1 chemistry density of unit cell × = × z m d a n d = density ( g/cm3
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1 Chemistry
Density of Unit Cell
3A
Z Mda N d = Density (g/cm3)
Z = number of atoms per unit cell a = Edge length in cm
Value of z for M = Molar mass in g/mol
simple cubic lattice = 1 NA = Avogadro constant in per mol
bcc lattice = 2
ccp or fcc lattice = 4
Radius Ratio : It is the ratio of radius of cation to the radius of anion.
radius ratio r
r
S.No. Radius Ratio C.N. Type of void or hole Example
1. 0.155 – 0.225 3 Trigonal planar B2O32 0.225 – 0.414 4 Tetrahedral ZnS3 0.414 – 0.732 6 Octahedral NaCl3 0.732 – 1.000 8 Cubic CsCl
Relationship between atomic radius and edge length.
Simple Cubic BCC FCC
2ar
34
ar 24
ar
In ccp or hcp packing two types of voids namely (i) tetrahedral (ii) octahedral are generated.
No. of octahendral voids present in a lattice = Number of close packed particles
No. of tetrahedral voids present in a lattice = 2 × number of close packed particles
In ionic solids, the larger ions (usually anions) form close packed structure and the smallerions (usually cations) occupy voids. If the cation is small enough then tetrahedral voids are occupied,if bigger, then octahedral voids. Not all octrahedral or teterahedral voids are occupied. The fractionof octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of thecompound.
2 Chemistry
Imperfections in Crystals
Schottky Defects
Cations and equal number of anions are missing from the lattice site of a crystal of the typeA+ B– e.g., there are 106 schottky pairs per cm3 at room temperature of NaCl. Schottkey defectlowers the density of crystal. Ionic substances showing schottky defects have almost similar sizesof cations and anions, e.g.; NaCl, KCl, KF, etc.
Frenkel Defects
Cation is missing from the lattice site but trapped within interstitial position, e.g., Ag Br,AgCl, Agl, AnS, etc. where cations and anions have much difference in their sizes, show this typeof defect.
F-Centre
Anions are missing from lattice sites and these anionic sites are occupied by unpaired electrons.Anionic sites occupied by unpaired electrons are called F-centres and are responsible for colourimparted to crystals.
Metal Deficiency Defect
Some metal oxides contain less amount of metal as compared to the stoichiometric proportion.In iron oxide of composition Fe0.95.O, some Fe2+ are replaced by definite number of Fe3+ resultingin the metal deficiency.
Packing Efficiency
It is the percentage of total space filled by particles.
Packing efficiency
hcp and ccp 74%
BCC 68%
Simple cubic lattice 52.4%
3 Chemistry
1. In a solid lattice, the cation has left a lattic site and is located in an interstitial position.The lattice defect is known as
(a) Interstitial defect (b) Vacancy defect
(c) Frenkel defect (d) Schottky defect
2. An octahedral void is surrounded by how many spheres
(a) 6 (b) 4
(c) 8 (d) 12
3. In NaCl crystal, number of Cl– ions around each Na+ ion will be
(a) 3 (b) 4
(c) 6 (d) 8
4. For an ionic crystal of general formula AB and coordination number 6, the radius ratio willbe
(a) greater than 0.732 (b) between 0.414 to 0.732
(c) between 0.225 to 0.414 (d) between 0.155 to 0.225
5. If pentavalent impurity is mixed in a crystal lattice of germanium, the semiconductorwill be
(a) p-type (b) n-type
(c) pnp (d) npn
6. A metallic crystal containing a sequence of layers AB AB AB....Any packing of spheresleaves out voids in the lattice. What percent of volume of thus lattice is empty space.
(a) 74% (b) 26%
(c) 52% (d) 68%
7. When electrons are trapped into anion vacancies, the defect is known as
(a) Schottky defect (b) Frenkel defect
(c) Non-stoichiometric (d) Metal excess defect and F-centres are formed
8. The empty space in CCP unit cell is
(a) 74% (b) 32%
(c) 47.6% (d) 26%
9. The packing efficiency of HCP unit is
(a) 26% (b) 74%
(c) 32% (d) 47.6%
4 Chemistry
10. The coordination number of metal crystallizing in hexagonal close packed structure is
(a) 4 (b) 6
(c) 12 (d) 8
11. A substance consisting of two elements, ‘P’ and ‘Q’, has atoms of ‘P’ occupying each cornerof the cube and atoms of ‘Q’ occupying the body centre. The composition of the substanceis
(a) PQ3 (b) P4Q3
(c) PQ (d) composition cannot be specified
12. A compound is formed by two elements ‘X’ and Y. Atoms of element Y (as anions) make ccparrangement and those of element X (as cations) occupy 50% of tetrahedral and octahedralvoids. The formula of compound is
(a) X3Y2 (b) X3Y3
(c) X2Y3 (d) XY2
13. On the basis of their magnetic properties, substances can be classified into various categorieslike paramagnetic, diamagnetic, ferromagnetic, antiferromagnetic and ferrimagnetic.Schematic alignment of magnetic moments in above-mentioned solids are :
(a) (b)
(c)
These representation respectively are of :(a) Ferromagnetic, ferrimagnetic, and antiferromagnetic(b) Ferrimagnetic; ferromagnetic, and antiferromagnetic(c) Ferromagnetic; antiferromagnetic, and ferrimagnetic(d) Ferrimagnetic, antiferromagnetic, and ferromagnetic
14. Which one of the following substances produce impurity defects when added to moltenNaCl?
(a) KCl (b) AgCl
(c) SrCl2 (d) AgBr
15. Germanium crystal doped with equal number of phosphorus and antimony atoms is
(a) an intrinsic semiconductor (b) p-type semiconductor
(c) an n-type semiconductor (d) a superconductor
16. If ‘a’ stands for the edge length of cubic systems sc, bcc and fcc, then ratio of radii of thespheres in these systems will be respectively,
(a) 1 : 3 : 2a a a (b)1 3 1: :2 4 2 2
a a a
(c) 1 1: 3 :2 2
a a a (d) 1 3 2: :2 2 2
a a a
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17. Which of the following statements is not correct?
(a) The number of Bravais lattices in which a crystal can be categorised is 14.
(b) The fraction of total volume occupied by the atoms in a primitive cell is 0.524.
(c) Molecular solids are generally volatile.
(d) Number of carbon atoms in an unit cell of diamond is ‘4’
18. Match box is an example of
(a) Tetragonal unit cell (b) Orthorhombic unit cell
(c) Monoclinic unit cell (d) Triclinic unit cell
19. CsBr crystallises in a body centered cubic lattice. The edge length of unit cell is 436.6 pm.Given that the atomic mass of Cs = 133 u and Br = 80 u, the Avogadro number being6.02 × 1023 mol–1, the density of CsBr is
(a) 8.25 g/cm3 (b) 4.25 g/cm3
(c) 42.5 g/cm3 (d) 0.425 g/cm3
20. The Ca2+ ions and F– ions are located in CaF2 crystal respectively at face centered cubiclattice points and in
(a) tetrahedral voids (b) half of tetrahedral voids
(c) octahedral voids (d) half of octahedral voids
21. Which is not the correct statement for ionic solids in which positive and negative ions areheld by strong electrostatic attractive forces?
(a) The radius ratio r+/r– increases as coordination number increases.
(b) As difference in the size of ions increases coordination number increases.
(c) When coordination number is eight, the r+/r– ratio lies between 0.225 to 0.414
(d) In ionic solids of type AX (ZnS, Wurzite) the coordination number of Zn2+ and S2–
respectively are 4 and 4.
22. The radii of Na+ and Cl– ions are 95 pm and 181 pm respectively. The edge length of NaClunit cell is
(a) 276 pm (b) 138 pm
(c) 552 pm (d) 415 pm
23. Potassium has a bcc structure with nearest neighbour distance of 4.52 Å. If atomic mass ofpotassium is 39 and then its density is
(a) 454 kg m–3 (b) 804 kg m–3
(c) 852 kg m–3 (d) 900 kg m–3
24. Which of the following fcc structures contains cations in alternate tetrahedral voids?
(a) ZnS (b) NaCl
(c) Na2O (d) CaF2
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25. Metallic lusture is explained by
(a) Diffusion of metal ions (b) Oscillation of positive ions
(c) Excitation of free electrons (d) Existence of bcc crystal lattice
26. Copper crystallises in fcc lattice with a unit length of 361 pm. What is the radius of copperatom in pm?
(a) 157 (b) 181
(c) 108 (d) 128
1. (c) 2. (a) 3. (c) 4. (c)
5. (b) 6. (a) 7. (d) 8. (d)
9. (b) 10. (c) 11. (a) 12. (a)
13. (c) 14. (c) 15. (c) 16. (b)
17. (d) 18. (b) 19. (b) 20. (b)
21. (c) 22. (c) 23. (d) 24. (b)
25. (c) 26. (d)
7 Chemistry
Molarity (M) is the number of moles of solute dissolved per litre of solution. Its unit is mol/L.
Number of moles of solute 1000Volume of solution in litre Volume of solution
B
B
WMM in ml
For liquids, where concentration is expressed in percentage and density of solution.
Percentage density 10Molar mass of solute
M
It is also defined as number of millimoles of solute dissolved in 1 mL of solution.
Molality (m) is the number of moles of solute dissolved per kg of solvent. Its unit is mol/kg
Number of moles of solute 1000
Mass of solvent in kilograms in gramsB
B A
WmM W
Normality (N) is the number of gram equivalents of solute dissolved per litre of solution. Itsunit is equivalents per litre.
Number of -equivalents 1000
Volume of solution in litres Equ. mass of solute Volume of solution inBWgN
mL
Mole FractionIt is the ratio of number of moles of one component to the total no. of moles of solution.
Number of moles of a component
Number of moles of solutionFor binary solutions A + B = 1
ppm (Parts Per Million)Mass of solute dissolved per million parts of the system.
6Number of parts by mass or volume of a component 10Total parts by mass or volume of the solution
ppm
Henry’s Law
The partial pressure of the gas in vapour phase (p) is directly proportional to the mole fractionof the gas () in solution and is expressed as :
8 Chemistry
p or p = KH ; where KH is Henry’s constant.
Raoult’s Law
For solution of volatile liquids, the partial vapour pressure of each component in solution isdirectly proportional to its mole fraction.
PA A
0A A AP P
0B B BP P
0 0Total A A B BP P P
0 0Total 1A B B BP P P
0 0 0Total A B A BP P P P X
Raoult’s Law for the solution of Non-Volatile Solute
Relative lowering of vapour pressure of a solution is equal to the mole fraction of solute whensolvent alone is volatile and the solute in non-volatile and non-electrolyte [when solution is dilute,i.e., nA + nB nA]
0soln
0A B B B
BA B B AA
P P W MM WP
··
IDEAL SOLUTIONS
The solution, where unlike interactions (A – B) are identical to like interactions (A – A) and(B – B) type, are known as ideal solutions.
Characteristics : (1) Vapour pressure of such solutions is the same as given by Raoult’s law.
(2) Hmix = 0 (3) Vmix = 0
Chorobenzene and bromobenzene, benzene and toluene, n-hexane and n-heptane form nearlyideal solutions.
NON-IDEAL SOLUTIONS
Non-ideal solutions have Hmix 0, Vmix 0 and do not obey Raoult’s law.
Non-Ideal Solution Showing + ve and – ve Deviations from Raoult’s Law
Showing Positive Deviation Showing Negative Deviation
Acetone + Carbon disulphide Chloroform + BenzeneAcetone + Ethyl alcohol Chloroform + AcetoneAcetone + Benzene Chloroform + Diethyl etherMethyl alcohol + WaterEthyl alcohol + eater Acetone + Aniline
P = P + PTotal
AB
PBP
A
Vap
our
pres
sure
P°A
A = 1B = 0
A = 0B = 1
Mole Fraction
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Carbon tetrachloride + Chloroform HCl + WaterCarbon tetrachloride + Benzene HNO3 + WaterCarbon tetrachloride + Toluene Acetic and + pyridine, phenol and aniline
Characteristics of Solution Showing –ve Deviation from Faoult’s Law
(a) Hmix < 0
(b) Vmix < 0
(c) Vapour pressure of solution is less than what isgiven by Raoult’s law.
0 0soln. A A B BP P p
0A A AP P
0B B AP P
Characteristics of Solution Showing +ve Deviation
(a) Hmix > 0
(b) Vmix > 0
(c) Vapour pressure of solution is more than what isgiven by Raoult’s law.
0 0soln A A B BP P p
0A A AP P
0B B BP P
Note : Curved lines show non-ideal behaviourand dotted line, ideal behaviour.
AZEOTROPES
(i) Azeotropic mixtures with minimum point i.e. whose boiling points is less than eitherof the two pure components. This is formed by that composition of non-ideal solution showingpositive deviation for which the vapour pressure is maximum. (95% ethanol + 5% H2O).
(ii) Azeotropic mixtures with maximum boiling point i.e. whose boiling is more thaneither of the two pure components. This is formed by that composition of a non-deal solutionshowing negative for which the vapour pressure is minimum. (68% HNO3 + 32% H2O), (20%HCl + 80% H2O).
COLLIGATIVE PROPERTIES
The properties of a solution which depends on number of moles of solute but are independentof its nature, are known as colligative properties. These are :
A = 1B = 0
A = 0B = 1
P°B
P°A
Vap
our
pres
sure
A = 1B = 0
A = 0B = 1
P°A
P°B
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1. Relative lowering of vapour pressure.
2. Elevation of Boiling point.
3. Depression of Freezing point.
4. Osmotic pressure.
Osmotic Pressure
Pressure applied on solution to stop osmosis is known as osmotic pressure
orB Bn T n RT CRTV V
Osmotic pressure of a solution is directly proportional to the number of moles of solute dissolvedper litre of solution at a given temperature.
Solutions having equal molar concentration and equal osmotic pressure at a given temperatureare called isotonic solutions, e.g., A 0.90% (mass/volume) solution of NaCl is isotonic with humanRBC. A solution of NaCl with concentration less than 0.90% (mass/volume) is hypotonic and RBCwill swells up and even burst in solution. A NaCl solution with concentration > 0.90% (mass/volume) is called hypertonic, RBC will shrink in such solution.
Elevation of Boiling Point
Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to thatof 1 atm or 1.013 bar. The elevation of b.p. is related to the molatily (m) of the solution as below:
1000Bb b b
B A
WT K m KM W
0b b bT T T Elevation of boiling point
Tb
1 atm = 1.013 bar
T°b TbTemperature
Kb = Molal boiling point elevation constant or Ebullioscopic constant.
Depression of Freezing Point
Freezing point of liquid is the temperature at which liquid and solid phases coexist and havethe same vapour pressure. Freezing point of the liquid solvent in depressed when a non-volatilesolute is added to it.
1000Bf f f
B A
WT K m KM W
of f fT T T Depression of freezing point
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Kf = Molal freezing point depression constant or cryoscopic constant.
Tf0Tf Temperature
Tf
Abnormal Molecular Mass : When the molecular mass of a substance as determined byusing colligative properties, does not come out to be the same as expected theoretically, it is saidto show abnormal molecular mass.
Abnormal molecular mass is obtained when the substance in the solution undergoesdissociation (e.g. NaCl in water) or association (e.g. organic acids or phenols in benzene).Dissociation results in the increases in the number of particles and hence increase in the value ofcolligative property and decrease in the molecular mass. Association results in the reverse.
Van’t Hoff factor (i) is given by
No. of particles after dissociation or association of soluteNumber of solute initially dissolved
iparticles
Experimental value of the colligative property
Calculated value of the colligative propertyi
1As molecular massColligative property
normal
observed
Calculated or normal molecular masshenceObserved molecular mass
MiM
Modified formulas for substance undergoing dissociation or association in the solution are
(i) Tb = i Kb m (ii) Tf = i Kf m
(iii) ni RTV
(iv)0
soln.0
AB
A
P p iP
When solute dissociates to give ‘n’ ions, the degree of dissociation () is related to Van’t Hofffactor as given below
11
in
If ‘n’ molecules of a solute undergo association in solution then degree of association () isgiven by
11
nin
12 Chemistry
1. An azeotropic mixture of two liquids has boiling point higher than either of them when it.
(a) Shows positive deviation from Raoult’s law.(b) Show no deviation from Raoult’s law.(c) Show negative deviation from Raoult’s law.(d) is saturated.
2. Consider 0.02 M aqueous solutions of
1. NaCl, 2. BaCl2 3. Urea
The relative lowering of vapour pressures in these solutions will be such that
(a) 2 < 1 < 3 (b) 2 > 3 > 1
(c) 1 > 2 > 3 (d) 3 < 1 < 2
3. Phenol associates in benzene as 6 5 6 5 212
C H OH C H OH If ‘x’ is the degree of associationof phenol, then the total number of moles of particles present at equilibrium is
(a) 1 – x (b) 1 + x
(c) 12x
(d) 12x
4. On mixing two pure liquids to form an ideal solution, the statement which is not correct
(a) no change in volume (b) no change in Gibbs energy
(c) no evolution or absorption of heat (d) entropy changes.
5. The dissolution of NH4Cl in water is endothermic even though dissolves in water spon-taneously. Which one of the following statement best explains this behaviour
(a) The bonds in solid are weak.
(b) The entropy driving force causes dissolution.
(c) Endothermic processes are energetically favourable.
(d) The dissolving process is unrelated to energy.
6. Which of following 0.1 M aqueous solutions will have the lowest freezing point
(a) K2SO4 (b) NaCl
(c) NH2CONH2 (d) C6H12O6
7. 0.2 molal acid, HX is 20% ionized in solution (Kb = 0.52 K/m) The boiling point of solutionis (standard boiling of water is 373K)
(a) 373.12 K (b) 373.10 K
(c) 373.24 K (d) 373.30 K
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8. At 80°C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is1000 mmHg. If the solution of ‘A’ and ‘B’ boils at 80°C and 1 atm pressure, the amount ofliquid ‘A’ in the mixture is (1 atm = 760 mm Hg)
(a) 48 percent mol (b) 50 percent mol
(c) 52 percent mol (d) 34 percent mol
9. A solution ‘X’ is prepared by mixing ethanol and water. The mole fraction of ethanol in themixture is 0.9, then more water is added to the solution ‘X’ such that the mole fraction ofwater in solution becomes 0.9, the boiling point of this solution is
(a) 380.4 K (b) 376.2 K
(c) 375.5 K (d) 354.7 K
10. During depression of freezing point in a solution, the following are in equilibrium
(a) Liquid solution, solid solvent (b) Liquid solvent, solid solute
(c) Liquid solute, solid solute (d) Liquid solute, solid solvent
11. 2g of C6H5 COOH dissolved in 25g benzene shows a depression in freezing point equal to1.62K. (Kf = 4.9 K kg mol–1) The value of van’t Hoff factor is :
(a) 0.5 (b) 1
(c) 2 (d) 1.5
12. Pick out the incorrect statement :
(a) The vapour pressure of solution containing non-volatile and non-dissociative soluteis less than that of pure solvent.
(b) Reverse osmosis occurs if a pressure smaller than the osmotic pressure of solutionis applied on solution side.
(c) Only solvent molecules solidify at the freezing point.
(d) Molar mass of NaCl determined by any of the colligative properties is found to beless than that of normal (expected) value.
13. Which one of the following colligative properties can provide molar mass of proteins withgreatest precision?
(a) Relative lowering of vapour pressure.
(b) Osmotic pressure
(c) Elevation of boiling point.
(d) Depression of freezing point.
14. Which of the following is not true for an ideal solution?
(a) Raoult’s law is obeyed in the entire concentration range of a binary solution.
(b) Hmix = 0
(c) Vmix = 0
(d) Smix = 0
14 Chemistry
15. If various terms in the below given expressions have usual meanings, the Van’t Hoff factor(i) cannot be calculated by which one of the following expressions :
(a) V i nRT (b) Tf = i kf . m
(c) Tb = i kb . m (c)
0solvent solution
0solvent
P P niN nP
16. A 5% solution of cane sugar (Mol. wt. = 342) is isotonic with 1% solution of ‘X’ under similarcondition. The mol. wt. of ‘X’ is :
(a) 136.2 u (b) 68.4 u
(c) 34.2 u (d) 171.2 u
17. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of solution containing1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is furtheradded to this solution, vapour pressure of solution increases by 10 mm Hg. Vapour pressure(in mm Hg) of X and Y in their pure states will be respectively :
(a) 500 and 600 (b) 200 and 300
(c) 300 and 400 (d) 400 and 600
18. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm.The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10moles of water at 298 K and 5 atm. pressure is
(a) 4 × 10–4 (b) 4 × 10–5
(c) 5.0 × 10–4 (d) 4.0 × 10–6
19. What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with30.0 mL of 0.10 M Ba(OH)2
(a) 0.40 M (b) 0.0050 M
(c) 0.12 M (d) 0.10 M
20. A 0.04 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at the temperature298 K. The apparent degree of dissociation of Na2SO4 is
(a) 25% (b) 50%
(c) 75% (d) 85%
21. To neutralize completely 20 ml of 0.1 M aq solution of phosphorous acid, H3PO3, the volumeof 0.1 M aqueous KOH solution required is
(a) 10 ml (b) 20 ml
(c) 40 ml (d) 60 ml
22. If two substances A and B have 0 0: 1 : 2A BP P and have mole fraction in solution 1 : 2,then mole fraction of A in vapours is
(a) 0.33 (b) 0.25
(c) 0.52 (d) 0.2
15 Chemistry
23. During osmosis, flow of water through a semipermeable membrane is
(a) from both sides of semipermeable membrane with unequal flow rates,
(b) from solution having lower concentration only,
(c) from solution having higher concentration only,
(d) from both sides of semipermeable membrane with equal flow rate.
