Chemical bonding and molecular structure

1) Explain the formation of a chemical bond.

Solution

A chemical bond is the force of attraction that holds various constituents (atoms or ions)

together in different chemical species.

Atoms combine in different ways to attain stable configuration of nearest inert gas. Atoms

combine by either transfer of electrons (ionic bond) or by sharing of electrons (covalent

bond). Due to the formation of chemical bond, the energy of the atoms is lowered,

resulting in maximum stability.

Types of bonds: Depending on the nature of formation, bonds are classified into two

kinds:

1) Ionic bond 2) Covalent bond

Ionic bond: The bond formed by the transfer of electrons from one atom to another atom

is called an ionic bond.

Example : NaCl

NaCl is formed by the complete transfer of one electron from Na to Cl.

Then Na+ and Cl

- ions get attracted to each other to form an ionic bond.

Sodium (Na):

Atomic number (Z) =11

Electronic configuration = 1s22s

22p

63s

1

Chlorine (Cl):

Atomic number (Z) =17

Electronic configuration = 1s22s

22p

63s

2 3p

5

After electron transfer :

Na+

Cl-

1s22s

22p

6 1s

22s

22p

6 3s

23p

6

(Electronic configuration of Neon) (Electronic configuration of Argon)

Na+

+ Cl- →NaCl

Covalent bond: The bond formed by sharing of electrons between the two bonded atoms is

called a covalent bond.

Example : H2

A hydrogen molecule (H2 ) is formed by the sharing of a pair of electrons between two

hydrogen atoms. The two hydrogen atoms in a hydrogen molecule are at such a distance

from each other that there is minimum repulsion and maximum stability.

Hydrogen atom (H):

Atomic number (Z)=1

Electronic configuration = 1s1

Formation of covalent bond in hydrogen molecule:

2) Write the Lewis dot symbol for the following elements : Mg, Na, B, O , N, Br.

Solution

Lewis dot symbol for Mg

Lewis dot symbol for Na:

Lewis dot symbol for B:

Lewis dot symbol for O:

Lewis dot symbol for N:

Lewis dot symbol for Br:

3) Write the Lewis symbols for the following atoms and ions:

S and S2-

; Al and Al3+

; H and H- .

Solution

Lewis dot symbols for S and S2-

:

Lewis dot symbols for Al and Al3+

:

Lewis dot symbols for H and H- :

4) Draw the Lewis structures for the following molecules and ions :H2S, SiCl4 ; BeF2, CO32-

,

HCOOH.

Solution

Lewis structure of H2S :

Lewis structure of SiCl4:

Lewis structure of BeF2 :

Lewis structure of CO32-

:

Lewis structure of HCOOH :

5) Define the octet rule. Write its significance and limitations.

Solution

Octet rule: Atoms combine either by the transfer of electrons or by their sharing to gain

eight electrons in their valence shell (octet). Atoms with eight electrons in their valence

shell are considered to be stable. This is called the octet rule.

Significance :

1) The octet rule explains the stability of most covalent compounds. By counting the

number of valence electrons in the central atom and the other atoms participating in the

bond, the stability can be explained. If the atoms in a molecule have an octet in their

valence shell, then the molecule is considered to be stable.

2) The octet rule is very useful in determining the structure of most organic compounds.

3) The octet rule is mostly applicable to the elements of the second period.

Limitations : Though the octet rule is applicable to explain the stability of covalent

molecules, it is not universal.

Compounds without an octet in the valence shell of their atoms are also stable.

For example :

1) Molecules like BCl3, in which the central atom boron has only six electrons in its

valence shell. It is 2 electrons short of a valence octet. Even then, BCl3 is a highly stable

molecule.

2) Molecules like NO and nitrogen dioxide, which ahve an odd number of electrons, do

not satisfy the octet rule. Even then they are stable.

3) Molecules like SF6, PF5 expand their octet by involving 3d orbitals in the bond

formation. They have more electrons than an octet in their valence shells. In SF6, sulphur

has 12 electrons in its valence shell, and in PF5 , phosphorus has 10 electrons in its valence

shell.

Though SF6 and PF5, have more electrons than octet, they are stable.

6) Write the factors favourable for the formation of an ionic bond.

Solution

Ionic bond: The bond formed by the complete transfer of an electron from one atom to

another atom, in which there exists an electrostatic force of attraction between the

positively charged and negatively charged ions, is called an ionic bond.

