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Density of Unit Cell

3 A

Z Md

a Nd = Density (g/cm3)

Z = number of atoms per unit cell a = Edge length in cmValue 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 ZnS

3 0.414 0.732 6 Octahedral NaCl

3 0.732 1.000 8 Cubic CsCl

Relationship between atomic radius and edge length.

Simple Cubic BCC FCC

2

ar

3

4

ar

2

4

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.

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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 Fe

2+ 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%

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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 formulaAB 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 formed8. 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%

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10. The coordination number of metal crystallizing in hexagonal close packed structure is

(a) 4 (b) 6

(c) 12 (d) 811. A substance consisting of two elements, P and Q, has atoms of P occupying each corner

of 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 mol1, 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 ofNa+ and Cl ions are 95 pm and 181pm respectively. The edge length ofNaClunit cell is

(a) 276 pm (b) 138 pm

(c) 552 pm (d) 415 pm23. Potassium has a bcc structure with nearest neighbour distance of 4.52 . If atomic mass of

potassium is 39 and then its density is

(a) 454 kg m3 (b) 804 kg m3

(c) 852 kg m3 (d) 900 kg m3

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 ofbcc crystal lattice26. Copper crystallises infcc lattice with a unit length of 361 pm. What is the radius of copper

atom 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)

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Molarity (M) is the number of moles of solute dissolved per litre of solution. Its unit is mol/L.

Number of moles of solute 1000

Volume of solution in litre Volume of solutionB

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 kgof solvent. Its unit is mol/kg

Number of moles of solute 1000

Mass of solvent in kilograms in grams

B

B A

Wm

M 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 Fraction

It 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 solution

For 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

Henrys 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 :

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p or p = KH ; where KH is Henrys constant.

Raoults 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

Raoults 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., n

A+ n

B nA]

0soln

0

A B B BB

A B B AA

P P W M

M 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 Raoults 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 Raoults law.

Non-Ideal Solution Showing + ve and ve Deviations from Raoults 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

+P

Total

AB

PBPA

V

apourpressure

PA

A = 1

B = 0A = 0B = 1

Mole Fraction

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Carbon tetrachloride + Chloroform HCl +Water

Carbon tetrachloride + Benzene HNO3 + Water

Carbon tetrachloride + Toluene Acetic and + pyridine, phenol and aniline

Characteristics of Solution Showing ve Deviation from Faoults Law

(a) Hmix < 0

(b) Vmix < 0

(c) Vapour pressure of solution is less than what isgiven by Raoults 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 Raoults law.

0 0soln A A B B

P P p

0A A AP P

0B B BP P

Note : Curved lines show non-ideal behaviour

and 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 showing

positive 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 than

either 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

PB

PA

Vapourpressure

A = 1

B = 0A = 0B = 1

PA

PB

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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 ofNaCl 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 ofb.p. is related to the molatily (m) of the solution as below:

1000Bb b b

B A

WT K m K

M W

0b b bT T T Elevation of boiling point

Tb

1 atm = 1.013 bar

Tb

Tb

TemperatureKb = 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 K

M W

of f fT T T Depression of freezing point

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Kf

= Molal freezing point depression constant or cryoscopic constant.

Tf0

Tf 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 said

to show abnormal molecular mass.Abnormal molecular mass is obtained when the substance in the solution undergoes

dissociation (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.

Vant Hoff factor (i) is given by

No. of particles after dissociation or association of solute

Number of solute initially dissolvedi

particles

Experimental value of the colligative property

Calculated value of the colligative property

i

1As molecular mass

Colligative property

normal

observed

Calculated or normal molecular masshence

Observed molecular mass

Mi

M

Modified formulas for substance undergoing dissociation or association in the solution are

(i) Tb = i Kb m (ii) Tf = i Kf m

(iii) n

i RTV

(iv)0

soln.

0

AB

A

P pi

P

When solute dissociates to give n ions, the degree of dissociation () is related to Vant Hofffactor as given below

1

1

i

n

If n molecules of a solute undergo association in solution then degree of association () isgiven by

1

1

ni

n

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1. An azeotropic mixture of two liquids has boiling point higher than either of them when it.

(a) Shows positive deviation from Raoults law.

(b) Show no deviation from Raoults law.