24. A solution of acetone in ethanol
(a) behaves like a near ideal solution,
(b) obeys Raoult’s law,
(c) shows a negative deviation from Raoult’s law,
(d) shows a positive deviation from Raoult’s law.
25. A 0.002 m aqueous solution of an ionic compound Co(NH3)5 (NO2) Cl freezes at – 0.00732°C.Number of moles of ions which 1 mol of ionic compound produces on being dissolved inwater will be (Kf = – 1.86°K/m)
(a) 3 (b) 4
(c) 1 (d) 2
26. A binary solution is prepared by mixing n-heptane and ethanol. Which one of the followingstatements are correct regarding the behaviour of solution?
(a) The solution formed is an ideal solution.
(b) The solution is non-ideal showing +ve deviation from Raoult’s law.
(c) The solution is non-ideal showing –ve deviation from Raoult’s law.
(d) None of these.
1. (c) 2. (c) 3. (d) 4 (b)
5. (b) 6. (a) 7. (a) 8. (b)
9. (b) 10. (a) 11. (a) 12. (b)
13. (b) 14. (d) 15. (a) 16. (b)
17. (d) 18. (a) 19. (b) 20. (c)
21. (c) 22. (d) 23. (a) 24. (d)
25. (d) 26. (b)
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Electrochemical Cells
Galvanic cell is an electrochemical cell that converts the chemical energy of the spontaneousredox reaction into electrical energy. The half-cell in which oxidation takes place is called anodeand it has negative potential with respect to the solution. The other half-cell in which reductiontakes place is called cathode and it has positive potential w.r.t. the solution. When the two half-cells are joined, the electrons start flowing from anode having negative polarity to cathode withpositive polarity.
The potential difference between the two electrodes is called cell potential. To maintain theflow of current in the external circuit, the two half-cells are joined by a salt bridge. The salt bridgemaintains the electroneutrality in both the half-cells.
Ecell = ECathode – EAnode
If the concentrations of the oxidised and reduced forms of the species are unity at 298 K, thencell potential is called standard cell potential (E°cell)
cell cathode anodeE E E
The potential of individual half-cell is determined with the help of a reference electrode, e.g.,standard hydrogen electrode (SHE) represented by Pt(s)/H2(g, 1bar)/H+ (aq, 1M), which is assigneda zero potential at all temperatures corresponding to the reaction : 2H+ (aq) + 2e– H2(g)
Significance of Sign with Standard Electrode Potential
2 20.76 and 0.34Zn Zn cu cuE V E V
–ve sign with
2Zn ZnE means that reduction of Zn2+ to Zn is non-spontaneous but oxidation
of Zn to Zn2+ occur spontaneously.
+ve sign with
2cu cuE means that reduction of Cu2+ is spontaneous but oxidation of Cu to Cu2+
is non-spontaneous.
Nernst Equation for an electrode Reaction : Mn+ + ne– M(s)
Reaction quotient
1
c nQ
MThe Nernst equation for this electrode reaction is :
2.303 logn n cM M M MRTE E Q
nF
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0.059 logn cM ME V Qn
Similarly Nernst equation for a cell reaction :
neaA bB cC dD
c d
c a bC DQA B
0.059 log
c d
cell cell a bC DE E V
n A BAt equilibrium, Ecell = 0 and Qc = Kc
0 00.059 0.0590 log logcell c cell cE V K E Kn n
Concentration cells have the same electrodes but the concentrations of the solutions of thesame electrolyte is different in the two half-cells. For example,
Cu(s)/Cu2+(C1)||Cu2+(C2)/Cu(s)
Cell Reactions : 2 2 2 22 1orRHC LHCCu Cu Cu C Cu C
2
20.059 log
2LHC
cellRHC
CuE VCu
Here RHC means right half-cell and LHC, left half-cell.
1
2
0.059 log2cell
CE VC
Cell potential and Gibbs energy of reaction (rG) are related as given below :
rG = – n Ecell F
Here rG is an extensive thermodynamic property and Ecell is an intensive parameter. ‘n’ isthe number of electrons involved in redox reaction. If the concentration of all the reacting speciesare taken as unity, then we have .r cellG nE F
1. Primary Batteries
(a) Dry Cell
Anode : Zn(s) Zn2+ + 2e–
Cathode : MnO2 + NH4+ + e– MnO (OH) + NH3
Mn is reduced from +4 oxidation state to +3 state. Ammonia produced forms a stable complex,[Zn (NH3)4]2+. This cell is not rechargeable because the products formed during discharging cannotbe converted into the reactants. The cell potential of nearly + 1.5 V decreases with time since theconc. of ions involved in the cell reaction changes during discharging.
(b) Mercury Cell
Anode : Zn(Hg) + 2OH– ZnO(s) + H2O + 2e–
18 Chemistry
Cathode : HgO(s) + H2O + 2e– Hg(l) + 2OH–
Overall reaction : Zn(Hg) + HgO(s) ZnO(s) + Hg(l)
Cell potential of nearly + 1.35 V remains constant during its life as the overall reaction doesnot involve any ion whose concentration can change during its life.
2. Lead Storage Battery
Lead Storage Battery is a secondary battery and is rechargeable. It consists of a lead anodeand a grid of lead packed with PbO2 as cathode. A 38% solution of H2SO4 is used as an electrolyte.During discharge, following reaction occurs at electrodes.
Anode : 2 –4 4 2Pb s SO aq PbSO s e
22 4 4 2
2 2 4 4 2
: 4 2 2: 2 2 2
Cathode PbO SO aq H aq e PbSO H O lOverallreaction Pb s PbO s H SO PbSO s H O
On charging the battery, the reaction is reversed.
Fuel cell is a galvanic cell which converts the energy of combustion of fuels like H2, CH4,C2H6, C3H8, CH3OH etc. directly into electrical energy.
In a fuel cell using H2 as a fuel, the following reactions are occurring at electrodes.
Cathode : O2 + 2H2O + 4e– 4 OH–
Anode :
2 2
2 2 2
2 4 4 4 ,2 2 r
H OH H O e G x kJ molH O H O
rG = – n Ecell F. Here n = 4
In a fuel cell using CH4 as a fuel, the reactions are :
Cathode : O2 + 4H+ + 4e– 2H2O] × 2
Anode :
4 2 2
4 2 2 2
2 8 8 , here 82 2 , r
CH H O CO H e nCH O CO H O G y kJ molrG = – n EcellF
Corrosion : Corrosion is an electrochemical phenomenon.
Anode : 222 2 4 ; 0.44Fe FeFe s Fe e E V
Cathode : 2 22 24 4 2 ; 1.23H O H OO g H e H o l E V
Overall reaction : 22 22 4 2 2 1.67cellFe s O g H aq Fe H O E V
H+ ions are furnished by H2CO3 formed by the dissolution of CO2 in water. More the concn.of H+, faster is the reaction at cathode. Corrosion occurs at faster rate in saline water because thesalts present in water perform the functions of salt bridge.
Atmospheric O2, further oxidises Fe2+ to Fe3+ to form hydrated Fe2O3 with further productionof H+ ions which facilitate corrosion.
We can prevent the corrosion by preventing the surface of the metal to come in contact withatmosphere by covering the metallic surface with paint or by the layer of other metals (Sn, Zn, etc.).
19 Chemistry
Conductivity : Conductivity (k) is related to resistance as given below :
Cell constantResistance
l AkR
Unit : ohm–1 cm–1 or S cm–1
Molar Conductivity (m) is given by kmc
where C = molar concentration and unit ofm= S cm2 mol–1.
Conductivity (k) decreases but molar conductivity(m) increases with the decrease in concentrations. Itslowly increases with the decrease in concentration forstrong electrolytes while the increase is very steep forweak electrolytes in very dilute solutions.
Molar conductivity at infinite dilution or zero concen-tration is called limiting molar conductivity denoted by°m. Limiting molar conductivity (°m) of strong electro-lyte is obtained by the extrapolation of curve at 0concbut °m for weak electrolyte is determined by the use ofKohlrausch law of independent migration of ions whichstates that limiting molar conductivity for an electrolyteis the sum of the contribution of molar conductivity of theions in which it dissociates.
0 0 0m v v where v+
and v– are number of cations and anions furnished by anelectrolyte.
(For Faraday’s laws of electrolysis please refer the NCERT Text Book for class XII Part I page 83).
Product of Electrolysis
Product of electrolysis depend on the different oxidising and reducing species present in theelectrolytic cell and their electrode potentials. Moreover some electrochemical process althoughfeasible but do not seem to take place because extra potential (overvoltage) has to be applied.
Electrolysis of Aqueous NaCl
Al cathode there is the competition between the following reduction reactions
Na+ + e– Na 2.71Na NaE V
2 21 22
H O e H OH 2 2
0.83H O HE V
The reduction reaction with higher E is preferred at cathode and therefore, reduction of wateris preferred at cathode.
Cathode : 2 21 22
H O e H OH
At the anode there is the competition between the following oxidation reactions :
21 ;2
Cl aq Cl e E = 1.36 V
Molar conductivity versus c for ocetic acid (weak electrolyte) and
potassium chloride (strong electrolyte) in aqueous solutions.
1/2
20 Chemistry
2 21 2 2 ;2
H O l O H e E = 1.23 V
At anode the reaction with lower E is preferred and therefore water should get oxidised inpreference to Cl–. However, on account of high over-potential of O2, the oxidation of Cl– ion ispreferred at anode.
Anode : 2
12aqCl Cl e
Note : During the electrolysis of aqueous solution of nitrates and sulphates, the oxidation of wateroccurs at anode in preference to the oxidation of NO3
– or SO4
2–. For example, during the electrolysisof aq. Na2SO4, Na+ is not reduced at cathode and SO4
2– is not oxidised at anode. In this case wateris oxidised at anode and water is also reduced at cathode.
21 Chemistry
1. Kohlrausch law states that at
(a) Infinite dilution, each ion makes definite contribution to the limiting molarconductivity of an electrolyte whatever be the nature of other ion of the electrolyte.
(b) Finite dilution, each ion makes definite contribution to the limiting molar conductivityof an electrolyte whatever be the nature of other ion of the electrolyte.
(c) Infinite dilution, each ion makes definite contribution to the limiting molarconductivity of an electrolyte depending upon the nature of other ion of the electrolyte.
(d) Infinite dilution, each ion makes definite contribution to electrolytic conductance ofan electrolyte whatever be the nature of the other ion of the electrolyte.
2. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10mA current. Thetime required to liberate 0.01 mol. of H2 gas at cathode is (1F = 96500 C/mol)
(a) 9.6 × 104 s (b) 19.3 × 104 s
(c) 28.95 × 104 s (d) 38.6 × 104 s
3. A dilute solution of NaBr is electrolysed using platinum electrodes. The products at anodeand cathode are :
(a) O2, H2 (b) Br2, H2
(c) Br2, Na (d) O2, Na
4. The standard emf of cell : Zn(s)/Zn2+ (aq, 0.01 M) || Fe2+ (aq, 0.001 M)/Fe(s) at 298K is0.3200 V, then the value of equilibrium constant for the cell reaction is :
(a)0.32
0.0295e (b)0.32
0.0591e
(c)0.32
0.029510 (d)0.32
0.059110
5. The half-cell reaction for corrosion :
2 212 2 1.232
H O e H O E V
2 2 0.44Fe e Fe s E V
Find the AG (in kJ) for the overall reaction :
(a) –76 (b) –161
(c) –152 (d) –322
6. Standard electrode potential data are useful for understanding the suitability of oxidant ina redox titration. Some half-cell reactions and their standard potentials are given below :
MnO4– (aq) + 8H+ (aq) + 5e– Mn2+ (aq) + 2H2O; E = 1.51V
Cr2O72– (aq) + 14 H+ (aq) + 6e– Cr3+ (aq) + 7H2O; E = 1.38V
22 Chemistry
Fe3+ (aq) + e– Fe2+ (aq) E = 0.77V
Cl2 (g) + 2e– 2Cl– (aq) E = 1.40V
Identify incorrect statement regarding the quantitative estimation of Fe(NO3)2 (aq)
(a) MnO4– can be used in aqueous HCl
(b) Cr2O72– can be used in aqueous HCl
(c) MnO4– can be used in aqueous H2SO4
(d) Cr2O72– can be used in aqueous H2SO4
7. The emf of concentration cell :
Zn (s)/Zn2+ (0.0 1M) || Zn2+ (0.10)/Zn is given by
(a) 0 1.102.303 log0.01cell cell
RTE EF
(b) 0 0.012.303 log2 0.10cell cellRTE E
F
(c) 0 0.012.303 log0.10cell cell
RTE EF
(d) 0.102.303 log
2 0.01cellRTE
F8. Standard Gibbs energy of formation fG in kJ mol–1 at 298 K for water is –237.2, The value
of E cell for the hydrogen-oxygen fuel cell is :
(a) 1.458 V (b) 1.0229 V
(c) 1.229 V (d) 2.229 V
9. The standard reduction potentials of the three metallic cations X, Y and Z are 0.52, –3.03and –1.18V respectively. The order of reducing powers of the corresponding metal is
(a) y > z > x (b) x > y > z
(c) z > y > x (d) z > x > y
10. The standard reduction potential for Fe2+/Fe and Sn2+/Sn are – 0.44 and – 0.14V respectively.For the cell reaction Fe2+ + Sn Fe + Sn2+, the
cellE is :
(a) + 0.30 V (b) – 0.58 V
(c) + 0.58 V (d) – 0.30 V
11. The correct order of molar conductivity at infinite dilution of LiCl, NaCl, and KCl is
(a) LiCl > NaCl > KCl (b) KCl > NaCl > LiCl
(c) NaCl > KCl > LiCl (d) LiCl > KCl > NaCl
12. Standard electrode potentials for Fe2+/Fe and Fe3+/Fe2+ are – 0.44 and + 0.77V respectively.Fe3+, Fe2+ and Fe blocks are kept together, then
(a) Fe3+ increases (b) Fe3+ decreases
(c) Fe2+/Fe3+ remains unchanged (d) Fe2+ decreases
23 Chemistry
13. On the basis of information available from reaction : 2 2 34 2 ,3 3
Al O Al O rG = – 827
kJ/mol of O2, the minimum emf required to carry out the electrolysis of Al2O3 is
(a) 2.14 V (b) 4.28 V
(c) 6.42 V (d) 8.56 V
14. Given
(i) Cu2+ + 2e– Cu; E = 0.337 V
(ii) Cu2+ + e– Cu+; E = 0.153 V
Electrode potential, E for the reaction :
Cu+ + e– Cu; will be
(a) 0.90V (b) 0.30 V
(c) 0.38 V (d) 0.52 V
15. Given :
E Fe3+/Fe = – 0.036, E Fe2+/Fe = – 0.439 V
The value of standard electrode potential for the change, Fe3+ + e– Fe2+ will be :
(a) – 0.072 V (b) 0.385 V
(c) 0.770V (d) – 0.27 V
16. In which of the following pairs the constants/quantities are not mathematically related toeach other?
(a) Gibb’s free energy and standard cell potential.
(b) Equilibrium constant and standard cell potential.
(c) Rate constant and activation energy.
(d) Rate constant and standard cell potential.
17. The Cell : Zn/Zn2+ (1M) || Cu2+ (1M)/Cu 1.10cellE V was allowed to be completely
discharged at 298 K. the relative concentration of Zn2+ to Cu2+
2
2ZnCu
is
(a) anti log (24.08) (b) 37.3
(c) 1037.3 (d) 9.65 × 104
18. The molar conductivities 0NaOAC
and 0HCl
at infinite dilution in water at 25°C are 91.0 and426.2S cm2/mol respectively. To calculate 0
HOAc, the additional value required is
(a) 2
0H O (b) 0
KCl
(c) 0NaOH (d) 0
NaCl
24 Chemistry
19. What will be the e.m.f. of the given cell?
2 1 2 2, bar ,1 ,s sPt H g p H aq M H g P bar Pt
(a) 1
2ln PRT
F P(b) 1
2ln
2PRT
F P
(c) 2
1ln PRT
F P(d) None of these
20. The charge required for the reduction of 1 mol of MnO4– to MnO2 is
(a) 1F (b) 3F
(c) 5F (d) 6F
1. (a) 2. (b) 3. (b) 4 (c)
5. (d) 6. (a) 7. (d) 8. (c)
9. (a) 10. (d) 11. (b) 12. (b)
13. (a) 14. (d) 15. (c) 16. (d)
17. (c) 18. (d) 19. (b) 20. (b)
25 Chemistry
Consider the Reaction : 2 N2O5(g) 4NO2(g) + O2 (g).
When a reaction proceeds, the concentration of a reactant decreases while that of a productincreases with time. The rate of change of concentration of a reactant is called rate of consumptionor disappearance of the reactant.
Average rate of consumption of 2 52 5
N ON O
t
since [N2O5] is a negative quantity, it is multiplied by – 1 to make the rate a positivequantity.
Similarly, rates of formation of NO2 and O2 are given by
Average rate of formation of
22
NONO
t
Average rate of formation of
22
OO
tThese rates can be equated if we divide the rate of change of concentration of a reactant or
a product by its stoichiometric coefficient appearing in balanced chemical equation, that is,
Average rate (rav)/or rate
2 5 2 21 12 4
N O NO Ot t t
Rate of a reaction at time ‘t’ is known as the instantaneous rate of reaction.
As t 0, rinst or rate
2 5 2 21 12 4
d N O d NO d Odt dt dt
For the Reaction
5 Br– (aq) + BrO3– (aq) + 6H+ (aq) 3Br2 (aq) + 3H2O(l)
We do not use the concentration term for water for expressing the rate of reaction sincechange in the concentration of water is negligibly small because the reaction is occurring in aqueousmedium. Hence, the rate of reaction is given as below :
3 21 1 1Rate5 6 3
BrO BrBr Ht t t t
Order and Molecularity of Reaction
Rate law : Consider a general reaction : a A + b B c C + d D
26 Chemistry
The rate law for this reaction is :
Rate = k [A]x [B]y where x and y may or may not be equal to the stoichiometric coefficients aand b of the reactants.
Rate law for any reaction cannot be predicted by merely looking at the balanced chemicalequation. It is determined experimentally.
x = order of reaction w.r.t. the reactant A
y = order of reaction w.r.t. the reactant B
and x + y = overall order of the reaction.
Units of rate constant depend upon the order or reaction.
Rate = k [A]x [B]y
x y
RatekA B
If x + y = n = order of reaction, then we have
1 11 .unit ofn n
nconcentration Concn Mk
time time sconcentrationFor zero order reaction (n = 0), unit of k is mol L–1 s–1 or Ms–1
For first order reaction (n = 1), unit of k is s–1
For second order reaction (n = 2), unit of k is L mol–1 s–1 or M–1 s–1.
Molecularity is the number of reacting species taking part in an elementary reaction, i.e., thenumber of species that collide simultaneously in order to bring about a chemical reaction.
For a bimolecular or trimolecular elementary reactions, the order of reaction is the same asits molecularity and order w.r.t. each reactant is equal to its stoichiometric coefficient.
For unimolecular elementary reactions, the order of reaction is one at high concentration orpressure. At low pressure the reaction becomes second order but then the reaction is no longer anelementary reaction. Order of reaction can be zero, 1, 2, 3 or even a fraction but molecularity cannever be zero or a non-integer. Order is applicable to elementary as well as complex reactions. Forcomplex reaction, molecularity has no meaning. For complex reaction, order of reaction is given byslowest step and generally molecularity of the slowest step is the same as the order of the overallreaction.
Zero Order Reaction : The rate of zero order reaction is independent of the concentrationchange.
0d RRate k R kdt
00 where R Initial concentrationR Rk
t R Concentration at time t
Half-life of a reaction is the time in which one-half of the initial concentration of the reactantis consumed.
0
01 2 1 2or2Rt t R
k
27 Chemistry
The decomposition of gaseous ammonia on hot platinum surface is a zero order reaction athigh pressure.
Rate = k [NH3]0 = k
First Order Reaction : 2N2O5 4NO2 is an example of first order reaction. For the reaction:
R P
d R k Rdt
0 1
2 12
1 1ln or lnR Rt t tk R k R
These relationships can be expressed in exponential form as given below
[R] = [R]0e–kt
Equation (ii) can be rewritten as :
0 1
2 12
2.303 2.303log or logR Rt t tk R k R
Expression for half-life for first order reaction :
1 20.693t
kHalf-life period of first order reaction is independent of initial concentration of the reactant.
log
[]
[]
R R0
28 Chemistry
Rseudo First Order Reaction is a reaction which is first order w.r.t. each of the tworeactants but becomes first order reaction under certain experimental conditions, i.e., if one of thereactants is taken in excess.