An ionic bond is formed between highly electronegative halogens and highly

electropositive alkali metals.

Conditions favourable for the formation of ionic bond:

1) An atom with low ionisation energy can lose an electron easily and can form a positive

ion (cation) easily.

2) An atom with high electron affinity can gain electron easily and form a negative

ion(anion) easily.

3) The negative and positive ions formed attain stable electronic configuration of the

nearest noble gas.

4) The negative and positive ions are stabilised by electrostatic attraction, resulting in the

formation of an ionic bond.

5) As alkali metals have low ionisation energy and halogen have high electron affinity,

usually an ionic bond is easily formed between them.

7) Discuss the shape of the following molecules using the VSEPR model: BeCl2 , BeCl3,

SiCl4.

Solution

The structure of a covalent molecule can be predicted by the number of electron pairs in

the valence shell of the central atom in the molecule based of VSEPR theory.

Structure of BeCl2 :

The central atom of BeCl2 is beryllium. In the valence shell of beryllium in BeCl2 there are

2 bond pairs of electrons. Therefore, the shape of the BeCl2 molecule is linear with a bond

angle of 180°.

Structure of BCl3 :

The central atom is BCl3 is boron. In the valence shell of boron, there are three bond pairs

of electrons. Therefore, the shape of the molecules is triangular, with a bond angle of 120°.

Structure of SiCl4 :

The central atom of SiCl4 is silicon. There are four bond pairs of electrons in the valence

shell of Si. Therefore, the structure of SiCl4 is tetrahedral with a bond angle of 109°28’.

Cl

Structure of AsF5 is arsenic. There are 5 bond pairs of electrons in the valence shell

of arsenic. Therefore, accoriding to VSEPR, the structure of AsF5 is trigonal bipyramidal

in the bond angles 90° and 120° .

Structure of H2S:

The central atom of H2S is sulphur. Sulphur has 2 lone pairs and 2 bond pairs of electrons

in its valence shell. According to VSEPR theory, the structure of the H2S molecule is bent

or angular due to lone pair-lone pair and bond pair-lone pair repulsions.

The lone apirs and the bond pairs of electrons are arranged tetrahedrally. The bond angle is

92°32’.

Structure of PH3 :

The central atom of the PH3 molecule is phosphorus. The phosphorus atom has four pairs

of electrons in its valence shell. Three electron pairs are bond pairs and one is lone pair.

According to VSEPR theory, the structure of PH3 molecule is trigonal pyramidal. The

bond angle is 93°36’ due to the repulasion between lone pair and bond pair of electrons,

and bond pair or electrons.

The arrangement of lone pair and bond pair of electrons around the phosphorus atom is

tetrahedral.

8) Although the geometries of NH3 and H2O molecules are distorted tetrahedral, the bond

angle in water is less than that of ammonia. Discuss.

Solution

The bond angle in a water molecule is less than that of an ammonia moleucle due to the

repulsion between lone pair-lone pair and lone pai-bond pair of electrons. In an ammonia

molecule, repulsion is only present between lone pair-bond pair of electrons. On the other

hand, in a water molecule, there is repulsion between lone pair-lone pair and lone pair-

bond pair of electrons. Therefore, in a water moleucle, the bond angle is less than that of

ammonia. The bond angle in a water moleucle is 104°.5’ , while in an ammonia molecule

the bond angle is 107°.

9) How do you express the bond length in terms of bond order?

Solution

Bond order is the number of bonds between the two atoms bound in a molecule. As the

bond order increases, the bond strength increases. This is because bond enthalpy increases

with an increase in the bond order.

10) Define bond length.

Solution

Bond length is defined as the equilibrium distance between the nuclei of two bonded

atoms in a molecule.

11) Explain the important aspects of resonance with reference to the CO32-

ion.

Solution

Lewis structure of some molecules and ions cannot represent its structural aspects.

Experimentally, it has been proved that, whenever a single Lewis structure cannot

represent the structure of a molecule accurately, then a number of structures with similar

energy, position of nuclei, bonding and non-bonding pairs of electrons are given. These

structures are called canonical structures or resonance structures. This is called resonance.

For example, the structure of a carbonate ion cannot be represented by a single Lewis

structure. It is proved experimentally that the structure of carbonate ion is a resonance

hybrid of the following three resonance structures I, II and III.