(c) Show negative deviation from Raoults 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 21

2C H OH C H OH If xis the degree of association

of phenol, then the total number of moles of particles present at equilibrium is

(a) 1 x (b) 1 + x

(c) 12

x(d) 1

2

x

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, HXis 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 80C, the vapour pressure of pure liquid A is 520 mm Hgand that of pure liquid B is1000 mmHg. If the solution of A and B boils at 80C 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. 2gof C6H5COOHdissolved in 25g benzene shows a depression in freezing point equal to1.62K. (K

f= 4.9 K kg mol1) The value of vant 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 ofNaCl 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) Raoults law is obeyed in the entire concentration range of a binary solution.

(b) Hmix = 0

(c) Vmix = 0

(d) Smix = 0

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15. If various terms in the below given expressions have usual meanings, the Vant 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 ni

N 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 liquidsXand Yform an ideal solution. At 300K, vapour pressure of solution containing1 mol ofXand 3 mol ofYis 550 mm Hg. At the same temperature, if 1 mol ofYis further

added to this solution, vapour pressure of solution increases by 10 mm Hg. Vapour pressure(in mm Hg) ofX 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 Henrys 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) 4 104 (b) 4 105

(c) 5.0 104 (d) 4.0 106

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 MBa(OH)2

(a) 0.40 M (b) 0.0050 M

(c) 0.12 M (d) 0.10 M

20. A 0.04 Msolution ofNa2SO4 is isotonic with a 0.010 Msolution 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 Maq 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 ofA in vapours is

(a) 0.33 (b) 0.25

(c) 0.52 (d) 0.2

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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 Raoults law,

(c) shows a negative deviation from Raoults law,

(d) shows a positive deviation from Raoults law.

25. A 0.002 m aqueous solution of an ionic compound Co(NH3)5 (NO2) Cl freezes at 0.00732C.Number of moles of ions which 1 mol of ionic compound produces on being dissolved inwater will be (Kf = 1.86K/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 Raoults law.

(c) The solution is non-ideal showing ve deviation from Raoults 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 reduction

takes 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 (Ecell)

cell cathode anodeE E EThe 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 assigned

a 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 ofCu2+ is spontaneous but oxidation ofCu to Cu2+

is non-spontaneous.

Nernst Equation for an electrode Reaction : Mn+ + ne M(s)

Reaction quotient

1

c nQ

M

The Nernst equation for this electrode reaction is :

2.303logn n cM M M M

RTE E Q

nF

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0.059logn cM ME V Qn

Similarly Nernst equation for a cell reaction :

neaA bB cC dD

c d

c a b

C DQ

A B

0.059

log

c d

cell cell a b

C DE E V

n A B

At equilibrium, Ecell = 0 and Qc = Kc

0 00.059 0.059

0 log logcell c cell cE V K E K n n Concentration cells have the same electrodes but the concentrations of the solutions of the

same 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 LHC Cu Cu Cu C Cu C

2

2

0.059log

2LHC

cell

RHC

CuE V

Cu

Here RHCmeans right half-cell and LHC, left half-cell.

1

2

0.059log

2cell

CE V

C

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 species

are 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 Vdecreases 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

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Cathode : HgO(s) + H2O + 2e Hg(l) + 2OH

Overall reaction : Zn(Hg) + HgO(s) ZnO(s) + Hg(l)

Cell potential of nearly + 1.35 Vremains 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 ofH2SO4 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 l

Overallreaction 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 rH OH H O e

G x kJ molH O H O

rG = nEcell 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 8

2 2 , r

CH H O CO H en

CH O CO H O G y kJ mol

rG = n EcellF

Corrosion : Corrosion is an electrochemical phenomenon.

Anode : 222 2 4 ; 0.44

Fe FeFe s Fe e E V

Cathode : 2 2

2 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.ofH+, 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 production

of 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.).

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Conductivity : Conductivity (k) is related to resistance as given below :

Cell constant

Resistance

l Ak

R

Unit : ohm1 cm1 or S cm1

Molar Conductivity (m) is given by k

mc

where C= molar concentration and unit of

m= S cm2 mol1.

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 bym. 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 the

ions in which it dissociates. 0 0 0m v v where v

+

and v are number of cations and anions furnished by anelectrolyte.

(For Faradays 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 Eis preferred at cathode and therefore, reduction of wateris preferred at cathode.