Note : Half-life period of nth order reaction in inversely proportional to initial concentrationof reactant raised to power (n – 1)
1 2 1
0
1nt
R
Dependence of rate of reaction on temperature is described by Arrhenius equation :
k = Ae–Ea/RT where
Ea = Activation energy and is given by energy difference between the activated complex andthe reactant molecules.
A = Arrhenius factor or pre-exponential factor or frequency factor that has the units of rateconstant.
Natural logarithm of both sides of Arrehenius equation gives :
ln lnaEk ART
The plot of ln k and 1T gives a straight line of slope .aE
RIf k1 and k2 are rate constants at temperature T1K and T2K then we have
2 2 1
1 2 1 2 1
1 1log2.303 2.303
a aE Ek T Tk R T T R T T
(Plot between ln and 1/T)k (Solid line denotes the reaction path withoutcatalyst and dotted line, with catalyst)
Effect of Catalyst : A catalyst provides an alternative reaction path or reaction mechanismby reducing the activation energy between reactants and products by lowering the potential energybarrier. A catalyst can catalyse those reactions for which rG < 0. It does not affect the equilibriumstate, increases the rates of forward as well as reverse reactions in the same proportion withoutaffecting the rH and the equilibrium is attained sooner.
29 Chemistry
Collision Theory of Chemical Reactions : The molecules are assumed to be hard spheres andthe reaction is postulated to occur when molecules collides with each other. The rate of reactionfor the following bimolecular elementry reaction : A + B Products, is given by
Rate = ZAB e–Ea/RT where
ZAB = Collision frequency of the reactants, A and B.
e–Ea/R = Fraction of molecules with energy equal to or greater than Ea.
The collisions in which molecules collide with sufficient kinetic energy (or threshold energy)and proper orientation are called effective collisions.
To account for effective collisions, another factor P, called the probability factor or steric factoris introduced, i.e.,
Rate = pZABe–Ea/RT
30 Chemistry
1. The specific rate constant of a first order reaction depends on the
(a) initial concentration of the reactants (s)
(b) time of the reaction
(c) temperature
(d) extent of the reaction
2. The rate of the reaction : N2 (g) + 3H2 (g) 2NH3 (g) can be expressed in terms of timederivative of concentration of N2, H2, or NH3. Identify the correct relationship amongst therate expressions.
(a) 32 21 13 2
d NHd N d HRate
dt dt dt
(b) 32 23 2d NHd N d H
Ratedt dt dt
(c)
32 21 13 2
d NHd N d HRate
dt dt dt
(d) 32 21 23
d NHd N d HRate
dt dt dt
3. In a first order reaction, the concentration of the reactant decreases from 400 mol dm–3 to25 mol dm–3 in 2 × 104 s. The rate constant in s–1 for this reaction is :
(a) 4 × 10–4 (b) 1.386 × 10–4
(c) 4 × 10–4 (d) 3.45 × 10–5
4. The rate constant for the reaction : 2N2O5 4NO2 + O2 is 3.0 × 10–5 s–1. If the rate ofreaction is 2.40 × 10–5 mol L–1 s–1, then [N2O5] in mol L–1 is :
(a) 1.4 (b) 1.2
(c) 0.04 (d) 0.8
5. Which one of the following statement for the order of reaction is not correct?
(a) order can be determined experimentally.
(b) Order of reaction is equal to the sum of the powers of the concentration terms in thedifferential rate law.
(c) Powers of the concentration terms may or may not be equal to the stoichiometriccoefficients of the reactants.
(d) Order cannot be fractional.
6. The reaction : R P follow first order kinetics. In 40 minutes the concentration of P changesfrom 0.1 to 0.025 M. The rate of reaction, when the concentration of P is 0.01 M is :
(a) 3.47 × 10–4 M min–1 (b) 3.47 × 10–5 M min–1
31 Chemistry
(c) 1.73 × 10–4 M min–1 (d) 1.73 × 10–5 M min–1
7. The rate constants k1, and k2 for two different reactions are 1016 2000eT
and 15 100010 eT
respectively. The temperature at which k1 = k2 is :
(a) 20002.303
K (b)10002.303
K
(c) 2000 K (d) 1000 K
8. For the reaction : R products, it is found that the rate of reaction increases by a factorof 6.25, when the concentration of R is increased by a factor of 2.5. The order of reactionwith respect to R is
(a) 2.5 (b) 2
(c) 1 (d) 0.5
9. The following data pertains to the reaction between A and B
S.No. [A]/mol L–1 [B]/mol L–1 Rate/mol L–1 s–1
(i) 1 × 10–2 2 × 10–2 2 × 10–4
(ii) 2 × 10–2 2 × 10–2 16 × 10–4
(iii) 2 × 10–2 4 × 10–2 15 × 10–4
The rate law for the reaction is :
(a) Rate = k [A]2 [B] (b) Rate = k [A] [B]2
(c) Rate = k [A]3 [B]0 (d) Rate = k [A]1/2 [B]2.5
10. The activation energy of a reaction is zero. The rate constant (k) of reaction of 280K is1.6 × 10–6 s–1. The value of k for this at 300 K is :
(a) zero (b) 3.2 × 10–5 s–1
(c) 1.6 × 10–5 s–1 (d) 1.6 × 10–6 s–1
11. If a reaction A + B C is exothermic to the extent of 30 kJ/mol and forward reaction hasan activation energy of 70 kJ/mol, the activation energy for the reverse reaction is :
(a) 30 kJ/mol (b) 40 kJ/mol
(c) 70 kJ/mol (d) 100 kJ/mol
12. Given the following diagram for the reaction A + B C + D
32 Chemistry
The enthalpy change and activation energy for the reverse reaction : C + D A + Brespectively are
(a) x, y (b) x, x + y
(c) – y, x + y (d) y, y + z
13. A reaction proceeds by a two step mechanism
1
12 2k
kA A (fast reaction)
2 productskA B (slow reaction)
(a) Rate = k [A2] [B] (b) Rate = k [A2]2 [B]
(c) Rate = k [A2]1/2 [B] (d) Rate = k [A2]1/2
14. If k1 = Rate constant at temperature T1 and k2 = Rate constant at temperature T2 for a firstorder reaction then which of the following velations is correct? (Ea : activation energy)
(a)
1 2 1
2 1 2
2.303log k T TEak RT T T
(b)
2 2 1
1 1 2log
2.303k T TEak RT T T
(c)
2 1 2
1 2 1log
2.303k T TEak RT T T
(d)
1 1 2
2 2 1log
2.303k T TEak RT T T
15. For a reaction between A and B, the initial rate of reaction is measured for various initialconcentration of A and B. The data provided are
S.No. [A] [B] Initial reaction rate
(i) 0.20 M 0.30 M 5.07 × 10–5
(ii) 0.20 M 0.10 M 5.07 × 10–5
(iii) 0.40 M 0.05 M 1.43 × 10–4
(a) One (b) Two
(c) One and a half (d) Three
16. For the reaction : N2 + 3H2 2 NH3, if
3 4 –1 12 10 ,
d NHmol L s
dt the value of 2d H
dtwould be
(a) 4 × 10–4 mol L–1 s–1 (b) 6 × 10–4 mol L–1 s–1
(c) 1 × 10–4 mol L–1 s–1 (d) 3 × 10–4 mol L–1 s–1
17. Half-life period of a first order reaction is 1386 sec. The specific rate constant of the reactionis :
(a) 0.5 × 10–2 s–1 (b) 0.5 × 10–3 s–1
(c) 5.0 × 10–2 s–1 (d) 5.0 × 10–3 s–1
33 Chemistry
18. For the reaction A + B products, it is observed that :
(i) On doubling the initial concentration of A only, the rate of reaction is also doubled.
(ii) On doubling the initial concentrations of both A and B there is a change by a factorof 8 in the rate of reaction.
The rate of this reaction is given by
(a) Rate = k [A] [B]2 (b) Rate = k [A]2 [B]2
(c) Rate = k [A] [B] (d) Rate = k [A]2 [B]
19. For a first order reaction A P, the temperature (T) dependent rate constant (k) was found
to follow the equation
1log 2000 6.0.kT
The pre-exponential factor A and the actua-
tion energy Ea respectively are
(a) 1.0 × 106 s–1 and 9.2 kJ mol–1 (b) 6.0 s–1 and 16.6 kJ mol–1
(c) 1.0 × 106 s–1 and 16.6 kJ mol–1 (d) 1.0 × 106 s–1 and 38.3 kJ mol–1
20. For the reaction 1 2 ,2
A B rate of disappearance of ‘A’ is related to the rate of appearanceof ‘B’ by the expression
(a)
d A d Bdt dt
(b) 4d A d B
dt dt
(c)
12
d A d Bdt dt
(d)
14
d A d Bdt dt
21. Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substancebecomes half in 40 sec and 20 sec through first order and zero order kinetics respectively.
Ratio 1
0
kk
of the rate constants for first order (k1) and zero order (k0) of the reaction is
(a) 0.5 mol–1 dm3 (b) 1.0 mol dm–3
(c) 1.5 mol dm–3 (d) 2.0 mol–1 dm3
22. The energies of activation for forward and reverse reactions for A2 + B2 2AB are 180kJ mol–1 and 200 kJ mol–1 respectively. The presence of a catalyst lowers the activationenergy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change for thereaction (A2 + B2 2AB) in the presence of catalyst in (kJ mol–1) will be
(a) 20 (b) 300
(c) 120 (d) 280
23. The reaction of hydrogen and iodine monochloride is given as H2(g) + 2 ICl(g) 2HCl(g)+ I2(g). This reaction is of first order with respect to H2(g) and ICl(g), following mechanismswere proposed.
Mechanism A : H2(g) + ICl(g) HCl(g) + HI(g)
34 Chemistry
Mechanism B :
2slowH g ICl g HCl g HI g
2fastHI g ICl g HCl g I g
which of the above mechanisms can be consistent with the given information about thereaction.
(a) A only (b) B only
(c) A and B both (d) Neither A nor B
24. The total number of - and - particles excited in the nuclear reaction 238 21492 82 isU Pb
(a) 6, 2 (b) 6, 10
(c) 4, 10 (d) 4, 2
25. The half life for radioactive decay of C–14 is 5730 years. An archeological artifact containingwood had only 25% of the C–14 found in living tree. The age of the sample is
(a) 5730 years (b) 14460 years
(c) 2865 years (d) 1432 years
26. The relative penetrating power of and neutron follows the order
(a) > > > n (b) n > > >
(c) > > n > (d) None of these
27. Given the hypothetical reaction mechanism
I II III IVA B C D E and the data as
Species formed Rate of its formation
B 0.002 mol/h per mole of A
C 0.030 mol/h per mol of B
D 0.011 mol/h per mol of C
E 0.420 mol/h per mol of D
The rate determining step is :
(a) Step I (b) Step II
(c) Step III (d) Step IV
35 Chemistry
1. (c) 2. (a) 3. (b) 4. (d)
5. (d) 6. (a) 7. (b) 8. (b)
9. (c) 10. (d) 11. (d) 12. (c)
13. (c) 14. (b) 15. (c) 16. (b)
17. (b) 18. (a) 19. (d) 20. (d)
21. (a) 22. (a) 23. (b) 24. (a)
25. (b) 26. (b) 27. (a)
36 Chemistry
Adsorption is essentially a surface phenomenon. The accumulation of molecular species atthe surface rather than in the bulk of a solid or liquid is termed adsorption. For example, silicagel adsorbs moisture.
Adsorbate is the substance whose concentration accumulates at the surface of a the solid ora liquid called adsorbent.
Desorption is the process of removing the adsorbate from the surface of the adsorbent.
If a substance is uniformly distributed throughout the bulk of the solid, the phenomenon isknown as absorption. In adsorption, the substance is concentrated on the surface and does notpenetrate through the surface to the bulk of adsorbent. For example, anhydrous calcium chlorideabsorbs moisture.
Both the adsorption and absorption can be also take place simultaneously. Then we use theterm sorption to describe both the processes. When a chalk stick is dipped in ink, colouredmolecules are adsorbed while solvent is absorbed. Adsorption is accompanied by the decrease inenthalpy as well as decrease in the entropy of the system. Since adsorption is a spontaneousprocess, therefore there is always decrease in the Gibbs energy. As the adsorption proceeds, Hbecomes less and less negative, ultimately |H| = |TS| and G = 0. At this state, equilibriumis attained.
Note : In exceptional cases, chemisorption may be endothermic. For example, H2 adsorbsendothermically on glass. The S in the process : H2(g) 2H (glass) is sufficiently positive makingG = H – TS negative. Similarly, highly hydrated solutes, when adsorbed on solids have positiveH but there is large positive S due to release of water molecules during adsorption.
[For comparation between physisorption (Van der Waals adsorption) and chemisorption, referthe NCERT text book part I pages 124-125].
Adsorption of Nitrogen on Iron
At 83 K nitrogen is physisorbed on iron surface as N2 molecules. At room temperature,practically there is no adsorption of nitrogen on iron. At 773 K and above, nitrogen is chemisorbedon iron surface as nitrogen atoms.
Factors Affecting Adsorption of Gases
(a) Nature of adsorbate : Higher the critical temperature of gas, the more easily it will bephysisorbed on the solid. A gas can be chemisorbed on the solid if it is capable of formingchemical bonds with the solid.
(b) Surface Area of the Adsorbent : Greater the surface area, the more the extent ofadsorption.
37 Chemistry
(c) Temperature : Adsorption is an exothermic process involving the equilibrium :
Gas (adsorbate) + Solid (adsorbent) Gas adsorbed on solid + Heat
Applying Le Chatelier principle, increase of temperature decreases the adsorption and viceversa.
(d) Pressure : Adsorption increases with pressure at constant temperature. The effect is largeif temperature is kept constant at low value.
(e) Activation of the Solid Adsorbent : This means increasing the adsorbing power of thesolid adsorbent. This can be done by subdividing the solid adsorbent or by removing thegases already adsorbed by passing superheated steam.
Freundlisch Adsorption Isotherm
(x/
These curves show that at a fixed pressure, there is the decrease in the extent of physicaladsorption with the increase in temperature. The relationship between the quantity of gas adsorbedby unit mass of adsorbent and pressure of the gas at constant temperature is expressed by thefollowing relation : x/m = KP1/n; (n > 1)
For adsorption from solution, the equilibrium concentration of solution is taken into account
x/m = KC1/n; (n > 1)
Homogeneous and Heterogeneous Catalysis
When the reactants and the catalyst are in the same phase, the process is called homogeneouscatalysis and if they are in different phases, then the processes are the examples of heterogeneouscatalysis. A catalyst enhances the rate of reaction without itself getting used up in the reaction.
Promoters are the substances that enhance the activity of a catalyst while poisons decreasethe activity of the catalyst.
2 2 33 2Fe s as CatalystMo s as promoterN H NH
In heterogeneous catalysis, the reactants in gaseous state or in solution are adsorbed on thesurface of solid catalyst. This results in the increase in concentration of reactants on the surfaceof solid Catalyst and hence, in the rate of reaction. Adsorption being an exothermic process, theenthalpy of adsorption is utilised in increasing the rate of reaction.
38 Chemistry
[For shape selective catalysis and enzyme catalysis please refer to the NCERT text book part 1pages 130-133].
Colloidal Solution
In Colloid or Colloidal dispersion, diameters of particles of the dispersed phase range from1nm to 1000 nm.
1. Classification Based on the Physical State of Dispersed Phase and Dispersion Medium
Types of Colloidal Systems
Dispersed Phase Dispersion Medium Name Examples
Solid Solid Solid sol Some coloured glassesSolid Liquid Sol Paints, muddy waterSolid Gas Aerosol Smoke, dustLiquid Solid Gel Cheese, butter, jelliesLiquid Liquid Emulsion Milk, hair creamLiquid Gas Aerosol Fog, mist, cloudGas Solid Solid foam Pumice stone, foam rubberGas Liquid Foam Froth, whipped cream
2. Classification Based on Nature of Interaction between Dispersed Phase and DispersionMedium
The colloidal sols are divided in two categories as lyophilic (solvent attracting) and lyophobic(solvent repelling).
Lyophilic Colloids/sols Lyophobic Colloids/sols
(i) These are the organic substances like gum, (i) These are inorganic substances like metals, theirstarch, gelatin etc. which, when mixed with sulphides etc. which do not form the colloidalthe liquid, directly form the colloidal sol. sol directly. These solutions are prepared
indirectly.(ii) They are reversible. (ii) They are irreversible.
(iii) Their viscosity is higher and surface tension (iii) Their viscosity and surface tension are nearlyis lower than that of the dispersion medium. same as that of the dispersion medium.
(iv) They are quite stable and are not easily pre- (iv) They are easily precipitated by adding a smallcipitated or coagulated. amount of a suitable electrolyte.
3. Classification Based on Type of Particles of the Dispersed Phase
(a) Multimolecular Colloids : A large number of atoms (as in case of gold sol) and smallermolecules (as in case of sulphur sol) having diameters less than 1nm aggregate via weakvan der walls forces to form particles in the colloidal range.
(b) Macromolecular Colloids : Macromolecules having colloidal dimensions form dispersionin a suitable solvent. Examples of macromolecules are starch, cellulose, proteins, enzymesand synthetic polymers like polythene, nylon 66, polystyrene, synthetic rubber, etc.
39 Chemistry
(c) Associated Colloids (Micelles) : Surface active agents such as soaps and syntheticdetergents behave as strong electrolytes at low concentrations, but at higher concentrationsexhibit colloidal properties due to the formation of aggregates called micelles or associatedcolloids.
Formation of micelles takes place only above a particular temperature called Krafttemperature (Tk) and above a particular concentration called critical micelle concentration(CMC). For soaps the CMC is 10–4 to 10–3 mol/L. These colloids have hydrocarbon part which ishydrophobic and anionic part which is hydrophilic.
Formation of Colloids
(a) Chemical methods like double decomposition, oxidation, reduction and hydrolyses areused to form the molecules which the aggregate forming the colloids.
Double decomposition : As2O3 + 3H2S As2S3(sol) + 3H2O
Oxidation : SO2 + 2H2S 3S(sol) + 2H2O
Reduction : 2AuCl3 + 3HCHO + 3H2O 2Au(sol) + 3HCOOH + 6HCl
Hydrolysis : FeCl3 + 3 H2O Fe2O3.xH2O + 6HCl
(b) Bredig’s Arc Method : Colloidal sols of metals such as Au, Ag, Pt, etc. are prepared bythis method.
(c) Preptisation : It is the process converting a precipitate into colloidal sols by shaking itwith a dispersion medium in presence of a small amount of electrolyte. The precipitateadsorbs on its surface the ions of the electrolyte common to the lattice of the precipitate andthen the precipitate breaks into charged colloidal particles.
Purification of Colloidal Solutions
While the traces of electrolyte is essential for the stability of the colloidal solution, largerquantities coagulate it. Hence, to reduce the concentration to a requisite minimum, dialysis,electrodialysis and ultrafiltration are used.
Properties Exhibited by Colloidal Solution
(a) Colligative Properties : The values of colligative properties are of small order as comparedto values shown by true solutions of the same concentration.
(b) Tyndall Effect : The Tyndall effect observed in colloidal solution is due to scattering oflight by colloidal particles in all directions provided (i) the diameters of dispersed particlesare not much smaller than the wavelength of light used and (ii) refractive indicates ofdispersed phase and dispersion medium differ greatly in magnitude.
(c) Brownian Movement : The zig-zag motion of colloidal particles is due to the unbalancedbombardment of the colloidal particles by the molecules of dispersion medium. This movementis responsible for the stability of sols.
(d) Charge on Colloidal Particles : Colloidal particles always carry an electric charge.
40 Chemistry
Positively Charged Sols Negatively Charged Sols
Hydrated metal oxides, e.g.,Al2O3.xH2O, Fe2O3.xH2O Metal sols, e.g., Ag, AuHaemoglobin (blood) Metallic sulphides, e.g., As2S3, Sb2S3, CdSOxide sols, e.g., TiO2 Sols of starch, gum, gelatin, clay, charcoal, etc.Basic dyes, e.g., methylene blue sol. Acid dyes, e.g., eosin, congo red sols.
The charge on sol particles is due to one or more reasons, viz, due to (i) electron captureby sol during electrodispersion of metal (ii) preferential adsorption of ions from the solutionand/or formulation of electrical double layer.
Having acquired a positive or negative charge by selective adsorption on the surface of acolloidal particle, the fixed layer of ions attract the counter ions from the medium forminga mobile or diffused layer. The potential difference between the fixed layer of ions anddiffused layer of counter ions is called electrokinetic or zeta potential.
(e) Electrophoresis : The movement of charged sol particles towards the oppositely chargedelectrode under the effect of electric field applied across the colloidal solution, is calledelectrophoresis.
If the movement of colloidal particles is prevented by some means, it is observed thatdispersion medium begins to move in an electric field. This phenomenon is calledelectroosmosis.