Resonance structures of a carbonate ion:

12) H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures

be taken as the canonical forms of the resonance hybrid representing H3PO3 ?

If not, give reason for the same.

Solution

No, the two given structures do not represent the resonance structures of phosphorous

acid. In phosphorous acid, there is a P-H bond, but in structure (2), it is missing. There is

double bond between P and O . This is also missing in structures (1) and (2). Phosphorus

forms five bonds, but in the given two structures , only four bonds have been shown to

phosphorus atom. Therefore , these two given structures cannot be considered as

resonance structures of H3PO3.

Structure of H3PO3

13) Write the resonance structures for SO3, NO2 and NO3-.

Solution

Sulphur trioxide (SO3)

Resonance structures of SO3 :

Resonance structures of NO2 :

Resonance structures of NO3-:

14) Use Lewis symbols to show the electron transfer between the following atoms to form

cations and anions.

a) K and S b) Ca and O c) Al and N

Solution

a) K and S

Lewis symbols for K and S atoms:

Lewis symbols for K+ and S

2- ions:

b) Ca and O

Lewis symbol for Ca and O atoms:

Lewis symbols for Ca2+

and O2- ions:

c) Al and N:

Lewis symbol for Al and N atoms:

Lewis symbols for Al3+

and N3-

ions:

15) Although both CO2 and H2O are triatomic molecules, the shape of an H2O molecule is

bent, while that of CO2 is linear . Explain this on the basis of dipole moment.

Solution

A water molecule is a bent molecule in which the two O-H bonds are oriented at an angle

of 104.5°. The bond moment of O-H bonds give the reultant net dipole moment of 1.85 D

for a water molecule. On the other hand, a CO2 molecule is a linear molecule with the two

C-O bonds at 180° in a straight line. The bond moment of one C-O bond is exactly

cancelled by the ond moment of another C-O bond. Thereofore, a CO2 molecule has zero

dipole moment and a linear structure.

16) Write the significance/applications of dipole moment.

Solution

Significance of dipole moment:

1. The percentage ionic character of a covalent bond can be determined from dipole

moment.

2. The symmetry of the moleucle can be decided using dipole moment value.

3. The bond angle of a covalent molecule can be determined using dipole moment value.

17) Define electro-negativity. How does it differ from electron gain enthalpy?

Solution

Electronegativity is the tendency of a bonded atom in a molecule to attract the shared pair

of electrons towards itself.

Electronegativity is the property of a onded atom , but not the property of an independent

individual atom. On the other hand, electron gain enthlpy is the energy released when an

electron is added to a gaseous neutral atom.

18) Explain with the help of a suitable example a polar covalent bond.

Solution

A polar covalent bond is a covalent bond formed by an unequal sharing of electrons

between two atoms participating in the bond due to the electronegativity difference

between them.

A polar covalent bond is formed between atoms of different elements.

For example, a polar covalent bond is formed between H and Cl in an HCl molecule. In an

HCl molecule, the polar covalent bond is formed by the unequal sharing of the bond pair

of electrons between H and Cl . The chlorine atom, due to high electronegativity , drags

the electron pair towards itself. Due to the unequal sharing of electrons between H and Cl,

H develops a partial and permanent positive charge, while Cl develops a partial and

permanent negative charge. Thus, a polar covalent bond is formed between H and Cl in an

HCl molecule.

Formation of polar covalent bond in HCl molecule:

19) Arrange the following bonds in order of increasing ionic character in the molecules:

LiF, K2O,N2, SO2 and ClF3.

Solution

Increasing order of ionic character in the molecules: N2 < ClF3 < SO2 < K2O < LiF.

20) The skeletal strucutre of CH3COOH as shown below is correct, but some of the bonds

are shown incorrectly. Write the correct Lewis strcuture for cetic acid.

Solution

21) Apart from tetrahedral geometry, another possible geometry for CH4 is square planar,

with the four H atoms at the corners of the square, and the C atom at its centre. Explain

why CH4 is not square planar?

Solution

Square planar geometry arises due to octahedral arrangement with four ond pairs and two

lone pairs of electrons. But in CH4, there are four bond pairs of electrons arrnaged in

tetrahedral manner with a bond angle of 109°28’, and there are no lone pairs of electrons.

Therefore, CH4 is tetrahedral and not square planar.

22)Explain why a BeH2 molecule has a zero dipole moment although the Be-H bonds are

polar?

