Chemical Thermodynamics 21
reversible path HH HH
26. (d) : C C +H–H H–C–C–H
HH
23. (d) : A B HH
DH Reaction = å BE reactant – å BE product
irreversible path = 4 × 414 + 615 + 435 – (6 × 414 + 347)
We know that for a cyclic process the net change in internal = 2706 – 2831
energy is equal to zero and change in the internal energy does
not depend on the path by which the final state is reached. = –125 kJ
24. (b) : For spontaneity, change in entropy (dS) must be positive, 27. (d) : It does not violate first law of thermodynamics but violates
means it should be greater than zero. second law of thermodynamics.
Change in Gibbs free energy (dG) must be negative means that
it should be lesser than zero. (dS)V , E > 0, (dG) T, P < 0. 28. (b) : For endothermic reaction, DH = +ve
Now, DG = DH – TDS
25. (c) : This is according to Hess's law. For nonspontaneous reaction, DG should be positive
Now DG is positive at low temperature if DH is positive.
DG is negative at high temperature if DS is positive.
29. (d) : DH = negative shows that the reaction is spontaneous.
Higher value for DH shows that the reaction is more feasible.
22 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER SOLUTIONS
6
1. K f for water is 1.86 K kg mol –1 . If your automobile radiator 7. If sodium sulphate is considered to be completely dissociated
holds 1.0 kg of water, how many grams of ethylene glycol into cations and anions in aqueous solution, the change in freezing
(C 2 H6 O 2 ) must you add to get the freezing point of the solution point of water (DTf ) , when 0.01 mol of sodium sulphate is dissolved
lowered to –2.8°C?
in 1 kg of water, is (Kf = 1.86 K kg mol– 1)
(a) 93 g (b) 39 g (a) 0.0186 K (b) 0.0372 K
(c) 27 g (d) 72 g (2012) (c) 0.0558 K (d) 0.0744 K (2010)
2. The density of a solution prepared by dissolving 120 g of 8. A binary liquid solution is prepared by mixing nheptane and
urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The ethanol. Which one of the following statements is correct
molarity of this solution is regarding the behaviour of the solution?
(a) 1.78 M (b) 1.02 M (a) The solution formed is an ideal solution.
(c) 2.05 M (d) 0.50 M (2012) (b) The solution is nonideal, showing +ve deviation from
Raoult’s law.
3. The degree of dissociation (a) of a weak electrolyte, Ax B y is
related to van’t Hoff factor (i) by the expression (c) The solution is nonideal, showing –ve deviation from
Raoult’s law.
(a) a = i - 1 (b) a= i - 1 (d) nheptane shows +ve deviation while ethanol shows
(x + y - 1) (x + y + 1) –ve deviation from Raoult’s law.
(2009)
(c) a = (x + y - 1) (d) a= (x + y + 1) (2011)
i - 1 i - 1
4. Ethylene glycol is used as an antifreeze in a cold climate. 9. Two liquids X and Y form an ideal solution. At 300 K, vapour
pressure of the solution containing 1 mol of X and 3 mol of
Mass of ethylene glycol which should be added to 4 kg of
Y is 550 mm Hg. At the same temperature, if 1 mol of Y is
water to prevent it from freezing at – 6°C will be : (K f for further added to this solution, vapour pressure of the solution
water = 1.86 K kg mol –1 , and molar mass of ethylene glycol
= 62 g mol –1 ) increases by 10 mm Hg. Vapour pressure (in mm Hg) of X
(a) 804.32 g (b) 204.30 g and Y in their pure states will be, respectively
(c) 400.00 g (d) 304.60 g (2011) (a) 200 and 300 (b) 300 and 400
5. A 5.2 molal aqueous solution of methyl alcohol, CH 3 OH, is (c) 400 and 600 (d) 500 and 600 (2009)
supplied. What is the mole fraction of methyl alcohol in the 10. The vapour pressure of water at 20°C is 17.5 mm Hg. If
solution? (b) 0.190 18 g of glucose (C 6H 12 O 6) is added to 178.2 g of water at
(a) 0.100 20°C, the vapour pressure of the resulting solution will be
(c) 0.086 (d) 0.050 (2011) (a) 17.325 mm Hg (b) 17.675 mm Hg
6. On mixing, heptane and octane form an ideal solution. At (c) 15.750 mm Hg (d) 16.500 mm Hg
373 K, the vapour pressure of the two liquid components
(2008)
(heptane and octane) are 105 kPa and 45 kPa respectively. 11. At 80°C, the vapour pressure of pure liquid A is 520 mm of
Vapour pressure of the solution obtained by mixing 25.0 g of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture
heptane and 35 g of octane will be (molar mass of heptane = solution of A and B boils at 80°C and 1 atm pressure, the
100 g mol– 1 and of octane = 114 g mol –1 ) amount of A in the mixture is (1 atm = 760 mm of Hg)
(a) 144.5 kPa (b) 72.0 kPa (2010) (a) 50 mol percent (b) 52 mol percent
(c) 36.1 kPa (d) 96.2 kPa (c) 34 mol percent (d) 48 mol percent
(2008)
Solutions 23
12. A 5.25% solution of a substance is isotonic with a 1.5% solution 21. Which one of the following statements is false?
of urea (molar mass = 60 g mol– 1) in the same solvent. If the
densities of both the solutions are assumed to be equal to (a) Raoult’s law states that the vapour pressure of a component
1.0 g cm– 3, molar mass of the substance will be
(a) 210.0 g mol– 1 (b) 90.0 g mol– 1 over a solution is proportional to its mole fraction.
(c) 115.0 g mol– 1 (d) 105.0 g mol– 1
(2007) (b) The osmotic pressure (p) of a solution is given by the
equation (p = MRT, where M is the molarity of the solution.
(c) The correct order of osmotic pressure for
0.01 M aqueous solution of each compound is
13. A mixture of ethyl alcohol and propyl alcohol has a vapour BaCl2 > KCl > CH3 COOH > sucrose.
(d) Two sucrose solutions of same molality prepared in
pressure of 290 mm at 300 K. The vapour pressure of propyl
different solvents will have the same freezing point
alcohol is 200 nm. If the mole fraction of ethyl alcohol is 0.6,
depression. (2004)
its vapour pressure (in mm) at the same temperature will be
(a) 360 (b) 350 22. Which of the following liquid pairs shows a positive deviation
(c) 300 (d) 700 (2007) from Raoult’s law?
14. The density (in g mL– 1) of a 3.60 M sulphuric acid solution that (a) Water hydrochloric acid
is 29% H2 S O4 (molar mass = 98 g mol– 1) by mass will be (b) Benzene methanol
(a) 1.45 (b) 1.64 (c) Water nitric acid
(c) 1.88 (d) 1.22 (2007) (d) Acetone chloroform (2004)
15. 18 g of glucose (C 6H 1 2O 6) is added to 178.2 g of water. The 23. To neutralise completely 20 mL of 0.1 M aqueous solution of
vapour pressure of water for this aqueous solution at 100°C is
phosphorous acid (H3 P O3 ) , the volume of 0.1 M aqueous KOH
(a) 759.00 torr (b) 7.60 torr solution required is
(c) 76.00 torr (d) 752.40 torr (2006) (a) 10 mL (b) 20 mL
(c) 40 mL (d) 60 mL (2004)
16. Density of a 2.05 M solution of acetic acid in water is 1.02 g/ 24. 6.02 × 10 20 molecules of urea are present in 100 ml of its solution.
mL. The molality of the solution is The concentration of urea solution is
(a) 1.14 mol kg– 1 (b) 3.28 mol kg– 1 (a) 0.001 M (b) 0.01 M
(c) 2.28 mol kg– 1 (d) 0.44 mol kg– 1 (2006) (c) 0.02 M (d) 0.1 M (2004)
17. Equimolal solutions in the same solvent have 25. Which one of the following aqueous solutions will exhibit
(a) same boiling point but different freezing point
(b) same freezing point but different boiling point highest boiling point?
(c) same boiling and same freezing points
(d) different boiling and different freezing points. (2005) (a) 0.01 M Na2 SO4 (b) 0.01 M KNO 3 (2004)
(c) 0.015 M urea (d) 0.015 M glucose
18. Two solutions of a substance (non electrolyte) are mixed in the 26. If liquids A and B form an ideal solution, the
(a) enthalpy of mixing is zero
following manner. 480 mL of 1.5 M first solution + 520 mL of (b) entropy of mixing is zero
(c) free energy of mixing is zero
1.2 M second solution. What is the molarity of the final mixture? (d) free energy as well as the entropy of mixing are each zero.
(2003)
(a) 1.20 M (b) 1.50 M
(c) 1.344 M (d) 2.70 M (2005)
19. Benzene and toluene form nearly ideal solutions. At 20°C, the 27. 25 mL of a solution of barium hydroxide on titration with a 0.1
vapour pressure of benzene is 75 torr and that of toluene is 22 molar solution of hydrochloric acid gave a litre value of 35 mL.
torr. The partial vapour pressure of benzene at 20°C for a solution The molarity of barium hydroxide solution was
containing 78 g of benzene and 46 g of toluene in torr is (a) 0.07 (b) 0.14
(a) 50 (b) 25 (c) 0.28 (d) 0.35 (2003)
(c) 37.5 (d) 53.5 (2005)
20. If a is the degree of dissociation of Na2 S O4 , the vant Hoff’s 28. In a 0.2 molal aqueous solution of a weak acid HX, the degree
factor (i) used for calculating the molecular mass is
of ionization is 0.3. Taking K f for water as 1.85, the freezing
(a) 1 + a (b) 1 – a point of the solution will be nearest to
(c) 1 + 2a (d) 1 – 2a (2005) (a) – 0.480°C (b) – 0.360 °C
(c) – 0.260 °C (d) + 0.480 °C (2003)
24 JEE MAIN CHAPTERWISE EXPLORER
29. In mixture A and B components show –ve deviation as 30. Freezing point of an aqueous solution is (–0.186)°C. Elevation
(a) DVm ix > 0
(b) DHm ix < 0 of boiling point of the same solution is K b = 0.512°C,
(c) A – B interaction is weaker than A – A and K f = 1.86°C, find the increase in boiling point.
B – B interaction (a) 0.186°C
(d) A – B interaction is stronger than A – A and
B – B interaction. (b) 0.0512°C
(2002)
(c) 0.092°C
(d) 0.2372°C (2002)
Answer Key
1. (a) 2. (c) 3. (a) 4. (a) 5. (c) 6. (b)
7. (c) 8. (b) 11. (a) 12. (a)
13. (b) 14. (d) 9. (c) 10. (a) 17. (c) 18. (c)
19. (a) 20. (c) 23. (c) 24. (b)
25. (a) 26. (a) 15. (d) 16. (c) 29. (b,d) 30. (b)
21. (d) 22. (b)
27. (b) 28. (a)
Solutions 25
1. (a) : Kf = 1.86 K kg mol –1 Here solute is methyl alcohol, solvent is water.
DTf = 0 – (–2.8) = 2.8°C
Mass of solvent = 1.0 kg Given n = 5.2, N = 1000
18
Mass of solute = ?
Molecular mass of solute = 62 \ Mole fraction = 5.2
DT f = Kf × m 5.2 + 101080
Weight of solute = 5.2 ´ 18 = 93.6 = 0.0855 » 0.086
m = Molecular mass of solute ´ 1000 93.6 + 1000 1093.6
Mass of solvent (g) 6. (b) : Given, p° heptane = 105 kPa
m = w / 62 ´ 1000 = w p°o ctane = 45 kPa
1000 62 wh eptane = 25 g
DT f = Kf × m wo ctane = 35 g
2.8 = 1.86 ´ w Þ w = 62 ´ 2.8 = 93 g nheptane = 25 = 0.25
62 1.86 100
2. (c) : Mass of solute taken = 120 g
Molecular mass of solute = 60 u noctane = 35 = 0.30
114
Mass of solvent = 1000 g
0.25
Density of solution = 1.15 g/mL xheptane = 0.25 + 0.30 = 0.45
Total mass of solution = 1000 + 120 = 1120 g
Volume of solution = Mass = 1120 mL xoctane = 0.30 = 0.54
Density 1.15 0.25 + 0.30
Mass of solute pT otal = x heptane p° heptane + x octane p° octane
= 0.45 × 105 + 0.54 × 45
Molarity = Molecular mass of solute ´ 1000
Volume of solution = 47.25 + 24.3 = 71.55 » 72 kPa
= 120 / 60 ´ 1000 = 2 ´ 1000 ´ 1.15 = 2.05 M 7. (c) :Depression in freezing point, DTf = i × Kf × m
1120 / 1.15 1120 For sodium sulphate, i = 3
m = 0.01 = 0.01 m.
3. (a) : Ax By xA y+ + yB x– 1 kg
Given, K f = 1.86 K kg mol –1
1 – a xa ya \ DT f = 3 × 1.86 × 0.01 = 0.0558 K
i = 1 – a + xa + ya = 1 + a(x + y – 1) 8. (b) : The solution containing nheptane and ethanol shows
nonideal behaviour with positive deviation from Raoult’s
\ a = (x i - 1 law. This is because the ethanol molecules are held together
+ y - 1) by strong Hbonds, however the forces between nheptane
and ethanol are not very strong, as a result they easily vapourise
4. (a) : DTf = K f × m = Kf ´ w2 ´ 1000 showing higher vapour presure than expected.
w1 ´ m2
w1 and w2 = wt. of solvent and solute
m 2 = molecular wt. of solute
DT f = 0 – (– 6) = 6
\ 6 = 1.86 ´ w2 ´ 1000 9. (c) : P T = p°X x X + p°Y x Y
4000 ´ 62 where, P T = Total presure
p°X = Vapour pressure of X in pure state
w2 = 6 ´ 62 ´ 4000 = 800 g p°Y = Vapour pressure of Y in pure state
1000 ´ 1.86 x X = Mole fraction of X = 1/4
x Y = Mole fraction of Y = 3/4
5. (c) : Mole fraction of solute = n (i) When T = 300 K, PT = 550 mm Hg
N + n
n = number of moles of solute
N = number of moles of solvent
26 JEE MAIN CHAPTERWISE EXPLORER
( ) ( ) \ 1 3 14. (d) : 3.6 M solution means 3.6 mole of H 2 SO 4 is present in
550 = p°X 4 + pY° 4 1000 mL of solution.
Þ p°X + 3 pY ° = 2200 .....(1) \ Mass of 3.6 moles of H2 SO4 = 3.6 × 98 g = 352.8 g
\ Mass of H2 SO4 in 1000 mL of solution = 352.8 g
(ii) When at T = 300 K, 1 mole of Y is added,
Given, 29 g of H2 S O4 is present in 100 g of solution
PT = (550 + 10) mm Hg \ 352.8 g of H2 S O4 is present in
\ x X = 1/5 and x Y = 4/5 100 ´ 352.8 = 1216 g of solution
29
( ) ( ) Þ 1 4
560 = pX° 5 + pY ° 5
or p°X + 4 pY ° = 2800 ....(2) Mass = 1216 = 1.216 g/mL
Now, density = Volume 1000
On solving equations (1) and (2), we get
p°Y = 600 mm Hg and p°X = 400 mm Hg 15. (d) : p° - ps = n
ps N
10. (a) : In solution containing nonvolatile solute, pressure is
directly proportional to its mole fraction. 760 - ps = 18/180 = 1/10
Ps olution = vapour pressure of its pure component ps 178.2 /18 9.9
× mole fraction in solution 1
\ Ps ol = P°X solvent Þ 760 - ps = 99 ps Þ 760 ´ 99 - 99 ps = ps
Let A be the solute and B the solvent
760 ´ 99
nB 178.2 Þ 100ps = 760 × 99 Þ p s = 100 = 752.4 torr
nA + nB
\ X B = = 18 18 16. (c) : Molality, m = M ´1000
180 + 17188. 2 1000d - MM 2
X B = 99.9.94 = 0.99 where M = molarity, d = density, M2 = molecular mass
Now Ps olution = P°X solvent = 17.5 × 0.99 m = 2.05 = 2.05
Ps olution = 17.325 1000 ´1.02 - 2.05 ´ 60 897
11. (a) : We have, PA ° = 520 mm Hg = 2.28 × 10 –3 mol g –1 = 2.28 mol kg –1
17. (c) : According to Raoult’s law equimolal solutions of all the
and PB° = 1000 mm Hg
Let mole fraction of A in solution = X A substances in the same solvent will show equal elevation in
and mole fraction of B in solution = X B boiling points as well as equal depression in freezing point.
Then, at 1 atm pressure i.e. at 760 mm Hg
18. (c) : Total millimoles of solute
PA° X A + PB° X B = 760 mm Hg = 480 × 1.5 + 520 × 1.2 = 720 + 624 = 1344
Total volume = 480 + 520 = 1000
PA° X A + PB° (1 - X A ) = 760 mm Hg Molarity of the final mixture = 1344 = 1.344 M
1000
Þ 520 X A + 1000 – 1000 X A = 760 mm Hg
Þ X A = 1 or 50 mol percent 19. (a) : According to Raoult's law, PB = P° B X B
2
P°B = 75 torr
12. (a) : Isotonic solutions have same osmotic pressure.
78 / 78 1 1
p1 = C1RT , p 2 = C 2 RT X B = (78/ 78) + (46 / 92) = 1 + 0.5 = 1.5
For isotonic solution, p1 = p2 1
\ C 1 = C2 PB = 75 ´ 1.5 = 50 torr
or, 1.5/ 60 = 5.25 / M 20. (c) : Na 2 SO 4 ƒ 2Na + + SO 42 –
VV
1 0 0
[Where M = molecular weight of the substance]
1 – a 2a a
or, 1.5 = 5.25 or M = 210 Vant Hoff factor (i) = 1- a + 2 a + a = 1 + 2a
60 M
1
13. (b) : According to Raoult’s law,
P = PA + PB = PA°xA + PB° xB 21. (d) : The extent of depression in freezing point varies with
or 290 = PA° ´ (0.6) + 200 ´ (1 - 0.6) the number of solute particles for a fixed solvent only and
or 290 = 0.6 × PA° + 0.4 × 200 it’s a characteristic feature of the nature of solvent also.
or PA o = 350 mm DT f = k f × m
For different solvents, value of k f is also different. So, for
two different solvents the extent of depression may vary even
if number of solute particles be dissolved in them.
Solutions 27
22. (b) : In solutions showing positive deviation, the observed 26. (a) : For ideal solutions, DH mix = 0, neither heat is evolved
vapour pressure of each component and total vapour pressure nor absorbed during dissolution.
are greater than predicted by Raoult's law, i.e.
pA > pºA x A; p B > p ºB x B ; p > p A + p B 27. (b) : Ba(OH) 2 HCl
In solution of methanol and benzene, methanol molecules M 1V 1 = M2 V 2
are held together due to hydrogen bonding as shown below:
M 1 × 25 = 0.1 × 35
CH 3 CH 3 CH 3 or, M1 = 0.12´5 35 = 0.14
O —– H O —– H O —– H 28. (a) : HX H + + X –
1 0 0
On adding benzene, the benzene molecules get in between 1 – 0.3 0.3 0.3
the molecules of methanol, thus breaking the hydrogen bonds.
As the resulting solution has weaker intermolecular attractions, Total number of moles after dissociation
the escaping tendency of alcohol and benzene molecules from
the solution increases . Consequently the vapour pres sure of = 1 – 0.3 + 0.3 + 0.3 = 1.3
the solution is greater than the vapour pres sure as expected
from Raoult’s law. Kf (observed) = no. of moles after dissociation
Kf (experimental) no. of moles before dissociation
23. (c) : H 3 PO 3 is a dibasic acid. or, Kf (o1b.8s5erved) = 11.3
N1 V1 (acid) = N 2V 2 (base) or, K f (observed) = 1.85 × 1.3 = 2.405
DT f = K f × molality = 2.405 × 0.2 = 0.4810
0.1 × 2 × 20 = 0.1 × 1 × V 2 Freezing point of solution = 0 – 0.481 = – 0.481°C
\ V2 = 0.10´.12´´ 12 0 = 40 mL 29. (b, d) : For negative deviation, from Raoult's law, DV mix < 0
and DH mix < 0. Here A – B attractive force is greater than
24. (b) : Moles of urea = 6.02 ´10 20 = 10-3 moles A – A and B – B attractive forces.
6.02 ´ 1023
Concentration (molarity) of solution
= no. of moles of solute = 10- 3 ´1000 = 0.0 1 M 30. (b) : DTb = K b W B ´ 1000
no. of litres of solution 100 B ´ WA
M
25. (a) : Elevation in boiling point is a colligative property which DTf = K f M BW´B W A ´1000
depends upon the number of solute particles.