(f) Coagulation or Precipitation of Sols : It is the process of changing the colloidal particlesin a sol into the insoluble precipitate by addition of some suitable electrolytes.
The coagulation of lyophobic sols can be carried out in the following ways :
(i) By electrophoresis
(ii) by boiling
(iii) by prolonged dialysis
(iv) by the addition of electrolytes.
Coagulation of Lyophilic Sols
This is done by adding (i) an electrolyte and (ii) a suitable solvent which can dehydrate thedispersed phase.
Hardy-Schulze Rule
A negative ion causes the precipitation of positively charged sol and vice versa. The greaterthe valence of the coagulating ion added, the greater its power to cause the precipitation. In thecoagulation of a negative sol, the flocculating power is in the order Al3+ > Ba2+ > Na+. Similarly,in the coagulation of a positive sol, the flocculating power is in the order :
4 3 26 4 4Fe CN PO SO Cl .
Coagulating Value
The minimum concentration of an electrolyte in mmol per litre required to cause the
41 Chemistry
precipitation of a sol in two hours, is called coagulating value. The smaller the amount required,the higher will be the coagulating power of an ion.
Lyophilic colloids are used to protect the lyophobic colloids. When a lyophilic sol is added, itsparticles from a layer around the lyophobic particles. This layer of lyophilic particles is extensivelysolvated.
To compare the protective action of different lyophilic colloids, Zsigmondy (1901) introduceda term called gold number. Gold number of a protective colloid is the minimum mass of it inmilligrams which must be added to 10 mL of a standard red gold sol so that no coagulation of thegold sol (i.e. the change of colour from red to blue) takes place when 1 mL of 10% sodium chloridesolution is rapidly added to it.
Evidently, smaller the gold number of a protective colloid, the greater is its protective action.The gold numbers of a few protective colloids are given below :
Sol Gold Number Reciprocal
Gelatin 0.005 – 0.01 200 – 100Casein 0.01 – 0.02 100 – 50Haemoglobin 0.03 – 0.07 33 – 14
Emulsions
Colloidal systems in which both dispersed phase and dispersion medium are liquids. Thesecan be of :
(i) Water in Oil Type (W/O Type) : Examples are milk and vanishing cream.
(ii) Oil in Water Type (O/W Type) : Examples are butter, cream and cold cream. Soaps anddetergents are most frequently used as emulsifiers for the stabilisation of the emulsion ofO/W type and long chain alcohols, heavy metal salts of fatly acids for the stabilisation ofthe emulsion of W/O type.
42 Chemistry
1. Rate of physisorption increases with the
(a) decrease in temperature (b) increase in temperature
(c) decrease in pressure (d) decrease in surface area
2. Adsorption of gases on solid surface is generally exothermic when
(a) entropy of gas decreases during adsorption,
(b) entropy of gas increases during adsorption,
(c) enthalpy of adsorption is positive,
(d) Gibbs energy increases.
3. Lyophilic sols are
(a) irreversible
(b) prepared from inorganic compounds like metal oxides and sulphides,
(c) coagulated by adding electrolytes,
(d) self stabilizing.
4. In a chemical reaction, a catalyst
(a) alters the amounts of products,
(b) lowers the activation energy,
(c) decreases rH for forward reaction,
(d) increases rH for the reverse section.
5. Which one of the following statement is correct?
(a) Lyophobic colloids do not easily coagulate on adding electrolytes.
(b) Lyophobic colloids are reversible in character.
(c) Lyophilic colloids are reversible in character.
(d) Lyophilic colloids are easily coagulated by electrolytes.
6. According to langumir adsorption isotherm, when the pressure of a gas is very large, theextent adsorption is
(a) directly proportional to pressure.
(b) Inversely proportional to pressure.
(c) Directly proportional to the square of pressure.
(d) Independent of pressure.
43 Chemistry
7. In the coagulation of arsenic sulphide solution, the flocculating powers of given ions aresuch that
(a) PO43– > SO4
2– > Cl– (b) Na+ > Ba2+ > Al3+
(c) Cl– > SO42– > PO4
3– (d) Al3+ > Ba2+ > Na+
8. Of the following which is not correct?
(a) As the adsorption proceeds, H becomes less and less negative, ultimately H becomesequal to TS and G becomes zero.
(b) The formation of micelles taken place above Kraft temperature Tk and above criticalmicelle concentration (CMC).
(c) The potential difference between the fixed layer and the diffused layer of oppositecharges around the collidal particle is called electrokinetic potential or zeta potential.
(d) Hydrated aluminium oxide, Al2O3 . xH2O sol consists of positively charged particles.When electric field is applied across the sol, charged sol particles move towards theoppositely charged electrode. This phenomenon is called electroosmosis.
9. Gold numbers of protective colloids a, b, c and d are 0.50, 0.01, 0.10 and 0.005 respectively.The correct order of their protective powers is
(a) d < a < c < b (b) c < b < d < a
(c) a < c < b < d (d) b < d < a < c
10. Which one of the following statements is correct? Peptisation is a process of
(a) precipitation of colloidal particles,
(b) purification of colloids,
(c) dispersing precipitate into colloidal solution,
(d) protection of colloidal solution.
11. Which of the following statements is incorrect regarding physisorption?(a) Enthalpy of adsorption ( Hadsorptions) is low and positive.(b) It occurs because of van der walls forces.(c) More easily liquefiable gases are adsorbed readily.(d) Under high pressure it results into multimolecular layer on adsorbent surface.
12. Freundlisch equation for adsorption of gases (in amount of ‘X’ g) on a solid (in amount of‘m’ g) at constant temperature can be expressed as :
(a) 1log log logx p k
m n(b)
1log log logx k pm n
(c) nx pm
(d) 1log logx p k
m n
13. If a liquid is dispersed in solid medium, then dispersion is called as :
(a) Sol (b) Emulsion
(c) liquid aesosol (d) Gel
44 Chemistry
14. Among the following, surfactant that will form mecelles in aqueous solution at the lowestmiscelle concentration at ambient conditions is
(a) CH3(CH2)15N+(CH3)3Br– (b) CH3(CH2)11OSO3–Na+
(c) CH3(CH2)6COO–Na+ (d) CH3(CH2)11N+(CH3)3Br–
15. Which among the following statements are correct with respect to adsorption of gases on asolid?
1. The extent of adsorption is equal to k P according to Freundlech isotherm.
2. The extent of adsorption is equal to k P1/n according to Freundlech isotherm.
3. The extent of adsorption is equal to (1 + bP)/aP according to Langmur isotherm.
4. The extent of adsorption is equal to aP/(1 + bP) according to Langmur isotherm.
(a) 1 and 3 (b) 1 and 4
(c) 2 and 3 (d) 2 and 4
16. Match list I (colloidal dispersion) with list II (nature of the dispersion) and select the correctanswer using the codes given below the lists.
List I List II(Colloidal Dispersion) (Nature of Dispersion)
A Milk 1. Solid in liquidB Clouds 2. Liquid in gasC Paints 3. Solid in solidD Jellies 4. Liquid in liquid
(a) A – 4, B – 2, C – 1, D – 5 (b) A – 1, B – 5, C – 3, D – 2
(c) A – 4, B – 5, C – 1, D – 2 (d) A – 1, B – 2, C – 3, D – 5
17. Which of the following statements about the zeolites is false?
(a) They are used as cation exchanges.
(b) They have open structure which enables them to take up small molecules.
(c) Zeolites are aluminosilicates having three dimensional network.
(d) Some of the SiO44– units are replaced by AlO4
5– and AlO69– ions in zeolites.
18. Which of the following forms catonic micelles above certain concentration
(a) sodium dodecyl sulphate (b) sodium acetate
(c) urea (d) cetyl trimethyl ammonium bromide.
19. Identify the correct statement regarding enzymes.
(a) Enzymes are specific biological catalysts that can normally function at very lowtemperature.
(b) Enzymes are normally heterogeneous catalysts that are very specific in action.
(c) Enzymes are specific biological catalysts that can not be poisoned.
(d) Enzymes are specific biological catalysts that possess well defined active sites.
45 Chemistry
20. Bredig’s arc are method can not be used to prepare colloidal solution of which of thefollowing?
(a) Pt (b) Fe
(c) Ag (d) Au
21. The volume of a colloidal particle Vc as compared to the volume of a solute particle in a truesolution Vs, could be
(a) 1c
s
VV
(b) 2310c
s
VV
(c) 310c
s
VV
(d) 310c
s
VV
22. The disperse phase in hydrated iron (III) oxide sol and colloidal gold are positively andnegatively charged respectively. Which of the following statements is not correct?
(a) Magnesium chloride solution coagulates the gold sol more readily than the hydratediron (III) oxide sol.
(b) Sodium sulphate solution causes coagulation in both sols.(c) Mixing the sols has no effect.(d) Coagulation of both sols can be brought about by electrophoresis.
23. An example of autocatalysis is
(a) oxidation of NO to NO2
(b) oxidation of SO2 to SO3
(c) decomposition of KClO3 to KCl and O2
(d) oxidation of oxalic acid by acidified KMnO4
24. Given below, catalyst and corresponding process/reaction are matched. The mismatch is
(a) [RhCl (PPh3)2] : hydrogenation. (b) TiCl4 + Al(C2H5) : polymerization.
(c) V2O5 : Haber-Bosch Process. (d) Nickel : hydrogenation.
25. Which of the following is true in respect to adsorption
(a) G < 0, S > 0, H < 0 (b) G < 0, S < 0, H < 0
(c) G > 0, S > 0, H < 0 (d) G > 0, S > 0, H > 0
1. (a) 2. (a) 3. (d) 4. (b)5. (c) 6. (d) 7. (d) 8. (d)9. (c) 10. (c) 11. (a) 12. (b)
13. (d) 14. (a) 15. (d) 16. (a)17. (d) 18. (d) 19. (d) 20. (b)21. (d) 22. (c) 23. (d) 24. (c)25. (b)
46 Chemistry
Minerals are naturally occurring chemical substances in earth’s crust obtainable by mining.Ores of a metal are minerals which are used as source of that metal profitably.
Aluminium is the most abundant metal and the oxygen is most abundant elementin the earth’s crust. Iron is the second most abundant metal in the earth’s crust. Silver,gold, Platinum, sulphur, oxygen and nitrogen are the elements that occur in native or free state.
Many gem-stones are impure form of Al2O3, the impurities range from Cr (in ruby)to Co (in sapphire). [For principal ores of Al, Fe, Cu and Zn, please refer Table 6.1 in the NCERTText Book Part I, Class XII, Page 148]. Concentration of ores depends upon the difference inphysical properties of the compound of metal present and that of the gangue.
Forth floation process is used for the concentration of sulphide ores. A suspension ofpowdered ore is made with water. To it collectors (e.g. pine oils, fatty acids, xanthates etc.) andfroth stabilisers (e.g., cresols, aniline which enhance the non-wettability of the mineral particles)and froth stabilisers (e.g., cresol and aniline) which stabilise the forth) are added. Sometimesdepressants are added to separate the sulphide ores. For example, NaCN is used as a depressantto selectively prevent ZnS from coming to froth but only allows PbS to come with froth. NaCNforms a layer of Na2[Zn(CN)4] on the surface of ZnS that prevent it from coming to froth.
Leaching is useful in case the ore is soluble in a suitable solvent.
242 3
2 3 2 2 3
Bauxite NaOH aq dilutionCO g
heat
Alumina
Al O s Na Al OH aq
Al O · xH O s Al O s
Conversion of Concentrated Ore to An Oxide
(a) Calcination : The hydrated or carbonate ores are heated in presence of limited supply ofair when the volatile matter escapes leaving behind the metal oxide.
2 3 2 2 3 2Fe O xH O Fe O s xH O g·
3 2ZnCO s ZnO s CO g
(b) Roasting : The sulphide ore is heated in a regular supply of air below the melting pointof metal to convert the metal into its oxide or sulphate. Sometimes a part of the sulphidemay act as reducing agent in the subsequent step.
47 Chemistry
2ZnS + 3O2 2ZnO + 2SO2
2PbS + 3O2 2PbO + 2SO2
PbS + 2O2 PbSO4
2Cu2S + 3O2 2Cu2O + 2SO2
Note : In auto reduction : For example, the partly converted Cu2O reduces Cu2S
Cu2S + 2Cu2O 6Cu + SO2PbS + PbSO4 2Pb + 2SO2
PbS + 2PbO 3Pb + SO2HgS + 2HgO 3Hg + SO2
Reduction of Oxide to Metal
Reduction means electron gain or electronation. For the reduction of metal oxides, heating isrequired. To understand the variation in temperature requirement for thermal reduction(pyrometallurgy) and to predict which element will suit as reducing agent for a metal oxide (MxOy),Gibbs energy interpretations are made at any specified temperature.
G = H – TS
For any reaction :
G0 = – 2.303 RT log K
when a reaction proceeds towards products, K will be positive which implies that G will benegative.
When the value of G is negative, only then the reaction proceed. If S is positive, on increasingthe temperature (T), the value of TS would increase i.e.; (H < TS) and then, TS will be –ve.
If the reactants and products of two reactions are put together in a system and the net Gof the two reactions is –ve, the overall reaction will occur spontaneously.
Ellingham Diagram
Gibbs energy (G) for formation of oxides per mol of O2 are plotted against temperature T.It is evident that elements for which Gibbs energy of formation of oxides per mol of oxygen is morenegative, can reduce the oxides of elements for which Gibbs energy of formation per mol of O2 isless negative, that is, the reduction of oxide represented by upper line is feasible by the elementrepresented by lower line.
Reduction by Carbon
Smelting : It is the process of extraction of metal from its roasted or calcined ore by heatingwith powdered coke in presence of a flux. In smelting, oxides are reduced to molten metal by carbonor carbon monoxide.
PbO + C Pb + CO and Fe2O3 + 3CO 2Fe + 3CO2
2 3 3 2
2 3
2 3acidic basic basic Acidic
SiO CaCO CaSiO COFlux impurities slag FeO SiO FeSiO slag
CaO SiO CaSiO slag
48 Chemistry
Extraction of Non-Metals or Metals by Oxidation
In simple electrolysis of molten salt, Mn+ are discharged at negative electrodes. Sometimes aflux is added for making molten mass more conducting as in the electrolysis of molten alumina,CaF2 or Na3[AlF6] is added to lower the melting point of mix and bring the conductivity.
Some extractions are based on oxidation for non-metals. For example, extraction of Cl2 frombrine is the oxidation of Cl– in aqueous medium.
2 2 22 2 2 ; 422aq aqCl H O l OH H g Cl g G KJ
Leaching of Ag or Au with CN– involves the oxidation of Ag Ag+ or Au Au+
2 24 8 2 4 2 4aq aqAu s CN H O O g Au CN OH
2
2 42 2Reducing agent
Au CN aq Zn s Au s Zn CN aq
–
REFINING OF METALS
Distillation
Useful for low boiling metals like zinc and mercury.
Liquation
Useful for low boiling metals like tin and lead.
Electrolytic Refining
Anode : Impure metal; cathode strip of pure metal. Soluble metal salt solution is used as anelectrolyte.
A large no. of metals such as Cu, Ag, Au, Pb, Ni, Cr, Zn, Al etc. are refined by this method.
Zone Refining
Zone refining is based on the principles that impurities are more soluble in the melt than inthe solid state of metal. Pure metals are crystallised out of the melt and impurities move intomolten zone. This is used for metals of very high purity, e.g., Ge, Si, B, Ga and In.
Vapour Phase Refining
Impure metal is converted into volatile compound which is then decomposed to get pure metale.g. Mond’s process for the purification of Ni metal involves formation of Nickel carbonyl which onfurther decomposition give pure nickel metal.
330 350 450 470
44 4K KImpure PureVapoursmetal
Ni CO Ni CO Ni CO g
49 Chemistry
Van Arkel method for Refining Zr or Ti
This method is used to remove all oxygen and nitrogen present in the form of impurity incertain metals like Zr or Ti.
2 4 218002 2Evacuated Tungston filament
Vessel Kimpure Deposited on FilamentZr I ArI Zr I
Chromatographic Methods
This method is based on the principle that different components of a mixture are differentlyadsorbed on a adsorbent. The adsorbed components are removed (eluted) by using suitable solvent(elutant).
50 Chemistry
1. Froth floation process may be used to increase concentration of the mineral in
(a) Bauxite (b) Calamine
(c) Haemetite (d) Copper pyrites
2. The slag obtained during the extraction of copper from copper pyrites is mainly of
(a) CuSiO3 (b) FeSiO3
(c) Cu2O (d) Cu2S
3. Heating the ore with carbon with the simultaneous removal of slag is called
(a) roasting (b) calcination
(c) smelting (d) leaching
4. Cryolite is used in electrolysis of alumina
(a) to increase the conductivity and decrease the melting point of mix,
(b) to decrease the conductivity and increase the melting point of mix,
(c) to increase the conductivity and melting point of mix,
(d) do decrease the conductivity and melting point of mix.
5. In the extraction of copper from sulphide ore, the metal is formed by reduction of Cu2O with
(a) FeS (b) CO
(c) Cu2S (d) SO2
6. The method of zone refining of metals is based on the principle of
(a) greater mobility of pure metal than that of impurity,
(b) higher melting point of impurity than that of pure metal,
(c) higher noble character of solid metal than that of impurity,
(d) greater solubility of impurity in molten state than in the solid.
7. Pyrolusite is an
(a) sulphide orde (b) oxide ore
(c) carbonate ore (d) phosphate ore
8. Pick out the incorrect statement
(a) Calamine and siderite are carbonates.
(b) Argentite and cuprite are oxides.
(c) Zinc blende and iron pyrites are sulphides.
(d) Malachite and azurite are the ores of copper.
51 Chemistry
9. ‘German silver’ does not have
(a) Cu (b) Zn
(c) Ni (d) Ag
10. The metal purified by fractional distillation is
(a) Zn (b) Cu
(c) Al (d) Si
11. Identify the reaction that does not take place in Blast furnace
(a) 2F e2O3 + 3C 4Fe + 3CO2 (b) CO2 + C 2CO
(c) CaCO3 CaO + CO2 (d) FeO + SiO2 FeSiO3
12. Native silver forms a water soluble complex with dilute solution of NaCN in presence of
(a) Nitrogen (b) Oxygen
(c) Carbon dioxide (d) Argon
13. Extraction of Zinc from zinc blende is achieved by
(a) electrolytic reduction,
(b) roasting followed by reduction with coke,
(c) roasting followed by reduction with other metal,
(d) roasting followed by self-reduction.
14. Blister copper is
(a) impure copper having 10% FeSiO3 (b) copper alloy
(c) pure copper (d) copper having about 1% impurity.
15. Which one of the following metals has greater tendency to form oxide?
(a) Al (b) Mg
(c) Cr (d) Fe
16. Bauxite ore is made up of Al2O3 + SiO2 + TiO2 + Fe2O3. This ore is treated with conc. NaOHsolution at 500K and 35 bar pressure for few hours and filtered hot. In the filtrate, thespecies present is/are
(a) Na[Al(OH)4] only (b) Na[Al(OH)4] and Na2SiO3 both
(c) Na2[Ti(OH)6] only (d) Na2SiO3 only
17. When copper pyrites is roasted in excess of air, a mixture of Cu2O + FeO is formed. FeOis present as impurities. This can be removed as slag during reduction of Cu2O. The fluxadded to from slag is
(a) SiO2 which is an acid flux. (b) Lime stone which is basic flux.
(c) SiO2 which is basic flux. (d) CuO which is basic flux.
52 Chemistry
18. In the process of extraction of gold
Roasted gold ore 22
OCN H O X OH
X Zn Y Au
Identify the complexes [X] and [Y]
(a) X = [Au(CN)2]–; Y = [Zn(CN)4]2– (b) X = [Au(CN)4]2–; Y = [Zn(CN)4]2–
(c) X = [Au(CN)2]–; Y = [Zn(CN)6]4– (d) X = [Au(CN)4]–; Y = [Zn(CN)4]2–
19. During electrolytic refining of copper, some metals present as impurity settle as anode mud.These are
(a) Sn and Ag (b) Pb and Zn
(c) Ag and Au (d) Fe and Au
20. Heating mixture of Cu2O and Cu2S will give
(a) Cu + SO2 (b) Cu + SO3
(c) CuO + CuS (d) CuO + SO3
21. Consider the following reactions at 1000°C
A 212
Zn s O g ZnO s rG = – 360 KJ mol–1
B 21graphite2
C O g CO g rG = – 460 KJ mol–1
Choose the correct statement at 1000°C
(a) Zinc can be oxidised by carbon monoxide.
(b) Zinc oxide can be reduced by graphite.
(c) Carbon monoxide can be reduced by zinc.
(d) All above statements are false.
22. Which of the following factors is of no significance for roasting sulphide ores to oxides andnot subjecting the sulphide ores to carbon reduction directly.
(a) CO2 is more volatile than CS2
(b) Metal sulphides are thermodynamically more stable than CS2.
(c) CO2 is thermodynamically more stable than CS2
(d) Metal sulphides are less stable than the corresponding oxides.