Solution

A BeH2 molecule is linear with 180° bond angle. The two Be-H bonds are polar and have

bond moment. But the bond moment of both the bonds are equal and opposite. Due to this,

the ond moment cancels each other and the resultant dipole moment becomes zero.

23) Which out of NH3 and NF3 has a higher dipole moment and why?

Solution

NH3 and NF3 both are oyramidal with one lone pair of electrons on a nitrogen atom.

Though fluorine is more electronegative than nitrogen, the dipole moment of NH3 is 4.9 x

10-30

cm, which is greater than the dipole moment of NF3(0.8 x 10-30

cm). This is becuase ,

in NH3, the orbital ipole due to the lone pair of electrons is in the same direction as the

resultant dipole moment of the N-H bonds, whereas in NF3 , the orbital dipole due to the

lone pair of electrons is in direction opposite to that of the resultant dipole moment of the

three N-F onds. Due to this , NF3 has a low dipole moment.

24) What is meant y hybridisation of atomic orbitals? Describe the shapes of sp, sp2

and

sp3 hybrid orbitals.

Solution

Hybridisation : The phenomenon of intermixing of the pure atomic orbitals of nearly same

or slightly different energies, so as to redistribute their energies, resulting in the formation

of a new set of hybrid orbitals of equivalent energies and shapes, is called hybridisation.

25) Describe the change in hybridisation (if any) of the Al atom in the following reaction.

AlCl3 + Cl- →AlCl4

-

Solution

When AlCl3 accepts a pair of electrons from Cl- ion to form AlCl4

-, then the hybridisation

of Al changes from sp2 to sp

3 . The structure of AlCl3 also changes from trigonal planar to

tetrahedral.

Type of

hybridistion

Shape Structure

sp Linear

Sp

2 Trigonal planar

Sp

3 Tetrahedral

26) Is there any change in the hybridisation of B and N atoms as a result of the following

reaction?

BF3 + NH3 →F3B.NH3

Solution

In the formation of ammonia boron trifluroide molecule the hybridisation of B changes

from sp2

to sp3

. There is no change in the hybridisation of N-atom as a result of the

formation of dative bond between the BF3 and NH3 molecules. The electron pair donated

by the nitrogen atom is accepted into the vacant sp3 hybrid orbital of the boron atom.

27) Draw diagrams showing the formation of a double bond and a triple bond between the

carbon atoms in C2H4 and C2H2 molecules.

Solution

Formation of double bond in C2H2 (Ethylene):

Formation of triple bond in C2H2(Acetylene):

28) What is the total number of sigma and Pi bonds in the following molecules?

a) C2H2 b) C2H4

Solution

a) C2H2 b) C2H4

σ bonds -3 σ bonds -5

π bonds- 2 π bonds- 1

29) Considering the X-axis as the intermolecular axis, which out of the following will not

form a sigma bond and why?

a) 1s and 1s b) 1s and 2px c) 2py and 2py d) 1s and 2s

Solution

c) 2py and 2py orbitls do not form a sigma bond as the inter-nuclear axis is the X –axis

and the py orbitals is perpenicular to it. Therefore, 2py and 2py can only for a π-bond by

lateral or sidewise overlap.

30) Which hybrid orbitals are used by the carbon atoms in the following molecules?

a) CH3 –CH3 b) CH3 –CH = CH2 (Propene)

c) CH3– CH2-OH d) CH3CHO e) CH3-COOH (ethanoic acid)

Solution

a) CH3 –CH3 (ethane)

In ethane, carbon uses sp3 hybrid orbitals.

b) CH3 –CH = CH2 (Propene)

In propene, CH3 –carbon uses sp3

hybrid orbitals, in –CH and –CH2 with double bond,

carbon atom uses sp2 hybrid orbitals.

c) CH3– CH2-OH (ethanol)

In ethanol , carbon uses sp3 hybrid orbitals.

d) CH3CHO (ethanal)

In ethanal, CH3-uses sp3 hybrid orbitals and –CHO carbon atom uses sp

2 hybrid orbitals.

e) CH3-COOH (ethanoic acid)

In ethanoic acid, CH3- carbon atom uses sp3

hybrid orbitals, and –COOH carbon atom

uses sp2 hybrid orbitals.

31) What do you understand by ond pairs and lone pairs of electrons? Illustrate by giving

one example of each type.