Greater the number of solute particles in a solution, higher DTb = Kb or DTb = 0.512 or, DTb = 0.0512°C
the extent of elevation in boiling point. D T f K f 0.186 1.86
Na 2 SO 4 ® 2Na + + SO 42 –
28 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER EQUILIBRIUM
7
1. How many litres of water must be added to 1 litre of an aqueous (a) 5.0 × 10– 8 g (b) 1.2 × 10– 10 g
solution of HCl with a pH of 1 to create an aqueous solution (c) 1.2 × 10– 9 g (d) 6.2 × 10 –5 g (2010)
with pH of 2? 8. In aqueous solution the ionisation constants for carbonic acid
(a) 9.0 L (b) 0.1 L are
(c) 0.9 L (d) 2.0 L (2013) K1 = 4.2 × 10– 7 and K2 = 4.8 × 10– 11
2. The equilibrium constant (K c ) for the reaction Select the correct statement for a saturated 0.034M solution of
N2 (g) + O 2(g) ® 2NO (g) at temperature T is 4 × 10– 4. The value of the carbonic acid.
Kc for the reaction, NO (g) ® 1 N 2(g) + 12 O2 (g) at the same (a) The concentration of H+ is double that of CO 32 –.
2 (b) The concentration of CO 32 – is 0.034 M.
temperature is (c) The concentration of CO 32 – is greater than that of HCO 3– .
(d) The concentration of H+ and HCO 3 – are approximately
(a) 2.5 × 102 (b) 4 × 10– 4
equal.
(c) 50.0 (d) 0.02 (2012)
3. The pH of a 0.1 molar solution of the acid HQ is 3. The value (2010)
of the ionization constant, K a of this acid is 9. The correct order of increasing basicity of the given conjugate
(a) 1 × 10 –3 (b) 1 × 10– 5 (2012) bases (R = CH3 ) is
(c) 1 × 10– 7 (d) 3 × 10 –1 (a) RCOO – < HC C– < NH2 – < R–
(b) RCOO – < HC C– < R – < NH2 –
4. A vessel at 1000 K contains CO 2 with a pressure of 0.5 atm. (c) R – < HC C– < RCOO – < NH2 – (2010)
Some of the CO 2 is converted into CO on the addition of (d) RCOO – < NH2 – < HC C– < R–
graphite. If the total pressure at equilibrium is 0.8 atm, the value
of K is (b) 3 atm 10. Solid Ba(NO 3) 2 is gradually dissolved in a 1.0 × 10 –4 M Na 2C O 3
(a) 1.8 atm solution. At what concentration of Ba 2+ will a precipitate begin
(c) 0.3 atm (d) 0.18 atm (2011) to form?(Ks p for BaCO 3 = 5.1 × 10 –9 )
(a) 4.1 × 10– 5 M (b) 5.1 × 10 –5 M
5. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10– 11. At (2009)
which pH, will Mg2 + ions start precipitating in the form of (c) 8.1 × 10– 8 M (d) 8.1 × 10– 7 M
Mg(OH) 2 from a solution of 0.001 M Mg2 + ions? 11. Four species are listed below :
(a) 8 (b) 9 (2010) (i) HCO 3– (ii) H3 O +
(c) 10 (d) 11 (iii) HSO 4– (iv) HSO 3F
6. Three reactions involving H2 P O4 – are given below: Which one of the following is the correct sequence of their
(i) H3 PO4 + H2 O H3 O + + H2 PO4 – acid strength?
HPO 42 – + H3 O +
(ii) H2 P O4 – + H2 O (a) iii < i < iv < ii
(iii) H2 PO4 – + OH– H3 PO4 + O 2–
(b) iv < ii < iii < i
In which of the above does H2 P O4 – act as an acid? (c) ii < iii < i < iv
(a) (i) only (b) (ii) only (d) i < iii < ii < iv (2008)
(c) (i) and (ii) (d) (iii) only (2010) 12. The pK a of a weak acid, (HA), is 4.80. The pK b of a weak base,
7. Solubility product of silver bromide is 5.0 × 10– 13 . The quantity BOH is 4.78. The pH of an aqueous solution of the
of potassium bromide (molar mass taken as 120 g mol– 1) to be
added to 1 litre of 0.05M solution of silver nitrate to start the corresponding salt, BA, will be
precipitation of AgBr is
(a) 9.22 (b) 9.58
(c) 4.79 (d) 7.01 (2008)
Equilibrium 29
13. For the following three reactions (i), (ii) and (iii), equilibrium 20. Phosphorus pentachloride dissociates as follows in a closed
constants are given reaction vessel,
(i) CO (g) + H2 O (g) CO 2(g) + H2 (g) ; K1 PCl5 (g) PCl3 (g) + Cl2 (g)
(ii) CH4 (g) + H2 O (g) CO (g) + 3H2 (g) ; K2 If total pressure at equilibrium of the reaction mixture is P and
(iii) CH4 (g) + 2H2 O (g) CO 2(g) + 4H2 (g) ; K3 degree of dissociation of PCl5 is x, the partial pressure of PCl3
will be
Which of the following relation is correct?
(a) K 3 ∙ K2 3 = K 12 (b) K1 K2 = K3 (a) æ x+ 1 ÷øö P (b) æ 2 x ÷øö P
çè çè 1 - x
(c) K2 K3 = K1 (d) K 3 = K 1K 2 (2008) x
14. The equilibrium constants K p 1 and K P2 for the reactions (c) æ x x- 1 ÷öø P (d) æ x ÷öø P (2006)
X 2Y and Z P + Q, respectively are in the ratio of çè çè 1 - x
1 : 9. If degree of dissociation of X and Z be equal then the 21. An amount of solid NH4 H S is placed in a flask already containing
ammonia gas at a certain temperature and 0.50 atm. pressure.
ratio of total pressures at these equilibria is
(a) 1 : 9 (b) 1 : 36 Ammonium hydrogen sulphide decomposes to yield NH3 and
H2 S gases in the flask. When the decomposition reaction reaches
(c) 1 : 1 (d) 1 : 3 (2008) equilibrium, the total pressure in the flask rises to 0.84 atm. The
15. In a saturated solution of the sparingly soluble strong electrolyte
AgIO3 (molecular mass = 283) the equilibrium which sets in is equilibrium constant for NH 4H S decomposition at this
temperature is
AgIO3 (s) Ag + (aq) + IO 3– ( aq).
If the solubility product constant K sp of AgIO 3 at a given (a) 0.30 (b) 0.18
temperature is 1.0 × 10– 8, what is the mass of AgIO3 contained (c) 0.17 (d) 0.11 (2005)
in 100 mL of its saturated solution? 22. Among the following acids which has the lowest pK a value?
(a) 1.0 × 10– 4 g (b) 28.3 × 10– 2 g (a) CH3 COOH (b) (CH3 ) 2C H – COOH
(c) 2.83 × 10 –3 g (d) 1.0 × 10– 7 g (c) HCOOH (d) CH3 CH2 COOH (2005)
(2007)
16. The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous 23. What is the conjugate base of OH– ?
bufferd solution of HA in which 50% of the acid is ionized is
(a) O2 (b) H2 O
(a) 7.0 (b) 4.5 (c) O– (d) O 2– (2005)
(c) 2.5 (d) 9.5 24. Hydrogen ion concentration in mol/L in a solution of pH = 5.4
(2007) will be
17. The first and second dissociation constants of an acid H2 A are (a) 3.98 × 108 (b) 3.88 × 10 6
1.0 × 10 –5 and 5.0 × 10– 10 respectively. The overall dissociation
(c) 3.68 × 10 –6 (d) 3.98 × 10– 6 (2005)
constant of the acid will be 25. For the reaction,
(a) 0.2 × 105 (b) 5.0 × 10– 5 2NO 2(g) 2NO (g) + O 2(g)
(c) 5.0 × 101 5 (d) 5.0 × 10– 15 (2007) (Kc = 1.8 × 10 –6 at 184°C)
18. Given the data at 25°C, (R = 0.0831 kJ/(mol.K)). When K p and K c are
Ag + I – ® AgI + e – ; E° = 0.152 V compared at 184°C it is found that
Ag ® Ag + + e– ; E° = – 0.800 V (a) K p is greater than Kc
(b) K p is less than Kc
What is the value of log Ks p for AgI? (c) K p = Kc
(d) whether Kp is greater than, less than or equal to Kc depends
æ 2.303 RT = 0.059 V ÷öø
çè F upon the total gas pressure.
(a) – 8.12 (b) + 8.612 (2005)
(c) – 37.83 (d) –16.13
(2006) 26. The exothermic formation of ClF 3 is represented by the equation:
19. The equilibrium constant for the reaction, Cl2 (g) + 3F2 (g) 2ClF3 (g); DH = –329 kJ
SO3 (g) 1 Which of the following will increase the quantity of ClF 3 in an
SO2 (g) + 2 O 2(g)
equilibrium mixture of Cl2 , F2 and ClF 3 ?
is Kc = 4.9 × 10– 2. The value of K c for the reaction (a) Increasing the temperature
2SO2 (g) + O 2(g) 2SO3 (g) will be (b) Removing Cl2
(a) 416 (b) 2.40 × 10– 3 (c) Increasing the volume of the container
(c) 9.8 × 10– 2 (d) 4.9 × 10– 2 (2006) (d) Adding F2 (2005)
30 JEE MAIN CHAPTERWISE EXPLORER
27. The solubility product of a salt having general formula MX2 , in CuSO4 solution deposits 1 gram equivalent of copper at
water is 4 × 10– 12. The concentration of M 2+ ions in the aqueous
the cathode. (2003)
solution of the salt is 36. The correct relationship between free energy change in a reaction
(a) 2.0 × 10– 6 M (b) 1.0 × 10– 4 M and the corresponding equilibrium constant Kc is
(a) DG = RT ln Kc (b) –DG = RT ln K c
(c) 1.6 × 10– 4 M (d) 4.0 × 10– 10 M (2005)
(c) DG° = RT ln K c (d) –DG° = RT ln Kc
28. Consider an endothermic reaction X ® Y with the activation (2003)
energies E b and E f for the backward and forward reactions, 37. The solubility in water of a sparingly soluble salt AB 2 is
respectively. In general 1.0 × 10 –5 mol L– 1. Its solubility product will be
(a) Eb < Ef (b) Eb > E f (a) 4 × 10– 15 (b) 4 × 10– 10
(c) Eb = Ef (c) 1 × 10 –15 (d) 1 × 10– 10 (2003)
(d) there is no definite relation between Eb and E f . (2005)
29. The molar solubility (in mol L– 1) of a sparingly soluble salt MX 4 38. For the reaction equilibrium,
is s. The corresponding solubility product is K sp. s is given in N 2O 4(g) 2NO 2(g)
terms of Ks p by the relation the concentrations of N 2 O 4 and NO 2 at equilibrium are
4.8 × 10– 2 and 1.2 × 10 –2 mol L– 1 respectively. The value of K c
(a) s = (K sp/ 128)1 /4 (b) s = (128Ks p) 1 /4
for the reaction is
(c) s = (256K sp) 1/5 (d) s = (K sp/ 256) 1/5 (2004)
(a) 3.3 × 10 2 mol L– 1
30. The equilibrium constant for the reaction, (b) 3 × 10– 1 mol L– 1
N 2(g) + O 2(g) 2NO (g) (c) 3 × 10– 3 mol L– 1
at temperature T is 4 × 10– 4. The value of K c for the reaction : (d) 3 × 10 3 mol L– 1 (2003)
NO (g) 12 N 2(g) + 12 O 2(g) at the same temperature is 39. Consider the reaction equilibrium:
(a) 2.5 × 102 (b) 50 2SO2 (g) + O 2(g) 2SO3 (g) ; DH ° = –198 kJ.
(c) 4 × 10 –4 (d) 0.02 (2004) On the basis of Le Chatelier's principle, the condition favourable
31. For the reaction, CO (g) + Cl2 (g) COCl2 (g), the Kp / K c is equal for the forward reaction is
to (2004)
(a) lowering of temperature as well as pressure
(a) 1/RT (b) RT (b) increasing temperature as well as pressure
(c) RT (d) 1.0 (c) lowering the temperature and increasing the pressure
(d) any value of temperature and pressure.
32. What is the equilibrium expression for the reaction (2003)
P4 (s) + 5O 2 (g) P4 O 10 (s) ? 40. In which of the following reactions, increase in the volume at
(a) Kc = [P[P44][OO102 ]]5 (b) Kc = [P4O10 ] constant temperature does not affect the number of moles at
5[P4 ][O2 ]
equilibrium?
(d) Kc = [O12 ]5
(c) K c = [O 2] 5 (2004) (a) 2NH3 ® N 2 + 3H2
(2004) (b) C( g) + (1/2) O 2(g) ® CO (g)
33. The conjugate base of H2 P O4 – is (c) H2 (g) + O 2(g) ® H2 O 2(g)
(d) None of these.
(a) PO4 3 – (b) P2 O 5 (2002)
(c) H3 P O4 (d) HPO4 2 –
34. When rain is accompanied by a thunderstorm, the collected rain 41. Change in volume of the system does not alter the number of
water will have a pH value moles in which of the following equilibria?
(a) slightly lower than that of rain water without thunderstorm
(b) slightly higher than that when the thunderstorm is not there (a) N 2(g) + O 2(g) 2NO (g)
(c) uninfluenced by occurrence of thunderstorm
(d) which depends on the amount of dust in air. (2003) (b) PCl5 (g) PCl3 (g) + Cl2 (g)
(c) N 2(g) + 3H2 (g) 2NH3 (g)
(d) SO2 Cl2 (g) SO2 (g) + Cl2 (g) (2002)
35. Which one of the following statements is not true? 42. For the reaction
(a) The conjugate base of H2 P O4 – is HPO4 2– .
(b) pH + pOH = 14 for all aqueous solutions. CO (g) + (1/2) O 2(g) = CO 2(g) , K p /Kc is
(c) The pH of 1 × 10– 8 M HCl is 8.
(d) 96,500 coulombs of electricity when passed through a (a) RT (b) (RT)– 1
(c) (RT) –1/2 (d) (RT)1 /2 (2002)
Equilibrium 31
43. Let the solubility of an aqueous solution of Mg(OH)2 be x then 45. 1 M NaCl and 1 M HCl are present in an aqueous solution. The
its Ks p is solution is
(a) not a buffer solution with pH < 7
(a) 4x3 (b) 108x5 (b) not a buffer solution with pH > 7
(c) a buffer solution with pH < 7
(c) 27x4 (d) 9x (2002) (d) a buffer solution with pH > 7.
44. Species acting as both Bronsted acid and base is (2002)
(a) (HSO4 ) –1 (b) Na2 C O 3 (2002)
(c) NH3 (d) OH– 1
Answer Key
1. (a) 2. (c) 3. (b) 4. (a) 5. (c) 6. (b)
7. (c) 8. (d) 11. (d) 12. (d)
9. (a) 10. (b)
13. (d) 14. (b) 17. (d) 18. (d)
19. (a) 20. (a) 15. (c) 16. (d) 23. (d) 24. (d)
25. (a) 26. (d) 21. (d) 22. (b) 29. (d) 30. (b)
31. (a) 32. (d) 27. (b) 28. (a) 35. (c) 36. (d)
37. (c) 38. (c) 41. (a) 42. (c)
43. (a) 44. (a) 33. (d) 34. (a)
39. (c) 40. (d)
45. (a)
32 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : Initial concentration of aq. HCl solution with pH 1 = 10– 1 M 7. (c) : Given, (K sp) A gBr = 5.0 × 10– 13
The required equation is,
Final concentration of this solution after dilution = 10– 2 M
MV = M 1 (V 1 + V2 ) KBr + AgNO 3 AgBr + KNO 3
10 –1 × 1 = 10 –2 (1 + V 2 ) Given, [AgNO 3 ] = 0.05 M
Þ [Ag +] = [NO– 3 ] = 0.05 M
0.1 = 1 + V2
0.01 [Ag+ ] [Br – ] = (K sp) AgBr
10 = 1 + V 2 Þ 0.05 × [Br– ] = 5 × 10 –13
Þ V2 = 9 L Þ [Br- ] = 5 ´ 10 -13 = 1 ´ 10-11 M
5 ´ 10-2
2. (c) : N 2(g) + O 2(g) 2NO (g) , K c = 4 × 10– 4 ... (i)
By multiplying the equation (i) by 1 . .. [K+ ] = [Br – ] = [KBr]
2
\ [KBr] = 1 × 10 –11 M
21 N 2(g) + 21 O 2(g) NO (g) ... (ii) Molarity = nKBr
VSolution (L)
Kc¢ = Kc = 4 ´ 10-4 = 2 ´ 10-2
By reversing the equation (ii), we get 1 ´ 10-11 = wKBr / 120 (Mol. wt. of KBr = 120)
1
NO (g) 12 N 2(g) + 21 O 2(g) Þ wK Br = 1 × 10 –11 × 120 = 120 × 10– 11
1 1 wK Br = 1.2 × 10– 9 g
Kc¢ ´ 10-2
Kc¢¢ = = = 50.0 8. (d) : H2 C O 3 H+ + HCO 3– ; K1 = 4.2 × 10 –7
2 HCO 3– H+ + CO 32 –; K2 = 4.8 × 10– 11
3. (b) : pH = 3
Molarity = 0.1 M . .. K 1 >> K 2 , so H2 CO 3 ionises more than HCO 3– and hence,
[H+ ] = KaC contribution of H+ is mostly due to ionisation of carbonic acid,
H+ = 10 –pH = 10 –3 thus the concentrations of H+ and HCO 3– are approximately
10-3 = Ka ´ 0.1 or 10-6 = Ka ´ 0.1 equal.
\ Ka = 10 –5
9. (a) : The order of acidity can be explained on the basis of the
4. (a) : CO 2 (g) + C (s) 2CO (g) acidity of the acids of the given conjugate base. Stronger is the
0.5 atm acid, weaker is the conjugate base. Since RCOOH is the
strongest acid amongst all, RCOO – is the weakest base. Due to
0.5 – P 2P sp hybridised carbon, acetylene is also acidic and hence a weak
base but stronger than RCOO – . As sp 3 carbon is less
Total pressure = 0.5 – P + 2P = 0.8 electronegative than sp 3 nitrogen, R – is more basic than NH2 – .
P = 0.3
KP = PC2O = (2P)2 = (0.6) 2 = 1.8 atm 10. (b) : K sp for BaCO 3 = [Ba 2+] [CO 23– ]
PCO 2 (0.5 - P) (0.5 - 0.3) given, [CO 23– ] = 1 × 10 –4 M (from Na 2 CO 3 )
Ks p = 5.1 × 10– 9
5. (c) : (K sp ) Mg(OH) 2 = [Mg 2+] [OH– ] 2 \ 5.1 × 10– 9 = [Ba 2+] × [10– 4]
1 × 10 –11 = [0.001][OH– ] 2 Þ [Ba 2+] = 5.1 × 10 –5 M
Thus, when [Ba 2+] = 5.1 × 10 –5 M, BaCO 3 precipitate will begin
Þ [OH- ]2 = 10 -11 = 10 -8 to form.
10-3
Þ [OH– ] = 10 –4
11. (d) : HSO 3F is the super acid. Its acidic strength is greater than
pOH = 4 any given species. The pKa value of other species are given
Thus, pH = 14 – 4 = 10 below :
6. (b) : In equation (ii), H2 P O4 – acts as a proton donor and thus,
HCO –3 ® 10.25
acts as an acid.