Cathode : 2 2

12

2H 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

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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

1

2aqCl Cl e

Note : During the electrolysis of aqueous solution of nitrates and sulphates, the oxidation of wateroccurs at anode in preference to the oxidation ofNO3

or SO4

2. For example, during the electrolysisof aq. Na2SO4, Na

+is not reduced at cathode and SO42 is not oxidised at anode. In this case water

is oxidised at anode and water is also reduced at cathode.

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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 ofNaBr 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 2

12 2 1.23

2H 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

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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 emfof concentration cell :

Zn (s)/Zn2+ (0.0 1M) || Zn2+ (0.10)/Zn is given by

(a) 01.10

2.303 log0.01cell cell

RTE E

F

(b) 00.01

2.303 log2 0.10

cell cell

RTE E

F

(c) 00.01

2.303 log0.10

cell cell

RTE E

F

(d) 0.10

2.303 log2 0.01

cell

RTE

F

8. Standard Gibbs energy of formation fG in kJ mol1 at 298Kfor 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, Yand Zare 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 0.14Vrespectively.

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+/Feand Fe3+/Fe2+ are 0.44 and + 0.77Vrespectively.Fe3+, Fe2+ and Fe blocks are kept together, then

(a) Fe3+ increases (b) Fe3+ decreases

(c) Fe2+/Fe3+ remains unchanged (d) Fe2+ decreases

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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 ofAl2O3 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 to

each other?

(a) Gibbs 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

2

Zn

Cu

is

(a) anti log (24.08) (b) 37.3

(c) 1037.3 (d) 9.65 104

18. The molar conductivities 0NaOACand0HCl

at infinite dilution in water at 25Care 91.0 and426.2S cm2/mol respectively. To calculate 0HOAc, the additional value required is

(a) 2

0H O (b)

0KCl

(c) 0NaOH (d) 0NaCl

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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) 12

lnPRT

F P(b) 1

2

ln2

PRT

F P

(c) 21

lnPRT

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)

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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 5N O

N Ot

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

t

These rates can be equated if we divide the rate of change of concentration of a reactant ora product by its stoichiometric coefficient appearing in balanced chemical equation, that is,

Average rate (rav)/or rate

2 5 2 21 1

2 4

N O NO O

t 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 1

2 4

d N O d NO d O

dt 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 1Rate

5 6 3

BrO BrBr H

t t t t

Order and Molecularity of Reaction

Rate law : Consider a general reaction : a A + b B c C + d D

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The rate law for this reaction is :

Rate = k [A]x[B]y where xand 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 yRate

k A B

Ifx + y = n = order of reaction, then we have

1 11 .unit of

n n

n

concentration Concn M k

time time sconcentration

For zero order reaction (n = 0), unit of k is molL1 s1 or Ms1

For first order reaction (n = 1), unit of k is s1

For second order reaction (n = 2), unit of k is L mol1 s1 or M1 s1.

Molecularity is the number of reacting species taking part in an elementary reaction, i.e., the

number 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 as

its 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 k

dt

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 2or

2

Rt t R

k

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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 Rk R

dt

0 12 1

2

1 1ln or ln

R Rt t t

k R k R

These relationships can be expressed in exponential form as given below

[R] = [R]0ekt

Equation (ii) can be rewritten as :

0 12 1

2

2.303 2.303log or log

R Rt t t

k R k R

Expression for half-life for first order reaction :

1 20.693

tk

Half-life period of first order reaction is independent of initial concentration of the reactant.

log

[

]

[

]

R R0

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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 ofnth order reaction in inversely proportional to initial concentrationof reactant raised to power (n 1)

1 2 1

0

1n

tR

Dependence of rate of reaction on temperature is described by Arrhenius equation :

k = AeEa/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 lnaE

k ART

The plot of ln k and1

Tgives a straight line of slope .a

E

R

If k1 and k2 are rate constants at temperature T1K and T2K then we have

2 2 1

1 2 1 2 1

1 1

log 2.303 2.303

a aE Ek T T

k R T T R T T

(Plot between ln and 1/T)k (Solid line denotes the reaction path without

catalyst 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

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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 eEa/RT where

ZAB = Collision frequency of the reactants, A and B.

eEa/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 = pZAB

eEa/RT

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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 1

