ans. (4)

EXAMPLEIn a crystal, the atoms are located at the position of :

(1) Zero P.E. (2) Infinite P.E.(3) Minimum P.E. (4) Maximum P.E.

Ans. (3)

EXAMPLEGraphite is an example of :

(1) Ionic solid(2) Covalent solid(3) Vander waal’s crystal(4) Metallic crystal Ans. (2)

EXAMPLEAmorphous solids :(1) Possess sharp melting points(2) Undergo clean cleavage when cut with knife(3) Do not undergo clean cleavage when cut with knife(4) Possess orderly arrangement over long distances

Ans. (3)

3. STUDY OF CRYSTALS Crystal : A crystal is a homogeneous portion of a

solid substance made by regular pattern ofstructural units bonded by plane surface makingdefinite angles with each other.

Space lattice : The arrangement of constituentslike atom, ions and molecules in different sites inthree dimensional space is called space lattice.

Unit cell : The smallest repeating unit in spacelattice which when repeats over and over again,results in a crystal of the given substance calledunit cell.

Face : The plane surface of the crystal are calledfaces.

Edge : An edge is formed by the intersection oftwo adjacent faces.

Interfacial angles : The angle between theperpendiculars two intersecting faces calledinterfacial angles.

(A) UNIT CELL IN TWO DIMENSIONS :

Now in order to uniquely explain a regulararrangement in two dimensions we need the help of threeparameters, two distance parameters and one angularparameter. Based upon their different relationships we candefine different cases

Case ‘A’ (a = b) angle = 90º


The unit cell in such a case is a square. Placing suchsquare side by side we will obtain the entire twodimensional arrangement.

Case ‘B’(a b) angle = 90ºThe unit cell formed in this case is a rectangle.

(B)UNIT CELL IN THREE DIMENSIONS :It has six parameters, 3-distance parameters and

3-angular parameter.

a, b, c are lengths of unit cell (also known as thecrystallographic axes). are known as thecrystallographic angles.

4. TYPES OF UNIT CELL :In every crystal class, the positioning of the lattice

points may be different. Based upon these differentpositions occupied by the lattice points, we have differenttypes of unit cells.

(i) Simple / Primitive type of unit cell : If latticepoints or the particles of the solid are present only at thecorners of the unit cell.

(ii) Body centred unit cell : lattice point are at thecorners as well as at the body centre.

(iii) Face centred unit cell : lattice points are atcorners as well as at each of the face centres.

(iv) End centred unit cell : lattice points are at thecorners as well as at centre of any of two opposite faces.

5. DIFFERENT CLASSES OF CRYSTALS :Based on different permutations of a, b, c and

we define different crystal classes. Each of thesearrangements corresponds to a unique and different typeof arrangement.These 14 different arrangements are calledthe 14 Bravais lattices. (In any lattice, the surrounding ofeach and every lattice point is exactly identical.)

a = b = cAll sides areof equallength; allangles are900

c a

b

Threeangles

changed

Cubic (3 unit cell) Trigonal (Rhombohedral) (1 unit cell)

One sideis

changed

a = b cOne side is ofd i f f e r e n tlength; al langles are 900

General case :All sides aredifferentlengths; allangles aredifferent

One side lengthchanged two angles

fixed at 900 one fixed at1200

Two side lengths madethe same; one angle

fixed at 1200

Length of another sideis changed

a b cThree sides areof differentlengths; al langles are 900

One anglechanged

Triclinic (1 unit cell) Hexagonal (1 unit cell)Tetragonal (2 unit cell)

Orthorhombic (4 unit cell) Monoclinic (2 unit cell)

c a

b

c a

b

c c a

b

c

ab a

c a

b

c a

b

c a

b

a = b = cAll sides are ofequal length;angles

Special Casea = b c

All sides are ofdifferent lengths;


14 BRAVIS LATTICE

2PbI,Zn,Cd,CdS,ZnO,GraphiteSimple120,90cbaHexagonal

)HgS(Cinnabar),3CaCO(Calcite,Bi,Sb,As,ICI,3NaNOSimple90cba

Trigonalor

bohedralhomR

3BO3H,7O2Cr2K,O2H5.4CuSOSimple90cbaTriclinic

sulphurMonoclinic,,4PbCrO,O2H10.4SO2NacentredEnd,Simple90cbaMonoclinic

sulphurbichomR,3CaCO,3PbCO,4BaSO,4SO2K,3KNOcentredEnd,centredFace

,centredBody,Simple90cba

bichomRor

bichomOrthor

4CaSO,2TiO,2SnO),Sn(tinWhitecentredBody,Simple90cbaTetragonal

blendeZinc,Diamond,Alums,KCl,NaCl,CucentredFace

,centredBody,Simple90cbaCubic

ExamplesFoundCellUnit

ofTypeAngleAxial

LengthAxial

ClassCrystal

GEOMETRY OF A CUBENumber of corners = 8 ; Number of faces = 6Number of edges = 12 ; Number of body centre = 1Number of body diagonals = 4 ; Number of face diagonals= 12

Contribution of an atom at different sites of cube :

A corner of a cube is common in 8 cubes. So 18

th part

of an atom is present at this corner of cube.

Note: In case of hexagonal unit cell corner atom is 1/6part.

A face of a cube is common in 2 cubes. So 12

th part of

an atom is present at the face of a cube.

An edge of a cube is common in four cubes, so 14

th

part of the atom is present at the edge of a cube

A body centre is not common in any another cube, so onecomplete atom is present at the cube centre.

6. MATHEMATICAL ANALYSIS OF CUBICSYSTEM (TYPES AND ANALYSIS) :

Simplest crystal is to be studied in cubic system. Threetypes of cubic systems are following.

(a) Simple Cubic (SC) :


(i) Atoms are arranged at the corners of the cube.(ii) Each corner atom is shared by eight surrounding

cubes.

Therefore, it contributes for 18 of an atom.

Z = 8 × 18 = 1

(iii) Radius of atom ‘r’ = 2a

a = edge length

Note: LANGTH OF FACE DIAGONAL AND BODYDIAGONAL :

Consider the triangle ABC, with the help ofpyathogorous theorem

2 2 2 2 2AC AB BC a a a? ? ? ? ? (length offace diagonal.)

Consider the triangle DAC, with the help ofpyathogorous theorem

2 2DC DA AC = 2 2a ( 2a) = 3a (length

of cube diagonal)

(b) Body Centered Cubic (BCC) : Atoms arearranged at the corners and at the centre of the cube.

(i) Eight Corner atoms contribute one atom per unitcell.

(ii) Centre atom contribute one atom per unit cell.(iii) So, total 1 + 1 = 2 atoms per unit cell.

Z = 8 × 18 + 1 = 2

Radius of atom r = 34

a

(c) Face Centered Cubic (FCC) : Atoms are arrangedat the corners and at centered of the each faces.


(i) The eight corners atoms contribute for 18 of an

atom and thus one atom per unit cell. Each of sixface centered atoms is shared by two adjacentunit cells and therefore one face centred atom

contribute half of its share. Means 6 × 12

= 3

atom per unit cell.(ii) So, total Z = 3 + 1 = 4 atoms per unit cell.

(iii) Radius of atom ‘r’ = 2 2a

EXAMPLE 5 :A compound formed by elements A and B has a cubicstructure in which A atoms are at the corners of the cubeand B atoms are at the face centres. Drive the formula ofthe compound.Sol.As A atoms are present at the 8 corners of the cube,therefore numbers of atoms of A in the unit cell

= 81

8 =1

As B atoms are present at the face centres of the 6faces of the cube, therefore, numbers of atoms of atoms

of B in the unit cell = 21

6 = 3

Ratio of atoms A : B = 1 : 3Hence, the formula of the compound is AB3

EXAMPLE 6 :A cubic solid is made up of two elements X and Y. AtomsY are present at the corners of the cube and atoms X atbody centre. What is the formula of the compound ?

SOLUTION:As atoms Y are present at the 8 corners of the cube,therefore, numbers of atoms of Y in the unit cell = 1/8 8= 1

As atoms X are present at the body centre. therefore,numbers of atoms of X in the unit cell = 1

ratio of atoms X : Y = 1 : 1Hence, the formula of the compound is XY

7. ARRANGEMENT OF THE ATOM / PARTICLESOF THE SOLIDS IN THREE DIMENSIONS

Now having gained a knowledge of some of the terms, letus study how the different arrangements in space arebrought about.

Firstly we will focus our attention on the solidscontaining only one type of lattice points.

The solids which contain only one type of lattice pointsare:

(i) metallic solids (eg. Iron)(ii) molecular solids (eg. dry ice)

(iii) covalent network solids (eg. diamond)(Ionic solids do not fall into this category as they

contain more than one type of particles,they will be studiedin the later parts of the chapter)

All the atoms or particles of the solids will berepresented by solid spheres, each of radius ‘r’.

We will be taking these spheres of radius ‘r’ andexplore how we can arrange these in threedimensions.Firstly we will begin with arrangement in onedimension.(A) ARRANGEMENT IN 1-D :

In one dimension it is possible to arrange the spheresin two possible ways.

1.

Not Stable [because the potential energy of the systemis not minimum]

2. a = 2r


Coordination number = 21-D close packing stable arrangementThis is the predominant way of packing in one

dimension and as such most of the space lattices will showsuch an arrangement in one dimension along the planes ofclose packing.(B) ARRANGEMENT IN TWO DIMENSION :

In two dimensions also there are two ways of packingthe spheres.

1. Square packing : If the one dimensional arraysare placed on top of one another, we get the square packingin two dimensions.

One sphere will be in contact with 4 other spheres.CN = 4area of square = a2 = 4r2

area of atoms in the square = 2 21 44

r r? ? ? ? ?

fraction of area occupied by spheres = 2

2 44rr

? ?? = 78%

2. Hexagonal close packing : (in 2-D) If in a twodimensional arrangement,every one dimensional array isplaced in the cavity of the just preceding array, we get thehexagonal packing in two dimensions. One sphere will bein contact with 6 other spheres. CN =6

area of hexagon = 6 × 3

4a2 = 6 ×

34

× 4r2 = 26 3 r

area of atoms = r2 + 13 × 6r2 = 3r2

fraction of area occupied = 2

23 3

66 3rr

? ?? –~ 91%

(C) ARRANGEMENT IN THREE DIMENSIONS :1. Simple cubical arrangement in three dimensions :(will be made up of 2-D sheets arranged one over other)

The simple cubical packing is obtained by arrangingthe square pack sheets of two dimension one over theother such that spheres of the second sheet are exactly(vertically) above the spheres of first sheet.

(Note that , hence crystal thus formed will

belong to the cubic crystal class, and as the lattice pointsare only at the corners, hence the unit cell will be simple,therefore what we get is the simple cubic)(i) Relation between ‘a’ and ‘r’a = 2r (because atoms along the edge are touching eachother)(ii) Effective no. of atoms per unit cell :

(Z) = 18 × 8 = 1

(iii) Packing fraction :

P.F. = volume occupied by per unit cell

volume of one unit cell

= 3

33

3 3 3

44 43

3(2 ) 3 8

r rra r r

? ?? ? ??

= 0.52 (or 52%)


(Note : This is not a very efficient way of packing asthe packing fraction is very low)(iv) Coordination Number :

It is defined as the number of atoms touching anyone particular atom. For simple cubic, coordination number= 6.

(v) Density of unit cell :

d = Massof unit cell

Volume = 3.A

Z MN a

? = 3.A

MN a

2. Body centred cubic :The body centred cubic is a unique way of packing,

i.e when the 2D arrays are placed on top of each other insuch a fashion that the spheres of the next plane are intothe cavities of the first plane of spheres.The third plane ofspheres is then exactly identical to the first plane ofspheres.

(i) a 2r (as atoms along the edge are not touchingeach other) they touch along the body diagonal,hence 3 a = 4r.

(ii) Effective number of atoms (Z) = 1 + 18 × 8 = 2.