Solution

Bond pairs- The electron pairs participating in bond formtion are called bond pairs.

Lone pairs: The electron pairs present in the valence shell of a onded atom, which do not

particiapate in bond formation, are called a lone pair of elctrons or non-bonding electrons.

Example: In a ammonia molecule, in the valence shell of the nitrogen atom, there are 3

bond pairs of electrons participating in 3 N-H bonds and one lone pair of electrons.

32) Distinguish between a sigma bond and a Pi bond.

Solution

S.No. Sigma bond (σ bond) S.No. Pi bond(π bond)

1. It is a head-on-head

overlap of atomic

orbitals along the inter-

nuclear axis during the

formation of a covalent

bond.

1. The bond formed by lateral or

side wise overlap of atomic

orbitals is called a Pi bond.

2. It is along the

inter-nuclear axis

2. It is above and below the inter-

nuclear axis.

3. It is a strong bond

as the extent of

overlap is more.

3. It is a weak bond as the extent

of overlap is less.

4. It is formed by

atomic orbitals and

hybrid orbitals.

4. It is not formed by hybrid

orbitals.

5. s-orbitals forms

only σ bonds.

5. s-orbitals cannot form π bonds.

33) Explain the formation of an H2 molecule on the basis of valence bond theory.

Solution

According to valence bond theory, a covalent bond is formed by an overlap of atomic

orbitals with unpaired electrons.

Formation of hydrogen molecule: A hydrogen atom has one unpaired electron in the 1s

orbital. An H2 molecule is formed by the overlap of the 1s orbital of one hydrogen atom

with the 1s orbital of another hydrogen atom, resulting in the formation of a σ bond and a

hydrogen molecule.

34) Write the importanat conditions required for the linear combination of atomic orbitals

to form molecular orbitals.

Solution

The conditions necessary for the linear combination of atomic orbitals to form

molecular orbitals are:

1) The atomic orbitals that combine to for molecular orbitals should have the same or

nearly the same energy.

Example: 1s orbital can combine with another 1s orbital, but not with 2s orbital has higher

energy.

2) The atomic orbital that combines must have the same symmetry about the molecular

axis.

Example: The 2pz orbital of one atom can only combine with the 2pz orbital of another

atom, but not with the 2px or 2py orbitals because of their different symmetry.

3) The extent of overlap between the combining atoms should be maximum. As greater the

extent of the overlap, the greater will be the electron density between the nuclei of a

molecular orbital.

35) Use molecular orbital theory to explain why the Be2 molecule does not exist.

Solution

The electronic configuration of Be is 1s2

2s2

. Each beryllium atom contains 4 electrons,

and , therefore, a Be2 molecule has 8 electrons. These 8 electrons are arranged in σ1s,

σ∗1s, σ2s, σ∗2s.

Be2 : σ1s, σ∗1s, σ2s2, σ∗2s.

Bond order of Be2 =1/2 (4-4)=0

∴ Be2 molecule is unstable due to zero bond order and does not exist.

36) Compare the relative stability of the following species and indicate their magnetic

properties: O2 , O2+

, O2-(superoxide) , O2

2-(peroxide).

Solution

O2 molecule: The electronic configuration of an O2 molecule is

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ∗2s)

2(σ2pz)

2 (π2px

2≡π2py)

2(π∗2px

1≡π∗2py

1)

The electronic configuration of an O2 molecule indicates that there are ten electrons in the

bonding molecular orbitals, and six electrons in the anti-bonding molecular orbitals.

∴Bond order=1/2 (10-6)=2

∴ The oxygen (O2)molecule is stable due to positive bond order. Due to the presence of

unpaired electrons in π∗O2px and π∗2py, the O2 molecule is paramagnetic.

O2+ ion:

Electronic configuration of O2+ ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2s)

2(π22pz)

2 (π2px

2≡π2py

2) (π∗2px

1)

The electronic configuration of an O2+ ion indicates that there are ten electrons in the

bonding molecular orbitals and five electrons in the anti-bonding molecular orbitals.

∴Bond order=1/2 (10-5)=2.5

O2+ has positive bond order. Therefore , it is stable. Due to the presence of one unaired

electron in O2+

, it is paramagnetic.

O2- ion:

Electronic configuration of O2- ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2pz)

2 (π2px

2≡π2py

2) (π2px

2≡π2py

1)

The electronic configuration of O2- ion indicates that there re 10 electrons in the bonding

molecular orbitals and 7 electrons in the anti-bonding molecular orbitals.