23. Which of the following statements about the advantage of roasting of sulphide ore beforereduction is not true.
(a) Roasting of the sulphide to the oxide is thermodynamically feasible.(b) Carbon and hydrogen are suitable reducing agents for metal sulphides.(c) The G0 for sulphide is greater than those for CS2 and H2S.(d) The G0 is negative for roasting sulphide ore to oxide.
53 Chemistry
24. Matte is a mixture of
(a) Cu2S + FeS (small amount) (b) FeS + Cu2S (small amount)
(c) Cu2O + FeO (small amount) (d) FeO + Cu2O (small amount)
25. During roasting copper pyrites are ultimately convented into a mixture of
(a) FeS + Cu2S (b) FeS + Cu2O
(c) FeO + Cu2S (d) FeS + Cu2S + FeO + Cu2O
1. (d) 2. (b) 3. (c) 4. (a)
5. (c) 6. (d) 7. (b) 8. (b)
9. (d) 10. (a) 11. (d) 12. (b)
13. (b) 14. (d) 15. (b) 16. (c)
17. (a) 18. (a) 19. (c) 20. (a)
21. (b) 22. (b) 23. (b) 24. (a)
25. (d)
54 Chemistry
General Characteristics of 15 Group Elements : [N, P, As, Sb, Bi]
Oxidation States
Negative Oxidation States : The elements of this group show an oxidation state of – 3.However, the tendency of these elements to show – 3 oxidation states decreases as we move downthe group from N to Bi due to a gradual decreases in the electronegativity and ionization enthalpy.
Positive Oxidation States : The elements of this group also show positive oxidation statessuch as + 3 and + 5. The stability of + 3 oxidation state increases while that of + 5 decreases aswe go done the group. This is due to inert pair effect. The + 5 oxidation state in Bi is less stablethan in Sb. The only well characterized Bi (V) compound is BiF5.
Nitrogen shows – 1, – 2, – 3, + 1, + 2, + 3, + 4 and + 5 oxidation states in NH2OH, N2H4, NH3,N2, N2O, NO, N2O3, N2O4 and N2O5 respectively.
Catenation : Nitrogen has little tendency for catenation since | |
– –N N. . . .
single bond is weak
(167 kJ mol–1) due to repulsion between non-bonded electron pairs owing to small N – N bondlength. As we move down the group, the element-element bond enthalpies decrease rapidly viz.N – N (167 kJ mol–1), P – P (201 kJ mol–1), As – As (146 kJ mol–1) and Sb – Sb (121 kJ mol–1) andtherefore, tendency for catenation decreases in the order P > N > As > Sb > Bi.
Elemental State
Because of small size and high electronegativity, nitrogen has a strong tendency to formmultiple (p – p) bonds with itself and with other elements like C and O having small size andhigh electronegativily but other elements do no form multiple bonds. Thus, nitrogen exists as adiatomic gas in which two nitrogen atoms are linked by a triple bond : N N : (one and two -bonds). Because of small bond length and high bond strength (946 kJ mol–1), nitrogen is inert atordinary temperature. Other elements of this group do not exist as diatomic molecules due to theirreluctance to form multiple bonds. Phosphorus, arsenic and antimony exist as discrete teratomictetrahedral molecules, i.e., P4. As4 and Sb4 in which the four atoms lie at the corners of a regulartetrahedron.
Formation of Hydrides
All the elements of group 15 form volatile hydride of the type EH3. They have a pyramidalstructure.
55 Chemistry
(i) Thermal stability of these hydrides decreases gradually form NH3 to BiH3 due to the decreasein bond dissociation enthalpy of E – H bond
NH3 > PH3 > AsH3 > SbH3 > BiH3
(ii) Reducing Character : Because of decrease in thermal stability, the tendency of thesehydrides to give hydrogen and thus act as reducing agents gradually increases in the order:
NH3 > PH3 > AsH3 > SbH3 > BiH3
(iii) Basic Character : The presence of the lone pair of electrons on the central atom E in EH3makes these hydrides as Lewis bases.
3 4Conjugate acidBase
EH H EH
As the size of the central atom increases, the stability of the conjugate acid decreases andhence the basic character decreases in the order :
NH3 > PH3 > AsH3 > SbH3 > BiH3Thus PH3 is weakly basic but AsH3, SbH3 and BiH3 are not at all basic.
(iv) Hydrogen Bonding : Due to small size and high electronegativity of nitrogen and presenceof a lone pair of electrons, NH3 forms H-bonds resulting in exceptionally high m.p. and b.p.
(v) Melting Points and Boiling Points : Due to H-bonding, the m.p. of NH3 is the highestamongst the hydrides of group 15 elements. As we move from NH3 to PH3, there is a sharpdecrease in the m.p. and b.p. of PH3 as compared to NH3 due to absence of H-bonding.However, the m.p. and b.p. of the hydrides of rest of the elements increase gradually as wemove down the group form PH3 to BiH3. This is due to increase in molecular size resultingin increase in Van der Wall forces of attraction holding the molecules together.Thus PH3 has the lowest and NH3 and SbH3 has the highest m.p. and b.p. respectively.
(vi) Bond Angles : The hydrides of group 15 have pyramidal shapes, i.e., the central atomundergoes sp3-hybridization. The HNH bond angle in NH3 is 107°. However, as we movedown the group the bond angles gradually decrease due to decrease in bond pair-bond pairrepulsion. NH3 (107.8°), PH3 (93.5°), AsH3 (91.8°), SbH3 (91.3°) and BiH3 (90°). Thus, thebond angles show the following trend :
NH3 > PH3 > AsH3 > SbH3 > BiH3.
Pentahalides
P, As and Sb also form pentachlorides. Nitrogen, however, does not form pentahalides due tothe absence of d-orbitals in its valence shell.
Phosphorus pentaiodide, PI5, does not exist probably due to steric factors. Bismuth forms onlypentafluoride. The non-existence of pentachloride, bromide and iodide of Bi is probably due to thestrongly oxidizing properties of Bi5+ due to inert pair effect.
These pentahalides have trigonal bipyramidal shapes in the vapour phase involving sp3d-hybridization. PF5 is not hydrolysed because the P – F bond is stronger that P – O covalent bond.PCl5 is not very stable due to its unsymmetrical (trigonal bipyramidal) shape where some bondangles are of 90° and the others are of 120°. It decomposes to give PCl5
PCl3 + Cl2. It is due to
this reason that PCl5 behaves as a good chlorinating agent. Solid PCl5 is an ionic compoundconsisting of [PCl4]+ and [PCl6]– ions. Solid PbBr5 exists as [PBr4]+ Br–.
56 Chemistry
Oxides
All the elements of this group form two types of oxides, i.e., M2O3 and M2O5 and are calledtrioxides and pentoxides.
The trioxides of N, P and As are acidic. Their acidic strength decreases in the order N2O3 >P2O3 > As2O3. The acidic strength of pentoxides also decreases in the order N2O5 > P2O5 > AS2O5> Sb2O5 > Bi2O5.
Dinitrogen (N2)
It is obtained
(i) by the reaction of aqueous solution of NH4Cl and NaNO2
NH4Cl + NaNO2 N2 + 2H2O + NaCl
(ii) by the thermal decomposition of ammonium dichromate, (NH4)2Cr2O7
4 2 7 2 2 2 32 4Heat
orange greenNH Cr O N H O Cr O
Few Typical Reaction of N2 are :
2 36 2HeatLi N Li N
2 3 23 HeatMg N Mg N
2 2 2HeatN O NO
2 2 33 2HeatN H NH
Ammonia (NH3)
Ammonium salts, when treated with caustic soda or lime, decompose to form NH3.
NH4Cl + NaOH NH3 + NaCl + H2O
Properties
Aqueous solution of ammonia is basic due to the formation of OH– ions :
3 2 4NH g H O l NH OH
It precipitates the hydroxides (hydrated oxides in case some metals like Al, Fe, etc.) of manymetals from their salt solutions. For example,
24 4 4 422white ppt
ZnSO NH OH Zn OH NH SO
Ammonia acts as a Lewis base due to the presence of a lone pair of electrons. It also acts asligand forming complexes.
223 3 44
blue deep blue
Cu aq NH aq Cu NH aq
3 3 2.2
white ppt colourlesssolution
AgCl s NH aq Ag NH Cl aq
57 Chemistry
Oxides of Nitrogen
Formula Oxidation state Common methods of preparation Physical appearance andof nitrogen chemical nature
N2O + 1 4 3 2 22HeatNH NO N O H O Colourless gas, neutral
NO + 2 2NaNO2 + 2FeSO4 + 3H2SO4 Colourless gas, neutral
Fe2(SO4)3 + 2NaHSO4 + 2H2O + 2NO
N2O3 + 3 250
2 4 2 32 2KNO N O N O blue solid, acidic
NO2 + 4 673
3 222 2 2KPb NO NO PbO brown gas, acidic
N2O4 + 4 2 2 42 CoolHeal
NO N O Colourless solid/liquid, acidic
N2O4 + 5 4HNO3 + P4O10 4HPO3 + 2N2O5 colourless solid, acidic
Structure of Oxides of Nitrogen
Formula Resonance Structures Bond Parameteres
58 Chemistry
Nitric Acid (HNO3)
On large scale it is manufactured by Ostwald’s process, i.e., by the catalytic oxidation of NH3by atmospheric oxygen.
3 2 24 5 4 6Pt Rh gaugeas CatalystNH O NO H O
2NO + O2 2NO2
3NO2 + H2O 2HNO3 + NO
Oxidising Action of Nitric Acid
3 3 2 24 2 2 2 2Cone
HNO e NO H O NO
3 3 28 6 6 4 2dil
HNO e NO H O NO
Nitric oxide predominate if the acid is dilute and nitrogen dioxide when the concentration ofacid is increased. Most metals form nitrates.
3Cu + 8HNO3 (dilute) 3cu (NO3)2 + 2NO + 4H2O
Cu + 4HNO3 (conc.) Cu (NO3)2 + 2NO2 + 2H2O
Zn + 4HNO3 (conc.) Zn (NO3)2 + 2NO2 + 2H2O
Fairly electropositive metals like Zn, Mg etc. reduce very dilute nitric acid further to give N2O,NH3 or H2, e.g.
3 3 4 3 224 10 4 3
diluteZn HNO Zn NO NH NO H O
3 3 2 224 10 4 5
Very diluteZn HNO Zn NO H O N O
23 3 22 2
Very diluteMg HNO Mg NO H
Concentrated nitric acid oxidises the non-metals. For example, I2 is oxidised to HIO3, carbonto CO2, Sulphur to H2SO4, and P4 to H3PO4.
Chemistry of Brown Ring Test is based on the ability of Fe2+ to reduce NO3– to NO which
reacts with the Fe2+(aq) to form a brown coloured complex, [Fe(H2O)5(NO)]2+ which decomposes onheating.
2 33 23 4 3 2Fe NO H NO Fe H O
222 2 26 5
Brown complex
Fe H O NO Fe H O NO H O
Phosphine (PH3)
It is prepared by the reaction of
(a) Ca3P2 with water or dilute HCl
59 Chemistry
Ca3P2 + 6H2O 3Ca(OH)2 + 2PH3
Ca3P2 + 6HCl 3CaCl2 + 2PH3
(b) White phosphorus with concentrated NaOH solution in the inert atmosphere of CO2.
3 10
4 2 3 2 23 3 3hypophosphite
P NaOH H O P H NaH PO
Here P4 molecule undergoes disproportionation reaction.
Properties
(i) Phosphine is weakly basic and gives phosphonium compounds with acids, e.g.,
PH3 + HCl PH4Cl
(ii) When absorbed in CuSO4 or HgCl2 solution, the corresponding phosphides are obtained.
3CuSO4 + 2PH3 Cu3P2 + 3H2SO4
Phosphorus trichloride (PCl3) : It is prepared from the following reactions :
P4 + 6Cl2 4PCl3
P4 + 8SOCl2 4PCl3 + 4SO2 + 2S2Cl2
Properties
(i) It reacts with organic compounds like RCOOH and R – OH containing – OH group :
3ROH + PCl3 3RCl + H3PO3
3RCOOH + PCl3 3RCOCl + H3PO3
(ii) It hydrolyses to form H3PO3 and HCl
PCl3 + 3H2O H3PO3 + 3HCl
Phosphorus Pentachloride
PCl5 molecule has trigonal bipyramidal structure, two axial P – Cl bonds are elongated andweaker than the other three equatorial P – Cl bonds.
Preparation
(i) By the reaction of white phosphorus with excess chlorine.
4 2 510 4
excessP Cl PCl
(ii) By the reaction of P4 with sulfuryl chloride.
P4 + 10 SO2Cl2 4PCl5 + 10 SO2
(iii) It hydrolyses to POCl3 and finally to H3PO4
PCl5 + H2O POCl3 + 2HCl
POCl3 + 3H2O H3PO4 + 3HCl
Oxides of Phosphorus
Phosphorus forms two types of oxides : P2O3 and P2O5. P2O3 is formed by heating phosphorus
60 Chemistry
in the limited supply of air and P2O5 is formed by heating phosphorus in the excess of air oroxygen.
4 2 2 33 2limited supply of air or oxygenP O P O
4 2 2 55 2excess of air or oxygenP O P O
The reluctance of P to enter into p – p multiple bonding leads to cage structures for theiroxides existing as the dimers. The phosphorus atoms are at the corners of the tetrahedron and sixoxygen atoms are along the edges forming six P – O – P single bonds. In P = O; P forms d
– p
bond with O atom.
P
P
P
PO
O O
OO O160pm
100°
123°
O
O
OOO
OO O
O OP
P
P
P102°123°
143 pm
( )bThe structures of (a) phosphorus (III) oxide, P4O6 and (b) phosphorus (V) oxide, P4O10.
Structure of Oxoacids of Phosphorus
O
P
HOHO OH
O
P
HOHO O
O
POH
OHH
O
POH
OHH
O
POH
H
H POOrthophosphoric acid
3 4 H P OPyrophosphoric acid
4 72 H POOrthophosphorus acid
3 3 H POHypophosphorus acid
3 2
HO
OP PO O
P
O
OH
O OH
O
(HPO )Cyclotrimatephosphoric acid
33
O
P
OHO O
O
P
OHO
O
P
OHO
(HPO )Polymetaphosphoric acid
n3
Hydrogen directly attached with phosphorus is reducing in nature while hydrogen of O – Hbonds is acidic.
Group - 16 Elements
Electron Gain Enthalpies
The elements of group 16 have relatively high electron gain enthalpy values which decreasesdown the group from S to PO. Oxygen has less negative electron gain enthalpy than sulphur. Thisis due to its small size of oxygen atom. As a result there is strong interelectronic repulsions in the
61 Chemistry
relative small 2p orbitals of oxygen and thus the incoming electron experience less attractioncompared to sulphur.
Element O S Se Po
Electron gain enthalpy (kJ mol–1) –140.9 –200.7 –195.1 –180.0
Oxidation State
The stability of –2 oxidation state decreases down the group. Due to very high electronegativityof oxygen it may shows –2 oxidation state (Exception O2F2 (+1), OF2 (+2), H2O2 (–1). Other elementsof the group exhibit +2, +4 and +6 oxidation states. The stability +6 oxidation state decreases andthat of +4 increases down the group (Inert pair effect).
Catenation : Oxygen has some but sulphur has greater tendency for catenation. Oxygenchains are limited to two atoms as in peroxides but sulphur chains contain upto four atoms as inpolysulphides.
2 42 2 2 2 2 3
, , ,or H Sor H O or H S or H S Polysulphides
Hydrogen peroxide
H O O H H S S H H S S H H S S S S H
Sulphur possesses the maximum tendency of catenation due to the highest bond strength ofS – S bond (O – O = 142, S – S = 226, Se – Se = 172 and Te – Te = 126 kJ mol–1). The tendencyfor catenation decreases markedly as we go down the group from S to Te. Oxygen possesses thisproperty to a very less extent. The decreasing order of catenation amongst group 16 elements is
S > Se > O > Te
Chemical Properties
Formation of Hydrides
All these elements form volatile, stable, bivalent hydrides of the formula H2E i.e., H2O, H2S,H2Se, H2Te and H2P0. The central atom (E) in these hydrides is sp3 hybridized. Due to the presenceof two lone pairs on the central atom, these have bent (V) shapes.
(i) Melting Points and Boiling Points : H2O has the highest and H2S has the lowest valuesof melting and boiling points. Their decreasing order is
H2O > H2Te > H2Se > H2S
(ii) Volatility : As H2O has the highest and H2S has the lowest boiling point, the volatilityincreases abruptly from H2O to H2S and then decreases from H2S to H2Te. Thus, H2O isleast volatile and H2S is most volatile hydrides of group 16 elements. Therefore, volatilityincreases in the order
H2O < H2Te < H2Se < H2S
(iii) Thermal Stability of the hydrides decreases from H2O to H2Te i.e. H2O > H2S > H2Se >H2Te. This is due to the reason that as the size of the central atom in H2E increases, theH – E bond becomes weaker and breaks easily on heating.
(iv) Reducing Character : Hydrides of all these elements except that of oxygen, i.e. H2O,behave as reducing agents. The reducing character of these hydrides increases as the thermalstability decreases from H2S to H2Te i.e.
62 Chemistry
H2S < H2Se < H2Te
(v) Bond Angles : The hydrides of group 16 are V-shaped in which the bond angles decreaseon moving down the group as the electronegativity of the central atom decreases from O toTe. Consequently, the bond pairs of electrons move away from the central atom and thebond pair- bond pair repulsion decreases. The bond angles, therefore, decrease in the order
2 2 2 2104 92 91 90H O H S H Se H Te
Formation of Halides
(a) Hexahalides : All the elements except oxygen form hexafluorides, i.e. SF6, SeF6 and TeF6.No other halogen forms stable hexahalides. The hexafluorides have octahedral (sp3d2) shapes.The stability of these hexafluorides decreases from SF6 to TeF6, i.e., SF6 is practically inert,SeF6 is slightly more reactive while TeF6 is hydrolysed by water. This is due to the reasonthat as the size of the atom increases, the polarity of the M – X bond increases and hencethe hexafluoride molecule becomes more susceptible to nucleophilic attack by water. Thus,the ease of hydrolysis varies as SF6 < SeF6 < TeF6.
(b) Tetrahalides : All the elements except oxygen form tetrafluorides (SF4, SeF4 and TeF4)and tetrachlorides (SCl4, SeCl4, TeCl4). These have trigonal bipyramid (sp3d) geometry. Asone equatorial position is occupied by a lone pair of electrons which repels the axial bondpairs thereby decreasing the angle from 180° to 173°, these halides have see-saw shape.
See-saw shape of SF molecule.4
Formation of Oxides
These elements form a variety of oxides in different oxidation states from + 2 to + 6.
The acidic character of oxides of these elements in the same oxidation state decreases as wemove down the group. Thus SO2 and SeO2 are acidic while TeO2 and PoO2 are amphoteric. Theorder varies as SO2 > SeO2 > TeO2 > PoO2 and SO3 > SeO3 > TeO3.
Reducing property of EO2 decreases from SO2 to TeO2. SO2 is reducing while TeO2 is anoxidising agent.
(a) Sulphur Dioxide, SO2 is gas and forms discrete molecules even in the solid state. It isacidic in nature and is also called anhydride of sulphurous acid (H2SO3). It can act as a reducingagent and also as an oxidising agent. It can also act as a bleaching agent in the presence of amoisture. It bleaches due to reduction and its bleaching action is temporary. (SO2 + 2H2O H2SO4+ 2H). The colour is, however, restored when the bleached article is exposed to air since the oxygen
63 Chemistry
of the air oxidises the colourless compound back to the original coloured substance. SO2 acts as aLewis base due to the presence of a lone pair of electrons on S atom. It also acts as a ligand andforms numerous coordination compounds.
SO2 molecule has a bent structure with a O – S – O bond angle of 119°. The -bonds betweenS and O are formed by sp2 – p overlap while one of the -bonds arises from p – p overlap andthe other from p – d overlap but even the both of the S – O bonds are identical (143 pm) dueto resonance.
Ozone (O3)
Ozone is an allotrope of dioxygen (O2) and is prepared by passing a silent electric dischargethrough dioxygen.
3O2 2O3 rH= + 142 kJ/mol
O3 is thermodynamically unstable with respect to oxygen (because rG> 0)
It liberates nascent oxygen (O3 O2 + O). Therefore, it oxidises PbS to PbSO4, and KI to I2.
PbS + 4O3 PbSO4 + 4O2
2I– + H2O + O3 2OH– + I2 + O2
The liberated Iodine is titrated against a standard solution of sodium thiosulphate toestimate O3.
I2 + 2S2O32– 2I– + S4O6
2–
Nitrogen oxides emitted from the exhaust system of supersonic aeroplanes deplete theconcentration of ozone layer in atmosphere.
Sulphurdioxide, SO2
It is prepared by
(i) treating a sulphite with dilute H2SO4 : SO32– + 2H+ SO2 + H2O
(ii) S + O2 SO2
(iii) 4FeS2 + 11O2 2Fe2O3 + 8SO2
Properties
It aqueous solution is called sulphurous acid (H2SO3)
SO2 + H2O H2SO3
It reacts with NaOH forming Na2SO3, which then reacts with more SO2 to form NaHSO3.