(iii) Packing fraction = 3

3

423

4 4 4

r

r

? ?

? ? 3 3 3? ? =

38

?–~ 0.68 = 68%

(iv) Coordination No. = 8 (the sphere at the bodycentre will be touching the spheres at the eightcorners)

(v) Density = 3A

ZMN a = 3

2

A

MN a

3. Close packing in three dimensions :(These are made up of two dimensional hexagonally

packed sheets)

In second layer we have two kinds of voids.(i) Voids of second layer below which there are

spheres of first layer (all voids of type ‘a’).(ii) Voids of second layer below which there are voids

of first layer (all voids of type ‘b’).For third layer, we have two possibilities.

(A) Hcp unit cell : a = 2r = bIf spheres of IIIrd layer are placed in voids of IInd

layer below which there are spheres of Ist layer (voids oftype ’a’) then Ist layer and IIIrd layer are identical so thisis called AB–AB pattern repeat or hexagonal close packing)


(i) Calculation of ‘c’ :For the estimation of ‘c’, consider the spheres marked

1,2,3,4 in the unit cell as shown.These four spheres forma regular tetrahedron. The length of the perpendicular from‘4’ to the equilateral triangle 1-2-3 will be equal to c/2.

cos 30º = 2ax x = 2cos30

a =

22 3

a = 3

a

Apply pythagoras theorem.

x2 + (c/2)2 = a2 c = 23

4r

Volume of the hexagon = Area of base × Height. = 6 34

a2 × c = 26 3 2(4 ) (4 )

4 3r r? = 24 2 . r3

(ii) Effective no. of atoms per unit cell :

Z = 16 × (no. of atoms at corner) +

12

× (no. of

atoms at face centres) + 1 × (no. of atoms inside thebody)

= 16 × 12 +

12

× 2 + 1 × 3 = 2 + 1 + 3 = 6

(iii) Packing fraction : = volume of the atomsvolume of unit cell

=

3

3

463

24 2

r

r

? ? 3 2

?? = 0.74 = 74%

(iv) Coordination No. : C.N. = 12

(v) Density = Mass

Volume = A

ZMN (volume) =

A

6MN (volume)

(B) ABC-ABC ARRANGEMENT :If the third layer spheres are placed in those voids of

second layer under which there are voids of the first layerof spheres (voids of type ‘b’), then the first and the thirdlayer of spheres will not be identical.Such an arrangementwill lead to an ABC-ABC type of arrangement.It is alsoknown as the cubical close packing (ccp) or also as theFace Centred Cubic structure (FCC).

A

B

A

A

ABABAB... or hexagonal close packing (hcp) of spheres

6 fold axis

BB

A

A

Coordination number ofhcp and ccp structure

3

7 8

9

2 1

X5

6

4

10

1211


(i) Relation between ‘a’ and ‘r’ :a 2r 2 4a r? (as the spheres touch along the face

diagonal)(ii) Effective no. of atoms :

Z = 18 × 8 +

12 × 6 = 4

(iii) Packing fraction :3

3

443 2 2

4 4 4

r

r

? ?? ?

? ? = 3 2

? =

0.74 = 74%(iv) Coordination number : 12

(v) Density = 3.A

Z MN a

? = 3

4

A

MN a

??

Note : In close packings, whenever two consecutivelayers are of different kinds(FCC, HCP) then packingefficiency will always be 74%

9. TYPES OF VOIDS FOUND IN CLOSEPACKINGS :

Eventhough the close packed structures have themaximum packing efficiency, there are indeed emptyspaces in between, let us analyse the types of such voidsand the maximum radius of a particle that can be placedin such voids.1. Triangular void (2-Dimensional 3 co-ordinate void)The triangular voids are found in the planes of the closepacked structures, whenever three spheres are in contactin such a fashion.

2R

30°

R

2R

C

B

A

C

B

A

ABCABCA... or cubic close packing (ccp) of spheres

3 fold axis

A

C

A

B

In the ABC – ABC pattern, the spheres of 4th layer arevertically above the spheres of Ist layer then theseconsecutive layers are different from each other, fourthlayer will be idential to first layer so it will be called ABC –ABC repeat pattern.


cos 30° = R

R r? R = Radius of the sphere,

32

= R

R r? r = maximum radius of a sphere that

can be placed inside the void.

23 =

R rR?

r = 0.155 R

2. Tetrahedral void (3-Dimensional 4 co-ordinate) :

a

b

c d

The tetrahedral void is formed whenever a sphere isplaced on top of the triangular arrangement as in case ofthe triangular void.

/2

A B

O

R+r

RR

sin 2?

= R

R r?

RR r? = sin 54° 44 r = R

1 1sin

2

? ?? ?

?? ??? ?? ?

r = 0.225 R

Location of tetrahedral voids in FCC unit cell :

The FCC unit cell has eight tetrahedral voids per unitcell. Just below every corner of the unit cell, there is one.As there are eight corners, there are eight tetrahedral voids.

The spheres 1, 2, 3, 4 form a tetrahedral void.

1 2

34

3. Octahedral void (3-Dimensional 6 coordinatevoid) The octahedral void is formed whenever two spheresare placed, one on top and the other below a squarearrangement of spheres

23

4 1

5

6

x = 2R

2(R + r) = 2

x = 2 2 R

r = ( 2 1)? R = 0.414 RR

r

Rr

x

Location of octahedral voids in a FCC unit cell :In a FCC unit cell, there are four octahedral voids.

They are present at all the edge centres and at the bodycentre. The contribution of the edge centre void per unit

cell is 14

.

Hence, total number of octahedral voids

= 1124

? ??? ?? ? + (1) = 4

edge centres body centre4. Cubical void (3-Dimensional 8-coordinate void)

The cubical void is generally not found in closedpacked structures, but is generated as a result of distortionsarising from the occupancy of voids by larger particles.

Along body diagonal

3 a = 2 (R + r)

2 3 R = 2 (R + r)

r = ( 3 1) R?

r = 0.732 R

rr

R

R

EXAMPLE 7 :Sodium has a bcc structure with nearest neighbourdistance 365.9 pm. Calculate its density (Atomic mass ofsodium = 23)


SOLUTION:For the bcc structure, nearest neighbour distance (d) is

related to the edge (a) as d = 3

2a

or a = 23 d =

21.732 365.9 = 422.5 pm

For bcc structure, Z = 2For sodium, M = 23

D = 30

Z Ma N

??

= 1

10 3 23 12 23gmol

(422.5 10 cm) (6.02 10 mol )

?

? ??

? ? ?

= 1.51 g /cm3

EXAMPLE 8 :X-ray diffraction studies shows that copper crystallizesin an fcc unit cell with cell edge of 3.608 10–8 cm. In aseparate experiment, copper is determined to have a densityof 8.92 g/cm3. Calculate the atomic mass of copper.

SOLUTION:

D = 30

Z Ma N

?? ; M =

30D a N

Z? ?

For fcc lattice, Z = 4 hence,3 8 3 23 1(8.92 g cm )(3.608 10 cm) (6.022 10 atoms mol )

4 atoms

? ? ?? ?

= 63.1 g mol–1

Atomic mass of copper = 63.1

EXAMPLE 9 :Density of Li atom is 0.53 g/cm3. The edge length of Li is3.5 Å Find out the number of Li atoms in a unit cell(N = 6.023 1023 , M = 6.94)

SOLUTION:The aim is to find Z in the formula

D = 30

Z Ma N

??

Z = 3

0D a NM

? ?

= 3 8 3 3 1

10.53 g cm (3.5 10 cm) (6.023 10 mol )

6.94g mol

? ? ?

?? ? ? ?

= 1.97 2

EXAMPLE 10:Transition metals, when they form interstitial compounds,the non-metals (H, B, C, N) are accomodated in:

(1) Voids or holes in cubic-packed structure(2) Tetrahedral voids(3) Octahedral voids(4) All of these Ans. (4)

EXAMPLE 11 :In a close pack array of N spheres, the number oftetrahedral holes are -

(1) 4N (2) N/2 (3) 2N (4) N Ans. (3)

10. IONIC SOLIDSIonic solids are characterised by the presence of atleasttwo types of particles, viz: the cation and the anion.

The Cations are generally found to be of smaller size,and the anions of larger sizes. The anions thus form thelattice by occupying the lattice positions and the cationsare found inside the voids in such structures.

The types of void occupied by the cation woulddepend upon the the ratio of its radius to that of the anion,popularly termed as the radius ratio. Hence, radius ratio =r+ / r–

C.No. Limiting radius Type of Voidratio Occupied

3 0.155 – 0.225 Triangular 4 0.225 – 0.414 Tetrahedral 6 0.414 – 0.732 Octahedral 8 0.732 – 0.999 Cubical

Examples of ionic crystals :(a) Rock Salt (NaCl) Coordination number (6 : 6)

NaCl crystallizes in the face centred cubic structure. Thechloride ions are present at all the lattice position and thesodium ions occupy all the octahedral voids.

Rock salt (NaCI) structure.Every sodium is in contact with six chloride ions,

and every chloride is in contact with six sodium ions


(b) CsCl C.No. (8 : 8)

Caesium chloride (CsCI) structure.The cesium ion is at the body centre and the chloride

ions are at the corners.(c) Zinc Blend (ZnS) C.No. (4 : 4)

(d) Fluorite structure (CaF2) C.No. (8 : 4)

Note: Some examples of metals with their lattice types and coordination number are given in the following table.

Li Be Body-centre Cubic [bcc]Face central cubic Hexagonal Closed packed [hcp]

Na Mg Al Simple Cubic

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

Cs Ba La Hf Ta W Re Os Ir Pt Au


Type

of Io

nic

Geo

met

ryCo

-ordin

ation

No.

of fo

rmul

a's

Exam

ples

Cry

stal

Num

ber

per U

.C.C

.

1.N

aCl (

1 : 1

)6

: 64N

a+ + 4

Cl–

Halid

es of

(Li, N

a, K,

Rb)

(R

ock

Salt

Type

)4N

aCl

Oxi

des a

nd su

lphi

des o

f(4

)II–

A (S

ome e

xcep

tion)

AgF

, AgC

l, A

gBr,

NH

4X

2.Cs

Cl Ty

pe (1

: 1)

8 : 8

1Cs+ +

1Cl

–H

alid

es o

f 'C

s'1C

sCl

TlCl

, TlB

r, Ca

S(1

)

3.Zn

S Typ

e (1 :

1)4

: 44Z

n+2 +

4S–2

BeS,

(Zinc

Blen

de Ty

pe)

4ZnS

BeO

, CaO

, AgI

,(S

phal

erite

)(4

)Cu

Cl, C

uBr,

CuI

4.Ca

F 2 Typ

e (1 :

2)4C

a+2 +

8F–

BaCl

2, Ba

F 2(F

luor

ite T

ype)

4CaF

2Sr

Cl2,

SrF 2

(4)

CaC

l 2, Ca

F 2

5.Na

2O Ty

pe (2

: 1)

8Na+ +

4O

–2Li

2O, L

i 2S(A

ntifl

ourit

e)4N

a 2ON

a 2O, N

a 2S

(4)

K2O

, K2S

6.Zn

S Ty

pe (1

: 1)

4 : 4

6Zn+2

+ 6

S–2Sa

me

as(W

urtz

ite)

6ZnS

spha

lerit

ean

other

geom

etry

(6)

of Z

nS

TY

PE

S O

F I

ON

IC C

RY

STA

L

Na+

Cl–

Cl–

Cs+ Zn

+2

S–2 F–

Ca+2 N

a+

O–2


11. CRYSTAL DEFECT (POINT DEFECTS) :Imperfection in solids can be because of :

(a) Conditions under which crystals have beendeveloped.