H3 O + ® –1.74
HSO4 – ® 1.92
Equilibrium 33
Lesser the pK a value, higher will be its acidic strength. Hence = 1.0 ´ 10–4 ´ 283 ´ 100 g/100mL
sequence of acidic strength will be 1000
HSO 3F > H3 O + > HSO 4 – > HCO 3– = 28.3 × 10 –4 g/100 mL
= 2.83 × 10– 3 g/100 mL
12. (d) : Given that pKa = 4.8 and pKb = 4.78
\ pH = 7 + 1/2 (pK a – pKb ) [A – ]
= 7 + 1/2 (4.80 – 4.78) = 7.01 16. (d) : For acidic buffer, pH = pK a + log [HA]
13. (d) : CO (g) + H2 O (g) ‡ˆˆˆ†ˆ CO 2(g) + H2 (g) When the acid is 50% ionised, [A– ] = [HA]
K1 = [[CCOO]2[]H[H2 O2 ]] .... (i) or pH = pKa + log1 or pH = pK a
Given pKa = 4.5 \ pH = 4.5
CH4 (g) + H2 O (g) ‡ˆˆˆ†ˆ CO (g) + 3H2 (g) \ pOH = 14 – 4.5 = 9.5
17. (d) : H 2 A H + + H A – ;
K2 = [C[CHO4]][[HH22 ] O3 ] .... (ii) [H+ ][H A– ]
K1 = [H2 A] = 1 × 10– 5
CH4 (g) + 2H2 O (g) ‡ˆˆˆ†ˆ CO 2(g) + 4H2 (g) H A – [H+ ][A2– ]
H + + A 2– ; K2 = 5 × 10– 10 = [HA – ]
K3 = [[CCHO42]][H[H22 O ]4] 2
.... (iv) K = [H+ ]2[A 2– ] K1× K 2 = 1× 10 –5 × 5 × 10 –10 = 5 × 10 –15
=
= 5 × 10 –15
[H 2 A]
From equations (i), (ii) and (iii) ; K 3 = K 1 × K2
14. (b) : X ‡ˆˆˆ†ˆ 2Y ; Z ‡ˆˆˆ†ˆ P + Q 18. (d) : AgI( s) + e– ® Ag (s) + I– , E° = – 0.152 V
Ag (s) ® Ag + + e– , E° = – 0.800 V
Initial mol. 1 0 1 0 0 AgI (s) ® Ag + + I – , E° = – 0.952 V
At equilibrium 1 – a 2a 1–a a a
æ 2a P1 ÷öø2 E°cell = 0.059 log K i.e. -0.952 = 0.059 log K sp
çè 1 + a P1 ÷øö n 1
PY 2
KP1 = PX = æ 1 - a K P2 = PZ or, log K = - 0.952 = - 16.135
0.059
sp
ç 1 + a 19. (a) : SO3 (g) ƒ SO2 (g) + 1/2 O 2(g)
è
PP PQ æa P2 ö æ a P2 ÷øö [SO2 ][O2 ]1 / 2 = Kc = 4.9 ´ 10 -2 ... (i)
PZ çè 1 + a ÷ø çè 1 + a ... (i) [SO3 ]
K P2 = =
æ 1 - a P2 öø÷
èç 1 + a SO3 (g) + 1/2 O 2(g) ƒ SO 3(g)
Þ K P 1 = 4a 2 P1 [SO3 ] = Kc¢ = 1 ... (ii)
1 - a 2 [SO2 ][O2 ]1/ 2 4.9 ´ 10 -2
4a 2 P1 For 2SO 2(g) + O 2(g) ƒ 2SO 3(g)
1 - a 2
Þ K P 1 = ...(ii) [SO ] 2 1
4.9 ´ 10 -4
K P1 3 = Kc¢ 2 =
K P2
Given is = 19 [SO2 ]2 [O2 ] 4.9 ´
...(iii) = 10000 = 416.49
24.01
Substituting values of from equation (i) and (ii) into (iii), we
get 20. (a) : Given PCl5 (g) ƒ PCl3 (g) + Cl2 (g)
4a 2 P1 t = 0 1 0 0
te q 1 – x x x
1 - a2 = 1 Þ 4 P1 = 1 Þ P1 = 1
a2 P2 9 P2 9 P2 36 Total number of moles = 1 – x + x + x = 1 + x
1 - a 2 Thus partial pressure of PCl3 = æ x ö P
çè 1 + x ÷ø
15. (c) : AgIO3 Ag+ + IO 3– [S = Solubility] 21. (d) : NH 4HS (s) ƒ NH3 (g ) + H 2S (g )
S S Initial pressure 0 0.5 0
Ks p = S2 At equi. 0 0.5 + x x
or, S 2 = 1.0 × 10 –8 or, S = 1.0 × 10– 4 mol/L
Total pressure = 0.5 + 2x = 0.84 \ x = 0.17 atm
= 1.0 × 10– 4 × 283 g/L
Kp = p NH3 × pH 2 S = (0.5 + 0.17)(0.17) = 0.11 atm2
= 1.0 ´ 10–4 ´ 283 g/L
1000 22. (b) : Higher the pK a value, weaker is the acid. Hence, strongest
acid has lowest pK a value.
34 JEE MAIN CHAPTERWISE EXPLORER
23. (d) : Conjugate base of OH– is O 2– . 33. (d) : Conjugate base is formed by the removal of H+ from acid.
OH– ƒ O 2– + H+ H2 P O4 – ® HPO4 2 – + H+
24. (d) : pH = – log[H+ ] 34. (a) : Due to thunderstorm, temperature increases. As temperature
[H+ ] = antilog (–pH) = antilog(–5.4) =3.98 × 10– 6 increases, [H+ ] also increases, hence pH decreases.
25. (a) : Kp = K c (RT)Dn 35. (c) : pH of an acid cannot exceed 7. Here we should also consider
Dn = 3 – 2 = 1 [H+ ] that comes from H2 O.
K p = K c (0.0831 × 457)1 Now [H+ ] = [H+ ] from HCl + [H+ ] f rom H 2O
\ Kp > Kc = 10– 8 + 10 –7 = 10– 8 + 10 × 10 –8 = 11 × 10– 8
\ pH = –log(11 × 10 –8 ) = 6.9587
26. (d) : Cl2 (g) + 3F2 (g) ƒ 2ClF3 (g) ; DH = –329 kJ
Favourable conditions: 36. (d) : DG = DG° + 2.303 RT logKc
(i) As the reaction is exothermic, hence decrease in At equilibrium, DG = 0
temperature will favour the forward reaction.
(ii) Addition of reactants or removal of product will farour DG° = –2.303 RT logKc
the forward reaction.
(iii) Here Dn = 2 – 4 = –2 (i.e., –ve) hence decrease in volume 37. (c) : AB2 A2 + + 2B –
or increase in pressure will favour the forward reaction.
S = 1.0 × 10 –5 mol L– 1
Ks p = [A2 +] [B– ] 2 = 1.0 × 10 –5 × (1.0 × 10 –5) 2
= 1.0 × 10– 15
27. (b) : MX 2(s) ƒ M 2 + + 2X– ( aq) 38. (c) : [N2O4 ] = 4.8´ 10-2 mol L-1
(aq) [NO2 ] = 1.2 ´ 10-2 mol L-1
s 2s
K sp = s ∙ (2s)2 = 4s 3 K c = [N O 2 ] 2 = 1.2 ´10-2 ´1.2 ´10 -2
4 × 10 –12 = 4s 3 or, s 3 = 1 × 10– 12 [N2O4 ] 4.8 ´ 10 -2
or, s = 1 × 10– 4 M Þ [M 2 +] = 1 × 10– 4 M = 0.3 × 10 –2 = 3 × 10 –3 mol L– 1
28. (a) : For endothermic reaction, DH = +ve 39. (c) : The conversion of SO2 to SO3 is an exothermic reaction,
hence decrease the temperature will favour the forward reaction.
DH = E f – Eb , it means Eb < Ef There is also a decrease in volume or moles in product side.
Thus the reaction is favoured by low temperature and high
29. (d) : M X4 (solid) M 4+ (aq) + 4X– (aq) pressure. (LeChatelier's principle).
s 4s
Solubility product, Ks p = s × (4s) 4 = 256 s5 .
\ s = 5 K sp = æ K sp ö 1/5 40. (d) : For those reactions, where Dn = 0, increase in volume at
256 çè 256 ÷ø constant temperature does not affect the number of moles at
equilibrium.
30. (b) : N 2(g) + O 2(g) 2NO (g)
Kc = [N[N2]O [O]2 2 ] = 4 ´ 10- 4 41. (a) : In this reaction the ratio of number of moles of reactants
to products is same i.e. 2 : 2, hence change in volume will not
NO (g) 12 N 2(g) + 12 O 2(g) alter the number of moles.
K c¢ = [ N 2 ]1/ 2 [O2 ]1 /2 = 1= 1 42. (c) : K p = Kc (RT)Dn ; Dn = 1 - æè1 + 1 ö = 1 - 3 = - 1
[ NO] Kc 4 ´ 10- 4 2ø 2 2
K p
\ K c = ( R T ) -1/ 2
= 2 1 = 1020 = 50
´ 10-2
43. (a) : Mg(OH)2 ® [Mg2 +] + 2[OH– ]
31. (a) : CO (g) + Cl2 (g) COCl2 (g) x 2x
Dn = 1 – 2 = –1
Ks p = [Mg 2+] [OH– ] 2
K p = K c (RT)Dn , \
K p RT ) -1 = R1T or, Ks p = (x) × (2x) 2 = x × 4x 2 = 4x 3
32. (d) : P4 (s) + 5O 2 (g) Kc
= ( 44. (a) : According to BronstedLowry concept, a Bronsted acid is
P4 O 10 (s) a substance which can donate a proton to any other substance
K c = [P4O10 ( s ) ] and a Bronsted base is a substance which can accept a proton
[P4 (s) ] [O 2 ( g ) ]5
from any other substance.
We know that concentration of a solid component is always (HSO4 ) – can accept and donate a proton.
(HSO4 ) – + H+ ® H2 S O4
taken as unity. (HSO4 ) – – H+ ® SO4 2 –
K c = [O12 ]5 45. (a) : HCl is a strong acid and its salt do not form bufter solution.
As the resultant solution is acidic, hence pH is less than 7.
Redox Reactions and Electrochemistry 35
CHAPTER REDOX REACTIONS
8 AND ELECTROCHEMISTRY
1. Consider the following reaction. (a) oxidises oxalic acid to carbon dioxide and water
(b) gets oxidised by oxalic acid to chlorine
xMnO 4– + yC 2 O 42 – + zH+ ® xMn2 + + 2yCO2 + 2z H2 O (c) furnishes H + ions in addition to those from oxalic acid
The values of x, y and z in the reaction are, respectively (d) reduces permanganate to Mn 2+ .
(a) 5, 2 and 8 (b) 5, 2 and 16 (2008)
(c) 2, 5 and 8 (d) 2, 5 and 16 (2013) 8. Given E° Cr3+/Cr = –0.72 V, E° Fe2+/Fe = –0.42 V
2. Given The potential for the cell
E°C r3 + /Cr = – 0.74 V; E° MnO 4– /Mn 2+ = 1.51 V Cr | Cr 3+ (0.1 M) | | Fe2 + (0.01 M) | Fe is
E°C r2 O 72 –/ Cr 3+ = 1.33 V; E° Cl/Cl– = 1.36 V (a) –0.26 V (b) 0.26 V
Based on the data given above, strongest oxidising agent will
(c) 0.339 V (d) –0.339 V
be (2008)
(a) MnO 4– (b) Cl – 9. The cell,
(c) Cr 3+ (d) Mn2 + (2013) Zn | Zn2 + (1 M) || Cu2 + (1 M) | Cu (E° cell = 1.10 V)
3. The standard reduction potentials for Zn 2+/ Zn, Ni 2+ /Ni, and was allowed to be completely discharged at 298 K. The relative
Fe2 +/ Fe are – 0.76, – 0.23 and – 0.44 V respectively. The æ [Zn2 + ] ö
reaction X + Y 2+ ® X 2+ + Y will be spontaneous when concentration of Zn2 + to Cu 2+ çè [Cu2+ ]÷ø is
(a) X = Ni, Y = Zn (b) X = Fe, Y = Zn (a) 9.65 × 10 4
(c) 37.3
(c) X = Zn, Y = Ni (d) X = Ni, Y = Fe (2012) (b) antilog(24.08)
(d) 10 37.3
4. The reduction potential of hydrogen halfcell will be negative (2007)
if 10. The equivalent conductances of two strong electrolytes at infinite
(a) p(H 2) = 1 atm and [H + ] = 2.0 M (2011) dilution in H2 O (where ions move freely through a solution) at
(b) p(H 2) = 1 atm and [H + ] = 1.0 M 25°C are given below:
(c) p(H 2) = 2 atm and [H + ] = 1.0 M L°C H 3C OONa = 91.0 S cm 2/ equiv.
(d) p(H 2) = 2 atm and [H + ] = 2.0 M L° HCl = 426.2 Scm2 /equiv.
5. The Gibb’s energy for the decomposition of Al 2 O 3 at 500°C What additional information/quantity one needs to calculate L°
is as follows : of an aqueous solution of acetic acid?
2/3Al2 O3 4/3Al + O 2 , Dr G = + 966 kJ mol –1 . (a) L° of chloroacetic acid (ClCH2 COOH)
(b) L° of NaCl
The potential difference needed for electrolytic reduction of
Al2 O 3 at 500°C is at least (c) L° of CH3 COOK
(d) The limiting equivalent conductance of H+ (l° H +).
(a) 5.0 V (b) 4.5 V
(2007)
(c) 3.0 V (d) 2.5 V (2010)
6. Given : E° Fe3+/Fe = – 0.036 V, E° Fe 2+ /Fe = – 0.439 V. The value 11. Resistance of a conductivity cell filled with a solution of an
of standard electrode potential for the change, electrolyte of concentration 0.1 M is 100 W. The conductivity
Fe3 + (aq) + e – ® Fe 2+ (aq) will be of this solution is 1.29 S m –1. Resistance of the same cell when
(a) – 0.072 V (b) 0.385 V filled with 0.2 M of the same solution is 520 W. The molar
(c) 0.770 V (d) – 0.270 V (2009) conductivity of 0.02 M solution of the electrolyte will be
7. Amount of oxalic acid present in a solution can be determined (a) 124 × 10– 4 S m2 mol– 1
by its titration with KMnO 4 solution in the presence of H2 S O4 .
The titration gives unsatisfactory res ult when carried out in (b) 1240 × 10– 4 S m2 mol– 1
the presence of HCl, because HCl
(c) 1.24 × 10 –4 S m2 mol– 1
(d) 12.4 × 10– 4 S m2 mol– 1 (2006)
36 JEE MAIN CHAPTERWISE EXPLORER
12. The molar conductivities L°N aOAc and L°H Cl at infinite dilution 19. The limiting molar conductivities L° for NaCl, KBr and KCl are
in water at 25°C are 91.0 and 426.2 S cm2 / mol respectively. To 126, 152 and 150 S cm2 mol– 1 respectively. The L° for NaBr is
(a) 128 S cm2 mol– 1 (b) 176 S cm2 mol– 1
calculate L° HOAc, the additional value required is (c) 278 S cm2 mol– 1 (d) 302 S cm2 mol– 1 (2004)
(a) L° H 2O (b) L° KCl
(c) L°N aOH (d) L°N aCl. (2006) 20. The standard e.m.f. of a cell, involving one electron change is
13. Which of the following chemical reactions depicts the oxidising found to be 0.591 V at 25°C. The equilibrium constant of the
behaviour of H2 S O4 ? reaction is (F = 96,500 C mol– 1, R = 8.314 JK– 1m ol– 1)
(a) 2HI + H2 SO4 ® I2 + SO2 + 2H2 O
(a) 1.0 × 10 1 (b) 1.0 × 105
(b) Ca(OH) 2 + H2 S O4 ® CaSO 4 + 2H2 O
(c) 1.0 × 101 0 (d) 1.0 × 103 0 (2004)
(c) NaCl + H2 S O4 ® NaHSO4 + HCl
(d) 2PCl5 + H2 S O4 ® 2POCl 3 + 2HCl + SO2 Cl2 . 21. Consider the following E° values.
(2006) E°Fe3+/Fe 2+ = +0 .77 V ; E° 2+ /Sn = - 0.14 V
Sn
14. Electrolyte Under standard conditions the potential for the reaction
KCl KNO3 HCl NaOAc NaCl Sn( s) + 2Fe3 +( aq) ® 2Fe2 +( aq) + Sn2 +( aq) is
(S cm2 mol– 1 ) 149.9 145.0 426.2 91.0 126.5 (a) 1.68 V (b) 1.40 V
Calculate molar conductance of acetic acid using appropriate (c) 0.91 V (d) 0.63 V (2004)
molar conductances of the electrolytes listed above at infinite 22. In a hydrogenoxygen fuel cell, combustion of hydrogen occurs
to
dilution in H2 O at 25°C. (a) generate heat
(b) create potential difference between the two electrodes
(a) 517.2 (b) 552.7 (c) produce high purity water
(d) remove adsorbed oxygen from electrode surface.
(c) 390.7 (d) 217.5 (2005) (2004)
15. Aluminium oxide may be electrolysed at 1000°C to furnish
aluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500
Coulombs). The cathode reaction is
Al 3+ + 3e- ® Al0 23. Among the properties (A) reducing (B) oxidising
To prepare 5.12 kg of aluminium metal by this method would (C) complexing, the set of properties shown by CN – ion towards
require metal species is
(a) 5.49 × 10 7 C of electricity (a) A, B (b) B, C
(b) 1.83 × 10 7 C of electricity (c) C, A (d) A, B, C. (2004)
(c) 5.49 × 104 C of electricity 24. Standard reduction electrode potentials of three metals A, B and
(d) 5.49 × 10 10 C of electricity (2005) C are +0.5 V, –3.0 V and –1.2 V respectively. The reducing
16. The highest electrical conductivity of the following aqueous power of these metals are
solutions is of (a) B > C > A (b) A > B > C
(a) 0.1 M acetic acid (c) C > B > A (d) A > C > B (2003)
(b) 0.1 M chloroacetic acid 25. For a cell reaction involving a twoelectron change, the standard
(c) 0.1 M fluoroacetic acid e.m.f. of the cell is found to be 0.295 V at 25°C. The equilibrium
(d) 0.1 M difluoroacetic acid. (2005) constant of the reaction at 25°C will be
E° (a) 1 × 10– 10 (b) 29.5 × 10– 2
17. The M 3+ / M 2 + values for Cr, Mn, Fe and Co are – 0.41, +1.57, (c) 10 (d) 1 × 101 0 (2003)
0.77 and +1.97 V respectively. For which one of these metals 26. For the redox reaction:
the change in oxidation state from +2 to +3 is easiest? Zn( s) + Cu 2+ (0.1 M) ® Zn + (1M) + Cu( s)
(a) Cr (b) Mn taking place in a cell, E° cell is 1.10 volt. E cell for the cell will be
(c) Fe (d) Co (2004) RT ( ) 2.303
F
18. In a cell that utilizes the reaction, = 0.0591
Zn (s) + 2H+ ( aq) ® Zn2 +( aq) + H2 (g)
addition of H2 SO4 to cathode compartment, will (a) 2.14 V (b) 1.80 V
(a) lower the E and shift equilibrium to the left
(b) lower the E and shift the equilibrium to the right (c) 1.07 V (d) 0.82 V (2003)
(c) increase the E and shift the equilibrium to the right
(d) increase the E and shift the equilibrium to the left. 27. When during electrolysis of a solution of AgNO 3, 9650 coulombs
(2004) of charge pass through the electroplating bath, the mass of silver
deposited on the cathode will be
(a) 1.08 g (b) 10.8 g
(c) 21.6 g (d) 108 g (2003)
Redox Reactions and Electrochemistry 37
28. The heat required to raise the temperature of body by 1ºC is 32. Conductivity (unit Siemen's S) is directly proportional to area
called of the vessel and the concentration of the solution in it and is
(a) specific heat inversely proportional to the length of the vessel then the unit
(b) thermal capacity of the constant of proportionality is
(c) water equivalent (a) S m mol– 1 (b) S m2 mol– 1
(d) none of these. (2002) (c) S– 2m 2 mol (d) S2 m2 mol– 2 (2002)
29. Which of the following reaction is possible at anode? 33. Which of the following is a redox reaction?
(a) NaCl + KNO 3 ® NaNO 3 + KCl
(a) 2 Cr3 + + 7H2 O ® Cr 2O 72 – + 14H+ (b) CaC2 O 4 + 2HCl ® CaCl2 + H2 C2 O 4
(c) Mg(OH) 2 + 2NH4 C l ® MgCl2 + 2NH4 OH
(b) F2 ® 2F– (d) Zn + 2AgCN ® 2Ag + Zn(CN)2
(c) (1/2) O 2 + 2H+ ® H2 O
(d) None of these. (2002) (2002)
30. What will be the emf for the given cell, 34. When KMnO 4 acts as an oxidising agent and ultimately forms
[MnO 4] – 1, MnO 2, Mn 2O 3, Mn 2+ then the number of electrons
Pt | H2 (P1 ) | H+ ( aq) | | H2 (P2 ) | Pt transferred in each case respectively is
(a) RFT log PP21 (b) RT log P1
(c) RFT log PP21 2 F P2
(a) 4, 3, 1, 5 (b) 1, 5, 3, 7
(d) none of these. (2002) (c) 1, 3, 4, 5 (d) 3, 5, 7, 1 (2002)
31. If f denotes reduction potential, then which is true? 35. EMF of a cell in terms of reduction potential of its left and right
electrodes is
(a) E° cell = fr ight – f left
(b) E° cell = fl eft + f right (2002) (a) E = E left – E right (b) E = El eft + E right
(c) E°c ell = fl eft – f right (c) E = Er ight – El eft (d) E = –(E right + El eft)
(d) E° cell = –(fl eft + fr ight)
(2002)
Answer Key
1. (d) 2. (a) 3. (c) 4. (c) 5. (d) 6. (c)
7. (d) 8. (b) 9. (d) 10. (b) 11. (a) 12. (d)
13. (a) 14. (c) 15. (a) 16. (d) 17. (a) 18. (c)
19. (a) 20. (c) 21. (c) 22. (b) 23. (c) 24. (a)
25. (d) 26. (c) 27. (b) 28. (b) 29. (a) 30. (b)
31. (a) 32. (b) 33. (d) 34. (c) 35. (c)
38 JEE MAIN CHAPTERWISE EXPLORER
1. (d) : 2MnO 4– + 5C 2 O 42 – + 16H + ® 2Mn 2+ + 10CO 2 + 8H 2 O Titration cannot be done in the presence of HCl because KMnO 4
\ x = 2, y = 5, z = 16 being a strong oxidizing agent oxidises HCl to Cl 2 and get
2. (a) : Greater the reduction potential of a substance, stronger itself reduced to Mn 2+ . So actual amount of oxalic acid in
is the oxidising agent. solution cannot be determined.
