Dr. Sangeeta Khanna Ph

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+1\+1 for School Exam Questions A4 size.doc

BASIC CONCEPTS IN CHEMISTRY 1. What is the difference between 5 g and 5.00 g? Sol. In 5g, one significant digit is present while 5.00 g contains three significant figures. Hence, 5.00 g is

more precise than 5g. 2. Define triple point. Sol. Triple point is that temperature at which the solid, liquid, and gaseous states all exist in equilibrium

with one another at the same temperature. 3. While examining a thermometer outside the house, due to sudden cold, the temperatures drops from

81° to 54° F. What is the corresponding temperature drop in degree Celsius? Sol. 81 °F temperature corresponding to 5/9 (81 – 32) °C = 27 °C While 54 °F corresponds to 5/9 (54 – 32) = 12 °C

The corresponding temperature drop in degree Celsius is (27 – 12) °C = 15 °C 4. At what temperature have the Celsius and Fahrenheit readings the same numerical value? Sol. x = 5/9 (x – 32) or x = -40°C = 40 °F 5. State the law of multiple proportion. Sol. According to Dalton in 1803, “If two elements A and B combine to give two or more compounds, then

weight of A which combines with a fixed weight of B, bear a simple ratio to one another. E.g.

)gm18(gm) (16)g2(

Water Oxygen Hydrogen

gm) 34() gm(32)g2(

peroxide HydrogenOxygen Hydrogen

Here the mass of oxygen (i.e. 16 gm and 32 gm) which combine with a fixed mass of hydrogen (2g) bears a simple ratio i.e. 16:32 or (1: 2)

6. Calculate the mass of 3 g molecules of nitric acid. Sol. 1 g molecule of HNO3 = 1 + 14 + 3 × 16 = 63 g gm molecule = 1 mole molecule

3 g molecule of HNO3 = 189 g 7. Define (a) molar mass (b) molar volume

Sol. (a) Molar mass Mass of one mole particles i.e. 6.023 × 1023 particles

(b) Molar volume The volume occupied by one mole of a gaseous substance. One mole molecules of all gases occupy 22.4 L at 273 K and 760 mmHg (STP).

8. How many moles are present in: (a) 24.5 g H2SO4 (b) 4.00 g O2?

Sol. (a) 98 g of H2SO4 = 1 mol

24.5 g of H2SO4 = 24.5/98 = 0.25 mol (b) 32 g of O2 = 1 mol

4.00 g of O2 = 1/8 = 0.125 mol 9. What is the average mass of a molecule of methane? Sol. 6.023 × 1023 molecules of CH4 weighs 16 g

1 molecule of methane will weigh 23106.023

g 16

= 2.66 × 10-23 g

10. A fluoride of oxygen was prepared by mixing oxygen and fluorine in the proper ratio at 60 K. This compound contains 32.1% F and 67.9% O. What is the empirical formula of the compound.

Sol.

Element % by mass Relative No of moles

Simple ratio of moles

Simplest whole no ratio

F 32.1 64.1

19

1.32 1

64.1

64.1

1 × 2 = 2

O 67.9 24.4

16

9.67 50.2

64.1

24.4

2.5 × 2 = 5

Hence, the empirical formula is O5F2



11. A compound contains 21.6% sodium, 33.3% chorine, 45.1% oxygen. Derive its empirical formula. Sol.

Element % by mass Relative No of moles Simple ratio of moles Simplest whole no ratio

Na Cl O

21.6 33.3 45.1

0.939 0.938 2.82

1 1 3

1 1 3

Hence, the empirical formula in NaClO3 12. In an experiment 5.0 g CaCO3 on heating gave 2.8 g CaO and 1120 ml of CO2 at STP. Show that

these results are in agreement with the law of conservation of mass.

Sol. CaCO3 CaO + CO2 5.0 g 2.8 g 1120 ml at STP 22400 ml of volume at STP is occupied by 1 mole of gas

1120 ml is occupied by 22400

1120 = 0.05 mol of CO2

1 mol of CO2 = 44 g

0.05 mol of CO2 = 44 × 0.05 = 2.2 g

1120 ml of CO2 at STP = 2.2 g Total mass of the products = 2.8 g + 2.2 g = 5.0 g Hence, law of conservation of mass is proved 13. A solution containing 8.5 g of silver nitrate was reacted with 2.9 g of sodium chloride. A white

precipitate of silver chloride was formed which weighed 7.15 g and residue containing sodium nitrate on evaporation to dryness yielded 4.25 g of sodium nitrate. Show that the data is in accordance with the law of conservation of mass.

Sol. AgNO3 + NaCl AgCl + NaNO3 8.5 g 2.9 g 7.15 g 4.25 g

As mass of the reactants (AgNO3 + NaCl = 11.4 g) and mass of the products (AgCl + NaNO3 = 11.4g) is same, the results are in accordance with the law of conservation of mass.

14. A certain elements ‘X’ forms three different binary compounds with chlorine, containing 59.68%, 68.95% and 74.75% chlorine, respectively. Show how these data illustrate the law of multiple proportions.

Sol.