(b) Impurities(c) Temperature (because of thermal conductivity

some atoms/ions can get displacedThese imperfections can be(a) Point defects – defects will be only at certain lattice

positions.(b) Line defects – If atoms/ions are misplaced/

missing/replaced by some other ions along a line(c) Plane (skew) defects – If atoms/ions are

misplaced/missing/replaced by some other ionsalong a line in a plane.

Types of point defects

(a) stoichiometric defects,(b) impurity defects and(c) non-stoichiometric defect

(a) Stoichiometric defect

These are the point defects that do not distrub thestoichiometry of the solid. They are also called intrinsicot thermodynamic defects. basically these are two types.Vacancy defects and interstitial defects.

(i) Schottky defect (Vacancy defect) : It is basically avacancy defect in ionic solids. In order to maintainelectrical nuetrality. The number of missing cations andanions are equal. Schottky defect decreases the densityof the substance, Number of such defects in ionic solidsis quite significant. For example, in NaCl there areapproximately 106 schottky pairs per cm3 at roomtemperature. In 1 cm3 there are about 1022 ions. Thisdefect can also develop when a substance is heated.

Schottky defect is shown by ionic substance in whichthe cation and anion are of almost similar sizes. Forexample,NaCl, KCl, CsCl and AgBr.

(ii) Frenkel defect (Interstitial defect) :

This defect is shown by ionic solids. The smaller ion(usually cation) is dislocated from its normal site to aninterstitial site. It creates a vacancy defect at its originalsite and an interstitial defect at its new location.

Frenkel defect is also called dislocation defect. It doesnot change the density of the solid. Frenkel defect is shownby ionic substance in which there is a large difference inthe size of ions, for example, ZnS, AgCl,AgBr and AgIdue to small size of Zn2+ and Ag+ ions.

Eg. ZnS, AgCl, AgBr, AgI etc.

It may be noted that AgBr shows both, Frenkel aswell as schottky defects.

(b) Impurity defects

If molten NaCl containing a little amount of SrCl2 iscrystallised, some of the sites of Na+ ions are occupiedby Sr2+. Each Sr2+ replaces two Na+ ions. It occupies thesite of one ion and the other site remains vacant. Thecationic vacancies thus produced are equal in number isthe solid solution. Other example is presence of CdCl2 inAgCl.

(c) Non-stoichiometric defectA large number of non-stoichiometric inorganic solids

are known which contain the constituent elements in non-stoichiometric ratio due to defects in their crystalstructures. These defects are of two types :

(i) metal excess defect and(ii) metal deficiency defect.


(i) metal excess defect

M+

X—

X—

X—

M+

M+ X—

M+

M+ X—

e— M+

M+ X—

X— M+

F - Center

Metal excess defectsdue to anion vacancies

(a) metal excess defect due to anionic vacancies :Alkali halides like NaCl and KCl show this type of defect.When crystals of NaCl are heated in an atmosphere ofsodium vapours, the sodium atoms are deposited on thesurface of the crystal. The Cl– ions diffuse to the surfaceof the crystal and combine with Na atoms to give NaCl.This happens by loss of electron by sodium atoms to formNa+

ions. The released electrons diffuse into the crystaland occupy anionic site. As a result the crystal and nowhas an excess of sodium. The anionic sites occupied byunpaired electrons are called F-centres (from german wordfarbenzenter for colour centre). They impart yellow colourto the crystals of NaCl. The colour results by excitationof these electrons when they absorb energy from the visiblelight falling on the crystals.Eg. : * The excess sodium in NaCl makes the crystalappears yellow.

* Excess potassium in KCl makes it violet.* Excess lithium in LiCl makes it pink.Greater the number of F-centres greater is the intensity

of colour. This type of defects are found in crystal whichare likely to possess schottky Defects.

(b) metal deficiency defect due to the presence ofextra cations at interstitial sites : Zinc oxideis white incolour at room temperature. On heating it loses oxygenand turns yellow.

ZnO heating? ? ? ? Zn2+ + 12 O2 + 2e–

M+

X—

X—

X—

M+

M+

M+ X—

M+M+

X—

X—

X—

X—M+

M+

M+

e—

Metal excess defectsdue to interstitial cation

Now there is excess of zinc in the crystal and itsformula becomes Zn1+xO. The excess Zn2+ ions move tointerstitial sites and the electrons to neghbouring interstitialsites.

(ii) metal deficiency defect : There are many solidswhich are difficult to prepare in the stoichiometriccomposition and contain less amount of the metal ascompared to the stoichiometric proportion. A typicalexample of this type is FeO which is mostly found with acomposition of Fe0.95O. It may actually range from Fe0.93Oto Fe0.96O. In crystals of FeO some Fe2+ cations aremissing and the loss of positive charge is made up by thepresence of required number of Fe3+ ions.

EXAMPLE 13 :CsCl has cubic structure. Its density is 3.99 g cm–3. Whatis the distance between Cs+ and Cl– ions ? (At mass of Cs= 133)

SOLUTION:CsCl has BCC structure. It has one formula unit in theunit cell So Z = 1

D = 30

Z Ma N

?? or a3 =

0

Z MD N

??

= 1

3 23 11 (133 35.5)g mol

3.99 g cm 6.02 10 mlo

?

? ?? ?

? ?

= 70.15 10–24 cm3

a = (70.15)1/3 10–8 cm = (70.15)1/3 102 pm (1 pm= 10–10 cm) = 4.124 102 pm = 412.4 pm

Interionic distance = 32

a =

1.7322 412.4 = 357 pm

EXAMPLE 14:In a crystal both ions are missing from normal sites inequal number. This is an example of -

(1) F-centres (2) Interstitial defect(3) Frenkel defect (4) Schotty defect Ans (4)

SOLUTION:Schottky defects are arised when one positive ion andone negative ion are missing from their respective positionsleaving behind a pair of holes. These are more common inionic compounds with high coordination number andhaving almost similar size of cations and anions.

EXAMPLE 15:Frenkel defect is noticed in -

(1) AgBr (2) ZnS (3) AgI (4) All Ans (4)


SOLUTION:Frenkel defect is arised when the cations are missing fromtheir lattice sites and occupy interstitial sites. As a resultof Frenkel defect, density remains unchanged but dielectricconstant increases.

EXAMPLE 16 :A solid AB has the NaCl structure, If radius of cation A+ is120 pm, calculate the maximum possible value of the radiusof the anion B–.

SOLUTION:We know for the NaCl structure, for maximum of radiusof B–, the ratio r+ / r– should be minimum for octahedralvoid i.e. 0.414.

Radius of cation/radius of anion = 0.414

–

A

B

rr

?

= 0.414

–Br = 0.414

Ar ? =

1200.414 = 290 pm.

EXAMPLE 17 :CsCI crystallises in a cubic that has a CI– at each cornerand Cs+ at the centre of the unit cell. If (rCs+) = 1.69 Åand rCI – = 1.81 Å, what is value of edge length a of thecube?

SOLUTION:

We assume that the closest Cs+ to CI– distance is the sumof the ionic radii of Cs and Cl.

= 1.69 + 1.81 = 3.50 Å

This distance is one-half of the cubic diagonal = 3

2a

3

2a

= 3.50 Å a = 4.04 Å

EXAMPLE 18 :Calculate the edge length of the unit cell of sodium chloridegiven density of NaCI is 2.17 × 103 kg m–3 and molecularweight 58.5 × 10–3 kg mol–1.

SOLUTION:NaCI is face-centred cubic lattice so that number of NaCImolecules in a unit cell (z) = 4.

We know density d = 30

zMa N

where a = length of the unit cell

Volume = a3 = 0

MzdN =

3

3 234 58.8 10

2.17 10 6.02 10

?? ?? ? ?

= 1.79 × 10–28 m3

a = 5.64 × 10–10 ma = 5.64 Å = 564 pm

EXAMPLE 19:In a CPS (close packed structure) of mixed oxides, it isfound that lattice has O2– (oxide ions), and one-half ofoctahedral voids are occupied by trivalent cations (A3+)and one-eighth of tetrahedral voids are occupied by divalentcations (B2+). Derive formula of the mixed oxide.

SOLUTION:

Number of octahedral voids per ion in lattice = 1

Hence, number of trivalent cations (A3+) = 1 × 1 12 2

?

number of tetrahedral voids per ion in lattice = 2

Hence, number of divalent cations (B2+) = 2 × 1 18 4

?

Thus, formula is A1/2 B1/4O or A2BO4.

EXAMPLE 20 :Calcium crystallises in a face-centred cubic unit cell witha = 0.556 nm. Calculate the density if

(i) it contained 0.1% Frenkel defects(ii) it contained 0.1% Schottky defects

SOLUTION:

Thus, density can be determined using d = 30

zMa N

(i) a = 0.556 nm = 0.556 × 10–9 m = 0.556 × 10–7

cm d (with Frenkel defects)

= 7 3 234 40

(0.556 10 ) 6.02 10??

? ? ? = 1.5463 g / cm3

(ii) d(with Schottky defect)

= 7 3 233.996 40

(0.556 10 ) 6.02 10??

? ? ? = 1.5448 g/cm3


12. PROPERTIES OF SOLIDS :(A) Electric Properties :

On the basis of electrical conductivity the solids canbe broadly classified into the three types:

(a) Metals (conductors) (b) Insulators

(c) Semi-conductors.

Following are salient features of electrical conductancein solids.

(i) The electrical conductivity of metallic conductorsis due to the motion of electrons or positive holes(electronic conductivity) or through the motionof ions (ionic conductivity)

(ii) The conductance through electrons is called n-type conduction and through positive holes iscalled p-type conduction.

(iii) The conductance in insulators and semiconductorsis mainly due to the presence of interstitialelectrons and positive holes in the solids due toimperfections.

(iv) The conductivity of semiconductors and insulatorsincreases with increase in temperature while thatof metals decreases.

(v) Electrical conductivity of metal is in the order of106–108 ohm–1 cm–1 while that of inculator is ofthe order of 10–12 ohm–1 cm–1. Semiconductorshave intermediate value in the range 10–9 ohm–1

cm–1.

B. MAGNETIC PROPERTIES :

C. DIELECTRIC PROPERTIES :A dielectric is a substance in which an electric field gives rise to no net flow of electric charge. This is due to the

reason that electrons in a dielectric are tightly held by individual atoms. However, under the effect of applied fielddisplacement of charges takes place, resulting in the creation of dipoles.

Properties Information Magnetic Example Application Alignment

1. Diamagnetic Repelled weakly in magnetic field. Benzene NaCl, InsulatorsSuch solids have only paired TiO2, V2O5, Zn,electrons. Cd, Hg, etc.

2. Paramagnetic Have unpaired electrons; weakly O2, VO, CuO, TiO, Electronicattracted in magnetic field. They FeSO4, etc. devicescannot be permanentlymagnetised.

3. Ferromagnetic Also, have unpaired electrons. Fe, Ni, Co, CrO2 CrO2 is usedStrongly attracted in magnetic in audio, videofield. Such solids can be tapes.permanently magnetised. Onheating to a temperature calledCurie Point, these solids changeto paramagnetic solid.

4. Antiferrro- In these solids unpaired electrons Cr2O3, CoO, Used in themagnetic align themselves in such a way Co3O4, Fe2O3 instruments of

that resultant magnetic moment MnO, MnO2 magneticis zero. susceptibility

measurement5. Ferrimagnetic Unpaired electrons align themselves Fe3O4, ferrites –

in such way that there is a net ORmagnetic moment.

and so on


Properties Information Dipolar Example Application Property

1. Piezoelectri- When a crystal of dielectrics is Development of Quartz andcity subjected to mechanical stress, charge in some Rochelle salt

then small magnitude current part of crystalis produced. It is called directPiezoelectric effect.