3 2

d NHd N d H Rate

dt dt dt

(b) 32 23 2d NHd N d H

Ratedt dt dt

(c) 32 21 1

3 2

d NHd N d H Rate

dt dt dt

(d) 32 21

23

d NHd N d H Rate

dt dt dt

3. In a first order reaction, the concentration of the reactant decreases from 400 moldm3

to25 mol dm3 in 2 104 s. The rate constant in s1 for this reaction is :

(a) 4 104 (b) 1.386 104

(c) 4 104 (d) 3.45 105

4. The rate constant for the reaction : 2N2O5 4NO2 + O2 is 3.0 105 s1. If the rate of

reaction is 2.40 105 mol L1 s1, then [N2O5] in mol L1 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 : RPfollow first order kinetics. In 40 minutes the concentration ofPchangesfrom 0.1 to 0.025 M. The rate of reaction, when the concentration ofP is 0.01 M is :

(a) 3.47 104

M min1

(b) 3.47 105

M min1

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(c) 1.73 104 M min1 (d) 1.73 105 M min1

7. The rate constants k1, and k2 for two different reactions are 1016

2000e

T

and15 100010 e

Trespectively. The temperature at which k1 = k2 is :

(a) 20002.303

K (b)1000

2.303K

(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 L1 [B]/mol L1 Rate/mol L1 s1

(i) 1 102 2 102 2 104

(ii) 2 102 2 102 16 104

(iii) 2 102 4 102 15 104

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]2.510. The activation energy of a reaction is zero. The rate constant (k) of reaction of 280K is

1.6 106 s1. The value of k for this at 300 K is :

(a) zero (b) 3.2 105 s1

(c) 1.6 105 s1 (d) 1.6 106 s1

11. If a reaction A + BCis 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

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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 2

k

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 TEa

k RT T T (b)

2 2 1

1 1 2

log2.303

k T TEa

k RT T T

(c)

2 1 2

1 2 1

log2.303

k T TEa

k RT T T (d)

1 1 2

2 2 1

log2.303

k T TEa

k RT T T

15. For a reaction between A and B, the initial rate of reaction is measured for various initialconcentration ofA and B. The data provided are

S.No. [A] [B] Initial reaction rate

(i) 0.20 M 0.30 M 5.07 105

(ii) 0.20 M 0.10 M 5.07 105

(iii) 0.40 M 0.05 M 1.43 104

(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) 4 104 mol L1 s1 (b) 6 104mol L1 s1

(c) 1 104 mol L1 s1 (d) 3 104mol L1 s1

17. Half-life period of a first order reaction is 1386 sec. The specific rate constant of the reactionis :

(a) 0.5 102 s1 (b) 0.5 103 s1

(c) 5.0 102 s1 (d) 5.0 103 s1

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18. For the reaction A + B products, it is observed that :

(i) On doubling the initial concentration ofA only, the rate of reaction is also doubled.

(ii) On doubling the initial concentrations of bothA andB 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 reactionAP, the temperature (T) dependent rate constant (k) was found

to follow the equation

1

log 2000 6.0.kT

The pre-exponential factorA and the actua-

tion energy Ea respectively are

(a) 1.0 106 s1 and 9.2 kJ mol1 (b) 6.0 s1 and 16.6 kJ mol1

(c) 1.0 106 s1 and 16.6 kJ mol1 (d) 1.0 106 s1 and 38.3 kJ mol1

20. For the reaction 1

2 ,2

A B rate of disappearance of A is related to the rate of appearance

of B by the expression

(a)

d A d B

dt dt(b)

4

d A d B

dt dt

(c)

1

2

d A d B

dt dt(d)

1

4

d A d B

dt dt

21. Under the same reaction conditions, initial concentration of 1.386 mol dm3 of a substancebecomes half in 40 sec and 20 sec through first order and zero order kinetics respectively.

Ratio1

0

k

k

of the rate constants for first order (k1) and zero order (k0) of the reaction is

(a) 0.5 mol1 dm3 (b) 1.0 mol dm3

(c) 1.5 mol dm3 (d) 2.0 mol1 dm3

22. The energies of activation for forward and reverse reactions for A2 + B2 2AB are 180kJ mol1 and 200 kJ mol1 respectively. The presence of a catalyst lowers the activationenergy of both (forward and reverse) reactions by 100 kJmol1. The enthalpy change for the

reaction (A2 + B2 2AB) in the presence of catalyst in (kJ mol1) 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)

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Mechanism B :

2slow

H 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 ofC14 is 5730 years. An archeological artifact containingwood had only 25% of the C14 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 IV A B C D E and the data as

Species formed Rate of its formation

B 0.002 mol/h per mole ofA

C 0.030 mol/h per mol ofB

D 0.011 mol/h per mol of C

E 0.420 mol/h per mol ofD

The rate determining step is :(a) Step I (b) Step II

(c) Step III (d) Step IV

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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 or

a 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 Sin the process : H2(g) 2H(glass) is sufficiently positive makingG = HTSnegative. 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 Kand 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.