\ MnO 4– is the strongest oxidising agent. 8. (b) : Cr ® Cr 3+ + 3e – E° red = –0.72 V
3. (c) : The elements with high negative value of standard Fe 2+ + 2e – ® Fe E° red = –0.42 V
reduction potential are good reducing agents and can be easily 2Cr + 3Fe 2+ ® 2Cr 3+ + 3Fe
oxidised.
Thus X should have high negative value of standard potential E° cell = E°c athode – E°a node = –0.42 – (–0.72)
than Y so that it will be oxidised to X2 + by reducing Y2 + to Y. E° cell = 0.3
X = Zn, Y = Ni According to Nernst equation,
Zn + Ni 2+ ® Zn 2+ + Ni
Alternatively, for a spontaneous reaction E° must be positive. Ecell = E °cell - 0n.0ce5ll9 log1 0 [[CFer23++ ] 2
]3
E° = E° reduced species – E°o xidised species Ecell = 0.3 - 0.065 9 log1 0 (0.1) 2
= – 0.23 – (– 0.76) (0.01) 3
Þ E° = + 0.53 V
4. (c) : 2H + (aq) + 2e – H 2 (g) Ecell =0.3 - 0.065 9 log10 104
Ec ell = 0.3 – 0.039
Ered = Er°ed - 0.0591 log pH 2 \ E cell = 0.261 V
n [H+ ]2
Ered = 0 - 0.0591 log (12) 2 9. (d) : Zn + Cu2 + Zn 2+ + Cu
2
Er ed will only be negative when p H2 > [H +] . So option (c) is Ecell = Ec °e ll – 0.059 log [Zn 2+ ]
correct. 2 [Cu 2+ ]
5. (d) : The ionic reactions are : When the cell is completely discharged, E cell = 0
2/3Al 23 + + 4e – 4/3Al 0 = 1.1 – 0.059 [Zn2+ ]
2 [Cu 2+ ]
2/3O 32 – O 2 + 4e – log
Thus, no. of electrons transferred = 4 = n [Zn2+ ] 2 × 1.1 Zn 2+
[Cu 2+ ] 0.059 Cu 2+
DG = –nFE = – 4 × 96500 × E or log = or, log = 37.3
or 966 × 10 3 = – 4 × 96500 × E
Þ E = - 966 ´ 10 3 = - 2.5 V or Zn 2+ = 103 7.3
4 ´ 96500 Cu 2+
6. (c) : Given, 10. (b) : According to Kohlrausch’s law, the molar conductivity
Fe 3+ + 3e – Fe ; E°1 = – 0.036 V at infinite dilution (L°) for weak electrolyte, CH 3 COOH is
Fe2 + + 2e – Fe ; E°2 = – 0.439 V Λ°CH3COOH = Λ°CH3 COONa + Λ°HCl - Λ° NaCl
Required equation is So, for calculating the value of Λ° C H3 COOH , value of Λ° N aCl
should also be known.
Fe3 + + e – Fe 2+ ; E°3 = ?
Applying DG° = –nFE°
\ DG°3 = DG°1 – DG°2
(–n3 FE°3 ) = (–n1 FE°1 ) – (–n2 FE°2 ) 11. (a) : k = 1 æ l ö i.e., 1.29 = 1 æ l ö
E°3 = 3E°1 – 2E°2 = 3 × (– 0.036) – 2 × (–0.439) R çè a ÷ø 100 çè a ÷ø
E° 3 = – 0.108 + 0.878 = 0.77 V l/a = 129 m –1
R = 520 W for 0.2 M, C = 0.02 M
7. (d) : Oxalic acid present in a solution can be determined by
its titration with KMnO 4 solution in the presence of H 2 SO 4 .
COOH lm = k´ 1000 = 1´129 ´ 1000 ´ 10-6 m 3
2KMnO4 + 3H2SO4 + 5 COOH molarity 520 0.02
K2SO4 + 2MnSO4 + 10CO2 + 5H2 O = 124 × 10 –4 S m 2 mol –1
Redox Reactions and Electrochemistry 39
12. (d) : CH 3C OONa + HCl ® CH 3 COOH + NaCl On adding H 2S O 4 the [H + ] will increase therefore E cell will
From the reaction, also increase and the equilibrium will shift towards the right.
L°CH3COONa + L°HCl = L°CH3C OOH + L° NaCl 19. (a) : L°NaCl = L° + + L°C l– ... (i)
or, L°CH3COOH = L°CH3C OONa + L°HCl - L° NaCl ... (ii)
Na ... (iii)
Thus to calculate the value of L°C H3C OOH one should know
the value of L° NaCl along with L°C H3C OONa and L° HCl. L°KBr = L° + + L° –
K Br
L°KCl = L° + + L°C l–
K
Equation (i) + (ii) – (iii)
13. (a) : In the reaction, L°NaBr = L° + + L° –
2 H I + H 2 S O 4 ® I 2 + S O 2 + 2 H 2 O
Na Br
+1–1 +1 +6 –2 0 +4 +1 –2
= 126 + 152 – 150 = 128 S cm 2 mol –1
I st half reaction : 2HI ® I 2
20. (c) : Ecell = E°cell - 0.0n591 log K c
–1 0
0 = 0.591- 0.015 91l og Kc
In this reaction oxidation number of I increases by one, thus Þ – 0.591 = – 0.0591 log K c
this is an oxidation reaction and HI behaves as a reducing Þ log Kc = 00..0559911 = 10
agent. \ K c = antilog 10 = 1 × 10 10
II nd half reaction : H 2 SO 4 ® SO 2
+6 +4
In this reaction oxidation number of S decreases by two,
thus this is a reduction reaction and H2 SO4 behaves as oxidising
agent.
14. (c) : L¥ AcOH = L¥ AcONa + L¥H Cl – L¥ NaCl 21. (c) : E°cell = E° 2+ + E°F e3+/Fe 2+
= 91.0 + 426.2 – 126.5
= 390.7 S cm 2 mol –1 Sn/Sn
= 0.14 + 0.77 = 0.91 V
15. (a) : From Faraday's 1st law, 22. (b) : Direct conversion of chemical energy to electric energy
W = Z × Q [W = weight, Z = electrochemical can be made considerably more efficient (i.e. upto 75%) than
equivalent, Q = quantity of electricity] the 40% maximum now obtainable through burning of fuel and
Now E = Z × F [E = equivalent weight , F = Faraday] using the heat to form steam for driving turbines. Furthermore,
the water obtained as a byproduct may be used for drinking by
or W = E ´ Q the astronauts.
F At anode : 2H2 (g) + 4OH– (aq) ® 4H2 O (l) + 4e–
At cathode : O 2(g) + 2H2 O (l) + 4e – ® 4OH– ( aq)
or Q = W ´ F or Q = W ´ F
E A 2H2 (g) + O 2(g) ® 2H2 O (l)
n
[A = Atomic weight , n = valency of ion] 23. (c) : CN – ions act both as reducing agent as well as good
complexing agent.
or Q = n ´ w ´ F
A
24. (a) : A B C
= 3 ´ 5.12 ´ 103 ´ 96500 = 5.49 × 10 7 C
27 E ° red +0.5 V –3.0 V –1.2 V
More is the value of reduction potential, more is the tendency
16. (d) : Higher the acidity, higher will be the tendency to release
protons and hence lighter will be the electrical conductivity. to get reduced, i.e. less is the reducing power.
Difluoroacetic acid will be strongest acid due to electron
withdrawing effect of two fluorine atoms so as it will show The reducing power follows the following order:
maximum electrical conductivity.
B > C > A.
17. (a) : Cr 2+ | Cr 3+ = +0.41 V 25. (d) : E°cell = 0.0n591 log K c
Mn 2+ | Mn 3+ = –1.57 V
Fe2 + | Fe3 + = – 0.77 V 0.295 = 0.025 91 log Kc
Co2 + | Co3 + = – 1.97 V or, 0.295 = 0.0295 logK c
More is the value of oxidation potential more is the tendency or, Kc = antilog 10 or K c = 1 × 10 10
to get oxidised.
As Cr will have maximum oxidation potential value, therefore 26. (c) : Ecell = E °cell - 0.0n591 log 1
its oxidation will be easiest. 0.1
Here n = 2, E ° cell = 1.10 V
18. (c) : Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Ecell = 1.10 - 0.025 91 log10
Ecell = E °cell - 0.025 9 log [ Zn2 + ] ´ pH 2 Ec ell = 1.10 – 0.0295 = 1.0705 V
[H+ ] 2
40 JEE MAIN CHAPTERWISE EXPLORER
27. (b) : The mass of silver deposited on the cathode or S = K AC [K = constant of proportionality]
L
= 10986´5906050 = 10.8 g
or K = SL
AC
S × m × m 3
28. (b) : It is also known as heat capacity. \ Unit of S × m = m2 ´ mol
K = m 2 × mmo 3l
29. (a) : Here Cr3 + is oxidised to Cr2 O 72 –.
30. (b) : 2H+ + 2e – H2 (P2 ) = S m2 mol– 1
H2 (P1 ) 2H+ + 2e– 33. (d) : The oxidation states show a change only in reaction (d).
Overall reaction : H2 ( P1 ) H2 (P2 ) – 2e –
0 +1 0 +2
P2 P2 P1 Zn + 2AgCN 2Ag + Zn(CN) 2
P1 P1 P2
E = E° - RT log = 0- RT og = RT log + 2 e –
nF nF nF
31. (a) : Ec ell = Er ight (cathode) – E left (anode). 34. (c) : +3 –4e – +7 –e – [MnO 4 ] –1
Mn2 O 3
[KMnO 4 ]
32. (b) : S µ A (A = area) –5e – –3e –
S µ C (C = concentration)
Mn2 + +4
MnO 2
S µ 1 (L = length) 35. (c) : E cell = Reduction potential of cathode (right) – reduction
L potential of anode (left)
Combining we get, S µ AC = E right – E left.
L
Chemical Kinetics 41
CHAPTER CHEMICAL KINETICS
9
1. The rate of a reaction doubles when its temperature changes (log 2 = 0.301)
from 300 K to 310 K. Activation energy of such a reaction (a) 230.3 minutes (b) 23.03 minutes (2009)
will be (R = 8.314 J K –1 mol –1 and log 2 = 0.301) (c) 46.06 minutes (d) 460.6 minutes
(a) 60.5 kJ mol –1 (b) 53.6 kJ mol –1 (2013) 7. For a reaction 12 A ® 2 B rate of disappearance of A is related
(c) 48.6 kJ mol –1 (d) 58.5 kJ mol –1 to the rate of appearance of B by the expression
2. For a first order reaction, (A) ® products, the concentration (a) - d[ A] = 4 dd[Bt ]
of A changes from 0.1 M to 0.025 M in 40 minutes. The rate dt
of reaction when the concentration of A is 0.01 M is (b) - d [ A] = 12 dd[Bt ]
(a) 3.47 × 10 –4 M/min (b) 3.47 × 10 –5 M/min dt
(c) 1.73 × 10 –4 M/min (d) 1.73 × 10 –5 M/min (2012)
3. The rate of a chemical reaction doubles for every 10°C rise (c) - d [ A] = 14 dd[Bt ]
dt
of temperature. If the temperature is raised by 50°C, the rate d[ A] = dd[Bt ]
dt
of the reaction increases by about (d) - (2008)
(a) 10 times (b) 24 times
(c) 32 times (d) 64 times (2011) 8. Consider the reaction, 2A + B ® products. When concentration
4. Consider the reaction : of B alone was doubled, the halflife did not change. When the
Cl2 (aq) + H 2 S (aq) S (s) + 2H +( aq) + 2Cl –( aq) concentration of A alone was doubled, the rate increased by two
The rate of reaction for this reaction is times. The unit of rate constant for this reaction is
rate = k[Cl 2 ][H 2 S] (a) s –1 (b) L mol– 1 s –1
Which of these mechanism is/are consistent with this rate
(c) no unit (d) mol L– 1 s –1. (2007)
equation? 9. The energies of activation for forward and reverse reactions for
A. Cl2 + H2 S H+ + Cl– + Cl+ + HS– (slow) A 2 + B 2 ƒ 2AB are 180 kJ mol– 1 and 200 kJ mol– 1 respectively.
Cl + + HS – H + + Cl – + S (fast) The presence of a catalyst lowers the activation energy of both
B. H2 S Û H + + HS – (fast equilibrium) (forward and reverse) reactions by 100 kJ mol– 1. The enthalpy
Cl 2 + HS – 2Cl – + H + + S (slow) change of the reaction (A 2 + B2 ® 2AB) in the presence of a
catalyst will be (in kJ mol– 1)
(a) A only (b) B only
(a) 20 (b) 300
(c) Both A and B (d) Neither A nor B (2010)
(c) 120 (d) 280 (2007)
5. The time for half life period of a certain reaction 10. The following mechanism has been proposed for the reaction
A Products is 1 hour. When the initial concentration
of the reactant A is 2.0 mol L– 1 , how much time does it take of NO with Br 2 to form NOBr.
for its concentration to come from 0.50 to 0.25 mol L– 1 if it
NO (g) + Br2 (g) NOBr2 (g)
is a zero order reaction? NOBr2 (g) + NO (g) ® 2NOBr( g)
If the second step is the rate determining step, the order of the
(a) 1 h (b) 4 h reaction with respect to NO (g) is
(c) 0.5 h (d) 0.25 h
(2010) (a) 1 (b) 0
6. The halflife period of a first order chemical reaction is (c) 3 (d) 2 (2006)
6.93 minutes. The time required for the completion of 99% 11. Rate of a reaction can be expressed by Arrhenius equation as :
of the chemical reaction will be k = Ae– E/RT . In this equation, E represents
42 JEE MAIN CHAPTERWISE EXPLORER
(a) the energy above which all the colliding molecules will (a) k is equilibrium constant.
(b) A is adsorption factor.
react (c) E a is energy of activation.
(d) R is Rydberg constant.
(b) the energy below which colliding molecules will not react
(c) the total energy of the reacting molecules at a temperature, T (2003)
(d) the fraction of molecules with energy greater than the 18. For the reaction system:
activation energy of the reaction. (2006) 2NO (g) + O 2(g) ® 2NO 2(g) ,
volume is suddenly reduced to half its value by increasing the
12. A reaction was found to be second order with respect to the
pressure on it. If the reaction is of first order with respect to O 2
concentration of carbon monoxide. If the concentration of carbon and second order with respect to NO 2, the rate of reaction will
(a) diminish to onefourth of its initial value
monoxide is doubled, with everything else kept the same, the
rate of reaction will be (b) diminish to oneeighth of its initial value
(a) remain unchanged (c) increase to eight times of its initial value
(b) tripled (d) increase to four times of its initial value. (2003)
(c) increased by a factor of 4
(d) doubled. (2006) 19. The rate law for a reaction between the substances A and B is
given by rate = k [A] n [B]m . On doubling the concentration of
13. t 1/4 can be taken as the time taken for the concentration of a A and halving the concentration of B, the ratio of the new rate
reactant to drop to 3/4 of its initial value. If the rate constant for
to the earlier rate of the reaction will be as
a first order reaction is k, the t1 /4 can be written as 1 (b) (m + n)
(a) 2m + n (d) 2( n – m)
(a) 0.10/k (b) 0.29/k (c) (n – m)
(c) 0.69/k (d) 0.75/k (2005) (2003)
14. A reaction involving two different reactants can never be 20. The formation of gas at the surface of tungsten due to adsorption
(a) unimolecular reaction is the reaction of order
(b) first order reaction (a) 0 (b) 1
(c) second order reaction (c) 2 (d) insufficient data. (2002)
(d) bimolecular reaction. (2005) 21. The differential rate law for the reaction,
H2 + I 2 ® 2HI is
15. The rate equation for the reaction 2A + B ® C is found to be :
rate = k[A] [B]. The correct statement in relation to this reaction (a) - d [H 2 ] = - d [dIt2 ] = - d [dHt I]
is that the dt
(a) unit of k must be s– 1
(b) t1 /2 is a constant (b) d [H2] = d [dIt2 ] = 12 d [dHt I]
(c) rate of formation of C is twice the rate of disappearance dt
of A
(d) value of k is independent of the initial concentrations of (c) 1 d [H2 ] = 12 d d[It2 ] = - d [dHt I]
A and B. 2 dt
(2004)
(d) -2 d [H2 ] = -2 d [dIt2 ] = d [dHt I] (2002)
dt
16. In a first order reaction, the concentration of the reactant, 22. For the reaction A + 2B ® C, rate is given by R = [A] [B]2 then
decreases from 0.8 M to 0.4 M in 15 minutes. The time taken the order of the reaction is
for the concentration to change from 0.1 M to 0.025 M is (a) 3 (b) 6
(a) 30 minutes (b) 15 minutes (c) 5 (d) 7 (2002)
(c) 7.5 minutes (d) 60 minutes. (2004) 23. Units of rate constant of first and zero order reactions in terms
17. In the respect of the equation k = Ae –Ea /RT in chemical kinetics, of molarity M unit are respectively
which one of the following statements is correct?
(a) s– 1, Ms –1 (b) s– 1, M
(c) Ms– 1, s –1 (d) M, s –1 (2002)
Answer Key
1. (b) 2. (a) 3. (c) 4. (a) 5. (d) 6. (c)
7. (c) 8. (b) 11. (b) 12. (c)
13. (b) 14. (a) 9. (a) 10. (d) 17. (c) 18. (c)
19. (d) 20. (a) 23. (a)
15. (d) 16. (a)
21. (d) 22. (a)
Chemical Kinetics 43
1. (b) : As r = k[A] n = 06.6.9933
Since reaction follows 1st order kinetics,
r2 = k 2
r1 k1
Since r2 = 2 (Given) t = 2.303 log [ A0 ]
r1 l [ A]
\ k 2 = 2 where [A0] = initial concentration
k1
and [A] = concentration of A at time t.
log 10 k2 = Ea é TT2 1-T2T 1 ûúù Q Reaction is 99% complete, \ [ A0 ] = 100
k1 2.303 R
ê [ A] 1
ë
log 2 = 2.303 ´ Ea ´ é 310 - 300 ù or t = 2.3003.6´936.93 ´ log(100)
8.314 10 -3 ëê 310 ´ 300 úû = 23.03 × 2 log(10) = 46.06 minutes.
Ea = 0.3010 ´ 2.303 ´ 8.314 ´ 10-3 ´ 93 ´ 10 3
10
7. (c) : For this reaction,
Ea = 53.6 kJ mol –1 Rate = - 1 d [ A] = 12 d[B ]
1/ 2 dt dt
2. (a) : For the first order reaction
k = 2.303 log a Þ - d [ A] = 1 d[B ]
t a - x dt 4 dt
a = 0.1 M, a – x = 0.025 M, t = 40 min 8. (b) : Rate = k [A] x [B] y
k = 2.303 log 0.1 = 2.303 log 4 = 0.0347 min –1 When [B] is doubled, keeping [A] constant halflife of the
reaction does not change.
40 0.025 40
[A] product 0.693
Thus, rate = k[A] Now, for a first order reaction t1 /2 = k
rate = 0.0347 × 0.01 M min –1 i.e. t 1/2 is independent of the concentration of the reactant.
= 3.47 × 10 –4 M min –1 Hence the reaction is first order with respect to B. Now
when [A] is doubled, keeping [B] constant, the rate also
Rate at 50°C = 50 doubles. Hence the reaction is first order with respect to A.
Rate at T1 °C
3. (c) : 2 10 = 25 = 32 times. \Rate = [A]1 [B]1 \ order = 2
4. (a) : The rate equation depends upon the rate determining Now for a nth order reaction, unit of rate constant is
step. The given rate equation is only consistent with the (L) n–1 (mol) 1–n s –1 when n = 2, unit of rate constant is
mechanism A. L mol– 1 s –1.
5. (d) : For a zero order reaction, t 1/2 is given as 9. (a) : DH R = E f - Eb = 180 - 200 = - 20 kJ mol –1
The correct answer for this question should be
t1/ 2 = [ A0 ] or k = [A0] –20 kJ mol– 1. But no option given is correct. Hence we can
2k 2t1/ 2
ignore sign and select option (a).