2. Anti In some solids, electric field Crystal suffers –piezoelectricity developes mechanical effect. elastic deforma-

tion in anelectric field

3. Ferroelectricity Piezoelectric crystals having BaTiO3, KH2PO4 Electromagneticpermanent dipoles are said to Rochelle salt appliances.possess ferroelectricity

4. Anti Piezoelectric crystals with zero PbZrO3ferroelectricity dipole are said to posses anti Lead zirconate

ferroelectrictity.5. Pyroelectricity Some polar crystals produce electric – Crystals of Used in fire alarms,

impulse on heating. tartaric acid. thermostat.

Used in mechanicalelectric transducer,e.g., in record playerTransmission ofdirect signals,sounding of seadepths.

?????????????


EXERCISE–0 (RECALL YOUR UNDERSTANDING)1. Define the term ‘space lattice.’2. What is the percentage of free space in bodycentred

cubic crystal ?3. What is the number of tetrahedral voids in a unit

cell of a cubic close-packed structure ?4. What is the percentage efficiency of packing in case

of a simple cubic lattice?5. Octahedral voids are larger than tetrahedral voids

or not. Give reason.6. Solid elements of group 13 or 15 impurities with

group 14 elements are found to exhibit unusualelectrical properties. Why ?

7. What is doping ? Why is it done ?8. What type of semiconductor is obtained when Si is

doped with boron?9. Given an example of antiferromagnetic solid.10. Classify the following solids on the basis of bonding

considerations :CO2, MgO, Al, H2, Si, Gd, Pb, AgCl.

11. Define the following terms in relation of crystallinesolids :(i) Unit cell (ii) Coordination numberGive one example in each case.

12. Explain how you can determine the atomic mass ofan unknown metal if you know its mass densityand the dimensions of unit cell of its crystal.

13. Calculate the packing efficiency of a metal crystalfor a simple cubic lattice.

14. The length of the unit cell edge of a body centredmetal crystal is 352 pm. Calculate the radius of anatom of a metal.

15. An element (atomic mass = 60) having facecentredcubic structure has a cell edge of 400 pm. What isits density ? [NA = 6.023 × 1023 ]

16. Explain the following terms with suitable examples(i) Schottky defect (ii) Interstitial defect

17. In Corundum, oxide ions are arranged in HCParrangement and the Al3+ ions occupy two third ofthe octahedral voids. What is the formula ofcorundum ?

18. An element crystallized in the simple cubic structure.Its density is 8 g cm–3 and its 200 g contains 24 ×1023 atoms. Calculate the edge length.

19. An alloy of Cu, Au and Ag is found to have Cuforming simple cubic strcture, Ag atom occupy edge

centres and Au is at body centre. What is the formulaof alloy ? Explain.

20. (a) Diamond and solid rhombic sulphur both arecovalent solids but the latter has very lowmelting point then former. Explain why ?

(b) Out of SiO2(s), Si(s), NaCl(s) and Br2(l) whichis best electrical conductor and why ?

(c) Sodium metal has bcc structure with edgelength 4.29 Å. What is the radius of an atom ?What is length of body diagonal of unit cell ?

21. Calcium metal crystallizes in face cented cubic latticewith length 556 pm. Calculate the density of metalif it contains, (i) 0.5% Frenkel defects (ii) 0.2%Schottky defects.

22. Account for the following :(i) Fe3O4 is ferrimagnetic at room temperature but

becomes paramagnetic at 850 K.(ii) Zinc oxide on heating becomes yellow.(iii) Frenkel defect does not change the density of

AgCl crystals.23. How do electrical resistivity of the following class

of materials vary with temperature ?Semiconductor,metallic conductor, insulator.

24. Explain the following terms with one suitableexample of each :(i) Ferromagnetism (ii) Paramagnetism

25. (i) What type of close packing is shown in Fig.?(ii) What is coordination number of central

sphere?(iii) What is free space in this packing ?(iv) How many atoms are present in single unit

cell having this type of packing ?

26. In a mixed oxide of a compound, 18 th of tetrahedral

voids are occupied by cation ‘A’. half of octahedralvoids are occupied by cation ‘B’ whereas oxide ionfroms cubic close packed structure.What will be the formula of oxide ? Explain.


27. How are the following properties of crystals affectedby Schottky and Frenkel defects :(i) Density (ii) Electrical conductivity

28. With the help of suitable diagrams, on the basis ofband theory, explain the difference between(i) a conductor and an insulator.(ii) a conductor and a semiconductor.

Introduction of Solids and Space-Lattice

1. In a face centered cubic unit cell, the contributionfrom the atom at the corner and face centre of thecube are respectively :

(a) 14

, 12

(b) 18 ,

14

(c) 14

, 12

(d) 18 ,

12

2. Which of the following are not properties of solid ?(a) They have definite mass, volume and shape.(b) Intermolecular force are weak.(c) Intermolecular distances are short.(d) They are incompresible and rigid.

3. Amorphous solids :(a) Possess sharp melting points(b) Undergo clean cleavage when cut with knife(c) Do not undergo clean cleavage when cut with

knife(d) Possess orderly arrangement over long

distances4. When molten form of crystalline solid is rapidly

cooled, it changes into -(a) crystalline solid (b) amorphous solid(c) insulator (d) superconductor

5. Which of the following is not a crystalline solid ?(a) Common salt (b) Sugar(c) Iron (d) Rubber

6. Solid CO2 is an example of(a) Ionic crystal (b) Covalent crystal(c) Metallic crystal (d) Molecular crystal.

7. An example of a metallic crystalline solid is(a) Si (b) C(c) P (d) W

8. Iodine crystal are :(a) Metallic (b) Ionic(c) Molecular (d) Covalent


9. NaCl crystal is made up of(a) NaCl molecules (b) Na+ and Cl– ions(c) Na and Cl atoms (d) Polymers of NaCl

10. Ionic solids are characterised by(a) Good conductivity in solid state(b) High vapour pressure(c) Low melting point(d) Solubility in polar solvents.

11. Which one of the following will have a low heat offusion ?(a) a covalent solid (b) an ionic solid(c) a metallic solid (d) a molecular solid

12. Diamond belongs to the crystal system :(a) cubic (b) triclinic(c) tetragonal (d) hexagonal

13. Wax is an example of -(a) lonic crystal (b) Covalent crystal(c) Molecular crystal (d) Metallic crystal

14. Which of the following is not the property ofsolids–(a) Solids are always crystalline in nature.(b) Solids have good density and less compressibility(c) Solids diffuse very slowly.(d) The volume of solids is fixed

15. A solid X melts slightly above 273K and is a poorconductor of heat and electricity. To which of thefollowing categories does it belong -(a) Ionic solid (b) Covalent solid(c) Metallic (d) Molecular

Lattice parameters and density

16. a b c, = = 90° 90° represents(a) tetragonal system (b) orthorhombic system(c) monoclinic system (d) triclinic system

17. Bravais lattices are of(a) 10 types (b) 8 types(c) 7 types (d) 14 types


18. If the radius of metal atom is 1.00 Å and its crystalstructure is simple cubic, then what is the volumeof one unit cell ?(a) 8 × 10–28 m3 (b) 4 × 10–30 cm3

(c) 8 × 10–30 m3 (d) 2 × 10–24 cm3

19. The intermetallic compound LiAg crystallizes incubic lattice in which both lithium and silver havecoordination number of eight. The crystal class is :(a) Simple cubic (b) Body centred cubic(c) Face centred (d) None of these

20. The crystal system for which a b c and = = = 900 is said to be :(a) triclinic (b) tetragonal(c) cubic (d) orthorhombic

21. Which of the following are the correct axial distanceand axial angles for rhombohedral system?(a) a = b = c, = = 900

(b) a = b c, = = = 900

(c) ab c, = = = 900

(d) ab c, 900

22. In a simple cubic cell, each point on a corner isshared by(a) 2 unit cells (b) 1 unit cell(c) 8 unit cells (d) 4 unit cells

23. In a face centred cubic cell, an atom at the cornercontributes to the unit cell(a) 1 part (b) 1/2 part(c) 1/4 part (d) 1/8 part

24. In face centred cubic cell, an atom at the facecentres is shared by(a) 4 units cells (b) 2 unit cells(c) One unit cell (d) 6 unit cells

25. In a face centred cubic cell, an atom at the facecontributes to the unit cell(a) 1 part (b) 1/2 part(c) 1/4 part (d) 1/8 part

26. In a body centred cubic cell, an atom at the bodycentre is shared by(a) 1 unit cell (b) 2 unit cell(c) 3 unit cells (d) 4 unit cells

27. Which of the following type of cubic lattice hasmaximum number of atoms per unit cell ?(a) Simple cubic (b) Body centred cubic(c) Face centred cubic (d) All have same

28. Which one of the following is a primitive unit cell ?(a) Simple cubic(b) Body-centred cubic(c) Face-centred cubic(d) Both body-centred and face-centred cubic

29. In a body centred cubic unit cell, a metal atom atthe centre of the cell is surrounded by how mayother metal atoms :(a) 8 (b) 6(c) 12 (d) 4

30. A compound is formed by elements A and B. Thiscrystallises in the cubic structure when atoms A areat the corners of the cube and atoms B are at thecentre of the body. The simplest formula of thecompound is(a) AB (b) AB2

(c) A2B (d) AB4

Closed packing (CROWDING)31. How many nearest neighbours are present in an atom

forming bcc lattice?(a) 4 (b) 6(c) 8 (d) 12

32. A metal crystallizes in a body centered cubic lattice(bcc) with the edge of the unit cell 5.2Å. Thedistance between the two nearest neighbour is(a) 10.4 Å (b) 4.5 Å(c) 5.2Å (d) 9.0Å

33. Sodium metal crystallises in bcc lattice with the celledge a = 42.29 Å. What is the radius of sodiumatom?(a) 1.86 Å (b) 1.90 Å(c) 18.3 Å (d) 1.12 Å

34. An element has bcc structure having unit cells 12.08× 1023. The number of atoms in these cells is(a) 12.08 × 1023 (b) 24.16 × 1023

(c) 48.38 × 1023 (d) 12.08 × 1022

35. The arrangment of the first two layers. one abovethe other, in hcp and ccp arrangements is :(a) Exactly same in both cases(b) partly same and partly different(c) Diffrent from each other(d) Nothing definite

36. Given below are schemes of placing 2-D closepacked sheets it generate 3-D closest packing.Which of these will not generate closest packing ?


(a) AB AB AB (b) ABCBCAB(c) ABCABBC (d) ACBCACB

37. The octahedral voids in a face-centred cubic (fccor ccp) structure are located at(a) at edge centres and 8 along diagonals(b) 12 at edge centres and one at a body centre(c) 8 along body diagonal and 6 at edge centres(d) All the edge centres only

38. The edge length of face centred cubic unit cell is508 pm. The nearest distance between two atomsis :(a) 360 pm (b) 288 pm(c) 618 pm (d) 398 pm

39. The maximum percentage of available volume thatcan be filled in a face centred cubic system by atomsis-(a) 74% (b) 68%(c) 34% (d) 26%

40. Fraction of empty space in ABAB type arrangementin 3D :(a) 0.74 (b) 0.26(c) 0.68 (d) 0.32

41. How many number of atoms are there in a cubebased unit cell having one atom on each corner andtwo atoms on each body diagonal of cube ?(a) 8 (b) 6(c) 4 (d) 9

42. The coordination number of hexagonal closestpacked(hcp) structure is(a) 12 (b) 10(c) 8 (d) 6

43. The more efficient mode of packing of identicalatoms in one layer is(a) Square close packing pattern(b) Hexagonal close packing pattern(c) Both (a) and (b)(d) None of the two

44. The ABAB....packing and ABC ABC...packing arerespectively called as(a) hexagonal close packing(hcp) and cubic close

packing (ccp)(b) ccp and hcp(c) body centred cubic (bcc) packing and

hexagonal close packing (hcp)(d) hcp and bcc

45. The space occupied in bcc arrangement is(a) 74% (b) 70%(c) 68% (d) 60.4%

46. The vacant space in bcc unit cell is(a) 32% (b) 10%(c) 23% (d) 46%

47. The available space occupied by spheres of equalsize in three dimensions in both hcp and ccparrangement is(a) 74% (b) 70%(c) 60.4% (d) 52.4%

48. The empty space in the hcp and ccp is about(a) 26% (b) 30%(c) 35% (d) 40%

49. Which one of the following is not a close packing ?(a) hcp (b) ccp(c) bcc (d) fcc

50. Close packing is maximum in the crystal lattice of(a) Simple cubic (b) Face centred(c) Body centred (d) None

51. Which of the following has hcp structure ?(a) Al (b) Mg(c) Cu (d) Ni

52. All noble gases crystallise in the ccp structure except(a) Helium (b) Neon(c) Argon (d) Krypton

53. If the coordination number of an element in itscrystal lattice is 8, then packing is :(a) fcc (b) hcp(c) bcc (d) None of the above

Voids and their geometries54. The void present in simple cubic crystal is :

(a) cubic (b) tetrahedral(c) octahedral (d) triangular

55. The empty space between the shaded balls andhollow balls as shown in the diagram is called

(a) hexagonal void(b) octahedral void(c) tetrahedral void(d) double triangular void


56. The coordination number of octahedral void is :(a) 8 (b) 6(c) 4 (d) 12

57. In a spinel structure O2– ions form an fcc array, ‘A’cation occupy one-eight of tetrahedral holes and ‘B’cation occupy one-half of octahedral holes. What isthe formula ?(a) A2BO4 (b) AB2O4

(c) A2B2O4 (d) A2B2O4

58. In a face centred lattice of X atoms, Y atoms occupy

12

of tetrahedral holes and 14

of octahedral holes.