2NaOH + SO2 Na2SO3 + H2O
Na2SO3 + H2O + SO2 NaHSO3
Reducing action of SO2 : It reduces Fe3+ ions to Fe2+ ions and acidified MnO4– to Mn2+.
2Fe3+ + SO2 + 2H2O 2Fe3+ + SO42– + 4H+
5SO2 + 2MnO4
– + 2H2O 5SO42– + 4H+ + 2Mn2+
64 Chemistry
Sulphuric Acid, H2SO4
It is manufactured by contact process which involves the following steps :
(i) Burning of sulphur to form SO2
(ii) Catalytic conversion of SO2 to SO3
2 52 2 32 2 196.6V OSO O SO rH kJ mol
(iii) Absorption of SO3 into H2SO4 (conc.) to form oleum, H2S2O7
Properties
(a) Acidic nature :
2 4 2 3 4 Virtually completeH SO H O H O HSO
22 4 2 3 4 about 10% completeH SO H O H O SO
(b) Dehydrating properties :
2 4
26 12 6 6 6H SO
H OC H O C
2 4
2
H SOH OHCOOH CO
(c) Oxidising properties : Hot conc. H2SO4 is less powerful oxidising agent that conc. HNO3
H2SO4 SO2 + H2O + O
2 4 4 2
2 4 2 42 2Cu O H SO CuSO H OCu H SO H O CuSO
2 4 2 2 490%
2 2acid
Zn H SO SO H O ZnSO
2 4 2 2 490%
5 4 4acid
Zn H SO H S H O ZnSO
2 4 2
diluteZn H SO H ZnSO
2 4 2 2 2 22
ConcentratedHBr I H SO H O SO Br I
(d) Preparation of more volatile acids from their corresponding salts : 2MX + H2SO4(conc.) 2HX + M2SO4
(M = Metal and X = F, Cl, NO3)
65 Chemistry
Oxoacids of Sulphur
O
S
OHO Sn–2
SOH
O
O(H S O ) and = 3 to 6n2 6n
O
S
HOHO O
S
HOHO O
O
S
OHO O
O
S
HOO O
O
S
OHO O
SO
OH
O
Sulphurous acid(H SO )2 3
Sulphurous acid(H SO )2 4
Peroxodisulphuric acid(H S O )2 2 8
Pyrosulphuric acid (oleum)H S O2 2 7
66 Chemistry
1. Which of the following are isoelectronic and isostructural? 23 3 3 3, , ,NO CO CIO SO
(a) 23 3,NO CO (b)
3 3,SO NO
(c) 23 3,ClO CO (d) 2
3 3,CO SO
2. The acid having O – O bond is
(a) H2S2O3 (b) H2S2O6
(c) H2S2O8 (d) H2S4O6
3. The shape of O2F2 is similar to that of
(a) C2F2 (b) H2O2
(c) H2F2 (d) C2H2
4. The ONO angle is maximum in
(a) 3NO (b)
2NO
(c) NO2 (d) 2NO
5. The correct order of bond angles (smallest first) in H2S, 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
6. The number of P – O – P bonds in the structures of P4O10 and P4O6 are respectively
(a) 6, 6 (b) 5, 5
(c) 5, 6 (d) 6, 5
7. Which of the following statement is not correct?
(a) Solid PCl5 exists as tetrahedral [PCl4]+ and octahedral [PCl6]– ions.
(b) Solid N2O5 exists as [NO2]+ [NO3]–
(c) Solid PBr5 exists as [PBr4]+ Br–
(d) Oxides of phosphorus P2O3 and P2O5 exist as monomers.
8. There is no S – S bond in
(a) 22 4S O (b) 2
2 3S O
(c) 22 5S O (d) 2
2 7S O
9. White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a
(a) dimerization reaction (b) disproportionation reaction
(c) condensation reaction (d) precipitation reaction
67 Chemistry
10. In pyrophosphoric acid, H4P2O7, number of – and d–p bonds respectively
(a) 8 and 2 (b) 6 and 2
(c) 12 and zero (d) 12 and 2
11. The following two reactions of HNO3 with Zn are given as (equations are not balanced)
3 3 22.Zn conc HNO Zn NO X H O A
3 3 22Zn dilute HNO Zn NO Y H O B
In reactions A and B, the compounds X and Y respectively are
(a) NO2 and NO (b) NO2 and NO2
(c) NO and NO2 (d) NO2 and NH4NO3
12. The reaction of P4 with X leads selectively to P4O6. The X is
(a) dny O2 (b) a mixture of O2 and N2
(c) Moist O2 (d) O2 in the presence of aqueous NaOH
13. The number of -bonds in P4O10 is
(a) 6 (b) 16
(c) 20 (d) 7
14. Which of the following has p – d bonding?
(a) 3NO (b) 2
3SO
(c) 33BO (d) 2
3CO
15. The percentage of p-character in the orbitals forming p – p bonds in p4 is
(a) 25 (b) 33
(c) 50 (d) 75
16. The correct order of increasing bond angles in the following triatomic species is
(a) 2 2 2NO NO NO (b) 2 2 2NO NO NO
(c) 2 2 2NO NO NO (d) 2 2 2NO NO NO
17. Reaction of HNO3 with I2, S8, P4 and C give respectively
(a) HIO3, H2SO4, H3PO4 and CO2 (b) HIO3, H2SO4, H3PO3 and CO2
(c) I2O5, H2SO4, H3PO4 and CO (d) I2O5, SO2, P2O5 and CO2
18. Two types of F – E – F angles are present in which of the following molecules [E = S,Xe and C]
(a) SF4 (b) XeF4
(c) SF6 (d) CF4
68 Chemistry
19. The number of hydrogen atoms attached to phosphorus atom in hypophosphorus acid is
(a) 0 (b) 2
(c) 1 (d) 3
20. Which of the following is not correct about structure of white phosphorus
(a) It has six P – P single bonds. (b) It has four P – P single bonds.
(c) Four lone pairs of electrons. (d) P – P – P angle of 60°
1. (a) 2. (c) 3. (b) 4. (d)
5. (c) 6. (a) 7. (d) 8. (d)
9. (b) 10. (d) 11. (d) 12. (b)
13. (b) 14. (b) 15. (d) 16. (d)
17. (a) 18. (a) 19. (b) 20. (b)
69 Chemistry
GENERAL CHARACTERISTICS OF 15 GROUP ELEMENTS
Electronegativity
Fluorine is the most electronegative element in the periodic table. With increase in atomicnumber down the group, the electronegativity decrease.
Electron gain enthalpy : Electron gain enthalpies of chlorine, bromine and iodine becomeless negative as the size of the atom increases. The electron gain enthalpy of fluorine is, however,less negative than that of Cl because of its small size as a result of which inter electronic repulsionspresent in its 2p-subshell are comparatively large. Thus, chlorine has the highest negative electrongain enthalpy.
Element F Cl Br I
Electron gain enthalpy (kJ mol–1) – 333 – 349 – 325 – 296
Oxidation states. All the halogens show an oxidation state of – 1. Fluorine being the mostelectronegative element always shows an oxidation state of – 1 while other halogens also showpositive oxidation states up to a maximum of + 7 (i.e. + 1, + 3, + 5, and + 7) due to the availabilityof vacant d-orbitals in the valence shell of these atoms. Some halogens also show + 4 and + 6oxidation states in oxides and oxo acids.
Colour
All the halogens have characteristic colours. F2 is light yellow, Cl2 is greenish yellow, Br2 isreddish brown and I2 is deep violet. The colour of halogens is due to the reason that their moleculesabsorb light in the visible region as a result of which electrons are excited to higher energy levels.
Bonds Dissociations Energy or Enthalpy of Dissociation
Bond dissociation energies of chlorine, bromine and iodine decrease down the group as the sizeof the atom increases. The bond dissociation energy of F2, is however, lower than those of Cl2 andBr2 because of large inter electronic repulsions among the lone pairs in F2 molecule.
X2 F2 Cl2 Br2 I2
Bond dissociation enthalpy (kJ mol–1) 158.8 242.6 192.8 151.1
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Oxidising Power : All the halogens act as strong oxidising agents since they have a strongtendency to attract electrons and have positive values of electrode potential (E°). The oxidisingpower, however, decreases as we move down the group from F to I i.e. F2 > Cl2 > Br2 > I2.
Since F2 is the strongest oxidising agent, it will oxidise all other halide ions to halogens insolution or even in the solid phase.
F2 + 2X– 2F– + X2 (X = Cl, Br to I)
Similarly, Cl2 will displace Br– and I– ions from their solutions while Br2 will displace I– ionsonly.
Cl2 + 2X– 2Cl– + X2 (X = Br or I); Br2 + 2I– 2Br– + I2
Hence F2 is the strongest and I2 is the weakest oxidising agent. This is also indicated by thedecrease in the electrode potential (E°) for the reaction X2(aq) + 2e– 2X– (aq) on moving downthe group.
Formation of Halides
Halogens combine with all the elements except He, Ne and Ar forming a large number ofbinary halides.
(a) Halides of Metals
(i) There is a regular gradation from ionic to covalent bonding as the atomic number of thehalogen increases for the same metal atom (M)
The ionic character of M – X bond and m.p. and b.p. of halides decrease in the orderM – F > M – Cl > M – Br > M – I.
(ii) Metals of low ionization enthalpies such as alkali metals form ionic halides whereas metalswith high ionization enthalpies such as transition metals form covalent halides. Molecularhalides show decrease in m.p. and b.p. as MI > MBr > MCl > MF.
(iii) Halides of metals in their higher oxidation states are more covalent than those formed inlower oxidation states. For example, SnCl4 is more covalent than SnCl2.
Similarly PbCl4, SbCl5 are more covalent than PbCl2 and SbCl3 respectively. Less ionichalides such as AgX show the solubility trends as AgI < AgBr < AgCl < AgF in water.
(b) Halides of Non-Metals
(i) Halides of non-metals are essentially covalent in nature.
(ii) The strength of M – X bond for a particular non metal (M) decreases in the order M – F> M – Cl > M – Br > M – I. Thus for hydrogen halides HX, the bond strength of H – X bonddecreases from HF to HI as the atomic size of the halogen increases from F to I i.e. H – F> H – Cl > H – Br > H – I.
(c) Reducing Character of Halides
The halide ions (X–) behave as reducing agents and their reducing power decreases in theorder :
I– > Br– > Cl– > F–
Aqueous solutions of hydrogen halides are known as hydrohalic acids.
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(i) Thermal stability of the hydrogen halides decrease from HF to HI i.e.
HF > HCl > HBr > HI
(ii) Acidic strength : The Acidic strength of hydrogen halides decreases from HI to HF i.e. HI> HBr > HCl > HF.
Oxoacids of Halogens
(i) Fluorine forms only one oxoacid HOF since it is the strongest oxidising agent Chlorine,bromine and iodine mainly form four series of oxoacids – namely halic (I) acid or hypohalousacid (HOX), halic (III) acid or halous acid (HOXO), halic (V) acid or halic acid (HOXO2)and halic (VII) acid or perhalic acid (HOXO3).
(ii) Acidic Character : All these acids are monoprotic containing an – OH group. The acidiccharacter of the oxoacids increases with increase in oxidation number of the halogen, i.e.,
HClO < HClO2 < HClO3 < HClO4
Oxoacids HOCl HOClO COClO2 HOClO3
pKa 7.5 2.0 – 1.2 – 10
(iii) Oxidising Power and Thermal Stability of Oxo Acids : The oxidising power of theseacids decreases as the oxidation number of the halogen increases, i.e., HClO > HClO2 >HClO3 > HClO4. This is due to the reason that as the oxidation number incrases, thehalogen-oxygen bond becomes more covalent. As a result, the thermal stability of anions ofoxo acids increases. Thus, hypohalites are stronger oxidising agents than perhalates.
Chlorine (Cl2)
It can be prepared from the following reactions
(i) By the oxidation of HCl with MnO2
MnO2 4HCl MnCl2 + Cl2 + H2O
Mixture of NaCl and conc. H2SO4 may be used in place of HCl.
(ii) By the oxidation of HCl with KMnO4
2KMnO4 + 16HCl 2KCl + 2MnCl2 + 8H2O + 5Cl2
(iii) By the electrolysis of brine (concentrated solution of NaCl)
Properties
(a) Reaction with metals and non-metals
2Al + 3ACl 2AlCl3, 2Fe + 3Cl2 2FeCl3
P4 + 6Cl3 4PCl3, S8 + Cl2 4S2Cl2
(b) Reaction with ammonia :
2 3 2 43 8 6
ExcessCl NH N NH Cl
2 3 33 3
ExcessCl NH NCl HCl
72 Chemistry
(c) Reaction with water and alkalies :
2 2Cl g water Cl
Cl (aq) + 2H O22 H O + Cl + HOCl+ –3
OH– OH–
2H O2 H O + OCl–2
Cl2 reacts with cold and dilute NaOH solution to form NaCl and NaClO :
2 22cold and dilute hypochloriteCl OH Cl OCl H O
Cl2 reacts with hot and concentrated NaOH solution to form NaCl and NaClO3
2 3 2.6 5 3
hot and concCl OH Cl ClO H O
(d) Oxidising and bleaching action of Cl2 :
Cl2 + H2O HCl + HClO
nascent oxygenHClO HCl O
It oxidises Fe2+ to Fe3+, SO32– to SO4
2–, SO2 to H2SO4, I2 to HIO3 and Br–/I– to Br2/I2.Bleaching action of Cl2 is due to oxidation in presence of moisture.
Cl2 + H2O 2HCl + O
Coloured substance + (O) oxidised from (colourless)
Interhalogen Compounds
The compounds of one halogen with the other are called interhalogen compounds. The stabilityof interhalogens increase as the size of central halogen increases (or with the increase in theelectronegativity difference of halogens).
For types of interhalogens ´ ´ ´3 5 7´, , andXX XX XX XX are formed where X is halogen of
larger size and X´ of smaller size. X´ is more electronegative than X. They are more reactive thanhalogens except F2 because X – X´ bond dissociation enthalpy is less than that of X – X bond exceptF – F bond. Their molecular geometries or structures can be predicted with the help of VSEPRtheory.
Groups 18 elements show least reactivity because :
(i) Noble gases except helium (1s2) have stable closed shell ns2np6 electronic configuration.
(ii) They have high ionisation enthalpies and more positive electron gain enthalpies.
Neil Bartlett prepared first noble gas compound, Xe+ [PtF6]–. True chemical compounds ofhelium, neon and argon are not yet known because of their high ionisation enthalpies.
Xenon-fluorine compounds, XeF2, XeF4 and XeF6 are powerful fluorinating agents and areformed by the direct reaction of xenon with fluorine under appropriate conditions.
73 Chemistry
Xenon fluorides react with F– acceptors like PF5, SbF5, etc. to form catonic species and F–
donor like alkali metal fluoride to form fluoroanions.
XeF2 + PF5 [XeF]+ [PF6]–
XeF4 + SbF5 [XeF3]+ [SbF6]–
XeF6 + MF M+ [XeF7]–
Hydrolysis
XeF2 is hydrolysed to give Xe and O2 :
XeF2 + 2H2O 2Xe + 4HF + O2
Hydrolysis of XeF4 and XeF6 with water gives XeO3.6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2
XeF6 + 3H2O XeO3 + 6HFPartial hydrolysis is XeF6 gives oxyfluorides, XeOF4 and XeOF2.
XeF6 + H2O XeOF4 + 2HFXeF6 + 2H2O XeO2F2 + 4HF
XeO3 is a colourless explosive solid and has a pyramidal molecular structure XeOF4 is acolourless volatile liquid and has square pyramidal molecular structure.
74 Chemistry
1. Which of the following is not true?
(a) Among halide ions, iodide ion is the most powerful reducing agent.
(b) Fluorine is the only halogen which does not show a variable oxidation state.
(c) HOCl is a stronger acid than HOBr
(d) HF is a stronger acid than HCl
2. Total number of lone pair of electrons in XeOF4 is
(a) 0 (b) 1
(c) 2 (d) 3
3. Among the following, the pair in which two species are not isostructural, is
(a) SiF4 and SF4 (b) IO3– and XeO3
(c) BH4– and NH4
+ (d) PF6– and SF6
4. Which one of the following arrangements represents the correct order of electron gainenthalpy (with negative sign) of the given atomic species.
(a) F < Cl < O < S (b) S < O < Cl < F
(c) O < S < F < Cl (d) Cl < F < S < O
5. In the case of alkalimetals, the covalent character decreases in the order :
(a) MF > MCl > MBr > MI (b) MF > MCl > MI > MBr
(c) MI > MBr > MCl > MF (d) MCl > MI > MBr > MF
6. Which is not the correct order for stated properly?
(a) Ba > Sr > Mg ; atomic radius
(b) F > O > N ; first ionisation enthalpy
(c) Cl > F > I ; negative electron gain enthalpy
(d) O > Se > Te ; electronegativity.
7. In which case the order of acidic strength is not correct?
(a) HI > HBr > HCl (b) HIO4 > HBrO4 > HClO4
(c) HClO4 > HClO3 > HClO2 (d) HF > H2O > NH3
8. In which of the following arrangements, the sequence is not strictly according to the propertywritten against it?
(a) B < C < 0 < N increasing first ionisation enthalpy
(b) HF < HCl < HBr < HI, increasing acid strength
(c) CO2 < SiO2 < SnO2 < PbO2, increasing oxidising power
(d) NH3 < PH3 < AsH3 < SbH3, increasing basic strength.
75 Chemistry
9. Which one of the following reactions of Xenon compounds is not feasible?
(a) Xe F6 + RbF Rb [XeF7]
(b) XeO3 + 6HF XeF6 + 3H2O
(c) 3XeF4 + 6H2O 2Xe + XeO3 + 12HF + 3/2 O2
(d) 2XeF2 + 2H2O 2Xe + 4HF + O2
10. Identify the incorrect statement among the following
(a) Ozone reacts with SO2 to give SO3
(b) Silicon reacts with NaOH (aq) in the presence of air to give Na2SiO3 and H2O
(c) Cl2 reacts with excess of NH3 to give N2 and HCl
(d) Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO3 and H2O.
11. Which is the most easily liquifiable rare gas?
(a) Ar (b) Ne
(c) Xe (d) Kr
12. Among the following molecules,
(i) XeO3 (ii) XeOF4 (iii) XeF6
those having same number of lone pairs on Xe are
(a) (i) and (ii) only (b) (ii) and (iii) only
(c) (ii) and (iii) only (d) (i), (ii) and (iii)
13. Which can do glass etching?
(a) HIO4 (b) SiF4
(c) HF (d) HNO3
14. High concentration of fluoride are poisonous and harmful to bones and teeth at levels over
(a) 1 ppm (b) 3 ppm
(c) 5 ppm (d) 10 ppm
15. In which of the following molecules, are all the bonds not equal
(a) NF3 (b) ClF3
(c) BF3 (d) AlF3
16. In BrF3 molecule, the lone pairs occupy equatorial positions to minimise
(a) lone pair–bond pair repulsions only
(b) bond pair–bond pair repulsions only
(c) lone pair–lone pair and lone pair–bond pair repulsions
(d) lone pair–lone pair repulsions only.
76 Chemistry
17. Which of the following contains maximum number of lone pairs of electron on the centralatom?
(a) ClO3– (b) XeF4
(c) SF4 (d) I3–
18. Which among the following factors is/are most important in making fluorine the strongestoxidising agent?
(a) Electron affinity (b) ionization energy
(c) hydration energy (d) low bond dissociation energy andhigh hydration energy
19. Which products are expected from the disproportionation reaction of hypochlorous acid?
(a) HClO3 and Cl2O (b) HClO2 and HClO4
(c) HCl and Cl2O (d) HCl and HClO3
20. Which of the following hydrogen halides is most volatile?
(a) HF (b) HCl
(c) HBr (d) HI
1. (d) 2. (b) 3. (a) 4 (d)
5. (c) 6. (b) 7. (d) 8. (d)
9. (b) 10. (b) 11. (c) 12. (d)
13. (c) 14. (b) 15. (b) 16. (c)
17. (d) 18. (c) 19. (d) 20. (b)
77 Chemistry
Melting and Boiling Points : These metals have very high melting and boiling points dueto stronger metallic bonding. The melting points of the transition elements first rise to a maximumand then fall as the atomic number increases Manganese and technetium have abnormally lowmelting points.
In a particular series, the metallic bond strength increases upto the middle with increasingnumber of unpaired electrons. Thus Cr, Mo and W have maximum number of unpaired d-electronsand have highest melting points in their respective series. Tungsten has the highest melting point(3683 K) among the d-block elements.
As there are no unpaired electrons in Zn, Cd and Hg, they have low melting points. Hg isliquid at ordinary temperature with lowest melting point (234 K) among the transition metals.
Ionization Enthalpies : The first ionization enthalpies of d-block elements are higher thanthose of s-block elements and are lesser than those of p-block elements. The ionization enthalpiesof 3d and 4 d-series are irregular but increase across the series while those 5d-series are muchhigher than 3d and 4 d-elements. This is because of the weak shielding of nucleus by 4 f electronsin 5d-transition series which result in greater effective nuclear charge acting on the outer valenceelectrons.