Given, t 1/2 = 1hr, [A0 ] = 2M
10. (d) : NO (g) + Br 2(g) ƒ NOBr2 (g)
\ k = 2 = 1 mol L-1 hr -1 NOBr 2(g) + NO (g) ® 2NOBr (g) [rate determining step]
2 ´ 1 Rate of the reaction (r) = K [NOBr2 ] [NO]
where [NOBr2 ] = KC [NO][Br 2]
Integrated rate law for zero order reaction is r = K ∙ K C ∙ [NO][Br 2 ][NO]
r = K¢ [NO]2 [Br2 ]
[A] = –kt + [A 0 ] The order of the reaction with respect to NO (g) = 2
Here, [A 0] = 0.5 M and [A] = 0.25 M
Þ 0.25 = –t + 0.5 Þ t = 0.25 hours 11. (b) : k = Ae –E/RT
6. (c) : Given, t1/2 = 6.93 min where E = activation energy, i.e. the minimum amount of
energy required by reactant molecules to participate in a
l = 0.693 (for 1 st order reaction) reaction.
t1 / 2
44 JEE MAIN CHAPTERWISE EXPLORER
12. (c) : Given r1 = dx = k [CO] 2 Therefore, the concentration of reactant will fall from 0.1 M
dt to 0.025 M in two half lives.
r2 = k[2CO]2 = 4k[CO] 2 i.e. 2t 1/2 = 2 × 15 = 30 minutes.
Thus, according to the rate law expression doubling the 17. (c) : In Arrhenius equation, k = Ae –E a /RT
concentration of CO increases the rate by a factor of 4. k = rate constant, A = frequency factor
13. (b) : t1 /4 = 2.303 log 4 = 0.29 T = temperature, R = gas constant, Ea = energy of activation.
k 3 k This equation can be used for calculation of energy of activation.
14. (a) : Generally, molecularity of simple reactions is equal to 18. (c) : Rate1 = k [NO]2 [O 2 ]
the sum of the number of molecules of reactants involved in When volume is reduced to 1/2, concentration becomes two
the balanced stoichiometric equation. Thus, a reaction times.
involving two different reactants can never be unimolecular. Rate2 = k [2NO]2 [2O 2]
But a reaction involving two different reactants can a first order
reaction. For example, for the following reaction Rate1 = k[ NO]2 [O2 ] or R at e1 = 1
Rate 2 k[2 NO]2 [ 2O 2 ] Rate2 8
RCl + H2 O ® ROH + HCl
Expected rate law : \ Rate2 = 8 Rate1
Rate = k[RCl][H2 O] expected order = 1 + 1 = 2 19. (d) : Rate1 = k [A]n [B] m
But actual rate law : On doubling the concentration of A and halving the
Rate = k¢[RCl] actual order = 1 concentration of B
Here water is taken in excess, hence its concentration may be Rate2 = k [2A]n [B/2]m
taken constant. Ratio between new and earlier rate
Here the molecularity of the reaction = 2 and the order of the k[2 A]n [B / 2]m m
reaction = 1. k[A]n [B ]m
( ) = = 2n ´ 1 = 2 n -m
2
15. (d) : 2 A + B ® C 20. (a)
rate = k [A] [B] 21. (d) : H2 + I2 ® 2HI
The value of k (velocity constant) is always independent of the When 1 mole of H2 and 1 mole of I 2 reacts, 2 moles of HI are
concentration of reactant and it is a function of temperature formed in the same time interval.
only.
For a second order reaction, unit of rate constant, Thus the rate may be expressed as
k is L mol– 1 sec– 1 for a second order reaction, -d [H2] = -d [I 2 ] = 1 d [ HI ]
1 dt dt 2 dt
ka
t1 / 2 = The negative sign signifies a decrease in concentration of the
i.e. t 1/2 is inversely proportional to initial concentration. reactant with increase of time.
2A + B ® C 22. (a) : Order is the sum of the power of the concentrations terms
Rate = - 1 d[ A] = - d[B] = d[C ] in rate law expression.
2 dt dt dt R = [A] × [B] 2
Thus, order of reaction = 1 + 2 = 3
i.e. rate of formation of C is half the rate of disappearance
of B. 23. (a) : Unit of K = (mol L– 1) 1 – n s –1,
16. (a) : The concentration of the reactant decreases from 0.8 M where n = order of reaction
to 0.4 M in 15 minutes, n = 0 Þ zero order reaction
i.e. t1 /2 = 15 minute. n = 1 Þ first order reaction
Surface Chemistry 45
CHAPTER SURFACE CHEMISTRY
10
1. The coagulating power of electrolytes having ions Na +, Al 3+ 5. In Langmuir’s model of adsorption of a gas on a solid surface
and Ba 2+ for arsenic sulphide sol increases in the order : (a) the rate of dissociation of adsorbed molecules from the
(a) Al 3+ < Na + < Ba 2+ (b) Al 3+ < Ba 2+ < Na + surface does not depend on the surface covered
(c) Na + < Ba 2+ < Al 3+ (d) Ba 2+ < Na + < Al 3+ (b) the adsorption at a single site on the surface may involve
(2013) multiple molecules at the same time
2. According to Freundlich adsorption isotherm, which of the (c) the mass of gas striking a given area of surface is
following is correct? proportional to the pressure of the gas
(a) x µ p1 (b) x µ p1/n (d) the mass of gas striking a given area of surface is
m m independent of the pressure of the gas.
(2006)
(c) x µ p0 6. The d+isperse phase in colloidal iron (III) hydroxide and
m colloidal gold is positively and negatively charged, respectively.
(d) All the above are correct for different ranges of pressure. Which of the following statements is NOT correct?
(2012) (a) Magnesium chloride solution coagulates, the gold sol more
3. Which of the following statements is incorrect regarding readily than the iron (III) hydroxide sol
physisorption? (b) Sodium sulphate solution causes coagulation in both sols
(a) It occurs because of van der Waals forces. (c) Mixing of the sols has no effect
(b) More easily liquefiable gases are adsorbed readily. (d) Coagulation in both sols can be brought about by
(c) Under high pressure it results into multi molecular layer
on adsorbent surface. electrophoresis
(2005)
(d) Enthalpy of adsorption (DHa dsorption ) is low and positive. 7. The volume of a colloidal particle, V c as compared to the volume
(2009)
of a solute particle in a true solution V s could be
4. Gold numbers of protective colloids A, B, C and D are 0.50, (a) ~ 1 (b) ~ 102 3
(c) ~ 10– 3 (d) ~ 103 (2005)
0.01, 0.10 and 0.005, respectively. The correct order of their 8. Which one of the following characteristics is not correct for
protective powers is physical adsorption?
(a) Adsorption on solids is reversible
(a) B < D < A < C (b) Adsorption increases with increase in temperature
(b) D < A < C < B
(c) C < B < D < A (c) Adsorption is spontaneous
(d) A < C < B < D (2008) (d) Both enthalpy and entropy of adsorption are negative.
(2003)
Answer Key
1. (c) 2. (d) 3. (d) 4. (d) 5. (c) 6. (c)
7. (d) 8. (b)
46 JEE MAIN CHAPTERWISE EXPLORER
1. (c) : For a negatively charged sol, like As2 S3 , greater the positive that the mass of the gas adsorbed per gram of the adsorbent is
charge on cations, greater is the coagulating power. related to the equilibrium pressure according to the equation:
2. (d) : According to Freundlich adsorption isotherm x = aP
m 1 + bP
x = kp1/n where x is the mass of the gas adsorbed on m gram of the
m
1/n can have values between 0 to 1 over different ranges of adsorbent, P is the pressure and a, b are constants.
pressure. 6. (c) : Opposite charges attract each other. Hence on mixing
3. (d) : Physical adsorption is an exothermic process (i.e., coagulation of two sols may be take place.
DH = –ve) but its value is quite low because the attraction of 7. (d) : For true solution the diameter range is 1 to 10 Å and for
colloidal solution diameter range is 10 to 1000 Å.
gas molecules and solid surface is weak van der Waals forces.
Vc = (4 / 3) prc3 = æ rc ö 3
4. (d) : The different protecting colloids differ in their protecting Vs (4 / 3)pr s 3 èç rs ø÷
powers. Zsigmondy introduced a term called Gold number
Ratio of diameters = (10/1) 3 = 10 3
to describe the protective power of different colloids. Smaller
the value of gold number greater will be protecting power of 8. Vc /V s ; 10 3
(b) : During adsorption, there is always decrease in surface
the protective colloid. Thus
energy which appears as heat. Therefore adsorption always takes
1
protective power of colloid µ Gold number place with evolution of heat, i.e. it is an exothermic process and
5. (c) : Assuming the formation of a monolayer of the adsorbate since the adsorption process is exothermic, the physical
on the surface of the adsorbent, it was derived by Langmuir adsorption occurs readily at low temperature and decreases with
increasing temperature. (Le Chatelier's principle).
CHEMISTRY 47
CHAPTER NUCLEAR CHEMISTRY*
11
1. Which of the following nuclear reactions will generate an 6. The halflife of a radioisotope is four hours. If the initial mass
isotope? of the isotope was 200 g, the mass remaining after 24 hours
(a) bparticle emission
(b) Neutron particle emission undecayed is
(c) Positron emission
(a) 1.042 g (b) 2.084 g
(c) 3.125 g (d) 4.167 g (2004)
(d) aparticle emission (2007) 7. Consider the following nuclear reactions:
2. A radioactive element gets spilled over the floor of a room. Its 92238M ® YX N + 2 42 He ; X N ® BA L + 2b +
Y
halflife period is 30 days. If the initial velocity is ten times the The number of neutrons in the element L is
permissible value, after how many days will it be safe to enter (a) 142 (b) 144
the room? (c) 140 (d) 146 (2004)
(a) 100 days (b) 1000 days 8. The halflife of a radioactive isotope is three hours. If the initial
(c) 300 days (d) 10 days (2007) mass of the iosotope were 256 g, the mass of it remaining
undecayed after 18 hours would be
3. In the transformation of 238 U to 234 U , if one emission is an a (a) 4.0 g (b) 8.0 g
92 92 (c) 12.0 g (d) 16.0 g
particle, what should be the other emission(s)? (2003)
(a) Two b – 9. The radionucleide 29304 Th undergoes two successive bdecays
(b) Two b – and one b+ followed by one adecay. The atomic number and the mass
(c) One b– and one g
(d) One b + and one b– number respectively of the resulting radionucleide are
(2006) (a) 92 and 234 (b) 94 and 230
4. A photon of hard gamma radiation knocks a proton out of 1224 Mg (c) 90 and 230 (d) 92 and 230 (2003)
nucleus to form 10. bparticle is emitted in radioactivity by (2002)
(a) the isotope of parent nucleus (a) conversion of proton to neutron
(b) the isobar of parent nucleus (b) form outermost orbit
(c) the nuclide 1213 Na (c) conversion of neutron to proton
(d) bparticle is not emitted.
(d) the isobar of 1213 Na (2005) 11. If halflife of a substance is 5 yrs, then the total amount of
5. Hydrogen bomb is based on the principle of substance left after 15 years, when initial amount is 64 grams
(a) nuclear fission (b) natural radioactivity
(c) nuclear fusion (d) artificial radioactivity. is
(2005) (a) 16 g (b) 2 g
(c) 32 g (d) 8 g (2002)
Answer Key
1. (b) 2. (a) 3. (a) 4. (c) 5. (c) 6. (c)
7. (b) 8. (a) 11. (d)
9. (c) 10. (c)
* Not included in the syllabus of JEE Main since 2008.
48 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : The atoms of the some elements having same atomic the atomic bomb initiates the fusion reaction between
number but different mass numbers are called isotopes.
3 H and 12 H releasing huge amount of energy.
1
ZA X –a A – 4 ZA X –b A Y 6. (c) : t1 /2 = 4 hours
Z – 2 Y; Z + 1 T 244 = 6 1 n
( ) n =t1/ 2 2
– 1 n = ; N = N0
ZA X 0 A – 1 ZA X +b+ A ( ) or, 1 6
2
Z X Z – 1Y N = 200 ´ = 3.125 g
2. (a) : Let A be the activity for safe working.
Given A0 = 10 A 7. (b) : 238 M ® 230 N + 2 42 He
A0 × N0 and A × N 92 88
230 N ® 28360 L + 2b +
88
t = 2.303 log N 0 = 2.303 log A0 Therefore, number of neutrons in element L
l N l A
= 230 – 86 = 144
= 2.303 log 10 A = 2.303´ 30 log 10 8. (a) : t1 /2 = 3 hours, n = T/t1 /2 = 18/3 = 6
( ) ( ) NN0 1 n N 1 6
0.693/ 30 A 0.693 = 2 Þ = 256 2
= 2.303 ´ 30 = 99.69 days » 100 days Þ N = 4.0 g
0.693
9. (c) : 90 Th 234 ¾-¾2 b¾® 92 X 234 ¾-¾a¾® 90 Th 230
3. (a) : 92 U 238 –a 90A 2 34 –b 91B 2 34 –b 92 U 234 Elimination of 1a and 2b particles give isotope.
Thus in order to get 92 U 234 as end product 1a and 2b particles 10. (c) : Since the nucleus does not contain bparticles, it is produced
by the conversion of a neutron to a proton at the moment of
should be emitted. emission.
0n 1 ® +1 p1 + –1 e 0
4. (c) : 24 Mg + g ® 1213Na + 11 p
12
5. (c) : Hydrogen bomb is based on the principal of nuclear n = T = 15 = 3
t1/ 2 5
fusion. In hydrogen bomb, a mixture of deuterium oxide and 11. (d) : t 1/2 = 5 years,
tritium oxide is enclosed in a space surrounding an ordinary 1ö n 1 ö 3
2ø 2 ø
atomic bomb. The temperature produced by the explosion of n = N 0 æè = 6 4 æ = 8 g
è
Classification of Elements and Periodicity in Properties 49
CLASSIFICATION OF ELEMENTS
CHAPTER AND PERIODICITY IN
PROPERTIES
12
1. The first ionisation potential of Na is 5.1 eV. The value of 7. The decreasing values of bond angles from NH3 (106°) to SbH3
electron gain enthalpy of Na + will be (101°) down group15 of the periodic table is due to
(a) + 2.55 eV (b) – 2.55 eV (2013) (a) increasing bondbond pair repulsion
(c) – 5.1 eV (d) – 10.2 eV (b) increasing porbital character in sp 3
(c) decreasing lone pairbond pair repulsion
2. Which of the following represents the correct order of (d) decreasing electronegativity. (2006)
increasing first ionization enthalpy for Ca, Ba, S, Se and Ar?
(a) Ca < Ba < S < Se < Ar 8. The increasing order of the first ionisation enthalpies of the
(b) Ca < S < Ba < Se < Ar (2013) elements B, P, S and F (lowest first) is (2006)
(c) S < Se < Ca < Ba < Ar (a) F < S < P < B (b) P < S < B < F
(d) Ba < Ca < Se < S < Ar (c) B < P < S < F (d) B < S < P < F
3. The increasing order of the ionic radii of the given isoelectronic 9. Which one of the following sets of ions represents a collection
species is
of isoelectronic species?
(a) S2 – , Cl – , Ca 2+ , K + (b) Ca 2+, K +, Cl – , S 2–
(a) K+ , Cl– , Ca 2+, Sc3 +
(c) K+ , S 2– , Ca 2+ , Cl – (d) Cl –, Ca 2+ , K + , S 2–
(b) Ba 2+, Sr 2+, K+ , S2 –
(2012)
(c) N 3– , O 2– , F– , S2 –
4. Which one of the following orders presents the correct sequence (d) Li+ , Na+ , Mg 2+, Ca 2+ (2006)
of the increasing basic nature of the given oxides? 10. In which of the following arrangements the order is NOT
according to the property indicated against it?
(a) Al 2O 3 < MgO < Na 2 O < K 2 O (2011) (a) Al 3+ < Mg2 + < Na+ < F increasing ionic size
(b) MgO < K 2 O < Al 2 O 3 < Na 2 O (b) B < C < N < O increasing first ionisation enthalpy
(c) Na2 O < K 2 O < MgO < Al 2 O 3 (c) I < Br < F < Cl increasing electron gain enthalpy
(d) K2 O < Na 2 O < Al 2 O 3 < MgO
5. The correct sequence which shows decreasing order of the (with negative sign)
ionic radii of the element is (d) Li < Na < K < Rb increasing metallic radius
(a) O 2– > F – > Na + > Mg 2+ > Al 3+ (2005)
(b) Al 3+ > Mg 2+ > Na + > F – > O 2–
(c) Na + > Mg 2+ > Al 3+ > O 2– > F – 11. Based on lattice energy and other considerations which one of
(d) Na + > F – > Mg 2+ > O 2– > Al 3+ (2010) the following alkali metal chlorides is expected to have the
6. Following statements regarding the periodic trends of chemical highest melting point?
reactivity of the alkali metals and the halogens are given. Which
of these statements gives the correct picture? (a) LiCl (b) NaCl
(c) KCl (d) RbCl (2005)
(a) The reactivity decreases in the alkali metals but increases 12. Lattice energy of an ionic compound depends upon
in the halogens with increase in atomic number down the (a) charge on the ion only
group. (b) size of the ion only
(b) In both the alkali metals and the halogens the chemical (c) packing of the ion only
reactivity decreases with increase in atomic number down (d) charge and size of the ion. (2005)
the group. 13. Which among the following factors is the most important in
(c) Chemical reactivity increases with increase in atomic making fluorine the strongest oxidising agent?
number down the group in both the alkali metals and (a) Electron affinity
halogens. (b) Ionization energy
(d) In alkali metals the reactivity increases but in the halogens (c) Hydration enthalpy
it decreases with increase in atomic number down the (d) Bond dissociation energy (2004)
group. (2006)
50 JEE MAIN CHAPTERWISE EXPLORER
14. The formation of the oxide ion O 2– (g) requires first an exothermic 16. Which one of the following ions has the highest value of ionic
and then an endothermic step as shown below.
O (g) + e– = O – (g) ; DH° = –142 kJmol– 1 radius?
O – (g) + e– = O 2– (g) ; DH° = 844 kJmol– 1
This is because (a) Li+ (b) B3 +
(a) oxygen is more electronegative
(b) oxygen has high electron affinity (c) O 2– (d) F– (2004)
(c) O – ion will tend to resist the addition of another electron
(d) O – ion has comparatively larger size than oxygen atom. 17. Which one of the following groupings represents a collection
(2004)
of isoelectronic species?
15. Which one of the following sets of ions represents the collection
of isoelectronic species? (At. nos.: Cs55, Br35)
(a) K+ , Ca 2+, Sc3 +, Cl– (a) Na+ , Ca 2+, Mg 2+ (b) N 3–, F– , Na+
(b) Na+ , Ca 2+, Sc3 +, F– (c) Be, Al3 +, Cl– (d) Ca 2+, Cs +, Br (2003)
(c) K+ , Cl– , Mg2 +, Sc3 + 18. According to the periodic law of elements, the variation in
(d) Na+ , Mg2 +, Al3 +, Cl– . properties of elements is related to their
(Atomic nos.: F = 9, Cl = 17, Na = 11, Mg = 12,
Al = 13, K = 19, Ca = 20, Sc = 21) (a) atomic masses
(2004)
(b) nuclear masses
(c) atomic numbers
(d) nuclear neutronproton number ratios. (2003)
19. Which is the correct order of atomic sizes?
(a) Ce > Sn > Yb > Lu
(b) Sn > Ce > Lu > Yb
(c) Lu > Yb > Sn > Ce
(d) Sn > Yb > Ce > Lu.
(At. Nos. : Ce = 58, Sn = 50, Yb = 70 and Lu = 71)
(2002)
Answer Key
1. (c) 2. (d) 3. (b) 4. (a) 5. (a) 6. (d)
7. (c) 8. (d) 9. (a) 10. (b) 11. (b) 12. (d)
13. (d) 14. (c) 15. (a) 16. (c) 17. (b) 18. (c)
19. (a)
Classification of Elements and Periodicity in Properties 51
1. (c) : Electron gain enthalpy = – Ionisation potential bond angles gradually decrease due to decrease in bond pair
= – 5.1 eV lone pair repulsion.
2. (d) : Ionization enthalpy decreases from top to bottom in a 8. (d) : Element: B S P F
group while it increases from left to right in a period. I.E. (eV): 8.3 10.4 11.0 17.4
3. (b) : For isoelectronic species as effective nuclear charge In general as we move from left to right in a period, the
increases, ionic radii decreases. Nuclear charge is maximum ionisation enthalpy increases with increasing atomic number.
of the specie with maximum protons. Order of nuclear charge: The ionisation enthalpy decreases as we move down a group.
Ca 2+ > K + > Cl – > S 2– P(1s 2 2s 2 2p 6 3s 2 3p 3 ) has a stable half filled electronic
Protons : 20 19 17 16 configuration than S (1s2 2s 2 2p 6 3s 2 3p 4) . For this reason,
ionisation enthalpy of P is higher than S.
Electrons : 18 18 18 18
Thus increasing order of ionic radii 9. (a) : K + = 19 – 1 = 18 e –
Ca 2+ < K + < Cl – < S 2– Cl – = 17 + 1 = 18 e –
4. (a) : While moving from left to right in periodic table basic Ca 2+ = 20 – 2 = 18 e –
character of oxide of elements will decrease.
Sc3 + = 21 – 3 = 18 e –
Thus all the species are isoelectronic.
\ 10. (b) : As we move from left to right across a period, ionisation
enthalpy increases with increasing atomic number. So the
And while descending in the group basic character of order of increasing ionisation enthalpy should be
corresponding oxides increases.