What is the possible formula of compound ?(a) X4Y6 (b) X4Y5

(c) X4Y8 (d) X4Y9

59. In a ccp structure of X atoms. Y atoms occupy halfof octahedral holes. If one Y atom and X atom fromeach unit cell is replaced by Z, then the formula ofcompound will be -(a) X4Y2Z2 (b) X3YZ2

(c) X3Y2Z2 (d) X3Y3Z60. A tetrahedral void in a crystal implies that

(a) shape of the void is tetrahedral(b) molecules forming the void are tatrahedral in

shape(c) the void is surrounded tetrahedrally by four

spheres(d) the void is surrounded by six spheres

61. In a closest packed lattice, the number of octahedralsites as compared to tetrahedral ones will be(a) Equal (b) Half(c) Double (d) None of these

62. The coordination number of a cation occupying anoctahedral hole is(a) 4 (b) 6(c) 8 (d) 12

63. The size of an octahedral void formed in a closestpacked lattice as compared to tetrahedral void is(a) Equal (b) Smaller(c) Larger (d) Not definite

64. The coordination number of a cation occupying atetrahedral hole is(a) 4 (b) 6(c) 8 (d) 12

65. Number of tetrahedral voids per atom in a crystallattice is :(a) 1 (b) 2(c) 4 (d) 8

66. A compound contains P and Q elements. Atoms Qare in ccp arrangement while P occupy all tetrahedralsites. Formula of compound is :(a) PQ (b) PQ2

(c) P2Q (d) P3Q67. A tetrahedral void in a crystal implies that

(a) Shape of the void is tetrahedral(b) Molecules forming the void are tetrahedral in

shape(c) the void is surrounded by four spheres.(d) the void is surrounded by six spheres

Ionic solids and Crystal defects

68. MgO exist in NaCl type lattice (rock salt structure).No. of nearest neighbour of Mg2+ ion are :(a) 12 O–2 ions (b) 8 O–2 ions(c) 6 O–2 ions (d) 4 O–2 ions

69. If the distance between Na+ and Cl– ions in NaClcrystal is 'a' pm, what is the length of the cell edge?(a) 2a pm (b) a/2 pm(c) 4a pm (d) a/4 pm

70. Which of the following expressions is correct inthe case of a sodium chloride unit cell (edge length= a)(a) rc + ra = a/2 (b) rc + ra = a(c) rc + ra = 2a (d) rc + ra = 21/2 a

71. The limiting radius ratio for tetrahedral shape is(a) 0 to 0.155 (b) 0.155 to 0.225(c) 0.225 to 0.414 (d) 0.414 to 0.732

72. For an octahedral arrangement the lowest radius ratiolimit is(a) 0.155 (b) 0.732(c) 0.414 (d) 0.225

73. If the radius ratio is in the range of 0.414 - 0.732then the co-ordination number will be :(a) 2 (b) 4(c) 6 (d) 8

74. In NaCl crystal r+/r– ratio is :(a) 0.4 (b) 0.98(c) 1.0 (d) 0.52


75. Which one of the following statements is incorrectabout rock salt type ?(a) It has fcc arrangement of Na+

(b) Na+ and Cl– ions have a co-ordination numberof 6 : 6

(c) A unit cell of NaCl consists of four NaCl units(d) All halides of alkali metals have rock-salt type

structure76. In sodium chloride, Cl– ions form ccp arrangement.

Which site a Na+ ions will occupy in thisstructure ?(a) Cubic (b) Tetragonal(c) Octahedral (d) Trigonal bipyramidal

77. The positions of Cl– ions in NaCl structure are(a) Corners of the cube(b) Centres of faces of the cube(c) Corners as well as centres of the faces of the cube(d) Edge centres of the cube

78. The number of NaCl units present in a unit cell ofNaCl are(a) 1 (b) 2(c) 4 (d) 8

79. The tetrahedral voids formed by ccp arrangementof Cl– ions in rock salt structure are(a) Occupied by Na+ ions(b) Occupied by Cl– ions(c) Occupied by either Na+ or Cl– ions(d) Vacant

80. The structure of MgO is similar to NaCl. The co-ordination number of Mg is(a) 2 (b) 6(c) 4 (d) 8

81. The co-ordination number of Cs+ and Cl– ions inCsCl structure is(a) 4 : 4 (b) 6 : 6(c) 8 : 8 (d) 4 : 8

82. A unit cell of CsCl consists of(a) one CsCl unit (b) two CsCl units(c) four CsCl units (d) eight CsCl units

83. The NaCl structure can be converted into CsClstructure(a) by application of high pressure(b) by heating to 760 K(c) both by heat and pressure(d) the conversion is not possible

84. TC has structure similar to CsCl. The co-ordination number of T+ is(a) 4 (b) 6(c) 10 (d) 8

85. The co-ordination number of Zn2+ and S2– ions inthe zinc blende (ZnS) type structure is(a) 4 : 4 (b) 6 : 6(c) 8 : 8 (d) 4 : 8

86. The co-ordination number of calcium fluoride(CaF2) type structure is(a) 1 : 2 (b) 4 : 4(c) 4 : 8 (d) 8 : 4

87. The number of molecules in an unit cell of fluoriteis(a) 2 (b) 4(c) 6 (d) 8

88. 4 : 4 Co-ordination is found in(a) ZnS (b) CuCl(c) Agl (d) All

89. Antifluorite structure is derived from fluoritestructure by(a) Heating fluorite crystal lattice.(b) Subjecting fluorite structure to high pressure.(c) Interchanging the positions of positive and

negative ions in the lattice.(d) None of these

90. A binary solid (A+B–) has a zinc blend structure withB– ions constituting the lattice and A+ ions occupying25% tetrahedral holes. The formula of solid is.(a) AB (b) A2B(c) AB2 (d) AB4

91. A solid AB has NaCl type structure. If the radius ofthe cation A+ is 100 pm, then the radius of the anionB– will be(a) 192.3 pm (b) 414 pm(c) 325 pm (d) 44.4 pm

92. The radius of Na+ is 95 pm and that of Cl– ion is181 pm. Hence the co-ordination number of Na+

will be(a) 4 (b) 6(c) 8 (d) unpredictable

93. Potassium fluoride has NaCl-type structure. Whatis the distance between K+ and F– ions if it's celledge 'a' cm ?(a) 2a cm (b) a/2 cm(c) 4a cm (d) a/4 cm


94. KF has NaCl structure. What is the distance betweenK+ and F– in KF if density is 2.48 g cm–3 ?(a) 268.8 pm (b) 537.5 pm(c) 155.3 × 10–24 cm (d) 5.375 cm

95. Potassium fluoride has NaCl– type structure. Whatis the distance between K+ and F– ions if cell edge is'a' cm ?(a) 2a cm (b) a/2 cm(c) 4a cm (d) a/4 cm

Crystal defects and Properties of solids96. In a solid lattice the cation has left a lattice site and

is located at an interstitial position, the lattice defectis(a) Interstitial defect (b) Vacancy defect(c) Frenkel defect (d) Schottky defect

97. At zero kelvin, most of the ionic crystals posses(a) Frenkel defect(b) Schottky defect(c) Metal excess deffect(d) No defect

98. As a result of Schottky defect(a) there is no effect on the density(b) density of the crystal increases(c) density of the crystal decreases(d) any of the above three can happen

99. Schottky as well as frenkel defects are observed inthe crystal of

(a) NaCl (b) AgBr(c) AgCl (d) MgCl2

100. Which one of the following is correct ?(a) Schottky defect lowers the density(b) Frenkel defect increases the dielectric constant

of the crystals(c) Stoichiometric defects make the crystals good

electrical conductors(d) All the three

101. Frenkel defect is generally observed in(a) AgBr (b) Agl(c) ZnS (d) All of these

102. Frenkel defect is found in crystals in which the radiusratio is :(a) low (b) 1.3(c) 1.5 (d) slightly less than unity

103. F-centres in an ionic crystal are(a) lattice sites containing electrons(b) interstitial sites containing electrons(c) lattice sites that are vacant(d) interstitial sites containing cations

104. The correct statement regarding F-centre is(a) Electrons are held in the voids of crystals(b) F-centre imparts colour to the crystal(c) Conductivity of the crystal increases due to

F-centre(d) All the three above


1. Tetragonal crystal system has the following unit celldimensions :(a) a = b = c, = = = 90°(b) a = b c, = = = 90°(c) a b c, = = = 90°(d) a = b c, = = 90° = 120°

2. Copper crystallises in a structure of face centerdcubic unit cell. The atomic radius of copper is 1.28Å. What is axial length on an edge of copper.(a) 2.16 Å (b) 3.62 Å(c) 3.94 Å (d) 4.15 Å

3. The number of atoms present in a unit cell of amonoatomic substance (element) of a simple cubiclattice, body-centred cubic and face centred cubicrespectively are

(a) 8, 9 and 14 (b) 1, 2 and 4(c) 4, 5 and 6 (d) 2, 3 and 5

4. Consider a Body Centered Cubic(bcc) arrangement,let de, dfd, dbd be the distances between successiveatoms located along the edge, the face-diagonal, thebody diagonal respectively in a unit cell.Their orderis given by:(a) de < dfd < dbd (b) dfd > dbd > de

(c) dfd > de > dbd (d) dbd > de > dfd,5. In a ccp structure, the (according to cubic 3D

arrangement) :(a) first and third layers are repeated(b) first and fourth layers are repeated(c) second and fourth layers are repeated(d) first, third and sixth layers are repeated.