∴Bond order=1/2 (10-7)=1.5

An O2- ion also has positive bond order. Therefore, it is stable . Due to the presence of one

unpaired electron in π∗2py1 orbital, it is paramagnetic.

O22-

ion:

Electronic configuration of O22-

ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2s)

2(σ2pz)

2(π2px

2≡π2py

2) (π∗2px

2≡π2py

2)

The electronic configuration of the O22

ion indicates that there are 10 electrons in the

bonding molecular orbitals and 8 electrons in the anti-bonding molecular orbitals.

∴Bond order=1/2 (10-8)=1

An O22-

ion has positive bond order. Therefore, it is stale. There are no unpaired electrons

in O22molecular orbitals. Therefore, it is diamagnetic .

Relative order of stability (Decreasing order): O2+

>O2 >O2->O 2

2-.

Higher the bond order more is the stability of the species.

37) Write the significance of a plus and a minus sign shown in representing orbitals.

Solution

The plus and minus sign shown in representing orbitals is the sign of the wave function

(ψ) of the electron wave or atomic orbital. When atomic orbitals join to form molecular

orbitals during bond formation, the lobes of the atomic orbitals with the same sign

combine to form bonding molecular orbitals, while in the formation of anti-bonding

orbitals, the lobes of the atomic orbitals with the opposite signs join together.

38) Describe the hybridisation in case of PCl5 . Why are the bonds longer as compared to

equatorial bonds?

Solution

In PCl5, the central atom phosphorus undergoes sp3d hybridisation in its excited state. The

five sp3d hybrid orbitals are arranged in a trigonal bipyramidal structure. These five sp

3d

hybrid orbitals with the pure p-orbital to five chlorine atoms froming 5P-Cl bonds. Two P-

Cl bonds lie perpendicular to the triangular plane at 90° and three P-Cl bonds lie in the

plane of the triangle making an angle of 120°.

P-Ground state

P- Excited state

PCl5

The axial bonds in PCl5 are longer than the equatorial bonds, since the axial bond pairs

suffer more repulsive interaction from the equatorial bond pairs.

39) Define a hydrogen bond. Is it weaker or stronger than vander waal’s forces?

Solution

A hydrogen bond is defined as the attractive force that binds the hydrogen atom of one

molecule with the electronegative atom (F, O or N) of another molecule.

A hydrogen bond is stronger than vander waal’s forces, as in a hydrogen bond, it is

electrostatic force of attraction between the opposite charges.

40) What is meant by the term bond order?

i)Calculate the bond order of : N2 ,O2.

Solution

Bond order is defined as one half the difference between the number of electrons present

in the bonding and the anti-bonding orbitals.

Bond order= ½ (Nb-Na)

Nb = Number of electrons in bonding molecular orbitals.

Na= = Number of electrons in anti- bonding molecular orbitals.

Bond order N2 :

Electronic configuration of N2 ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2s)

2(σ2pz)

2(π2px

2≡π2py

2)

Nb= 10

Na =4

∴Bond order=1/2 (Nb-Na)

= ½ (10-4)

=6/2=3

Bond order of N2 molecule is 3.

Bond order O2:

Electronic configuration of O2 ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2s)

2(σ2pz)

2(π2px

2≡π2py

2) (π2px

2≡π2py

2) (π∗2px

2≡π∗2py

1)

Nb =10

Na = 6

Bond order= ½ (Nb-Na)

= ½ (10-6)

=6/2 =2

∴Bond order of O2 molecule is 2.

ii) Calculate the bond order of : O2+

and O2-

Solution

Bond order of O2+

:

Electronic configuration of O2+

ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ2s)

2(σ2pz)

2(π2px

2≡π2py

2) (π2px

2≡π2py

2) (π∗2px

1)

Na =5

Nb =10

Bond order = ½ (10-5)

= 5/2 =2.5

∴Bond order of O2+

ion is 2.5.

Bond order of O2- ion:

Electronic configuration of O2- ion:

(σ1s)2 (σ∗1s)

2 (σ2s)

2 (σ∗2s)

2(σ2pz)

2(π2px

2≡π2py

2) (π∗2px

2≡π∗2py

1)

Na =10

Nb =7

Bond order = ½ (10-7)

=3/2 =1.5

∴Bond order of O2- ion is 1.5.