\ B < C < N < O.
\ Correct order is But N(1s 2 2s 2 2p 3 ) has a stable half filled electronic
Al 2O 3 < MgO < Na 2 O < K 2 O configuration. So, ionization enthalpy of nitrogen is greater
than oxygen.
5. (a) : All the given species are isoelectronic. Among So, the correct order of increasing the first ionization enthalpy
isoelectronic species, anions generally have greater size than
cations. is B < C < O < N.
Also greater, the nuclear charge (Z) of the ion, smaller the 11. (b) : In case of halides of alkali metals, melting point decreases
size. Thus the order is : going down the group because lattice enthalpies decreases
O 2– > F – > Na + > Mg 2+ > Al 3+
6. (d) : All the alkali metals are highly reactive elements since as size of alkali metal increases. But LiCl has lower melting
they have a strong tendency to lose the single valence s point in comparison to NaCl due to covalent nature. Thus,
electron to form unipositive ions having inert gas configuration. NaCl is expected to have the highest melting point among
This reactivity arises due to their low ionisation enthalpies given halides.
and high negative values of their standard electrode potentials. 12. (d) : The value of lattice energy depends on the charges present
However, the reactivity of halogens decreases with increase on the two ions and the distance between them.
in atomic number due to following reasons:
(a) As the size increases, the attraction for an additional 13. (d) : The bond dissociation energy of F – F bond is very low.
The weak F – F bond makes fluorine the strongest oxidising
electron by the nucleus becomes less. halogen.
(b) Due to decrease in electronegativity from F to I, the
bond between halogen and other elements becomes 14. (c) : The addition of second electron in an atom or ion is
weaker and weaker. always endothermic.
7. (c) : NH 3 PH 3 AsH 3 SbH 3 15. (a) : Isoelectronic species are those which have same number
Bond angle 106.5° 93.5° 91.5° 91.3° of electrons.
The bond angle in ammonia is less than 109° 28¢ due to K+ = 19 – 1 = 18 ; Ca 2+ = 20 – 2 = 18
repulsion between lone pairs pres ent on nitrogen atom and Sc3 + = 21 – 3 = 18 ; Cl – = 17 + 1 = 18
bonded pairs of electrons. As we move down the group, the Thus all these ions have 18 electrons in them.
52 JEE MAIN CHAPTERWISE EXPLORER
16. (c) : This can be explained on the basis of z/e electrons but different nuclear charge. Number of electrons
in N 3– = 7 + 3 = 10.
{ } nuclear charge , whereas z/e ratio increases, the size
no. of electrons Number of electrons in F – = 9 + 1 = 10
decreases and when z/e ratio decreases the size increases. Number of electrons in Na + = 11 – 1 = 10
For Li +, z = 3 = 1.5 18. (c) : According to modified modern periodic law, the properties
e 2 of elements are periodic functions of their atomic numbers.
For B3 + , z = 5 = 2.5 19. (a) : Generally as we move from left to right in a period,
e 2 there is regular decrease in atomic radii and in a group as the
atomic number increases the atomic radii also increases. Thus
For O 2– , z = 8 = 0.8 the atomic radius of Sn should be less than lanthanides.
e 10 La > Sn. But due to lanthanide contraction, in case of lanthanides
there is a continuous decrease in size with increase in atomic
For F –, z = 9 = 0.9 number. Hence the atomic radius follow the given trend :
e 10 Ce > Sn > Yb > Lu.
Hence, O 2– has highest value of ionic radius.
17. (b) : Isoelectronic species are the neutral atoms, cations or
anions of different elements which have the same number of
General Principles and Processes of Isolation of Metals 53
CHAPTER GENERAL PRINCIPLES
AND PROCESSES OF
13
ISOLATION OF METALS
1. Which method of purification is represented by the following (a) Magnetite (b) Cassiterite
equation? (c) Galena (d) Malachite.
Ti (s) + 2I 2(g)
TiI 4(g) Ti (s) + 2I 2(g) (2004)
(a) Cupellation (b) Poling 5. When the sample of copper wth zinc impurity is to be purified
(c) Van Arkel (d) Zone refining
(2012) by electrolysis, the appropriate electrodes are
cathode anode
2. Which of the following factors is of no significance for roasting (a) pure zinc pure copper
sulphide ores to the oxides and not subjecting the sulphide (b) impure sample pure copper
ores to carbon reduction directly? (c) impure zinc impure sample
(a) CO 2 is more volatile than CS 2 . (d) pure copper impure sample.
(b) Metal sulphides are thermodynamically more stable than
(2002)
CS2 . 6. Cyanide process is used for the extraction of
(c) CO 2 is thermodynamically more stable than CS 2 . (a) barium (b) aluminium
(d) Metal sulphides are less stable than the corresponding (c) boron (d) silver.
oxides. (2002)
(2002)
(2008)
3. During the process of electrolytic refining of copper, some metals 7. The metal extracted by leaching with a cyanide is
present as impurity settle as ‘anode mud’. These are (a) Mg (b) Ag
(a) Sn and Ag (b) Pb and Zn (c) Cu (d) Na.
(c) Ag and Au (d) Fe and Ni. 8. Aluminium is extracted by the electrolysis of
(2005) (a) bauxite
(b) alumina
4. Which one of the following ores is best concentrated by froth (c) alumina mixed with molten cryolite
flotation method? (d) molten cryolite.
(2002)
Answer Key
1. (c) 2. (a) 3. (c) 4. (c) 5. (d) 6. (d)
7. (b) 8. (c)
54 JEE MAIN CHAPTERWISE EXPLORER
1. (c) : Van Arkel method which is also called as vapourphase properties with the frothing agent and water. Here galena (PbS)
refining is used for preparing ultrapure metals like titanium, is the only sulphide ore.
zirconium, thorium and uranium. 5. (d) : The impure metal is made anode while a thin sheet of
2. (a) : Oxidising roasting is a very common type of roasting pure metal acts as cathode. On passing the current, the pure
in metallurgy and is carried out to remove sulphur and arsenic metal is deposited on the cathode and equivalent amount of
in the form of their volatile oxides.CS 2 is more volatile than the metal gets dissolved from the anode.
CO 2. So option (a) is of no significance for roasting sulphide 6. (d) : Gold and silver are extracted from their native ores by
ores to their oxides. The reduction process is on the MacArthur forrest cyanide process.
thermodynamic stability of the products and not on their 7. (b) : Silver ore forms a soluble complex with NaCN from which
volatility.
silver is precipitated using scrap zinc.
3. (c) : In the electrolytic refining of copper the more Ag 2S + 2NaCN ® Na[Ag(CN)2 ] ¾¾Zn¾®
electropositive impurities like Fe, Zn, Ni, Co, etc. dissolve
in the solution and less electropositive impurities such as Na2 [Zn(CN)4 ] + Ag ¯
Ag, Au and Pt collect below the anode in the form of anodic
sod. argentocyanide
mud. (soluble)
4. (c) : Frothflotation method is used for the concentration of 8. (c) : Aluminium is obtained by the electrolysis of the pure
alumina (20 parts) dissolved in a bath of fused cryolite (60 parts)
sulphide ores. The method is based on the preferential wetting
and fluorspar (20 parts).
Hydrogen 55
CHAPTER HYDROGEN
14
1. Very pure hydrogen (99.9%) can be made by which of the (b) CO and H 2 are fractionally separated using differences
following processes? in their densities.
(a) Mixing natural hydrocarbons of high molecular weight. (c) CO is removed by absorption in aqueous Cu 2 Cl2 solution.
(b) Electrolysis of water. (d) H2 is removed through occlusion with Pd.
(c) Reaction of salt like hydrides with water.
(2008)
(d) Reaction of methane with steam. (2012)
3. Which one of the following processes will produce hard water?
2. In context with the industrial preparation of hydrogen from (a) Saturation of water with CaCO 3. (2003)
water gas (CO + H 2 ), which of the following is the correct (b) Saturation of water with MgCO 3 .
statement? (c) Saturation of water with CaSO 4 .
(d) Addition of Na2 SO4 to water.
(a) CO is oxidised to CO 2 with steam in the presence of a
catalyst followed by absorption of CO 2 in alkali.
1. (b) 2. (a) Answer Key
3. (c)
56 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : Dihydrogen of high purity is usually prepared by the CO + H2 O Fe2O3, CoO CO2 + H 2
electrolysis of water using platinum electrodes in presence 673 K
of small amount of acid or alkali.
CO 2 NaOH Na2CO 3
Dihydrogen is collected at cathode. alkali
CO 2 is absorbed in alkali (NaOH).
The entire reaction is called water gas shift reaction.
2. (a) : Carbon monoxide is oxidised to carbon dioxide by passing 3. (c) : Permanent hardness is introduced when water passes over
the gases and steam over an iron oxide
rocks containing the sulphates or chlorides of both of calcium
or cobalt oxide or chromium oxide catalyst at 673 K resulting and magnesium.
in the production of more H 2 .
sBlock Elements 57
CHAPTER sBLOCK ELEMENTS
15
1. Which of the following on thermal decomposition yields a (c) prevent action of water and salt
basic as well as an acidic oxide? (d) prevent puncturing by undersea rocks. (2003)
(a) KClO 3 (b) CaCO 3 7. In curing cement plasters water is sprinkled from time to time.
(c) NH4 NO 3 (d) NaNO 3
(2012) This helps in
2. The set representing the correct order of ionic radius is (a) keeping it cool
(a) Li + > Be 2+ > Na + > Mg 2+ (b) developing interlocking needlelike crystals of hydrated
(b) Na + > Li + > Mg 2+ > Be 2+ silicates
(c) Li + > Na + > Mg 2+ > Be 2+ (c) hydrating sand and gravel mixed with cement
(d) Mg 2+ > Be 2+ > Li + > Na + (2009) (d) converting sand into silicic acid. (2003)
3. The ionic mobility of alkali metal ions in aqueous solution is 8. The solubilities of carbonates decrease down the magnesium
maximum for group due to a decrease in
(a) K+
(c) Li+ (b) Rb+ (2006) (a) lattice energies of solids
(d) Na + (b) hydration energies of cations
(c) interionic attraction
4. Beryllium and aluminium exhibit many properties which are (d) entropy of solution formation. (2003)
similar. But, the two elements differ in 9. The substance not likely to contain CaCO 3 is
(a) exhibiting maximum covalency in compounds (a) a marble statue (b) calcined gypsum
(b) forming polymeric hydrides
(c) forming covalent halides (c) sea shells (d) dolomite. (2003)
(d) exhibiting amphoteric nature in their oxides. (2004) 10. A metal M readily forms its sulphate MSO 4 which is water
5. One mole of magnesium nitride on the reaction with an excess soluble. It forms its oxide MO which becomes inert on heating.
of water gives
It forms an insoluble hydroxide M(OH) 2 which is soluble in
(a) one mole of ammonia NaOH solution. Then M is
(b) one mole of nitric acid (a) Mg (b) Ba
(c) two moles of ammonia (c) Ca (d) Be. (2002)
(d) two moles of nitric acid. (2004) 11. KO 2 (potassium super oxide) is used in oxygen cylinders in space
6. Several blocks of magnesium are fixed to the bottom of a ship and submarines because it
to
(a) keep away the sharks (a) absorbs CO 2 and increases O 2 content (2002)
(b) make the ship lighter (b) eliminates moisture
(c) absorbs CO 2
(d) produces ozone.
Answer Key
1. (b) 2. (b) 3. (b) 4. (a) 5. (c) 6. (b)
7. (b) 8. (b) 9. (b) 10. (d) 11. (a)
58 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : magnesium makes the ship lighter when it is fixed to the bottom
of the ship.
2. (b) : Moving from left to right in a period, the ionic radii 7. (b) : Water develops interlocking needlelike crystals of hydrated
silicates. The reactions involved are the hydration of calcium
decrease due to increase in effective nuclear charge as the
aluminates and calcium silicates which change into their
additional electrons are added to the same shell, however from
colloidal gels. At the same time, some calcium hydroxide and
top to bottom the ionic radii increase with increasing atomic
aluminium hydroxides are formed as precipitates due to
number and presence of additional shells. Also Li and Mg
hydrolysis. Calcium hydroxide binds the particles of calcium
are diagonally related and hence the order is silicate together while aluminium hydroxide fills the interstices
Na + > Li + > Mg 2+ > Be 2+ . rendering the mass impervious.
3. (b) : The alkali metal ion exist as hydrated ions M+ ( H2 O )n in 8. (b) : The stability of the carbonates of the alkaline earth metals
the aqueous solution. The degree of hydration, decreases with increases on moving down the group. The solubility of carbonate
ionic size as we go down the group. Hence Li+ ion is mostly of metals in water is generally low. However they dissolve in
hydrated e.g. [Li(H2 O )6 ] +. Since the mobility of ions is inversely water containing CO2 yielding bicarbonates, and this solubility
proportional to the size of the their hydrated ions, hence the decreases on going down in a group with the increase in stability
increasing order of ionic mobility is of carbonates of metals, and decrease in hydration energy of the
Li+ < Na + < K+ < Rb+ cations.
4. (a) : Beryllium has the valency +2 while aluminium exhibits 9. (b) : The composition of gypsum is CaSO 4 ∙2H2 O . It does not
its valency as +3. have CaCO 3 .
5. (c) : Mg 3 N 2 + 6H 2O ® 3Mg(OH) 2 + 2NH 3 10. (d) : Be forms water soluble BeSO4 , water insoluble Be(OH) 2
6. (b) : Magnesium, on account of its lightness, great affinity for and BeO. Be(OH)2 is insoluble in NaOH giving sodium beryllate
Na2 BeO 2.
oxygen and toughness is used in ship. Being a lighter element,
11. (a) : 4KO 2 + 2CO 2 ® 2K2 CO 3 + 3O 2
pBlock Elements 59
CHAPTER pBLOCK ELEMENTS
16
1. Which of the following exists as covalent crystals in the solid 7. Which one of the following reactions of xenon compounds
state? is not feasible?
(a) Phosphorus (b) Iodine (a) XeO3 + 6HF ® XeF6 + 3H2O
(c) Silicon (d) Sulphur (2013) (b) 3XeF4 + 6H2O ® 2Xe + XeO3 + 12HF + 1.5O2
2. Which of the following is the wrong statement? (c) 2XeF2 + 2H2O ® 2Xe + 4HF + O2 (2009)
(a) Ozone is diamagnetic gas. (d) XeF6 + RbF ® Rb[XeF7]
(b) ONCl and ONO – are not isoelectronic. 8. In which of the following arrangements, the sequence is not
(c) O 3 molecule is bent. strictly according to the property written against it?
(d) Ozone is violetblack in solid state. (2013) (a) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power
3. Boron cannot form which one of the following anions? (b) HF < HCl < HBr < HI : increasing acid strength
(c) NH3 < PH3 < AsH3 < SbH3 :increasing basic strength
(a) BF6 3 – (b) BH 4– (d) B < C < O < N : increasing first ionization enthalpy
(c) B(OH) 4– (d) BO 2–
(2009)
(2011)
4. Which of the following statement is wrong? 9. Which one of the following is the correct statement?
(a) The stability of hydrides increases from NH 3 to BiH 3 in (a) B2 H 6∙ 2NH 3 is known as ‘inorganic benzene’.
group 15 of the periodic table. (b) Boric acid is a protonic acid.
(b) Nitrogen cannot form dppp bond. (c) Beryllium exhibits coordination number of six.
(c) Single N — N bond is weaker than the single P – P bond. (d) Chlorides of both beryllium and aluminium have bridged
(d) N 2O 4 has two resonance structure. chloride structures in solid phase.
(2011) (2008)
5. Which of the following statements regarding sulphur is 10. Among the following substituted silanes the one which will
incorrect? give rise to cross linked silicone polymer on hydrolysis is
(a) S2 molecule is paramagnetic. (a) R 3S iCl (b) R4 Si (2008)
(b) The vapour at 200°C consists mostly of S 8 rings. (c) RSiCl 3 (d) R 2 SiCl 2
(c) At 600°C the gas mainly consists of S 2 molecules.
(d) The oxidation state of sulphur is never less than +4 in 11. The stability of dihalides of Si, Ge, Sn and Pb increases steadily
its compounds. in the sequence
(2011) (a) PbX 2 = SnX 2 = GeX 2 = SiX 2
(b) GeX2 = SiX2 = SnX2 = PbX 2
6. The bond dissociation energy of B – F in BF3 is 646 kJ mol– 1 (c) SiX2 = GeX 2 = PbX 2 = SnX2 (2007)
whereas that of C – F in CF4 is 515 kJ mol –1. The correct (d) SiX 2 = GeX 2 = SnX 2 = PbX 2.
reason for higher B – F bond dissociation energy as compared
to that of C – F is
(a) smaller size of Batom as compared to that of Catom 12. Identify the incorrect statement among the following.
(b) stronger s bond between B and F in BF3 as compared to (a) Br 2 reacts with hot and strong NaOH solution to give NaBr
that between C and F in CF4 and H2 O .
(c) significant pppp interaction between B and F in BF3 (b) Ozone reacts with SO2 to give SO3 .
whereas there is no possibility of such interaction between (c) Silicon reacts with NaOH( aq) in the presence of air to give
C and F in CF4.
Na 2 SiO 3 and H2 O.
(d) lower degree of pppp interaction between B and F in (d) Cl2 reacts with excess of NH3 to give N 2 and HCl.
BF3 than that between C and F in CF4.
(2009) (2007)
60 JEE MAIN CHAPTERWISE EXPLORER
13. Regular use of the following fertilizers increases the acidity of 22. The correct order of the thermal stability of hydrogen halides
soil? (H – X) is
(a) Ammonium sulphate (a) HI > HBr > HCl > HF
(b) Potassium nitrate (b) HF > HCl > HBr > HI
(c) Urea (c) HCl < HF > HBr < HI
(d) Superphosphate of lime (2007) (d) HI > HCl < HF > HBr (2005)
14. A metal, M forms chlorides in +2 and +4 oxidation states. Which 23. In silicon dioxide
(a) each silicon atom is surrounded by four oxygen atoms and
of the following statements about these chlorides is correct? each oxygen atom is bonded to two silicon atoms
(b) each silicon atom is surrounded by two oxygen atoms and
(a) MCl2 is more volatile than. MCl4 . each oxygen atom is bonded to two silicon atoms
(b) MCl2 is more soluble in anhydrous ethanol than MCl4 . (c) silicon atom is bonded to two oxygen atoms
(c) MCl2 is more ionic than MCl4 . (d) there are double bonds between silicon and oxygen atoms.
(d) MCl2 is more easily hydrolysed than MCl4 . (2006) (2005)
15. What products are expected from the disproportionation reaction
of hypochlorous acid? 24. Which of the following oxides is amphoteric in character?
(a) HClO3 and Cl2 O (b) HClO 2 and HClO4 (a) CaO (b) CO 2
(c) SiO 2 (d) SnO2
(c) HCl and Cl2 O (d) HCl and HClO 3 (2006) (2005)
16. Which of the following statements is true? 25. The soldiers of Napolean army while at Alps during freezing
winter suffered a serious problem as regards to the tin buttons
(a) H3 P O3 is a stronger acid than H2 SO3 . of their uniforms. White metallic tin buttons got converted to
(b) In aqueous medium HF is a stronger acid than HCl. grey powder. This transformation is related to
(a) an interaction with nitrogen of the air at very low
(c) HClO4 is a weaker acid than HClO3 . (2006) temperatures
(d) HNO 3 is a stronger acid than HNO 2. (b) a change in the crystalline structure of tin
(c) a change in the partial pressure of oxygen in the air
17. Heating an aqueous solution of aluminium chloride to dryness (d) an interaction with water vapour contained in the humid
air. (2004)
will give
(a) AlCl3 (b) Al 2 Cl6 (2005)
(c) Al2 O 3 (d) Al(OH)Cl2
18. The number and type of bonds between two carbon atoms in 26. Aluminium chloride exists as dimer, Al 2C l6 in solid state as well
as in solution of nonpolar solvents such as benzene. When
calcium carbide are
(a) one sigma, one pi dissolved in water, it gives
(b) one sigma, two pi (a) Al 3+ + 3Cl– (b) [Al(H2 O )6 ] 3+ + 3Cl–
(c) two sigma, one pi (c) [Al(OH)6 ] 3 – + 3HCl (d) Al 2 O 3 + 6HCl. (2004)
(d) two sigma, two pi. (2005) 27. Among Al2 O 3, SiO 2, P2 O 3 and SO 2 the correct order of acid
strength is
19. The structure of diborane (B2 H 6 ) contains
(a) four 2c2e bonds and two 3c2e bonds (2005) (a) SO2 < P2 O 3 < SiO 2 < Al 2O 3 (2004)
(b) SiO 2 < SO2 < Al 2 O 3 < P2 O 3
(b) two 2c2e bonds and four 3c2e bonds (c) Al2 O 3 < SiO 2 < SO2 < P2 O 3
(c) two 2c2e bonds and two 3c3e bonds (d) Al 2 O 3 < SiO 2 < P2 O 3 < SO2 .