6. Volume of HCP unit cell is :

(a) 24 2 r3 (b) 8 2 r3

(c) 16 2 r3 (d) 24 2 r3

7. In a multi layered close-packed structure(a) there are twice as many tetrahedral holes as

there are close-packed atoms(b) there are as many tetrahedral holes as there are

closed packed atoms(c) there are twice as many octahedral holes as

there are close-packed atoms(d) there are as many tetrahedral holes as there are

octahedral holes8. If cations C form a hexagonal close packing & anion

(a) occupy 23 rd of octahedral holes, then the general

formula of the compound is :(a) C3A4 (b) C6A5

(c) C3A2 (d) C2A3

9. How many rectangular faces are in an orthorhombicunit cell ? (Given : a b c & = = = 90º)(a) 4 (b) 2(c) 6 (d) No faces rectangular

10. In a lattice of X and Y atoms, if X atoms are presentat corners and Y atoms at the body centre & one Xatom is removed from a corner from each unit cell,then the formula of the compound will be :(a) X7Y (b) X8Y7

(c) X7Y8 (d) X7Y7

11. Graphite belongs to(a) cubic system(b) tetragonal system(c) rhombohedral system(d) hexagonal system

12. The characteristic features of solids are -(a) Definite shape(b) Definite size(c) Definite shape and size(d) definite shape, size and rigidity

13. Copper crystallises as fcc unit cell. If atomic radiusof copper is 1.28Å, then what is the edge length ofthe unit cell ?(a) 2.16Å (b) 3.63Å(c) 3.97Å (d) 4.15Å

14. The vacant space in B.C.C. unit cell is :(a) 32% (b) 10%(c) 23% (d) 46%

15. When electrons are trapped into the crystal in anionvacancy, the defect is known as :(a) Schottky defect (b) Frenkel defect(c) Stoichhiometric defect (d) F–centres

16. Which of the following describes the hexagonalclose packed arrangment of spheres ?(a) ABCABA (b) ABCABC(c) ABABA (d) ABBABB

17. In a face centerd lattice of X and Y, X atoms arepresent at the corners while Y atoms are at facecenters. Then the formula of the compound is :(a) XY3 (b) X2Y3

(c) X3Y (d) XY18. When NaCl crystal is heated in sodium vapors, then

it attains yellow colour. It is due to :(a) electrons trapped in cation vacancies.(b) F-centres, which is electron trapped in anion

vacancy created by Cl–.(c) F-centres, which is cation trapped in cation

vacancy created by Na+.(d) interstitial defect caused by external impurity.

19. The radius of Ag+ ion is 126 pm & that of – ion is216 pm. What is the possible Ag lattice ?(a) I– form simple cube and Ag+ occupy the cubic

void.(b) I– form fcc and Ag+ occupy the octahedral

voids.(c) I– form fcc and Ag+ are present on the edge

centres.(d) I– form hcp and Ag+ occupy half of tetrahedral

holes.20. 1 g of X has atoms arranged in cubic packing so as

to give best packing efficiency. The possiblearrangement is :(a) simple cubic(b) face centered cubic(c) body certered cubic(d) hexagonal close packing

21. If R is the radius of the sphere in the close packedarrangment and r is the radius of the tetrahedral void,then(a) R = 0.225 r (b) r = 0.225 R(c) r = 0.414 R (d) R = 0.414 r


22. In 3D close packed structures, for every 100 atoms,it contain :(a) 50 octahedral voids(b) 100 tetrohedral voids(c) 200 octohedral voids(d) 100 octahedral voids

23. For an ionic solid of general formula AB andcoordination number 8 for both A and B, the valueof radius ratio will be :(a) less than 0.225(b) in between 0.225 and 0.414(c) between 0.414 and 0.732(d) greater than 0.732

24. The ionic radii of Rb+ and I– are 1.46 and 2.16 Årespectively. The most probable type of structureexhibited by it is(a) CsCI type (b) NaCl type(c) ZnS type (d) CaF2 type

25. A binary solid (A+B–) has a rock salt structure if theedge length is 400 pm and radius of cation (A+) is75 pm, the radius of anion (B–) is(a) 100 pm (b) 125 pm(c) 250 pm (d) 325 pm

26. A solid X+Y– has a bcc structure. If the distance ofclosest approach between the two atoms is173 pm, the edge length of the cell is

(a) 200 pm (b)32

pm

(c) 142.2 pm (d) 2 pm

27. If the distance between Na+ and Cl– ions in NaClcrystal is 'a' pm, what is the length of the cell edge?(a) 2a pm (b) a/2 pm(c) 4a pm (d) a/4 pm

28. The lattice of CaF2 is called fluorite structure. SrCl2has fluorite structure. Which of the followingstatements is true for SrCl2 ?(a) Sr2+ are at the corners and face centres of the

cubic arrangment(b) Sr2+ are arranged in bcc lattice(c) a– are arranged in fcc lattice(d) Cl– occupy octahedral holes in the lattice.

29. In the crystal lattice of diamond, carbon atomsadopt:(a) fcc arrangement along with occupancy of 50%

tetrahedral holes

(b) fcc arrangement along with occupancy of 25%tetrahedral holes

(c) fcc arrangement along with occupancy of 25%octahedral hole

(d) bcc arrangement30. The most unsymmetrical system is :

(a) Cubic (b) Hexagonal(c) Triclinic (d) Orthorhombic

31. Schottky defect occurs mainly in electrovalentcompounds where :(a) positive ions and negative ions are of different

size(b) positive ions and negative ions are of same size(c) positive ions are small and negative ions are

big(d) posiive ions are big and negative ions are small.

32. Which one of the following statements is incorrectabout salt type NaCl ?(a) It has fcc arrangment of Na+

(b) Na+ and Cl– ions have a co-ordination numberof 6 : 6

(c) A unit cell of NaCl consists of four NaCl units(d) All halides of alkali metals have rock-salt type

structure.33. The ionic radii of Rb+ and I– are 1.46 and 2.16 Å.

The most probable type of structure exhibited bythis –(a) CsCl type (b) NaCl type(c) ZnS type (d) CaF2 type

34. A solid AB has NaCl type structure. If the radius ofthe cation A is 100 pm, then the radius of the anionB will be :(a) 241 pm (b) 414 pm(c) 225 pm (d) 44.4 pm

35. If the radius of K+ and F– are 133 pm and 136 pmrespectively, the distance between K+ and F– in KFis :(a) 269 pm (b) 134.5 pm(c) 136 pm (d) 3 pm

36. In zinc blende structure the coordination number ofZn2+ ion is(a) 2 (b) 4(c) 6 (d) 8

37. Cesium chloride on heating to 760 K changes into(a) CsCl(g)(b) NaCl structure(c) antifluorite structure(d) ZnS structure


38. Diamond is hard because :(a) All the four valence electrons are bonded to

carbon atoms by covalent bonds(b) It is a giant molecule(c) It is made up of carbon atoms(d) It cannot be burnt

39. An ionic compound is expected to have tetrahedralstructure if r+/r– lies in the range of :(a) 0.155 to 0.225 (b) 0.732 to 0.414(c) 0.414 to 0.732 (d) 0.225 to 0.414

40. Malleability and ductility of metals can be acounteddue to :(a) the capacity of layers of metal ions to slide over

the other(b) the interaction of electrons with metal ions in

the other(c) the presence of electrostatic forces(d) the crystalline structure of metal

41. In NaCl & AgCl, AgNa rr , so(a) NaCl has higher tendency to show Schottky

defect than AgCl.(b) NaCl has lower tendency to show Schottky

defect than AgCl.(c) NaCl has lower tendency to show Frenkel

defect than AgCl.(d) Both have equal tendency to show Frenkel

defect and Schottky defect.42. KCl crystallises in the same type of lattice as does

NaCl. Given that rNa+ / rCl

– = 0.55 and rK+/rCl

– =0.74. Calculate the ratio of the side of the unit cellof KCl to that of NaCl :(a) 1.123 (b) 0.891(c) 1.414 (d) 0.414

43. A solid compound contains X, Y and Z atoms in acubic lattice with X atom occupying the corners, Yatoms in the body centred positions and Z atoms atthe centres of faces of the unit cell. What is theempirical formula of the compound :(a) XY2Z3 (b) XYZ3

(c) X2Y2Z3 (d) X8YZ6

44. Which of the following statements are correct incontext of point defects in a crystal ?(a) AgCl has anion Frenkel defect and CaF2 has

Schottky defects(b) AgCl has cation Frenkel defects and CaF2 has

anion Frenkel defects

(c) AgCl as well as CaF2 have anion Frenkel defects(d) AgCl as well as CaF2 has Schottky defects

45. In the schottky defect(a) cations are missing from the lattice sites and

occupy the interstitial sites(b) Cations and anions are missing(c) anions are missing and electrons are present

their place(d) equal number of extra cations and electrons are

present in the interstitial sites.46. As a result of schottky defect,

(a) there is no effect on the density(b) density of the crystal increases(c) density of the crystal decreases(d) any of the above three can happen.

47. F-centres in an ionic crystal are(a) lattice sites containing electrons(b) interstitial sites containing electrons(c) lattice sites that are vacant(d) interstitial sites containing cations

48. Doping of silicon with P or Al increases theconductivity. The difference in the two cases is(a) P is non-metal whereas Al is a metal(b) P is a poor conductor while Al is a conductor(c) P gives rise to extra electrons while Al gives

rise to holes(d) P gives rise to holes while Al gives rise to extra

electrons.49. The cubic unit cell of Al(molar mass 27 g mol–1)

has an edge length of 405 pm. Its density is 2.7 gcm–3. The cubic unit cell is :(a) face centered (b) body centered(c) primitive (d) edge centered

50. An element (with atom. mass = 250 g) crystallises in asimple cubic.If the density of the unit cell is 7.2 g cm–

3, what is the radius of the element ?(a) 1.93 10–6 cm (b) 1.93 10–8 cm(c) 1.93 10–8 Å (d) 1.93 10–8 cm

51. In AgBr, there can occur(a) only schottky defect(b) only Frenkel defect(c) both (a) and (b)(d) None of these

52. Which of the following shows ferrimagnetism ?(a) TiO2 (b) CrO2

(c) MnO (d) Fe3O4


53. Metallic lustre is explained by :(a) diffusion of metal ions(b) oscillation of loose electrons(c) excitation of free protons(d) existence of bcc lattice

54. The hardest substance amongst the following is :(a) Be2C (b) Graphite(c) Titanium (d) SiC

55. Which has no axis of rotation of symmetry ?(a) Hexagonal (b) Orthorhombic(c) Cubic (d) Triclinic

56. In a ccp structure of X atoms, Y atoms occupy allthe octahedral holes. If 2 X atom are removed fromcorners and replaced by Z, then the formula of thecompound will be :(a) X15Y16Z (b) X7Y8Z(c) X7.5 Y8Z (d) X8Y8Z3

57. The number of atoms in 100 g of a fcc crystal withdensity = 10.0 g/cm3 and cell edge equal to 200 pmis equal to :(a) 5 1024 (b) 5 1025

(c) 6 1023 (d) 2 1025

58. The maximum proportion of available volume thatcan be filled by hard spheres in diamond is(a) 0.52 (b) 0.34(c) 0.32 (d) 0.68

59. In the sphalerite (ZnS) structure, S2– ions form aface-centred cubic lattice. Then Zn2+ ions arepresent on the body diagonals at a distance fromeach corner

(a) 13 rd of the distance

(b) 14

th of the distance

(c) 16 th of the distance

(d) 18 th of the distance

60. Consider a simple cubic arrangement let d1,d2,d3,be the distances between successive atoms alongthe edge, the face diagonal and body diagonal in aunit cell. The correct order is :(a) d1 > d2 > d3 (b) d3 > d1 > d2

(c) d3 > d2 > d1 (d) d2 > d3 > d1

61. In a solid, S2– ions are packed in fcc lattice. Zn2+

occupy half of the tetrahedral voids in an alternatingarrangement. Now if a plane is cut (as shown) thenthe cross-section would be :

(a) (b)

(c) (d)

62. In a bcc- arrangement which of the marked planeshave maximum spatial density of atoms ?

(a) 1 (b) 2(c) 3 (d) 4

63. The C–C and Si–C interatomic distances are 154pm and 188 pm. The atomic radius of Si is :(a) 77 pm (b) 94 pm(c) 114 pm (d) 111 pm

64. S1 : Cubic system have four possible type of unitcells.S2 : H2O is diamagnetic substance and it is weaklyattracted in magnetic field.S3 : Graphite is a covalent solid with vanderwaal’sforces as well.(a) F F T (b) F T F(c) T F F (d) F F F

65. S1 : Distance between Na+ & Cl¯ in NaCl crystal ismore than half of edge length.S2 : The no. of triagular viods in the givenarrangement in the enclosed region is 3.S3 : In ZnS structure, 2 Zn2+ & 2 S2– ions arepresent in each unit cell.