Electrode Potentials and Reducing Character : Quantitatively the stability of transitionmetal ions in different oxidation states in solution can be determined on the basis of electrodepotential data. The lower the electrode potential (i.e., more negative the standard reduction potential)of the electrode, more stable is the oxidation state of the transition metal ion in aqueous solution.Electrode potential values depend upon energy of sublimation of the metal, the ionization enthalpyand the hydration enthalpy.
Oxidation States : All transition elements except the first and the last member in eachseries show variable oxidation states. This is because difference of energy in the (n–1) d and nsorbitals is very little. Hence electrons from both the energy levels can be used for bond formation.
(i) The highest oxidation states of transition metals are found in fluorides and oxides sincefluorine and oxygen are the most electronegative elements. The highest oxidation state shown byany transition elements is + 8. Osmium shows highest oxidation state of + 8 in OsO4. Oxygen issuperior to fluorine in showing higher oxidation state because it has the ability to form multiplebonds with metals.
(iv) Lower oxidation states (zero or + 1) are stabilized by ligands which can accept electronsfrom the metal through -bonding (Such as CO), i.e., with -acceptor ligands.
(v) In going down a group, the stability of higher oxidation states increases while that of loweroxidation states decreases.
(vii) The relative stability of different oxidation states of transition metal atoms can bedetermined with the help of standard electrode potential data. For example, E° Values for the
78 Chemistry
couples Cr3+/Cr2+ = 0.41 V, Mn3+/Mn2+ = + 1.57V suggests that Cr2+ is unstable and is oxidised toCr3+ (which is more stable) and acts as a reducing agent whereas Mn3+ is unstable and is reducedto Mn2+ (which is more stable) and acts as an oxidising agent. It may be noted that both Cr2+ andMn3+ are d4 species.
Catalytic Properties : Many transition metals (like Co, Ni, Pt Fe, Mo etc.) and their compoundsare used as catalysts because of the following reasons.
(i) Transition metal ions change their oxidation states e.g.Fe3+ Catalyses the reaction between 1– and S2O2–
4 ions2Fe3+ + 2I– 2Fe2+ + I2
2Fe2+ + S2O82– 2Fe3+ + 2SO4
2–
(ii) Because of variable oxidation states, they easily combine with one of the reactants to formintermediate which reacts with the second reactant to form the final products.
(iii) They have tendency to adsorb reactants on the surface. This has the effect of increasing theconcentration of reactants and weakening of bonds in reactants.
Coloured Ions: Most of the transition metal compounds are coloured both in the solid stateand in aqueous solution. This is because of the presence of incompletely filled d-orbitals.
When white light falls on these compounds, some wavelength is absorbed for promotion ofelectrons from one set of lower energy orbitals to another set of slightly higher energy within thesame d-subshell. This is called d-d transition. The remainder light is reflected which has a particularcolour.
Colours of Cr2O72–, CrO4
2–, MnO–4, MnO4
2–, Cu2O, AgBr, AgI are due to charge transfertransitions.
Compounds of s and p-block elements are generally white as high energy is required forpromotion of s and p-electrons of incomplete subshells.
Magnetic Properties : Due to the presence of unpaired electrons in the (n-–1) d-orbitals, themost of the transition metal ions and their compounds are paramagnetic i.e., they are attracted bythe magnetic field. As the number of unpaired electrons increases from 1 to 5, the magneticmoment and hence paramagnetic character also increases. Those transition elements which havepaired electrons are diamagnetic i.e., they are repelled by the magnetic field.
The magnetic moment of species depends upon the sum of orbital and spin contributions foreach unpaired electron present. In transition metal ions, the orbital magnetic moment is largelysuppressed or quenched by the electrostatic field of other atoms, ions or molecules surrounding themetal ion. Thus the effective magnetic moment arises mainly from the spin of electrons.
A paramagnetic substance is characterised by its effective magnetic moment () which iscalculated by using spin-only formula 2 . . µ n n B M
where n is the number of unpaired electrons and B.M. stands for Bohr magneton.
Complex Formation : Transition metal ions form a large number of complexes in which thecentral metal ion is linked to a number of ligands. This is because of the following characteristicproperties of transition metal ions :
(i) They have high nuclear charge and small size i.e., charge/size ratio (charge density is large.
(ii) Thus in complexes, metal ions behave as Lewis acids and the ligands as Lewis bases.
79 Chemistry
Formation of Interstitial Compounds : Transition metals form a number of interstitialcompounds in which small non-metal atoms such as H, C, B, N and He occupy the empty spaces(interstitial sites) in their lattices and also form bonds with them.
Alloy Formation : Due to similarity in atomic sizes, atoms of one transition metal can easilytake up positions in the crystal lattice of the other in the molten state and are miscible with eachother forming solid solutions and smooth alloys on cooling. Alloys are generally harder, have highermelting points and more resistant to corrosion than the individual metals.
Some Important Compounds of Transition Elements
(1) Potassium Dichromate (K2Cr2O7)
Preparation : It is prepared from chromite ore (Fe2Cr2O4 or FeO.Cr2O3) through the followingreactions :
4FeCr2O4 + 16NaOH + 7O2 8Na2CrO4 + 2Fe2O3 + 8H2O
2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O
2Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl
(ii) In the solution, dichromate ions (Cr2O72–) exist in equilibrium with chromate ions
(CrO42–) as follows :
OH2 22 7 2 4H YellowOrange
Cr O H O 2CrO 2H –
+
In alkaline solution, equilibrium shifts in the forward direction and the solution is yellow. Inthe acidic medium, equilibrium shifts in the backward direction and the solution is orange. Thisconversion takes place with change in pH of the solution
2. Potassium Permanganate (KMnO4)
Preparation : It is prepared from pyrolusite ore (MnO2) through the following reactions :
2MnO2 + 4KOH + O2 2K2MnO4 + 2H2O(green)
MnO42– undergoes disproportionation in a neutral or acid medium to give KMnO4.
3MnO42– + 4H+ 2MnO4
– + MnO2 + 2H2O.(green) (purple)
(b) Electrolytic oxidation : This is the most preferred method. The manganate solution iselectrolysed. The electrode reactions are as follows :
At anode : 2 – –4 4
Green PurpleMnO MnO e At cathode : 2H+ + 2e– H2.
The solution is filtered and evaporated to get deep purple black crystals of KMnO4.
OXIDISING ACTION OF K2Cr2O7 AND KMnO4
Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O (acidic medium), E = 1.33 V
MnO4– + e– MnO4
2– (highly alkaline medium), E = 0.56 V
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MnO4– + 2H2O + 3e– MnO2 + 4OH– (alkaline medium), E = 1.69 V
MnO4– + 8H+ + 5e– Mn2+ + 4H2O (acidic medium), E = 1.52 V
Acidified K2Cr2O7 Oxidises
(i) Iodides to iodine : 2l I2 + 2e–
(ii) Surphides to sulphur : H2S S + 2H+ + 2e–.(iii) tin (II) to tin (IV) : Sn2+ Sn4+ + 2e–
(iv) Iron (II) to iron (III) : Fe2+ Fe3+ + e–
Acidified KMnO4 Oxidises
(i) Iodides to iodine : 2I– I2 + 2e–
(ii) Iron (II) to iron (III) : Fe2+ Fe3+ + e–
(iii) Hydrogen sulphides to sulphur : H2S 2H+ + S + 2e–
(iv) Oxalate iron or oxalic acid to CO2 : C2O42– 2CO2 + 2e–
(v) Sulphites to sulphate : SO32– + H2O SO4
2– + 2H+ + 2e–
(vi) Nitrite to nitrate : NO2– + H2O NO3
– + 2H+ + 2e–
(vii) hydrogen peroxide to O2 : H2O2 2H+ + O2 + 2e–
Neutral or Faintly Alkaline KMnO4 Oxidises
(i) Iodide to iodate : I– + 6OH– IO3– + 3H2O + 6e–
(ii) Thiosulphate to sulphate : S2O32– + 10.OH– 2SO4
2– + 5H2O + 8e–
Lanthanoids and Actinoids
(a) Lanthanoids : The elements with atomic numbers 58 to 71 i.e., cerium to letetium (whichcome immediately after lanthanum Z = 57) are called lanthanoids. These elements involvethe filling of 4f-orbitals. Their general electronic configuration is [Xe]4f1–14 5d0–16s2
Promethium (Pm), At. No. 61 is the only synthetic radioactive lanthanide.Lanthanoid contraction is due to imperfect shielding of 4f electrons. This causes the radiiof the members of third transition series to be very similar to those of the correspondingmembers of second transition series.Oxides and hydroxides of lanthanoids are basic in nature. Trivalent lanthanoid ions arecoloured (except La3+ and Lr3+) due f – f transition which occurs by the absorption of visiblelight.
(b) Actionoids : The elements with atomic numbers 90 to 103 i.e., thorium to lawrencium(which come immediately after actinium, Z = 89) are called actinoids or actinides.These elements involve the filling of 5 f-orbitals. Their general electronic configuration is[Rn] 5f1–14 6d0–1 7s2.5f orbitals in actinoids penetrate less into the inner core of electrons than 4f orbitals inlanthanoids, i.e., 5f electrons have poorer shielding effect than 4f electrons. Hence actioidcontraction from element to element is more than lanthanoid contraction.5f orbitals are not as buried as 4f orbitals. Hence 5f electrons participate in bonding to afar greater extent in actinoids compared with lanthanoids resulting in the participation of5f electrons of actinoids in bond formation.
81 Chemistry
UNIT – 16
1. MnO42– (1 mole) in neutral aqueous medium disproportionates to
(a) 2/3 mole of MnO4– and 1/3 mole of MnO2
(b) 1/3 mole of MnO4– and 2/3 mole of MnO2
(c) 1/3 mole of Mn2O7 and 1/3 mole of MnO2
(d) 2/3 mole of Mn2O7 and 1/3 mole of MnO2
2. In which of the following pairs of ions the lower oxidation state in aqueous solution is morestable than the other?
(a) Tl+, Tl3+ (b) Cu+, Cu2+
(c) Cr2+, Cr3+ (d) V2+, VO2+
3. The basis character of the transition metal monoxide follows the order
(a) VO > CrO > TiO > FeO (b) CrO > VO > FeO > TiO
(c) TiO > FeO > VO > CrO (d) TiO > VO > CrO > FeO
4. When MnO2 is fused with KOH, a coloured compound is formed. The product and itscolour is
(a) K2MnO4, purple green (b) KMnO4, purple
(c) K2MnO4 brown (d) Mn2O4, black.
5. Lanthanoids are
(a) 14 elements in the sixth period (atomic number 90 to 103) that are filling 4f sublevel.
(b) 14 elements in the seventh period (atomic number 90 to 103) that are filling 5fsubshell.
(c) 14 elements in the sixth period (atomic number 58 to 71) that are filling the 4fsubshell.
(d) 14 elements in the seventh period (atomic number 58 to 71) that are filling the 4fsubshell.
6. The product of oxidation of I– with MnO4– in alkaline medium is
(a) IO3– (b) I2
(c) IO– (d) IO4–
7. Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statementsabout cerium is incorrect?
(a) The common oxidation sates of cerium are +3 and +4.
82 Chemistry
(b) The +3 oxidation state of cerium is more stable than +4 oxidation state.
(c) The +4 oxidation state of cerium is not known in solutions.
(d) Cerium (IV) acts as an oxidizing agents.
8. Lanthanoids and actinoids resemble in
(a) electronic configuration (b) Oxidation state
(c) ionization energy (d) formation of complexes.
9. For decolorisation of 1 mole of KMnO4, the moles of H2O2 required is
(a) 1/2 (b) 3/2
(c) 5/2 (d) 7/2
10. The value of ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM.The correct one is
(a) d4 (in strong ligand field) (b) d4 (in week ligand field)
(c) d3 (in weak as well as strong fields)
(d) d5 (in strong ligand field)
11. Which of the following factor may be regarded as the main causes of lanthanoide contraction?
(a) Poor shielding of one 4f-electron by another in the subshell.
(b) Effective shielding of one 4f-electrons by another in the subshell.
(c) Poorer shielding of 5d-electrons by 4f-electrons.
(d) Greater shielding of 5d-electrons by 4f-electrons.
12. The aqueous solution containing which one of the following ions will be coloured :
(a) CuCl (b) K3[Cu(CN)4]
(c) CuF2 (d) [Cu(CH2CN)4]BF4
13. The spin only magnetic moment value in Bohr magnition units of Cr(CO)6 is
(a) 6 (b) 2.84
(c) 4.90 (d) 5.92
14. For the reduction of NO3– ion in an aqueous solution E° is 0.96V. Values of E° for some
metal ions are given below.
V3+(aq) + 2e– V(s)
Fe3+(aq) + 3e– Fe(s)
Au3+ (aq) + 3e– Au(s)
Mg2+ (aq) + 2e– Hg(l)
The metal/metals that can be oxidised by NO3– in aqueous medium are
(a) V and Au (b) Fe and Au
(c) Hg and Au (d) V, Fe and Hg
83 Chemistry
15. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is
(a) 4 (b) 6
(c) 2 (d) 7
16. The number of unpaired electrons in gaseous species of Mn3+, Cr3+ and V3+ respectively are__________ and most stable species is __________.
(a) 4, 3, 2 and V3+ is most stable
(b) 3, 3 and 2 and Cr3+ is most stable
(c) 4, 3 and 2 and Cr3+ is most stable
(d) 3, 3 and 3 and Mn3+ is most stable.
17. Which is not correct statement about the chemistry of 3d and 4f series elements?
(a) 3d elements show more oxidation states than 4f series elements
(b) The energy difference between 3d and 4s orbitals is very little.
(c) Europium (II) is more stable than Ce(II)
(d) The paramagnetic character in 3d series elements increases from scandium to copper.
18. A mixture of salts (Na2SO3 + K2Cr2O7) in a test tube is treated with dil H2SO4 and resultinggas is passed through lime water. Which of the following observations is correct about thistest?
(a) Solution in test tube becomes green and water turns milky.
(b) Solution in test tube is colourless and lime water turns milky.
(c) Solution in test tube becomes green and lime water remains clear.
(d) Solution in test tube remains clear and lime water also remains clear.
19. Which one of the elements with the outer orbital configuration may exhibit the largestnumber of oxidation states?
(a) 3d54s1 (b) 3d54s2
(c) 3d24s2 (d) 3d34s2
20. Out of TiF62–, CoF6
3–, Cu2Cl2 and NiCl42– the colourless species are
(Z of Ti = 22, Co = 27, Cu = 29, Ni = 28)
(a) Cu2Cl2 and NiCl42– (b) TiF62– and Cu2Cl2
(c) CoF63– and NiCl42– (d) TiF6
2– and CoF63–
21. Knowing that chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, whichof the following statement is incorrect?
(a) Because of the large size of the Ln(III) ions the bonding in its compounds ispredominantly ionic in character.
(b) The ionic sizes of Ln(III) decrease in general with increasing atomic number.
(c) Ln(III) compounds are generally colourless
(d) Ln(III) hydroxides are mainly basic in character.
84 Chemistry
22. Large number of oxidation states are exhibited by the actinoids than those by the lanthanoids,the main reason being
(a) more energy difference between 5f and 6d than between 4f and 5d orbitals.
(b) more reactive nature of the actionoids than the lanthanoids.
(c) 4f orbitals are more diffused than 4f orbitals
(d) lower energy difference between 5f and 6d than between 4f and 5d orbitals.
23. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4solution in the presence of H2SO4. The titration gives unsatisfactory result when carried outin the presence of HCl because HCl
(a) reduces permanganate to Mn(II)
(b) Oxidises oxalic acid to carbon dioxide and water
(c) gets oxidised by oxalic acid to chlorine
(d) furnishes H+ ions in addition to these from oxalic acid.
24. The correct order of decreasing second ionization enthalpy of Ti(22), Cr(24) and Mn(25) is
(a) Cr > Mn > V > Ti (b) V > Mn > Cr > Ti
(c) Mn > Cr > Ti > V (d) Ti > V > Cr > Mn
25. Identify the incorrect statement among the following:
(a) 4f and 5f orbitals are equally shielded
(b) d-block elements show irregular and erratic chemical properties among themselves
(c) La and Lu have partially filled d orbitals and no other partially filled orbitals
(d) The chemistry of various lanthanoids is very similar.
1. (a) 2. (a) 3. (d) 4. (a)
5. (c) 6. (a) 7. (c) 8. (b)
9. (c) 10. (a) 11. (a) 12. (c)
13. (a) 14. (d) 15. (b) 16. (c)
17. (d) 18. (a) 19. (b) 20. (b)
21. (c) 22. (d) 23. (a) 24. (a)
25. (a)
85 Chemistry
Consider the following coordination compound
[Co(H2O)Cl(en)2]Cl2
Here,
Co (Cobalt) : is the central metal atom/ion
(en)2, (H2O), (Cl–) are neutral didentate, neutral unidentate and anionic unidentate ligandsrespectively.
Ligands and central metal atom/ion are bonded through coordinate bonds to from coordinationsphere, which is enclosed within square brackets.
Coordination number of a metal atom is number of bonds between metal and ligands. e.g.,in [Co(en)2(H2O)Cl]Cl2, coordination number of Co is six.
Cl– is a counter ion which is outside the coordination sphere.
IUPAC Nomenclature of Coordination Compounds
(i) Name of the cation is written before the name of the anion.
(ii) Within a complex ion or coordination entity, names of ligands are cited before the metalatom/ion.
(iii) Identical ligands are represented by prefixing di, tri, tetra etc. to their name.
(iv) Different ligands are reported in alphabetical order.
(v) When the name of ligands includes a numerical prefix like mono or di. Their total numberis represented by prefixing bis for di, tris for tri, tetrakis for tetra and so on and the ligandto which it refers is enclosed in parenthesis.
(vi) Name of anionic ligand ends in o
Cl– = Chlorido Br– = Bromido
C2O42– = Oxalato H– = Hydrido
CN– = Cyanido EDTA4– = Ethylenediaminetetracetato
(vii) The names of neutral and cationic ligands are the same except NH3(ammine), CH3NH2
(methylamine/methanamine), H2O (aqua), CO (carbonyl), NO (nitrosyl) and CH2 NH2
CH2 NH2
(ethylenediamine/ethane-1,2-diamine).
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(viii) When the charge on coordination complex is negative name of central metal atom/ion endsin ‘ate.’
Iron (ferrum) – Ferrate Cobalt = Cobaltate
Silver (argentum) – argentate Nickel = Nickelate
Gold (aurum) – aurate
(ix) Oxidation state of metal atom is expressed in Roman numeral and is written following itsname and enclosed in parenthesis.
(x) Name of coordinate entity is a single word.
For e.g. [Co(H2O) Cl(en)2] Cl2 will be named as aquachloridobis (ethane-1, 2-diamine) cobalt(III) chloride.
Isomerism in Coordination Compounds
(i) Ionization Isomerism : This isomerism results when compounds with same molecularformula give different ions in the solution e.g., {Co (NH3)5Br]2+ SO4
2– and [Co(NH3)5(SO4)]+
Br–. Here counter ion itself is a potential ligand.
(ii) Solvate Isomerism : It arises when H2O molecule acting as ligand becomes water ofhydration outside the coordination sphere, i.e., here water forms a part of the coordinationentity or is outside it. For example, there are three hydrate isomers of CrCl3.6H2O as shownbelow :
[Cr(H2O)6]Cl3, [Cr(H2O)5Cl] Cl2.H2O, and [Cr(H2O)4Cl2] Cl.2H2O.
(iii) Linkage Isomerism : This isomerism results when a ligand attaches with central atom/ion in two different ways. Some common examples of such ligands are : nitrito–N–(–NO2)and nitrito–N–(–ONO), cyano (–CN) and isocyano (–NC), thiocyanato (–SCN), andisothiocyanato (–NCS), cyanato (–OCN) and isocyanato (–NCO).
Some example of linkage isomers are : [Co(NO2) (NH3)5]Cl2 and [Co(ONO)(NH3)5]Cl2;[Cr(H2O)5(SCN)]2+ and [Cr(H2O)5 (NCS)]2+
(iv) Coordination Isomerism : This type of isomerism is shown by compounds in which bothcation as well as anion are complexes (coordination entities) and ligands may interchangetheir position between the two complex ions. For example,
[CO(NH3)6] [Cr(C2O4)3] and [Cr(NH3)6] [Co(C2O4)3]
(v) Geometrical Isomerism
(a) Isomerism in complexes with coordination number 4 : Tetrahedral complexesdo not show geometrical isomerism because the relative position of the ligands withrespect to each other will be the same. The square planar complexes of the typeMA2X2, MA2XY, MA2BX and MABXY show this kind of isomerism, e.g., [Pt(NH3)2Cl2].