(d) four 2c2e bonds and four 3c2e bonds
20. The molecular shapes of SF4 , CF4 and XeF4 are 28. The states of hybridisation of boron and oxygen atoms in boric
(a) the same with 2, 0 and 1 lone pairs of electrons on the
acid (H3 B O 3) are respectively
central atom respectively (a) sp2 and sp2 (b) sp 2 and sp 3
(b) the same with 1, 1 and 1 lone pair of electrons on the central (c) sp 3 and sp 2 (d) sp 3 and sp 3 . (2004)
atoms respectively 29. Which one of the following statements regarding helium is
incorrect?
(c) different with 0, 1 and 2 lone pairs of electrons on the
central atom respectively (a) It is used to fill gas in balloons instead of hydrogen because
(d) different with 1, 0 and 2 lone pairs of electrons on the it is lighter and noninflammable.
central atom respectively (2005) (b) It is used as a cryogenic agent for carrying out experiments
21. The number of hydrogen atom(s) attached to phosphorus atom at low temperatures.
in hypophosphorous acid is (c) It is used to produce and sustain powerful superconducting
(a) zero (b) two magnets.
(c) one (d) three. (2005) (d) It is used in gascooled nuclear reactors. (2004)
pBlock Elements 61
30. Glass is a (c) PCl5 and HCl are formed and the mixture cools down
(a) microcrystalline solid
(b) supercooled liquid (d) PH3 ∙ Cl2 is formed with warming up. (2003)
(c) gel
(d) polymeric mixture. (2003) 37. Which one of the following statements is correct?
(a) Manganese salts give a violet borax test in the reducing
31. Graphite is a soft solid lubricant extremely difficult to melt. flame.
The reason for this anomalous behaviour is that graphite (b) From a mixed precipitate of AgCl and AgI, ammonia
(a) is a noncrystalline substance solution dissolves only AgCl.
(b) is an allotropic form of diamond (c) Ferric ions give a deep green precipitate on adding
(c) has molecules of variable molecular masses like polymers potassium ferrocyanide solution.
(d) has carbon atoms arranged in large plates of rings of (d) On boiling a solution having K+ , Ca 2+ and HCO 3– ions we
strongly bound carbon atoms with weak interplate bonds. get a precipitate of K2 Ca(CO3 ) 2 .
(2003) (2003)
32. Which one of the following pairs of molecules will have 38. Alum helps in purifying water by
(a) forming Si complex with clay particles
permanent dipole moments for both members? (b) sulphate part which combines with the dirt and removes
it
(a) SiF4 and NO 2 (b) NO 2 and CO 2 (2003) (c) coagulating the mud particles
(c) NO 2 and O 3 (d) SiF4 and CO 2 (d) making mud water soluble.
(2002)
33. Which one of the following substances has the highest proton
affinity?
(a) H2 O (b) H2 S (2003) 39. In case of nitrogen, NCl3 is possible but not NCl5 while in case
(c) NH3 (d) PH3 of phosphorus, PCl3 as well as PCl5 are possible. It is due to
(a) availability of vacant d orbitals in P but not in N
34. Which one of the following is an amphoteric oxide?
(b) lower electronegativity of P than N
(a) ZnO (b) Na2 O (2003)
(c) SO2 (d) B2 O 3 (c) lower tendency of Hbond formation in P than N
35. Concentrated hydrochloric acid when kept in open air sometimes (d) occurrence of P in solid while N in gaseous state at room
produces a cloud of white fumes. The explanation for it is that temperature.
(a) concentrated hydrochloric acid emits strongly smelling HCl
gas all the time (2002)
(b) oxygen in air reacts with the emitted HCl gas to form a
40. In XeF 2, XeF 4, XeF 6 the number of lone pairs on Xe are
respectively
cloud of chlorine gas (a) 2, 3, 1 (b) 1, 2, 3
(c) strong affinity of HCl gas for moisture in air results in
(c) 4, 1, 2 (d) 3, 2, 1. (2002)
forming of droplets of liquid solution which appears like 41. Which of the following statements is true?
a cloudy smoke (a) HF is less polar than HBr.
(d) due to strong affinity for water, concentrated hydrochloric (b) Absolutely pure water does not contain any ions.
acid pulls moisture of air towards itself. This moisture (c) Chemical bond formation takes place when forces of
forms droplets of water and hence the cloud. attraction overcome the forces of repulsion.
(d) In covalency transference of electron takes place.
(2003) (2002)
36. What may be expected to happen when phosphine gas is mixed 42. When H2 S is passed through Hg2 S we get
with chlorine gas?
(a) The mixture only cools down (a) HgS (b) HgS + Hg 2S
(b) PCl3 and HCl are formed and the mixture warms up
(c) Hg 2S + Hg (d) Hg 2 S. (2002)
Answer Key
1. (c) 2. (None) 3. (a) 4. (a) 5. (d) 6. (c)
7. (a) 8. (c) 9. (d) 10. (c) 11. (c) 12. (d)
13. (a) 14. (c) 15. (d) 16. (d) 17. (b) 18. (b)
19. (a) 20. (d) 21. (b) 22. (b) 23. (a) 24. (d)
25. (b) 26. (b) 27. (d) 28. (b) 29. (a) 30. (b)
31. (d) 32. (c) 33. (c) 34. (a) 35. (c) 36. (c)
37. (b) 38. (c) 39. (a) 40. (d) 41. (c) 42. (c)
62 JEE MAIN CHAPTERWISE EXPLORER
1. (c) Beryllium exhibits coordination number of four as it has only
2. (None) : All the statements are correct. four available orbitals in its valency shell.
3. (a) : Due to nonavailability of dorbitals, boron is unable to Also,
expand its octet. Therefore, the maximum covalency of boron
cannot exceed 4. 3B2H6 + 6NH 3 3[BH2(NH3)2] + BH4 –
or
4. (a) : Thermal stability decreases gradually from NH 3 to BiH 3 . B2H6 ∙ 2NH3
So the stability also decreases. or
NH 3 PH 3 AsH 3 SbH 3 BiH 3 B2H6 ∙ 2NH 3 Heat, 450 K 2B3N3H6 + 12H2
Decomposition 1300°C 440°C 2 8 0 °C 150°C room temp. Borazine
t e mpe ra t ure Borazine has structure similar to benzene and therefore,
The size of the central atom increases from N to Bi therefore, it is called inorganic benzene. Hence option (d) is correct.
the tendency to form a stable covalent bond with small atom
like hydrogen decreases and therefore, stability decreases. 10. (c) : RSiCl 3 on hydrolysis gives a cross linked silicone.
The formation can be explained in three steps
5. (d) : Sulphur exhibits –2, +2, +4, +6 oxidation states but +4
and +6 are more common. Cl OH
R – Si – OH
6. (c) : In BF3, B is sp 2 hybridised and has a vacant 2porbital (i) R – Si – Cl 3H2O
–3HCl OH
which overlaps laterally with a filled 2porbital of F forming
strong pppp bond. However in CF4, C does not have any Cl
vacant porbitals to undergo pbonding.
Thus B.E.B – F > B.E.C – F. R R R
7. (a) : The reaction is not feasible because XeF6 formed will (ii) HO – Si – OH + H O – Si – OH + H O – Si – OH
OH OH OH
R R R
further produce XeO3 by getting hydrolysed. –H2O HO – Si – O – Si – O – Si – OH
OH OH OH
XeF6 + 3H2O XeO3 + 3H2F2.
8. (c) : In group15 hydrides, the basic character decreases on R R R
going down the group due to decrease in the availability of
the lone pair of electrons because of the increase in size of (iii) HO – Si – O – Si – O – Si – OH
elements from N to Bi. Thus, correct order of basicity is
NH3 > PH3 > AsH3 > SbH3. OH OH OH –3H2 O
H O H O H O
9. (d) : Boric acid is a weak monobasic acid HO – Si – O – Si – O – Si – OH
(K a = 1.0 × 10 –9 ). It is a notable part that boric acid does not R R R
act as a protonic acid (i.e., proton donor) but behaves as a
Lewis acid by accepting a pair of electrons from OH – ions. R R R
B(OH) 3 + 2H 2 O ® [B(OH) 4 ] – + H 3O + – O – Si – O – Si – O – Si – O –
BeCl 2 like Al 2C l 6 has a bridged polymeric structure in solid
phase generally as shown below. O O O
Cl – O – Si – O – Si – O – Si – O –
R R R
Cl – Be Be – Cl Cross linked silicone
11. (c) : Due to the inert pair effect (the reluctance of ns 2 electrons
Cl of outermost shell to participate in bonding) the stability of
M2 + ions (of group IV elements) increases as we go down the
Polymeric structure of BeCl 2 Cl group.
Cl Cl
Al Al 12. (d) : 3Br2 + 6NaOH ® 5NaBr + NaBrO 3 + 3H2 O
O 3 + SO2 ® O 2 + SO3
Cl Cl Cl Si + 2NaOH + O 2 ® Na 2 SiO 3 + H2 O
Polymeric structure of Al2C l 6
pBlock Elements 63
Cl2 reacts with excess of ammonia to produce ammonium the four orbitals of each of the boron atom overlap with two
chloride and nitrogen.
terminal hydrogen atoms forming two normal B – H sbonds.
2NH3 + 3Cl 2 N 2 + 6HCl One of the remaining hybrid orbital (either filled or empty) of
6NH3 + 6HCl 6NH4 Cl
8NH3 + 3Cl 2 N 2 + 6NH4 Cl one of the boron atoms, 1s orbital of hydrogen atoms (bridge
atom) and one of hybrid orbitals of the other boron atom overlap
to form a delocalised orbital covering the three nuclei with a
pair of electrons. Such a bond is known as three centre two
13. (a) : (NH 4) 2S O 4 + 2H 2O (2H + + SO 42 – ) + 2NH 4 OH electron (3c – 2e) bonds.
Strong acid Weak base . H. . H. H
H . . H . . H
(NH4 ) 2S O4 on hydrolysis produces strong acid H2 S O4 , which B
increases the acidity of the soil. B . B B
. H . .H H . . H
14. (c) : The elements of group 14 show an oxidation state of +4 .
and +2. The compounds showing an oxidation state of +4 H. H
are covalent compound and have tetrahedral structures . e. g.
SnCl4 , PbCl4 , SiCl 4, etc. whereas those which show +2 oxidation Structure of diborane
state are ionic in nature and behave as reducing agent. e.g.
SnCl 2 , PbCl 2 , etc. 20. (d) : SF4 (sp 3 d, trigonal bipyramidal with one equatorial position
Further as we move down the group, the tendency of the element occupied by 1 lone pair), CF4 (sp 3, tetrahedral, no lone pair),
to form covalent compound decreases but the tendency to XeF4 (sp 3d 2, square planar, two lone pairs).
form ionic compound increases.
F F
15. (d) : 3HClO4 (aq) ® HClO 3(aq) + 2HCl( aq)
It is a disproportionation reaction of hypochlorous acid where F F F
the oxidation number of Cl changes from +1 (in ClO– ) to +5 S C Xe
(in ClO 3 –) and –1 (in Cl– ).
F F F F F F
16. (d) : Higher is the oxidation state of the central atom, greater F
is the acidity.
Hence, HClO 4 is a stronger acid than HClO3 . (SF4 ) (CF4 ) (XeF 4)
HNO 3 is a stronger acid than HNO 2 .
Now, greater is the electronegativity and higher is the oxidation 21. (b) : Hypophosphorous acid
state of the central atom, greater is the acidity. Hence H2 SO3 O
is a stronger acid than H3 PO3 .
Due to higher dissociation energy of H – F bond and molecular P
association due to hydrogen bonding in HF, HF is a weaker HO H H
acid than HCl. Number of hydrogen atom(s) attached to phosphorus atom = 2.
17. (b) : Aluminium chloride in aqueous solution exists as ion 22. (b) : As the size of the halogen atom increases from F to I,
H – X bond length in HX molecules also increases from
pair. H – F to H – I (H – F < H – Cl < H – Br < H – I).
The increase in H – X bond length decreases the strength of
2AlCl3 + aq. ® [AlCl2 ( H2 O )4 ] + ( aq) + [AlCl4 ( H2 O )2 ] – ( aq) H – X bond from H – F to H – I (H – F > H – Cl > H – Br
> H – I). The decrease in the strength of H – X bond is evident
The crystallisation of AlCl3 from aqueous solution, therefore, from the fact that H – X bond dissociation energies decrease
yields an ionic solid of composition [AlCl 2 (H 2 O) 4 ] + from H – F to H – I. Due to successive decrease in the strength
[AlCl4 (H2 O )2 ] – ∙ xH2 O. This compound decomposes at about of H – X bond from H – F to H – I, thermal stability of HX
molecules also decreases from HF to HI (HF > HCl > HBr >
190°C to give the nonionic dimer Al2 C l6 . HI).
[AlCl 2( H 2 O) 4 ] + [AlCl 4( H 2 O) 2 ] – ∙ xH 2O heat 23. (a) : Silicon dioxide exhibits polymorphism. It is a network
190°C solid in which each Si atom is surrounded tetrahedrally by four
oxygen atoms.
Al 2 Cl 6 + H 2 O
18. (b) : Calcium carbide is ionic carbide having [ .. C C .. ] 2– . O O
– O – Si – O – Si – O –
Ca 2+ [ .. C 1 s C .. ] 2–
O O
2 p
19. (a) : According to molecular orbital theory, each of the two 24. (d): CaObasic, CO 2 and SiO 2 acidic, SnO 2 amphoteric, as it
boron atoms is in sp3 hybrid state. Of the four hybrid orbitals, reacts both with acids and bases.
three have one electron each while the fourth is empty. Two of
64 JEE MAIN CHAPTERWISE EXPLORER
SnO 2 + 4HCl ® SnCl4 + 2H2 O 33. (c) : Ammonia is a Lewis base, accepting proton to form
SnO 2 + 2NaOH ® Na 2 SnO 3 + H2 O
ammonium ion as it has tendency to donate an electron pair.
25. (b) : Grey tin is very brittle and easily crumbles down to a +
H H
powder in very cold climates. H – N .. + H + H – N ® H
Grey tin White tin
(cubic) (tetragonal) H H
The change of white tin to grey tin is accompanied by increase
in volume. This is called tin disease or tin plague. 34. (a) : ZnO is an amphoteric oxide and dissolves readily in acids
forming corresponding zinc salts and alkalies forming zincates.
26. (b) : Al 2C l 6 + 12H 2 O 2[Al(H 2 O) 6] 3+ + 6Cl – ZnO + H2 S O4 ® ZnSO4 + H2 O
27. (d) : Acidity of the oxides of non metals increases with the zinc sulphate
electronegativity and oxidation number of the element.
Al2 O 3 < SiO 2 < P2 O 3 < SO2 ZnO + 2NaOH ® Na2 Z nO 2 + H2 O
Al2 O 3 is amphoteric. SiO 2 is slightly acidic whereas P2 O 3 and
SO2 are the anhydrides of the acids H3 P O3 and H2 S O3 . sodium zincate
Hs p 3 .O. .. 35. (c) : HCl gas in presence of moisture in air forms droplets of
liquid solution in the form of cloudy smoke.
28. (b) : B sp2
H .. 36. (c) : Phosphine burns in the atmosphere of chlorine and forms
sp3 O. . .. H phosphorus pentachloride.
O. . sp3 PH 3 + 4Cl2 ® PCl5 + 3HCl
29. (a) : Helium is twice as heavy as hydrogen, its lifting power is 37. (b) : The solubility product of AgCl, AgBr and AgI at the room
92 percent of that of hydrogen. temperature are 2.8 × 10– 10, 5.0 × 10– 13 and 8.5 × 10– 17 respectively.
Helium has the lowest melting and boiling points of any element Thus, AgI is the least soluble silver halide.
which makes liquid helium an ideal coolant for many extremely The lattice energies of AgBr and AgI are even higher because
lowtemperature applications such as superconducting magnets, of greater number of electrons in their anions. Consequently,
and cryogenic research where temperatures close to absolute they are even less soluble than AgCl. Due to greater solubility
zero are needed. of AgCl than AgI, ammonia solution dissolves only AgCl and
forms a complex.
AgCl + 2NH4 O H ® [Ag(NH3 ) 2 ]Cl + 2H2 O
Diammine silver chloride
30. (b) : Glass is a transparent or translucent amorphous supercooled 38. (c) : The negatively charged colloidal particles of impurities get
solid solution (supercooled liquid) of silicates and borates, neutralised by the Al 3+ ions and settle down and pure water can
having a general formula R 2O ∙MO∙6SiO2 where R = Na or K
and M = Ca, Ba, Zn or Pb. be decanted off.
39. (a) : 7N = 1s 2 2s 2 3p 3
31. (d) : Graphite has a twodimensional sheet like structure and
each carbon atom makes a use of sp2 hybridisation. 15P = 1s 2 2s 2 2p 6 3s 2 3p 3
The above layer structure of graphite is less compact than that In phosphorus the 3dorbitals are available.
of diamond. Further, since the bonding between the layers
involving only weak van der Waal's forces, these layers can slide 40. (d) : XeF2 sp 3 d 3 lone pairs
over each other. This gives softness, greasiness and lubricating XeF4 sp 3d 2 2 lone pairs
character of graphite. XeF6 sp 3 d 3 1 lone pair
32. (c) : NO 2 and O 3 both have unsymmetrical structures, so they 41. (c) : Due to the higher electronegativity of F, HF is more polar
have permanent dipole moment. than HBr pure water contains H+ and OH– ions. In covalency,
sharing of electrons between two nonmetal atoms takes place.
42. (c)
d and fBlock Elements 65
CHAPTER d and f BLOCK ELEMENTS
17
1. Which of the following arrangements does not represent the 7. In context with the transition elements, which of the following
correct order of the property stated against it? statements is incorrect?
(a) Sc < Ti < Cr < Mn : number of oxidation states (a) In addition to the normal oxidation states, the zero
(b) V 2+ < Cr 2+ < Mn 2+ < Fe 2+ : paramagnetic behaviour
(c) Ni 2+ < Co2 + < Fe 2+ < Mn 2+ : ionic size oxidation state is also shown by these elements in
(d) Co 3+ < Fe 3+ < Cr 3+ < Sc 3+ : stability in aqueous solution. complexes.
(b) In the highest oxidation states, the transition metals show
(2013) basic character and form cationic complexes.
2. Four successive members of the first row transition elements (c) In the highest oxidation states of the first five transition
are listed below with atomic numbers. Which one of them is elements (Sc to Mn), all the 4s and 3d electrons are used
expected to have the highest E° M 3+ /M 2+ value? for bonding.
(a) Co (Z = 27) (b) Cr (Z = 24) (2013) (d) Once the d 5 configuration is exceeded, the tendency to
(c) Mn (Z = 25) (d) Fe (Z = 26) involve all the 3d electrons in bonding decreases.
3. Iron exhibits +2 and +3 oxidation states. Which of the following (2009)
statements about iron is incorrect?
(a) Ferrous compounds are relatively more ionic than the 8. Knowing that the chemistry of lanthanoids (Ln) is dominated
by its +3 oxidation state, which of the following statements
corresponding ferric compounds.
(b) Ferrous compounds are less volatile than the is incorrect?
corresponding ferric compounds. (a) Because of the large size of the Ln(III) ions the bonding
(c) Ferrous compounds are more easily hydrolysed than the in its compounds is predominantly ionic in character.
corresponding ferric compounds. (b) The ionic sizes of Ln(III) decrease in general with
(d) Ferrous oxide is more basic in nature than the ferric oxide. increasing atomic number.
(2012) (c) Ln(III) compounds are generally colourless.
4. The outer electronic configuration of Gd (Atomic No : 64) (d) Ln(III) hydroxides are mainly basic in character.
is (2009)
(a) 4f 3 5d 56 s 2 (b) 4f 8 5d 0 6s 2 9. In which of the following octahedral complexes of
(c) 4f 45 d 46 s 2 (d) 4f 7 5d1 6 s 2 (2011) Co (At. no. 27), will the magnitude of D oct be the highest?
5. In context of the lanthanoids, which of the following statement (a) [Co(NH 3 ) 6] 3+ (b) [Co(CN) 6 ]3 –
is not correct? (c) [Co(C 2 O 4) 3 ] 3– (d) [Co(H 2O ) 6] 3+ (2008)
(a) There is a gradual decrease in the radii of the members 10. Larger number of oxidation states are exhibited by the actinoids
with increasing atomic number in the series. than those by the lanthanoids, the main reason being
(b) All the members exhibit +3 oxidation state.