(a) F F T (b) F T F(c) T F F (d) F F F

66. Which of the following is rue about hexagonal closepacking ?


(a) It has a coordination number of 6(b) It has 26% empty space(c) It is ABCABC ..... type of arrangement(d) It is as closely packed as body centred cubic

packing.67. Which of the following is not true ?

(a) In NaCl crystals, Na+ ions are present in all theoctahedral voids

(b) In ZnS (zinc blende),n2+ ions are present inalternate tetrahedral voids.

(c) In CaF2, F– ions occupy all the tetrahedralvoids.

(d) In Na2O, O2– ions occupy half the octahedralvoids.

68. Crystal systems in which two axial lengths are equalis:(a) tetragonal (b) orthaorhombic

(c) monoclinic (d) triclinic69. Which of the following do not have rock-salt type

structure :

(a) NH4Cl (b) CaO(c) AgCl (d) AgI


These questions consist of two statements each,printed as Assertion and Reason. While answering theseQuestions you are required to choose any one of thefollowing four responses.A. If both Assertion & Reason are True & the

Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason

is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False.1. Assertion : In CsCl crystal, the Co-ordination

number of Cs+ ions is 6.Reason : Cl– ion in CsCl, adopt body centred cubicarrangement.

2. Assertion : Metals are generally good conductorsof electricity.Reason : Electrical conductivity of metals is due toschottky defect.

3. Assertion : Na2O adopts structure similar to thatof CaF2 but positions of positive and negative ionsare reversed.Reason : The structure of Na2O is also calledantifluorite structure.

4. Assertion : The number of tetrahedral void is doublethe number of octahedral voids.Reason : The size of the tetrahedral void is half ofthat of the octahedral void.

5. Assertion : Graphite is an example of tetragonalcrystal system.Reason : For a tetragonal system a=bc, ==90°,=120°.

6. Assertion : BCC & HCP has same packingefficiency.

Reason : Both have same number of atoms perunit Cell and same arrangement. [AIIMS-2011]

7. Assertion : Due to Frenkel defect the density ofsolid decrease.

Reason : In Frenkel defect cation or anion is presentleaving lattice sites vacant.

8. Assertion : By increasing the temperature C.N.of CsCl changes from 8 : 8 to 6 : 6.

Reason : Temperature also influences the C.N. ofsolid.

9. Assertion : Solid having more F-centers haveintense colour.

Reason : Excess of Na+ in NaCl solid having F-centers make it appears to pink.

10. Assertion : Close packing of atoms in cubicstructure is in order of : fcc > bcc > sc

Reason : Packing efficiency

= 3vol. of sphere in an unit cell? ?

? ?? ?a

11. Assertion : In NaCl structure, the interionicdistance is a/2. (a = Unit cell edge legth)Reason : NaCl form face centered Cubic Unit Cell.

12. Assertion : The co-ordination number of CaF2 is8 : 4.Reason : Ca2+ ions occupy ccp lattice whileF– ions occupy 50 % octahedral voids & 50%tetrahedral voids.


13. Assertion : The number of spheres are equal tothe number of octahedral void as well as tetrahedralvoid.

Reason : Octahedral void & tetrahedral void hasequal size.

14. Assertion : In Schottky defect, density of crystaldecreases.

Reason : Equal number of cations & anions aremissing in Schottky defect.

15. Assertion : If a tetrad axis is passed through theunit cell of NaCl & all ions are removed which aretouching to tetrad axis than the formula of NaClbecomes Na3 Cl4.

Reason : Only one Na+ is removed not the Cl–.

16. Assertion : Crystalline solids are anisotropic.

Reason : The constituent particles are very closelypacked.

17. Assertion : A particle present at the corner of theface centred unit cell has 1/8th of its contributionof the unit cell.Reason : In any space lattice, the corner of the unitcell is always shared by the eight unit cells.

18. Assertion : Crystal density is maximum in closedpacking geometry.Reason : In three dimension closed packinggeometry, Co-ordination number and packingefficiency is always maximum.

19. Assertion : ABAB.... pattern of close packing givesccp arrangement.Reason : In fcc arrangement each sphere associatedwith two tetrahedral voids.

20. Assertion : In crystal lattice, the size of the cationis larger in a tetrahedral hole than in an octahedralhole.Reason : The cations occupy more space thanatoms in crystal paking.


1. In a face centred cubic lattice, unit cell is sharedequally by how many unit cells ? [AIPMT 2005]

(a) 4 (b) 2

(c) 6 (d) 8

2. What is the co-ordination number of body centredcube ? [RPMT 2005]

(a) 8 (b) 6

(c) 4 (d) 12

3. If ‘Z’ is the number of atoms in the unit cell thatrepresents the closest packing sequence .... ABC,ABC....., the number of tetrahedral voids in the unitcell is equal to [AIIMS 2005]

(a) Z (b) 2 Z

(c) Z/2 (d) Z/4

4. CsBr crystallises in a body centred cubic lattice.The unit cell length is 436.6 pm.Given that the atomicmass of Cs = 133 and that of Br = 80 amu andAvogadro number being 6.02 1023 mol–1 thedensity of CsBr is : [AIPMT 2006]

(a) 8.25 g /cm3 (b) 4.25 g /cm3

(c) 42.5 g / cm3 (d) 0.425 g /cm3

5. The appearance of colour in solid alkali metal halidesis generally due to : [AIPMT 2006]

(a) F-centres (b) Schottky defect

(c) Frenkel defect (d) Interstitial positions

6. Total volume of atoms present in a face centredcubic unit cell of a metal is : (r is atomic radius)

[AFMC 2006]

(a) 163 r3 (b)

203 r3

(c) 243 r3 (d)

123 r3

7. The Ca2+ and F– ions are located in CaF2 crystal,respectively at body centred cubic lattice points andin: [AIIMS 2006](a) tetrahedral voids (b) half of tetrahedral voids(c) octahedral voids (d) half of octahedral voids

8. The fraction of the total volume occupied by theatoms present in a simple cube is : [AIPMT 2007]

(a) 4?

(b) 6?

(c) 3 2?

(d) 4 2?


9. If NaCl is doped with 10–4 mol% of SrCl2, theconcentration of cation vacancies will be: (NA = 6.02 1023 mol–1) [AIPMT 2007]

(a) 6.02 1014 mol–1 (b) 6.02 1015 mol–1

(c) 6.02 1016 mol–1 (d) 6.02 1017 mol–1

10. An fcc unit cell of aluminium contains the equivalentof how many atoms ? [AFMC 2007]

(a) 1 (b) 2

(c) 3 (d) 4

11. If AgI crystallises in zinc blende structure with I–

ions at lattice points. What fractions of tetrahedralvoids is occupied by Ag+ ions ? [AIIMS 2007]

(a) 25 % (b) 50 %

(c) 100 % (d) 75 %

12. Which of the following statements is not correct ?

[AIPMT 2008]

(a) The fraction of the total volume occupied bythe atoms in a primitive cell is 0.48.

(b) Molecular solids are generally volatile

(c) The number of carbon atoms in a unit cell ofDiamond is 8

(d) The number of Bravais lattices in which a crystalcan be categorized is 14

13. If ‘a’ stands for the edge length of the cubic systems:simple cubic, body centred cubic and face centredcubic, then ratio of the radii of the spheres in thesesystems will be respectively : [AIPMT 2008]

(a) 12

a : 34

a : 1

2 2 a

(b) 12

a : 3 a : 12 a

(c) 12

a : 32

: 22

a

(d) 1a : 3 a : 2 a

14. A metallic crystal has the bcc type stacking pattern.What percentage of volume of this lattice in emptyspace ? [AIPMT 08,Gujrat CET 08](a) 68 % (b) 32 %

(c) 26 % (d) 74 %

15. Edge length of a cube is 400 pm, its body diagonalwould be : [AFMC 2008](a) 566 pm (b) 600 pm(c) 500 pm (d) 693 pm

16. How many unit cells are present in a cube shapedideal crystal of NaCI of mass 1.00 g ?

[RPMT 2008]

(a) 2.57 × 1021 (b) 5.14 × 1021

(c) 1.28 × 1021 (d) 1.71 × 1021

17. Assertion : No compound has both Schottky andFrenkel defects. [AIIMS 2008]

Reason : Both defects change the density of thesolid.

(a) If both assertion and reason are true and reasonis a correct explanation of assertion.

(b) If both assertion and reason are true but reasonis not a correct explanation of assertion.

(c) If assertion is true but reason is false.


(e) If assertion is false but reason is true.

18. Copper crystallises in a face-centred cubic latticewith a unit cell length of 361 pm. What is the radiusof copper atom in pm ? [AIPMT 2009]

(a) 128 (b) 157

(c) 181 (d) 108

19. Lithium metal crystallises in a body centred cubiccrystal. If the length of the side of the unit cell oflithium is 351 pm, the atomic radius of the lithiumwill be : [AIPMT 2009]

(a) 240.8 pm (b) 151.8 pm

(c) 75.5 pm (d) 300.5 pm

20. Sodium crystallises in bcc arrangements with theinterfacial sepration between the atoms at the edge53 pm. The density of the solid is : [AIIMS 2009]

(a) 1.23 g/cc (b) 485 g/cc

(c) 4.85 g/cc (d) 123 g/cc

21. AB crystallizes in a body centred cubic lattice withedge length 'a' equal to 387 pm. The distancebetween two oppositively charged ions in the latticeis: [AIPMT 2010]

(a) 335 pm (b) 250 pm

(c) 200 pm (d) 300 pm


22. A solid compound XY has NaCl structure. If theradius of the cation is 100 pm, the radius of theanion (Y–) will be : [AIPMT 2011](a) 275.1 pm (b) 322.5 pm

(c) 241.5 pm (d) 165.7 pm23. A metal crystallizes with a face-centered cubic

lattice. The edge of the unit cell is 408 pm. Thediameter of the metal atom is: [AIPMT 2012](a) 288 pm (b) 408 pm(c) 144 pm (d) 204 pm

24. The number of octahedral void(s) per atom presentin a cubic close-packed structure is :

[AIPMT 2012](a) 1 (b) 3(c) 2 (d) 4

25. Structure of a mixed oxide is cubic close-packed(c.c.p). The cubic unit cell of mixed oxide iscomposed of oxide ions. One fourth of thetetrahedral voids are occupied by divalent metal Aand the octahedral voids are occupied by amonovalent metal B. The formula of the oxide is :

[AIPMT 2012](a) ABO2 (b) A2BO2

(c) A2B3O4 (d) AB2O2

26. A metal has a fcc lattice. The edge length of the unitcell is 404 pm. The density of the metal is 2.72 g cm–

3. The molar mass of the metal is : (NA Avogadro'sconstant = 6.02 1023 mol–1) [NEET 2013](a) 30 g mol–1 (b) 27 g mol–1

(c) 20 g mol–1 (d) 40 g mol–1

27. The number of carbon atoms per unit cell of diamondunit cell is : [NEET 2013](a) 8 (b) 6(c) 1 (d) 4

28. A given metal crystallises out with a cubic structurehaving edge length of 361 pm. If there are four metalatoms in one unit cell, what is the radius of oneatom? [AIPMT-2015](a) 40 pm (b) 127 pm(c) 80 pm (d) 108 pm

29. If the distance between Na+ and Cl– ions in sodiumchloride crystal is y pm, the length of the edge ofthe c unit cell is : [AIIMS-2015](a) 4y pm (b) y/4 pm(c) y/2 pm (d) 2y pm

30. Lithium has bcc structure. Its density is 530 kgm–3

and its atomic mass is 6.94 g mol–1. Calculate theedge length of a unit cell of Lithium metal. (NA =6.02 × 1023 mol–1) [NEET-2016](a) 264 pm (b) 154 pm(c) 352 pm (d) 527 pm

31. The ionic radii of A+ and B– ions are 0.98 × 10–10 mand 1.81 × 10–10 m. The coordination number ofeach ion in AB is [NEET-2016]

(a) 2 (b) 6

(c) 4 (d) 8

32. Which of the incorrect statement? [NEET-2017]

(a) FeO0.98 has non stoichiometric metal deficiencydefect.