(b) Isomerism in complexes with coordination number 6 : Octahedral complexesof the type : MA4X2, MA3X3, MA4XY exist as cis-and trans-isomers. Common examplesare : [Co(NH3)4Cl2], [Co(NH3)4Cl(NO2)]
87 Chemistry
Complexes of the type : [M(AA)2X2] and [M(AA)2XY] also show cis-trans isomers(Here (AA) refers to symmetrical bidentate ligands and X and Y refer to anionicligands. Common examples are : [Co(en)2 Cl2]+, [Ni(OX)2Cl2]+ etc.
(ii) Optical Isomerism : Chiral molecules i.e., molecules which do not have plane of symmetry,exhibit optical isomerism. The optically active isomers called enantiomers or enantiomorphsare non superimposable on mirror images of each other and rotate the plane of polarisedlight equally but in opposite directions.
If the substance rotates the light to the right, i.e., in clockwise direction, it is dextrorotatoryor the d- form or (+) form. If the light is rotated in the anti-clockwise direction, the substanceis said to be levorotatory or l- form or (–) form. The optical isomers of a compound haveidentical physical and chemical properties.
(b) Isomerism in complexes with coordination number 6 : Octahedral complexes ofthe type [M(AA)X2Y2], [M(AA)2X2] and [M(AA)2XY] where (AA) is symmetricalbidentate ligand and X and Y are monodentate ligands, can be resolved into a pairof enantiomers. As seen earlier, the complexes of these types show geometricalisomerism i.e., exist as cis-and trans-isomers. The trans isomer does not show opticalisomerism since it is symmetrical while only cis-isomer show optical activity as it isunsymmetrical. Thus total three isomers may be possible in such cases.
Some common examples of the above type of complexes are :
[CO(en)(NH3)2Cl2]+, [Co(en)2Cl2]+
(ii) Valence bond theory (VBT) : This theory, put forward by Linus Pauling (1931), is usedto explain the shapes of complexes and relation between the observed magnetic behaviourand the bond type. It involves the hybridization of empty non equivalent orbitals of themetal atom/ion, each of which then can accept a lone pair of electrons donated by theligands. The shape is octahedral, tetrahedral, square, planar, trigonal bipyramidal (or squarepyramidal) or linear when hybridization is d2sp3, sp3, dsp2, dsp3 or sp respectively.
88 Chemistry
Complexes of some transition metal ions/atoms on the basis of V.B. Theory.
Complex Metal Electron distribution Type of No. ofIon 3s 3p 3d 4s 4p 4d Hybridisation unpaired
[V(H2O)63]+ V3+ 2 6 2 d2sp3 2Cr(NH3)6
3+ Cr3+ 2 6 3 d2sp3 3Mn(CN)6
3– Mn3+ 2 6 4(le– pairs up) d2sp3 2Fe(CN)63– Fe3+ 2 6 5 (2e– pair up) d2sp3 1Fe(CN)6
4– Fe2+ 2 6 6 (2e– pair up) d2sp3 0[FeF6]3– Fe3+ 2 6 5 sp3d2 5[Co(NH3)6]3+ Co3+ 2 6 6 (2e– [pair up) d2sp3 0[CoF6]3– Co3+ 2 6 6 sp3d2 4[Ni(CN)4]2– Ni2+ 2 6 8 (1e– pairs up) dsp2 0[Ni(CO)4] Ni 2 6 8 2 (Jump to 3d) sp3 0[Cu(NH3)2]+ Cu+ 2 6 10 sp 0[Cu(NH3)4]2+ Cu2+ 2 6 8 0 1 0 dsp2 1
(1e– promoted to 4p)[Cu(CN)4]3– Cu+ 2 6 10 sp3 0[Zn(NH3)4]2+ Zn2+ 2 6 10 sp3 0
(iii) Crystal Field Theory (CFT) : Crystal field theory is based on the assumption that themetal ion and the ligands act as point charges and the interaction between them is purelyelectrostatic, i.e., metal-ligands bonds are 100% ionic. The five d-orbitals is an isolatedgaseous metal atom/ion are degenerate. This degeneracy is maintained in a sphericallysymmetrical field of negative charges.
However, when this negative field is due to real ligands in a complex, the degeneracy of thed-orbitals is lifted due to asymmetrical field. This results in splitting of the d-orbital energies.The pattern of splitting depends upon (a) nature of the crystal field such as octahedral,tetrahedral of square planar and (b) basic strength of ligands.
1. Crystal field effects in octahedral coordination entities : In an octahedral complex,the six ligands approach the central metal atom lying at the origin symmetrical along the
Ene
rgy
d-orbitals in free ionAverage energy of
d-orbitals in sphericalcrystal field
d , d , d (t )Splitting of d-orbitals in
an octahedral crystal field
xy yz xz 2g
Splitting of five -orbitals in an octahedral crystal field.d
d – , d (e )2 gx y z2 2
0 D or 10 q
0.6 0
–0.4 0
89 Chemistry
cartesian axes. Initially, there is an increase in the energy of d-orbitals relative to that ofthe free ion (just like that in a spherical field). Next, the orbitals lying along the axes (dxy,dyz and dxz) which are lowered in energy relative to the average energy in the sphericalcrystal field as shown below :
This splitting of five degenerate d-orbitals of the metal ion into two sets of d-orbitals withdifferent energies is called crystal field splitting. The two sets of d orbitals, i.e., 2 2 2– andx y zd dand , andxy yz zxd d d are commonly called eg and t2g
orbitals respectively. The crystal field splittingis the energy difference between t2g
and eg orbitals and is frequently measured in terms of aparameter 0 where the subscript (0) stands for octahedral. (This is also measured in terms ofanother parameter called Dq. The magnitude of splitting is arbitrarily fixed at 10Dq so that(0 = 10Dq). The above energy level diagram shows that an electron will prefer to go into morestable t2g orbital is stabilized by –0.40 (–4Dq) energy while each electron occupying the eg orbitalshall be destabilized by an amount 0.60(6Dq) energy. This gain in energy achieved by preferentialfilling of stable t2g orbitals over the energy of a randomly filled d-orbital is called the crystal fieldstabilization energy (CFSE).
The net CFSE is equal to (6x – 4y) Dq where x and y are the number of electrons in eg andt2g orbitals respectively. The actual configuration adopted is decided by the relative value of 0 andP (where P is the pairing energy). The energy required to cause pairing of electrons in the sameorbital is called pairing energy. If 0 < P, we have a weak field and high spin or spin free complexand the fourth electron will enter one of the eg orbitals giving the configuration t2g
3eg1. If now a fifth
electron is added to a weak field coordination entitity, the configuration becomes t2g3e2
g. Thepairing of electrons will take place only if the gain in stability in terms of 0 is large enough toovercome the loss in stability due to electron pairing. When 0 > P, we have a strong field and lowspin or spin paired complex and pairing will occur in the t2g level with eg level remaining unoccupiedin d1 to d6 systems.
90 Chemistry
1. Which of the following is an organometallic compounds?
(a) Ti(C2H4)4 (b) Ti(OC2H5)
(c) Ti(O COCH3)4 (d) Ti (OC6H5)4
2. The oxidation state of Fe in the brown ring complex [Fe(H2O)5 (NO)] SO4 is
(a) +1 (b) +2
(c) +3 (d) +4
3. In the compound, lithiumtetrahydrido aluminate (III), the ligands is
(a) H+ (b) H
(c) H– (d) None of these
4. IUPAC name of [Pt(NH3)3 Br(NO2) Cl] Cl is
(a) Triamminechloroidobromidonitrito-N-platinum (IV) chloride)
(b) Triamminechloridobromidonitrito-N-platinum (IV) chloride
(c) Triamminebromidochloridonitrito-N-platinum (IV) chloride
(d) Triamminenitrito-N-chloridobromidoplatinum (IV) chloride.
5. The geometry of [Ni(CO)4] and [Ni(PPh3)2Cl2] are
(a) both square planar
(b) tetrahedral and square planar respectively
(c) both tetrahedral
(d) square planar and tetrahedral respectively.
6. Which one of the following complexes will have four isomers?
(a) [Co(en) (NH3)2Cl2]Cl (b) [Co(PPh3)2 (NH3)2Cl2]Cl
(c) [Co(en)3]Cl3 (d) [Co(en)2Cl2] Br (en = ethylenediamine)
7. The correct structure of Fe(CO)5 is
(a) octahedral (b) tetrahedral
(c) square pyramidal (d) trigonal bipyramidal.
8. Which of the following has magnesium?
(a) Chlorophyll (b) Haemocyanin
(c) Carbonic anhydrase (d) Vitamin B12
9. Which of the following shall form an octahedral complex?
(a) d4 (low spin) (b) d8 (high spin)
(c) d6 (low spin) (d) All of these
91 Chemistry
10. Which of the following is expected to be a paramagnetic complex?
(a) [Ni(H2O)6]2+ (b) [Ni(CO)4]
(c) [Zn(NH3)4]2+ (d) [Co(NH3)6]3+
11. Which of the following will give maximum numbers of isomers?
(a) [Co(NH3)4Cl2]+ (b) [Ni(en)(NH3)4]2+
(c) [Ni(C2O4) (en)2]2– (d) [Cr(SCN)2(NH3)4]+
12. Which of the following organometallic compound is and bonded?
(a) [Fe(n5 – C5H5)2] (b) K[PtCl3(n2–C2H4)]
(c) [Co(CO)5(NH)3]2+ (d) Fe(CH3)3
13. The complex which has no ‘d’ electrons in the central metal atom is
(a) [MnO4]– (b) [Co(NH3)6]3+
(c) [Fe(CN)6]3– (d) [Cr(H2O)6]3+
14. Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1complex of BF3 and NH3
(a) N : tetrahedral, sp3; B : tetrahedral, sp3
(b) N : pyramidal, sp3; B : pyramidal sp3
(c) N : pyramidal, sp3; B : planar, sp2
(d) N : pyramidal, sp3; B : tetrahedral, sp3.
15. Among the following the most stable complex is
(a) [Fe(H2O)6]3+ (b) [Fe(CN)6]3–
(c) [Fe(C2O4)3]3– (d) [FeCl6]3–
16. Among [Ni(CO)4], [Ni(CN)4]2–, [NiCl4]2– species, the hybridisation states at Ni atom arerespectively
(a) sp3, dsp2, dsp2 (b) sp3, dsp2, sp3
(c) sp3, sp3, dsp2 (d) dsp2, sp3, sp3
17. Considering H2O as a weak field ligand, the number of unpaired electrons in [Mn(H2O)6]2+
will be (atomic number of Mn = 25]
(a) 3 (b) 5
(c) 2 (d) 4
18. In Fe(CO)5, the Fe – CO bond possesses
(a) ionic character (b) – character only
(c) – character only (d) both and characters
19. How many EDTA (ethylenediaminetetraacetate) ions as ligands are required to make anoctahedral complex with a Ca2+ ion.
(a) one (b) two
92 Chemistry
(c) three (d) four
20. [C o(N H 3)4(NO2)2Cl] exhibits
(a) ionisation isomerism, geometrical isomerism and optical isomerism
(b) linkage isomerism, geometrical isomerism, and optical isomerism
(c) linkage isomerism, ionisation isomerism and optical isomerism
(d) linkage isomerism, ionisation isomerism and geometrical isomerism.
21. One mole of coordination compound, Co(NH3)5Cl3 gives three moles of ions on dissolutionin water. One mol of the same complex reacts with two moles of AgNO3 solution to yield twomoles of AgCl(s). The structure of the complex is :
(a) [Co(NH3)3Cl3].2NH3 (b) [Co(NH3)4Cl2] Cl.NH3
(c) [Co(NH3)4Cl] Cl2.NH3 (d) [Co(NH3)5Cl] Cl2
1. (a) 2. (b) 3. (c) 4. (c)
5. (c) 6. (d) 7. (d) 8. (a)
9. (c) 10. (a) 11. (d) 12. (b)
13. (a) 14. (a) 15. (c) 16. (b)
17. (b) 18. (d) 19. (a) 20. (d)
21. (d)
93 Chemistry
1. The total volume of atoms present in face-centered cubic unit cell of a metal is (r is atomicradius)
(a)320
3r (b)
3243r
(c)312
3r (d)
3163r
2. An ionic compound has unit cell consisting of A ions at the corners of a cube and B ions atthe centres of the faces of the cube. The empirical formula of the compound would be
(a) A3B (b) AB3
(c) A2B (d) AB
3. What type of crystal defect is indicated in the diagram below?
Na+ Cl– Na+ Cl– Na+ Cl–– –Cl Cl Na Na
– – –Na Cl Cl Na Cl
– –Cl Na Cl Na Na
(a) Frenkel and schottky defect (b) Schottky defect
(c) Interstitial defect (d) Frenkel defect.
4. Which among the following will show maximum osmotic pressure?
(a) 1M NaCl (b) 1M MgCl2(c) 1M (NH4)3 PO4 (d) 1M Na2SO4
5. If is the degree of dissociation of Na2SO4, the vant Hoff ’s factor (i) used for calculatingthe molecular mass is
(a) 1 + (b) 1 –
(c) 1 + 2 (d) 1 + 2
6. At certain hill station pure water boils at 99.725°C. If Kb for water is 0.513°C Kg mol–1, theboiling point of 0.69 m solution of urea will be
(a) 103°C (b) 100.079°C
(c) 100.359°C (d) unpredictable.
94 Chemistry
7. If all the following four compounds were sold at same price, which would be the cheapestfor preparing an antifreeze solution for a car radiator?
(a) CH3OH (b) C2H5OH
(c) C2H4(OH)2 (d) C3H5(OH)3
8. The molar solution of sulphuric acid is equal to
(a) 1 N solution of H2SO4 (b) 2N solution of H2SO4
(c) N/2 solution of H2SO4 (d) 3N solution of H2SO4
9. A tooth paste has 0.2 g L–1 fluoride. Its concentration in ppm will be :
(a) 250 (b) 200
(c) 400 (d) 1000
10. Solution boils at a temperature T1 and the solvent at a temperature T2, the elevation ofboiling point is given by
(a) T1 + T2 (b) T1 – T2
(c) T2 – T1 (d) T1 + T2
11. If salt bridge is removed from two half cells, the voltage
(a) Drops to zero (b) Does not change
(c) Increase gradually (d) Increases rapidly
12. A smuggler could not carry gold by depositing iron on the gold surface because
(a) gold is denser (b) iron rusts
(c) gold has higher reduction potential than iron
(d) gold has lower reduction potential than iron.
13. The charge required for the reduction of 0.4 mol of K2Cr2O7 to Cr3+ ions is
(a) 0.6 × 96500 C (b) 2.4 × 96500 C
(c) 6 × 96500 C (d) 12.4 × 96500 C
14. A cell is constituted as follows Pt, H2(1 atm.) |HA1||HA2|H2(1 atm), Pt. The pH of twoacids solutions HA1 and HA2 are 3 and 5 respectively. The emf of the cell is
(a) 0.059 V (b) 0.0295 V
(c) 0.118 V (d) –0.118 V
15. The values of m° for KCl and KNO3 are 149.86 and 154.96 –1 cm2 mol–1. Also oCl – is
71.44 ohm–1 cm2 mol–1. The value of °(NO3–) is
(a) 76.54 ohm–1 cm2 mol–1 (b) 133.08 ohm–1 cm2 mol–1
(c) 37.7 ohm–1 cm2 mol–1 (d) unpredictable.
16. Which of the following is a primary cell?
(a) Lead storage battery (b) Nickel–Cadmium cell
(c) Mercury cell (d) Both (a) and (b)
95 Chemistry
17. The units of rate constant for first order reaction is
(a) s–1 (b) mol L–1 s–1
(c) mol s–1 (d) L mol–1 s–1
18. A first order reaction is 75% complete after 32 minutes. When was 50% of the reactioncompleted?
(a) 4 minutes (b) 8 minutes
(c) 16 minutes (d) 32 minutes
19. Which of the following represents zero order reaction :
(a)t1/2
[R]0
(b)t1/2
[R]0
(c)t1/2
[R]0
(d)t1/2
[R]0
20. The rate constant of the reaction at temperature 200K is one-tenth of the rate constant at400K. What is the activation energy of the reaction?
(a) 1842.4R (b) 921.2R
(c) 460.6R (d) 230.3R
21. For a first order reaction, the plot of log K against 1/T is a straight line. The slope of theline is equal to
(a) aE–R (b)
a
2.303–E R
(c) aE–2.303 (d) aE–
2.303 R
22. Select the rate law that corresponds to the data shown for the following reaction A + B C.
Exp. No. (A)/M (B)/M Initial rate/Ms–1
1. 0.012 0.035 0.12. 0.024 0.070 0.83. 0.024 0.035 0.14. 0.012 0.070 0.8
96 Chemistry
(a) Rate = k [B]3 (b) Rate = k [B]4
(c) Rate = k [A] [B]3 (d) Rate = k [A]2 [B]2
23. A redioactive isotope has a half life of 10 days. If today 125 mg is left over, what was itsoriginal weight 40 days earlier?
(a) 2g (b) 600 mg
(c) 1g (d) 1.5g
24. The relationship between standard reduction potential of a cell and requilibrium constantis given by
(a) ocell c
nE log k0.059
(b) ocell c
0.059E log kn
(c) ocell cE 0.059 n log k (d) o c
celllog k
En
25. The specific conductance of 0.1M HNO3 is 6.3 × 10–2 ohm–1 cm–1. The molar conductance ofthe solution is
(a) 630 ohm–1 cm2 mol–1 (b) 315 ohm–1 cm2 mol–1
(c) 100 ohm–1 cm2 mol–1 (d) 63 ohm–1 cm2 mol–1
26. When mercuric iodide is added to an aqueous solution of KI, the
(a) freezing point is decreases (b) freezing point in raised
(c) boiling point does not change (d) freezing point does not change
27. Cloud or fog is colloidal system in which the dispersed phase and dispersion medium arerespectively
(a) Gas, Liquid (b) Liquid, Gas
(c) Liquid, Liquid (d) Solid, Liquid
28. When a river enters into the sea, a delta is formed. Formation of delta is due to
(a) peptization (b) coagulation
(c) Emulsification (d) Dialysis.
29. Gold number is the number of milligrams of the protective colloid which prevents thecoagulation of 10 mL of a gold hydrosol on adding 1 mL of a ____ solution of sodium chloride.
(a) 1% (b) 5%
(c) 25% (d) 10%
30. Which of the following enzymes catalyzes the conversion of proteins into amino acids?
(a) Pepsin (b) Amylase
(c) Nuclease (d) Carbonic anhydrase
31. During the process of electrolytic refining of copper some metals present as impurity settleas anode mud? These are
(a) Sn and Ag (b) Pb and Zn
(c) Ag and Au (d) Fe and Ni
97 Chemistry
32. Extraction of gold is done by the following process
Au + CN– X
(X) + Zn Y + A
The (X) and (Y) are
(a) [Au(CN)2]–, [Zn(CN)4]2– (b) [Au(CN)2]+, [Zn(CN)4]2–
(c) [Au(CN)4]–, [Zn(CN)2] (d) [Au(CN)4]–, [Zn(CN)4]–
33. Which method of purification is represented by the following equations?
773K 1675K2 4 2Ti 2I TiI Ti 2I
(a) Cupellation (b) Poling
(c) Van Arkel method (d) Zone refining
34. Ellingham diagram is used to explain
(a) Kinetics of metallurgical operations
(b) Thermodynamics of metallurgy
(c) Transition temperature
(d) None of these.
35. The correct order of the thermal stability of hydrogen halides (H–X)
(a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI
(c) HCl > HF > HBr > HI (d) HI > HCl > HF > HBr
36. The number of bonds in P4O10 is
(a) 6 (b) 16
(c) 20 (d) 7
37. Nitrogen forms N2 but phosphorus is converted into P4. The reason for this is
(a) Triple bond is present between phosphorus atoms.
(b) p – p bonding is weak
(c) p – p bonding is strong
(d) Multiple bond is formed easily.
38. (NH4)2 Cr2O7 on heating libertes a gas. The same gas will be obtained by heating :
(a) NH4 NO2 (b) NH4NO3
(c) H2O2 with NaNO2 (d) Mg3N2 with H2O
39. The formation of O2+[PtF6]– is the basis for the formation of Xenon fluorides. This is because
(a) O2 and Xe have comparable electronegativies
(b) O2 and Xe have comparable ionization ethalpies
98 Chemistry
(c) Both O2 and Xe are gases
(d) O2 and Xe have comparable sizes.
40. In the following reaction :
P4 + 3NaOH + 3H2O PH3 + 3NaH2PO2
(a) Phosphorus is oxidised (b) Phosphorus is oxidised and reduced
(c) Phosphorus is reduced (d) Phosphorus is neutralised
1. (d) 2. (b) 3. (b) 4. (c)
5. (c) 6. (b) 7. (a) 8. (b)
9. (b) 10. (b) 11. (a) 12. (c)
13. (b) 14. (a) 15. (a) 16. (c)
17. (a) 18. (c) 19. (a) 20. (b)
21. (d) 22. (a) 23. (a) 24. (b)
25. (a) 26. (b) 27. (b) 28. (b)
29. (d) 30. (a) 31. (c) 32. (a)
33. (c) 34. (b) 35. (b) 36. (b)
37. (b) 38. (a) 39. (b) 40. (b)