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(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 Catalyst

Mo 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.

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Positively Charged Sols Negatively Charged Sols

Hydrated metal oxides, e.g.,

Al2O

3.xH

2O, Fe

2O

3.xH

2O Metal sols, e.g., Ag, Au

Haemoglobin (blood) Metallic sulphides, e.g., As2S3, Sb2S3, CdS

Oxide 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 and

diffused layer of counter ions is called electrokinetic or zeta potential.(e) Electrophoresis : The movement of charged sol particles towards the oppositely charged

electrode 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

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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, its

particles 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 0.01 200 100

Casein 0.01 0.02 100 50

Haemoglobin 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/WType) : Examples are butter, cream and cold cream. Soaps and

detergents 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.

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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.

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7. In the coagulation of arsenic sulphide solution, the flocculating powers of given ions aresuch that

(a) PO4

3 > 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, Hbecomes less and less negative, ultimately Hbecomesequal 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 ofm g) at constant temperature can be expressed as :

(a) 1

log log logx

p km n

(b) 1

log log logx

k pm n

(c) nx

pm

(d) 1

log logx

p km n

13. If a liquid is dispersed in solid medium, then dispersion is called as :

(a) Sol (b) Emulsion

(c) liquid aesosol (d) Gel

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20. Bredigs 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) 1cs

V

V(b) 2310c

s

V

V

(c) 310cs

V

V(d) 310c

s

V

V

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 ofNO to NO2

(b) oxidation ofSO2 to SO3

(c) decomposition ofKClO3 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)

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Minerals are naturally occurring chemical substances in earths 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 earths crust. Iron is the second most abundant metal in the earths crust. Silver,gold, Platinum, sulphur, oxygen and nitrogen are the elements that occur in native or free state.

Many gem-stones are impure form ofAl2O3, the impurities range from Cr (in ruby)

to Co (in sapphire). [For principal ores ofAl, 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. Sometimes

depressants 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

BauxiteNaOH aq dilution

CO 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.

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2ZnS + 3O2 2ZnO + 2SO2

2PbS + 3O2 2PbO + 2SO2

PbS + 2O2 PbSO42Cu2S + 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 ofG is negative, only then the reaction proceed. If Sis positive, on increasingthe temperature (T), the value of TSwould increase i.e.; (H < TS) and then, TSwill 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 ofO2 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 3

acidic basic basic Acidic

SiO CaCO CaSiO CO

Flux impurities slag FeO SiO FeSiO slag

CaO SiO CaSiO slag

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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 ofAg or Au with CN involves the oxidation ofAg 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 agentAu 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. Monds process for the purification of Ni metal involves formation of Nickel carbonyl which onfurther decomposition give pure nickel metal.

330 350 450 47044 4

K K

Impure PureVapoursmetal

Ni CO Ni CO Ni CO g

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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 21800

2 2Evacuated Tungston filament

Vessel K impure 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).

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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 ofCu2O 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.

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9. German silver does not have

(a) Cu (b) Zn

(c) Ni (d) Ag10. 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 ofAl2O3 + 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.

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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 ofCu2O and Cu2S will give

(a) Cu + SO2 (b) Cu + SO3

(c) CuO + CuS (d) CuO + SO3

21. Consider the following reactions at 1000C

A 21

2Zn s O g ZnO s rG

= 360 KJ mol1

B 21

graphite2

C O g CO g rG = 460 KJ mol1

Choose the correct statement at 1000C

(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.

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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)

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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 NtoBi 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 mol1) 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 mol1), P P(201 kJ mol1), As As (146 kJ mol1) and Sb Sb (121 kJ mol1) 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 a

diatomic 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 mol1), 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.

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(i) Thermal stability of these hydrides decreases gradually form NH3 toBiH3 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 Ein 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. ofPH3 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 atom

undergoes sp3-hybridization. The HNHbond 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.