(b) Density decreases in case of crystals withSchottky’s defect.

(c) NaCl(s) is insulator, silicon is semiconductor,silver is conductor, quartz is piezo electriccrystal.

(d) Frenkel defect is favoured in those ioniccompounds in which sizes of cation and anionsare almost equal.

33. Iron exhibits bcc structure at room temperature.Above 900ºC, it transforms to fcc structure. Theratio of density of iron at room temperature to thatat 900ºC (assuming molar mass and atomic radii ofiron remains constant with temperature) is :

[NEET-2018]

(a) 32

(b) 12

(c) 3 34 2

(d) 4 33 2

34. What is observe when ZnO is heated

[AIIMS-2018 (M)]

(a) yellow (b) Violet

(c) Green (d) Blue

35. f-centre is [AIIMS-2018 (M)]

(a) anion vacancy occupied by unpaired electron

(b) anion vacancy occupied by electron

(c) cation vacancy occupied by electron

(d) anion present in interstitial site


36. Example of Molecular solid is :[AIIMS-2018 (M)]

(a) SO2(s) (b) SiC(c) C (graphite) (d) NaCl

37. Assertion : Metal deficiency defect can be seen inFeO.Reason : Li compound (LiCl) have violet colourdue to F center. [AIIMS-2018 (M)](a) If both assertion and reason are true and reason



(c) If assertion is true but reason is false.(d) If both assertion and reason are false.

38. In HCP of A, 13 of tetrahedral are occupied by B.

What is the formula for compound:[AIIMS-2018 (E)]

(a) A2B3 (b) A3B2

(c) AB3 (d) A2B


ANSWER KEY

EXERCISE–0 (RECALL YOUR UNDERSTANDING)1. Space latice is the three demensional regular arrange-

ment of atoms, ions or molecules represented bypoints called lattice points.

2. In body-centred cubic crystal free space is 32%.3. There are 8 tetrahedral voids in a unit cell of ccp

structure.4. Packing efficiency in a simple cubic lattice = 52.4%.5. The size of octahedral void is larger than tetrahedral

void because cation can occupy octahedral void ifsize of cation is 0.414 times as large as anionwhereas in tetrahedral voids it should be 0.225 timesas large as anion.

6. It is because their presence leads to excess of elec-trons or leads to formation of positive holes whichincrease electrical conductivity.

7. Doping is a process of adding impurities in a crystallattice. It is done by adding calculated amount ofimpurity. Addition of one boron atom per 105 sili-con atoms increases the conductivity of silicon by103 at room temperature.

8. p-type semiconductor9. MnO is antiferromagnetic solid.10. Molecular solids CO2, H2

Ionic solids MgO, AgClCovalent solids SiMetallic solids Gd, Al, Pb

11. (i) Unit cell is the smallest portion of a crystal latticewhich, when repeated in different directions,generates the entire lattice.e.g., primitive unit cells : simple cubiccentred unit cells : bcc, fcc, ecc(ii) The number of spheres which are touching agiven sphere is called the coordination number.e.g., In simple cubic arrangement coordinationnumber is : 6.In bcc arrangement coordination number is : 8.In hcp or fcc arrangement coordination number is :12.

12. M = 3 30

0 10D a NZ

13. In a simple cubic unit cell : Suppose the edgelength of the unit cell = a and radius of thesphere = r.

Number of spheres per unit cell = 18 × 8 = 1.

Volume of the sphere = 43 pr3

Volume of the cube = a3 = (2r)3 = 8r3

Fraction occupied, i.e., packing fraction =

3

3

438

r

r

= 6

= 0.524.

% of occupied volume = 52.4%.14. 152.416 pm15. 6.23 g cm–3

16. (i) Schottky defect : It occurs when the equalnumber of vacant cation and anion sites are foundin lattice structure. It decreases the density of thesubstance.(ii) Interstitial defect : When some constituentparticles (atom or molecules) occupy an interstitialsite, the crystal is said to have interstitial defect.This defect increases the density of the substance.For example, non-ionic solids.

17. O2– ions are arranged in hcp, number of O2– ionsare equal to = 6Al3+ ions occupy 2/3 of the octahedral voids.

No. of Al3+ ions = 2 63 2 6

3

Al O6 Al2/3O1

Formula of corundum is Al2O3.18. 218 pm

19. Cu forms sc Cu = 8 × 18 = 1Ag at edge centres

= 12 × 14

= 3 Au at body centre = 1

Formula = Cu Ag3 Au20. (a) It is due to strong covalent bonds in diamod

whereas there are weak van der Waal’s forces ofattraction in S8 molecules.


(b) Silicon is best conductor out of these because itis semiconductor whereas others are insulator insolid state.

(c) In bcc, 4r = 3a

r = 3 1.732 4.294 4

a Å ; r = 7.43

4Å = 1.857 Å

Body diagonal = 3a = 1.732 × 4.29 Å = 7.43 Å.21. (i) Frenkel defects have no effect on the density

because number of atoms per unit cell remains thesame.Given : For fcc, Z = 4, Atomic weight of Ca = 40g mol–1

a = 0.5506 nm = 0.556 × 10–7 cm, NA = 6.023 ×1023

Asked : d = ?

Formulae : d = 3A

Z MN a

d = 7 3 234 40

(0.556 10 ) 6.022 10

= 2160 160

103.50.17 6.022 10

d = 1.5458 g cm–3

Ans.: 1.5458 g cm–3

(ii) Due to 0.2% Schottky defect, No. of atoms

0.2100 × 4 = 3.992

d = 3A

Z Ma N

= 7 3 23

4 3.992(0.556 10 ) 6.022 10

= 1.5427 g cm–3.22. (i) At 850 K ,Fe3O4 (magnetite) loses its alignment

of magnetic moment is parallel and anti-paralleldirection in unequal numbers and becomeparamagnetic.(ii) Zinc oxide is white in colour at room temperature.On heating it loses oxygen and turns yellow. (due toformation of F - centre)

ZnO heating Zn2+ + 12

O2 + 2e–

(iii) Due to Frenkel defect, no ions are missing fromthe crystal as a whole. Thus, there is no change indensity.

23. In semiconductor, electrical resistivity decreaseswith increase in temperature. In metals, electricalresistivity increases with increases in temperatureand in insulators electrical resistivity decreases withincrease in temperature.

24. (i) Ferromagnetism : Materials which are stronglyattracted by magnetic fields are called ferromagneticmaterials and the property thus exhibited is calledferromagnetism. only three elements (Fe, Co, Ni)show ferromagnetism at room temperature.(ii) Paramagnetism : Materials which are weaklyattracted by magnetic fields are called paramagneticmaterials and the property thus exhibited is calledparamagnetism. Paramagnetic substances containunpaired electron. e.g. TiO, CuO, O2 and VO2 etc.

25. (i) It is a hexagonal close packing(ii) The coordination number is 12(iii) Free space = 26%(iv) Z = 6, There are six atoms in single unit cellhaving hexagonal close packing.

26. Cations A = 18 × 8 = 1; Cations B =

12

× 4 = 2

O2– = 8 × 18 + 6 ×

12

= 4

[ O2– forms cubic close packed structure]AB2O4 is formula of mixed oxide.

27. (i) Schottky defect decreases the density of thecrystal.Frenkel defect does not change the density of thecrystal.(ii) Electrical conductivity of the crystal is increasedin both Schottky and Frenkel defects.

28. (i) In conductors the energy gap are very close oreven overlapped by the conduction bands. Due tothis reason electron can flow very easily fromvalence to conduction band under an electronic fieldand shows conductivity.In insulator, the energy gap between the valenceband and conduction band is too high so that electroncan not flow from valence band to conduction band.Therefore, no conduction is possible.(ii) In semiconductor the energy gap betweenvalence band and conductor band is small whichmake some of the electron enable to jump fromvalence to conduction band and shows conductivity


EXERCISE–1 (CHECK YOUR UNDERSTANDING)1. (d) 2. (b) 3. (c) 4. (b) 5. (d) 6. (d) 7. (d) 8. (c) 9. (b) 10. (d)

11. (d) 12. (a) 13. (c) 14. (a) 15. (d) 16. (c) 17. (d) 18. (c) 19. (b) 20. (d)21. (a) 22. (c) 23. (d) 24. (b) 25. (b) 26. (a) 27. (c) 28. (a) 29. (a) 30. (a)31. (c) 32. (b) 33. (c) 34. (b) 35. (a) 36. (c) 37. (b) 38. (a) 39. (a) 40. (b)41. (d) 42. (a) 43. (b) 44. (a) 45. (c) 46. (a) 47. (a) 48. (a) 49. (c) 50. (b)51. (b) 52. (a) 53. (c) 54. (a) 55. (b) 56. (b) 57. (b) 58. (b) 59. (b) 60. (c)61. (b) 62. (b) 63. (c) 64. (a) 65. (b) 66. (c) 67. (c) 68. (c) 69. (a) 70. (a)71. (c) 72. (c) 73. (c) 74. (d) 75. (d) 76. (c) 77. (c) 78. (c) 79. (d) 80. (b)81. (a) 82. (a) 83. (a) 84. (d) 85. (a) 86. (d) 87. (b) 88. (d) 89. (c) 90. (c)91. (a) 92. (b) 93. (b) 94. (a) 95. (b) 96. (c) 97. (d) 98. (c) 99. (b) 100. (d)

101. (d) 102. (a) 103. (a) 104. (d)

EXERCISE–2 (CHALLENGE YOUR UNDERSTANDING)1. (b) 2. (b) 3. (b) 4. (c) 5. (b) 6. (a) 7. (a) 8. (c) 9. (c) 10. (c)

11. (d) 12. (d) 13. (b) 14. (a) 15. (d) 16. (c) 17. (a) 18. (b) 19. (b) 20. (b)21. (b) 22. (d) 23. (d) 24. (b) 25. (b) 26. (a) 27. (a) 28. (a) 29. (a) 30. (c)31. (b) 32. (d) 33. (b) 34. (a) 35. (a) 36. (b) 37. (b) 38. (a) 39. (d) 40. (a)41. (a) 42. (a) 43. (b) 44. (b) 45. (b) 46. (c) 47. (b) 48. (c) 49. (a) 50. (b)51. (c) 52. (d) 53. (b) 54. (d) 55. (d) 56. (a) 57. (a) 58. (b) 59. (b) 60. (c)61. (b) 62. (a) 63. (d) 64. (a) 65. (d) 66. (b) 67. (d) 68. (a) 69. (d)

EXERCISE–3 (AIIMS SPECIAL)1. (d) 2. (c) 3. (b) 4. (c) 5. (d) 6. (d) 7. (d) 8. (a) 9. (c) 10. (a)

11. (b) 12. (c) 13. (d) 14. (a) 15. (d) 16. (b) 17. (c) 18. (a) 19. (d) 20. (d).

EXERCIS-4 (ARCHIVES FROM VARIOUS MEDICAL ENTRANCE)1. (c) 2. (a) 3. (b) 4. (b) 5. (a) 6. (a) 7. (a) 8. (b) 9. (d) 10. (d)

11. (b) 12. (a) 13. (a) 14. (b) 15. (d) 16. (a) 17. (d) 18. (a) 19. (b) 20. (a)21. (a) 22. (c) 23. (a) 24. (a) 25. (d) 26. (b) 27. (a) 28. (b) 29. (d) 30. (c)31. (b) 32. (a) 33. (c) 34. (a) 35. (a) 36. (a) 37. (b) 38. (b)

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