(c) Because of similar properties the separation of lanthanoids (a) more reactive nature of the actinoids than the lanthanoids
is not easy. (b) 4f orbitals more diffused than the 5f orbitals
(d) Availability of 4f electrons results in the formation of (c) lesser energy difference between 5f and 6d than between
compounds in +4 state for all the members of the series. 4f and 5d orbitals
(2011) (d) more energy difference between 5f and 6d than between
6. The correct order of E° M2 + /M values with negative sign for 4f and 5dorbitals.
the four successive elements Cr, Mn, Fe and Co is
(2008)
(a) Cr > Mn > Fe > Co (b) Mn > Cr > Fe > Co 11. The actinoids exhibit more number of oxidation states in general
(c) Cr > Fe > Mn > Co (d) Fe > Mn > Cr > Co than the lanthanoids. This is because
(2010)
66 JEE MAIN CHAPTERWISE EXPLORER
(a) the 5f orbitals extend further from the nucleus than the 4f 19. Heating mixture of Cu2 O and Cu2 S will give
orbitals
(a) Cu + SO2 (b) Cu + SO3
(b) the 5f orbitals are more buried than the 4f orbitals
(c) there is a similarity between 4f and 5f orbitals in their (c) CuO + CuS (d) Cu 2S O 3 (2005)
angular part of the wave function 20. The correct order of magnetic moments (spin only values in
(d) the actinoids are more reactive than the lanthanoids. B.M.) among is
(2007) (a) [MnCl4 ]2 – > [CoCl4 ] 2– > [Fe(CN)6 ] 4–
(b) [MnCl4 ] 2– > [Fe(CN)6 ] 4 – > [CoCl4 ]2 –
12. Identify the incorrect statement among the following: (c) [Fe(CN)6 ]4 – > [MnCl4 ]2 – > [CoCl4 ] 2–
(a) 4fand 5forbitals are equally shielded. (d) [Fe(CN)6 ] 4 – > [CoCl4 ] 2– > [MnCl4 ] 2 –.
(b) dBlock elements show irregular and erratic chemical
properties among themselves. (Atomic nos.: Mn = 25, Fe = 26, Co = 27)
(c) La and Lu have partially filled dorbitals and no other (2004)
partially filled orbitals.
(d) The chemistry of various lanthanoids is very similar. 21. Cerium (Z = 58) is an important member of the lanthanoids.
(2007)
Which of the following statements about cerium is incorrect?
(a) The common oxidation states of cerium are +3 and +4.
(b) The +3 oxidation state of cerium is more stable than +4
13. The “spinonly” magnetic moment [in units of Bohr magneton, oxidation state.
(m B) ] of Ni2 + in aqueous solution would be (atomic number of (c) The +4 oxidation state of cerium is not known in solutions.
Ni = 28)
(d) Cerium (IV) acts as an oxidising agent. (2004)
(a) 2.84 (b) 4.90 22. Excess of KI reacts with CuSO 4 solution and then Na 2S 2 O 3
solution is added to it. Which of the statements is incorrect for
(c) 0 (d) 1.73 (2006)
this reaction?
14. Nickel (Z = 28) combines with a uninegative monodentate ligand
X – to form a paramagnetic complex [NiX 4] 2 – . The number of (a) Cu2 I2 is formed. (b) CuI 2 is formed.
unpaired electron(s) in the nickel and geometry of this complex (c) Na2 S2 O 3 is oxidised. (d) Evolved I 2 is reduced.
ion are, respectively
(a) one, tetrahedral (b) two, tetrahedral (2004)
(c) one, square planar (d) two, square planar.
(2006) 23. Of the following outer electronic configurations of atoms, the
highest oxidation state is achieved by which one of them?
(a) (n – 1)d 8 ns2 (b) (n – 1)d 5n s 1
15. Which of the following factors may be regarded as the main (c) (n – 1)d3 n s 2 (d) (n – 1)d 5n s2 . (2004)
cause of lanthanide contraction?
(a) Poor shielding of one of 4felectron by another in the 24. For making good quality mirrors, plates of float glass are used.
subshell.
(b) Effective shielding of one of 4felectrons by another in the These are obtained by floating molten glass over a liquid metal
subshell.
(c) Poorer shielding of 5d electrons by 4felectrons. which does not solidify before glass. The metal used can be
(d) Greater shielding of 5d electrons by 4felectrons.
(2006) (a) mercury (b) tin
(c) sodium (d) magnesium. (2003)
25. Which one of the following nitrates will leave behind a metal
on strong heating?
(a) Ferric nitrate
16. The lanthanide contraction is responsible for the fact that (b) Copper nitrate
(a) Zr and Y have about the same radius (c) Manganese nitrate
(b) Zr and Nb have similar oxidation state (d) Silver nitrate. (2003)
(c) Zr and Hf have about the same radius 26. The radius of La 3+ (Atomic number of La = 57) is 1.06 Å. Which
(d) Zr and Zn have the same oxidation state. (2005) one of the following given values will be closest to the radius
17. Calomel (Hg 2C l2 ) on reaction with ammonium hydroxide gives of Lu3 + (Atomic number of Lu = 71)?
(a) HgNH2 Cl (b) NH2 – Hg – Hg – Cl (a) 1.60 Å (b) 1.40 Å
(c) Hg 2 O (d) HgO (2005) (c) 1.06 Å (d) 0.85 Å (2003)
18. The oxidation state of chromium in the final product formed by 27. What would happen when a solution of potassium chromate is
the reaction between KI and acidified potassium dichromate treated with an excess of dilute nitric acid?
solution is (a) Cr3 + and Cr2 O 72 – are formed.
(b) Cr 2 O 72 – and H2 O are formed.
(a) +4 (b) +6 (c) CrO 42 – is reduced to +3 state of Cr.
(c) +2 (d) +3 (d) CrO 4 2– is oxidised to +7 state of Cr.
(2005) (2003)
d and fBlock Elements 67
28. The number of delectrons retained in Fe2 + (At. no. Fe = 26) 32. How do we differentiate between Fe3 + and Cr3 + in group III?
ions is (a) By taking excess of NH4 O H solution.
(b) By increasing NH4 + ion concentration.
(a) 3 (b) 4
(c) By decreasing OH– ion concentration.
(c) 5 (d) 6 (2003)
29. The atomic numbers of vanadium (V), chromium (Cr), (d) Both (b) and (c). (2002)
manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 33. The most stable ion is
(a) [Fe(OH)3 ] 3 –
26. Which one of these may be expected to have the highest (c) [Fe(CN)6 ] 3– (b) [Fe(Cl)6 ] 3–
(d) [Fe(H2 O )6 ] 3+
second ionisation enthalpy?
(a) V (b) Cr (2002)
(c) Mn (d) Fe (2003) 34. Arrange Ce3 + , La 3+ , Pm 3+ and Yb 3+ in increasing order of their
30. A reduction in atomic size with increase in atomic number is a ionic radii.
characteristic of elements of (a) Yb 3+ < Pm 3+ < Ce3 + < La 3+
(a) high atomic masses (b) Ce3 + < Yb 3+ < Pm 3+ < La 3+
(b) d block (c) Yb 3+ < Pm 3+ < La 3+ < Ce3 +
(c) f block (d) Pm 3+ < La 3+ < Ce3 + < Yb 3+ (2002)
(d) radioactive series. (2003)
35. Most common oxidation states of Ce (cerium) are
31. A red solid is insoluble in water. However it becomes soluble (a) +2, +3 (b) +2, +4
if some KI is added to water. Heating the red solid in a test tube (c) +3, +4 (d) +3, +5 (2002)
results in liberation of some violet coloured fumes and droplets
of a metal appear on the cooler parts of the test tube. The red 36. Which of the following ions has the maximum magnetic
solid is moment?
(a) (NH4 ) 2 Cr 2O 7 (b) HgI2 (a) Mn2 + (b) Fe2 +
(c) HgO (d) Pb3 O 4. (2003) (c) Ti2 + (d) Cr 2+ (2002)
Answer Key
1. (b) 2. (a) 3. (c) 4. (d) 5. (d) 6. (b)
7. (b) 8. (c) 9. (b) 10. (c) 11. (a) 12. (a)
13. (a) 14. (b) 15. (a) 16. (c) 17. (a) 18. (d)
19. (a) 20. (a) 21. (c) 22. (b) 23. (b) 24. (a)
25. (d) 26. (d) 27. (b) 28. (d) 29. (b) 30. (c)
31. (b) 32. (d) 33. (b) 34. (a) 35. (c) 36. (a)
68 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : Number of unpaired electrons in Fe2 + is less than Mn 2+ , 13. (a) : 28 Ni ® [Ar] 3d 8 4s 2 4p
so Fe 2+ is less paramagnetic than Mn 2+ . 3d 4s
2. (a) Number of unpaired electrons (n) = 2
m = n(n + 2) = 2(2 + 2) = 8 » 2.84
3. (c) : Ferrous oxide is more basic, more ionic, less volatile
and less easily hydrolysed than ferric oxide. 3d 4s 4p
4. (d) : The electronic configuration of 14. (b) : Ni ®
64G d = [Xe] 54 4f 7 5d1 6s 2
Ni2 + ®
5. (d) : Availability of 4f electrons does not result in the formation
of compounds in +4 state for all the members of the series. [NiX 4] 2 – ® × × × × × × × ×
6. (b) : E° Mn 2+ /Mn = – 1.18 V sp 3 hybridisation
E° Cr 2+ /Cr = – 0.91 V
E° Fe 2+ /Fe = – 0.44 V Number of unpaired electrons = 2
E°C o 2+ /Co = – 0.28 V Geometry = tetrahedral.
7. (b) : When the transition metals are in their highest oxidation 15. (a) : As we proceed from one element to the next element in
state, they no longer have tendency to give away electrons, the lanthanide series, the nuclear charge, i.e. atomic number
thus they are not basic but show acidic character and form increases by one unit and the addition of one electron occurs
anionic complexes. at the same time in 4fenergy shell. On account of the very
diffused shapes of forbitals, the 4felectrons shield each other
8. (c) : Ln 3+ compounds are generally coloured in the solid state quite poorly from the nuclear charge. Thus, the effect of nuclear
as well as in aqueous solution. Colour appears due to presence charge increase is somewhat more than the changed shielding
of unpaired felectrons which undergo ff transition. effect. This brings the valence shell nearer to the nucleus and
hence the size of atom or ion goes on decreasing as we move
9. (b) : Strong field ligand such as CN –, usually produce low in the series. The sum of the successive reactions is equal to
spin complexes and large crystal field splittings. H 2 O is a the total lanthanide contraction.
weaker field ligand than NH 3 and C 2 O 42 – therefore
Do ct [Co(H 2 O) 6] 3+ < D oct[ Co(C 2 O 4) ] 3– < [Co(NH 3 )6 ] 3+ 16. (c) : In each vertical column of transition elements, the elements
Common ligands in order of increasing crystal field strength of second and third transition series resemble each other more
are given below : closely than the elements of first and second transition series
I – < Br – < Cl – < F – < OH – < C 2O 4 2– < H 2O < NH 3 < en < NO 2 – on account of lanthanide contraction. The pairs of elements
< CN – such as ZrHf, MoW, NbTa, etc; possess almost the same
properties.
10. (c) : Actinoids show different oxidation states such as +2,
+3, +4, +5, +6 and +7. However +3 oxidation state is most 17. (a) : Calomel on reaction with ammonium hydroxide turns
common among all the actinoids.
The wide range of oxidation states of actinoids is attributed black. The black substance is a mixture of mercury and mercuric
to the fact that the 5f, 6d and 7s energy levels are of comparable
energies. Therefore all these three subshells can participate. amino chloride.
11. (a) : As the distance between the nucleus and 5f orbitals Hg2 C l2 + 2NH4 OH ® NH 2
(actinides) is more than the distance between the nucleus and H{g Cl + H g + NH 4C l + 2H 2 O
4f orbitals (lanthanides) hence the hold of nucleus on valence
electron decreases in actinides. For this reason the actinoides Black
exhibit more number of oxidation states in general.
12. (a) : The decrease in the force of attraction exerted by the
nucleus on the valency electrons due to presence of electrons
in the inner shells is called shielding effect. An 4f orbital is
nearer to the nucleus than 5f orbitals. Hence shielding of 4f is
more than 5f.
d and fBlock Elements 69
18. (d) : +3 28. (d) : 26F e = 1s 2 2s2 2p 6 3s 2 3p 6 3d 6 4s2
Fe2 + = 1s 2 2s 2 2p6 3s 2 3p6 3d 6
+6 2Cr 3+ + 7H2 O + 3I2 The number of d electrons retained in Fe2 + = 6.
Cr2 O 7 2 – + 14H+ + 6I–
19. (a) : Cu 2 S + 2Cu 2 O ® 6Cu + SO2 29. (b) : The second ionisation potential values of Cu and Cr are
This is an example of autoreduction. sufficiently higher than those of neighbouring elements. This is
because of the electronic configuration of Cu + which is 3d 10
20. (a) : (completely filled) and of Cr + which is 3d 5 (halffilled), i.e.,
for the second ionisation potentials, the electron is to be removed
3d 4s 4p from very stable configurations.
[MnCl4 ] 2–® - - - - - 1.. 442.. 4.. 43.. 30. (c) : With increase in atomic number i.e. in moving down a
group, the number of the principal shell increases and therefore,
sp 3 the size of the atom increases. But in case of f block elements
there is a steady decrease in atomic size with increase in atomic
Number of unpaired electrons = 5 number due to lanthanide contraction.
As we move through the lanthanide series, 4f electrons are being
[CoCl 4] 2–® 1.. 442.. 4.. 43.. added one at each step. The mutual shielding effect of f electrons
is very little. This is due to the shape of the f orbitals. The
sp3 nuclear charge, however increases by one at each step. Hence,
the inward pull experienced by the 4f electrons increases. This
Number of unpaired electrons = 3 causes a reduction in the size of the entire 4f n shell.
[Fe(CN) 6 ] 4– ® 1.. .4. 4..2 4.. 4..3 ..
d 2s p3
Number of unpaired electrons = 0
Magnetic moment = n n + 2 31. (b) : The precipitate of mercuric iodide dissolves in excess of
where n = number of unpaired electrons.
i.e. greater the number of unpaired electrons, greater will be the potassium iodide forming a complex, K 2 HgI 4 .
paramagnetic character.
HgI2 + 2KI ® K 2 HgI 4
HgI 2 on heating liberates I 2 gas.
21. (c) : +4 oxidation state of cerium is also known in solutions. HgI 2 Hg + I 2
–1 0 violet vapours
22. (b) : 4KI + 2CuSO4 I 2 + Cu 2I 2 + 2K 2S O 4 32. (d) : NH4 + ions are increased to suppress release of OH– ions,
hence solubility product of Fe(OH) 3 is attained. Colour of
0 +2 +2.5 –1 precipitate is different.
I 2 + 2Na 2 S2 O 3 Na 2 S4 O 6 + 2NaI
(n – 1)d ns 33. (b) : A more basic ligand forms stable bond with metal ion, Cl–
is most basic amongst all.
23. (b) :
(n – 1)d5 ns 2 can achieve the maximum oxidation state of +7. 34. (a) : According to their positions in the periods, these values
are in the order:
24. (a) : Mercury is such a metal which exists as liquid at room
temperature. Yb 3+ < Pm 3+ < Ce3 + < La 3+
25. (d) : When heated at red heat, AgNO 3 decomposes to metallic At. Nos. 70 61 58 57
silver.
2AgNO 3 ® 2Ag + 2NO 2 + O 2 Ionic radii (pm) 86 98 103 106
26. (d) : Due to lanthanide contraction, the ionic radii of Ln3 + Ionic size decreases from La 3+ to Lu 3+ due to lanthanide
(lanthanide ions) decreases from La 3+ to Lu3 +. Thus the lowest
value (here 0.85 Å) is the ionic radius of Lu3 +. contraction.
27. (b) : Dilute nitric acid converts chromate into dichromate and 35. (c) : The common stable oxidation state of all the lanthanides
is +3. The oxidation states of +2 and +4 are also exhibited and
H2 O . these oxidation states are only stable in those cases where stable
4 f 0 , 4 f 7 or 4 f 14 configurations are achieved. Ce4 + is stable
2K2 C rO 4 + 2HNO 3 ® K2 C r2 O 7 + 2KNO 3 + H2 O due to 4 f 0 configuration.
or, 2CrO 42 – H + Cr 2 O 72 – + H 2 O 36. (a) : Mn2 + (3s 23 p 63 d5 ) has the maximum number of unpaired
electrons (5) and therefore has maximum moment.
yellow orange
70 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER COORDINATION COMPOUNDS
18
1. Which of the following complex species is not expected to 9. The coordination number and the oxidation state of the element
exhibit optical isomerism? E in the complex [E(en)2 (C2 O 4) ]NO 2 (where (en) is ethylene
diamine) are, respectively
(a) [Co(en)(NH 3 ) 2 Cl 2 ] + (b) [Co(en) 3 ] 3+
(c) [Co(en) 2C l 2 ] + (d) [Co(NH 3 ) 3 Cl 3 ] (2013) (a) 6 and 3 (b) 6 and 2
2. Which among the following will be named as dibromidobis (c) 4 and 2 (d) 4 and 3 (2008)
(ethylene diamine) chromium(III bromide? 10. Which of the following has a square planar geometry?
(a) [Cr(en) 2B r 2] Br (b) [Cr(en)Br 4] – (a) [PtCl4 ] 2 – (b) [CoCl4 ] 2 –
(c) [Cr(en)Br 2] Br (d) [Cr(en) 3 ]Br 3 (2012) (c) [FeCl4 ] 2– (d) [NiCl4 ] 2–.
3. The magnetic moment (spin only) of [NiCl 4] 2– is (At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)
(a) 1.82 BM (b) 5.46 BM (2007)
(c) 2.82 BM (d) 1.41 BM (2011) 11. How many EDTA (ethylenediaminetetraacetic acid) molecules
4. Which of the following facts about the complex [Cr(NH3 ) 6 ]Cl 3 are required to make an octahedral complex with a Ca 2+ ion?
is wrong?
(a) Six (b) Three
(a) The complex involves d 2 sp 3 hybridisation and is (c) One (d) Two (2006)
octahedral in shape. 12. In Fe(CO)5 , the Fe – C bond possesses (2006)
(b) The complex is paramagnetic. (a) pcharacter only (b) both s and p characters
(c) The complex is an outer orbital complex.
(d) The complex gives white precipitate with silver nitrate (c) ionic character (d) scharacter only.
solution. (2011) 13. The IUPAC name for the complex [Co(NO 2) (NH3 ) 5] Cl2 is
5. Which one of the following has an optical isomer? (a) nitritoNpentaamminecobalt(III) chloride
(b) nitritoNpentaamminecobalt(II) chloride
(a) [Zn(en) 2] 2+ (b) [Zn(en)(NH 3 )2 ]2 + (2010) (c) pentaammine nitritoNcobalt(II) chloride (2006)
(c) [Co(en) 3] 3+ (d) [Co(H 2O ) 4 (en)] 3+ (d) pentaammine nitritoNcobalt(III) chloride.
6. A solution contains 2.675 g of CoCl 3 ∙6NH 3 (molar mass
= 267.5 g mol –1 ) is passed through a cation exchanger. The 14. The value of the ‘spin only’ magnetic moment for one of the
chloride ions obtained in solution were treated with excess of following configurations is 2.84 BM. The correct one is
AgNO 3 to give 4.78 g of AgCl (molar mass = 143.5 g mol– 1 ). (a) d 4 ( in strong ligand field)
The formula of the complex is (At. mass of Ag = 108 u) (b) d 4 ( in weak ligand field)
(a) [CoCl(NH 3 ) 5 ]Cl 2 (b) [Co(NH 3 )6 ]Cl 3 (c) d 3 ( in weak as well as in strong fields)
(c) [CoCl 2 (NH 3) 4] Cl (d) [CoCl 3( NH 3 ) 3 ] (2010) (d) d 5 (in strong ligand field) (2005)
7. Which of the following pairs represents linkage isomers? 15. Which one of the following cyano complexes would exhibit the
(a) [Cu(NH 3 ) 4] [PtCl 4 ] and [Pt(NH 3 ) 4 ][CuCl 4 ] lowest value of paramagnetic behaviour?
(b) [Pd(PPh 3 ) 2 (NCS) 2 ] and [Pd(PPh 3 ) 2 (SCN) 2 ] (a) [Cr(CN)6 ] 3– (b) [Mn(CN)6 ] 3–
(c) [Co(NH 3 )5 (NO 3 )]SO 4 and [Co(NH 3 )5 (SO 4 )]NO 3 (c) [Fe(CN)6 ] 3 – (d) [Co(CN)6 ] 3 – (2005)
(d) [PtCl 2 (NH 3 ) 4 ]Br 2 and [PtBr 2( NH 3 ) 4] Cl 2 (2009)
16. Which of the following compounds shows optical isomerism?
8. Which of the following has an optical isomer?
(a) [Cu(NH3 ) 4] 2 + (b) [ZnCl4 ] 2–
(a) [Co(NH 3 ) 3 Cl] + (b) [Co(en)(NH 3 ) 2 ]2 + (2009) (c) [Cr(C2 O 4) 3 ]3 – (d) [Co(CN)6 ] 3 – (2005)
(c) [Co(H 2O ) 4 (en)] 3+ (d) [Co(en) 2 (NH 3 ) 2 ] 3+
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