Mass of X Mass of Cl Mass of Cl per gram of X Dividing by 1.48 Multiplication

40.32

31.05

25.25

59.68

68.95

74.75

1.48

2.22

2.96

1

1.5

2

2

3

4

For a given mass of X (i.e., 1 g). The ratio of number of Cl is a simple integral ratio (2 : 3 : 4). Hence, the law is proved.

15. How many molecules of H2 are present in 7.5 g of H2? How many H atoms?

Sol. Number of mole of H2 contained in 7.5 g of H2 are mol 75.3mol 2g

g 5.7

1-

Number of molecules of H2 contained in 3.75 mol are 3.75 × 6.023 × 1023 = 2.26 × 1024 molecules Number of atoms of H present in 3.75 mol of H are 2 × 3.75 × 6.023 × 1023 = 4.52 × 1024 atoms of H 16. What is the weight of one molecule of

(a) CH3CH2OH (b) C60H122 (c) C1200 H2000 O1000

Sol. (a) molecular mass of CH3CH2OH = 46 6.023 × 1023 molecules are present in 46 g C2H5OH

1 molecule has a mass of 2310023.6

46

= 7.64 × 10–23 g

(b) Molecular mass of C60H122 = 842

Mass of one molecule of C60H122 = 23100234.6

842

= 1.39 × 10–21 g



(c) Molecular mass of C1200H2000O1000 = 32400

Mass of one molecule of C1200 H2000 O1000 = g1038.510023.6

32400 2023

17. Calculate the number of atoms of helium in each of the following:

(i) 52 moles of He (ii) 52 amu of He (iii) 52 g of He

Sol. (a) Number of atoms in 52 moles of He = 52 × 6.023 × 1023 = 3.13 × 1025

(b) Mass of 1 He atom = 4 amu

4 amu = 1 He atom

or 52 amu = 13 He atom

(c) 52 g He = 13 mole He

Number of atoms contained in 52 g of He (13 mole He)

= 13 × 6.023 × 1023

= 7.83 × 1024 atom He

18. How many molecules of water of hydration are present in 36 mg of Mohr’s Salt

(FeSO4(NH4)2SO46H2O)?

Sol. Molecular mass of Mohr’s salt = 392

Number of moles of Mohr’s salt contained in 36 mg

= 1-

3

mol g 392

g1036 = 9.18 × 10-5 mol

1 Mole of Mohr’s salt contains 6 × 6.023 × 1023 molecules of water of hydration

9.18 × 10-5 moles of Mohr’s salt contain 6 × 9.18 × 10-5 × 6.023 × 1023

= 3.32 × 1020 molecules of water of hydration.

19. Calculate how many methane molecule and how many hydrogen and carbon atoms are there in 25.0

g of methane.

Sol. Number of methane molecules = 16

0.25 × 6.023 × 1023 = 9.40 × 1023 molecules CH4

Number of C atoms = 9.40 × 1023

Number of H atoms = 4 × 9.40 × 1023

= 3.76 × 1024

20. Calculate the number of moles, molecules and atoms in 11.2 litre of H2 at STP.

Sol. 22.4 L of H2 at STP = 1 mol H2

11.2 L of H2 at STP = 2H mol 5.04.22

2.11

Number of molecules contained in 0.5 mol

H2 = 0.5 × 6.02 × 1023 = 3.01 × 1023

Number of atoms in 0.5 moles H2

= 2 × 0.5 × 6.02 × 102 = 6.02 × 1023 atom H

21. Nitrogen and oxygen combine to form many oxides. In some oxides, 14 g of nitrogen combine with

either 16 g oxygen or 32 of oxygen. In some other oxides, 28 g of nitrogen combines with 16, 48 or

80 g of oxygen.

(a) What are the formulae of oxides?

(b) What law of chemical combination does the formation of these compounds illustrate? How?

Sol. (a) The atomic masses of nitrogen and oxygen are 14 and 16 respectively. Hence, their relative

masses in different compound are:-

N : O N : O N : O

Mass 14 g 16 g 14g 32g 28g 16g

Mole ratio = 14

14

16

16

14

14

16

32

14

28

16

16

= 1 = 1 =1 =2 = 2 =1

Compound NO NO2 N2O



N : O N : O

mass 28 g 48 g 28 g 80 g

mole ratio 14

28

16

48

14

28

16

80

= 2 =3 = 2 =5

N2O3 N2O5

Hence, the formula of different oxides are N2O, NO, N2O3, NO2 and N2O5.

(b) Since, in this example a fixed mass of N combines with different masses of oxygen that bear a

simple ratio to one another, hence the law of multiple proportion is followed.

22. (a) If the elemental composition of butyric acid is found to be 54.2% C, 9.2% H and 36.6% O,

determine the empirical formula.

(b) The molar mass of butyric acid was determined by experiment to be 88 g mol-1. What is the

molecular formula?

Sol. (a)

Element mass in per 100 g number of moles Divide by 2.2 g

C 54.2 52.412

2.54 2

H 9.2 20.91

2.9

4

O 36.6 16

6.36 = 2.29 1

Hence, the empirical formula is C2H4O.

(b) EFW = 24 + 4 + 16 = 44

Molar mass = 88 g mol–1

n = 244

88

Hence, molecular formula is (C2H4O)2 or C4H8O2

23. What is a limiting reagent?

Sol. The limiting reagent is defined as the reactant which is completely consumed during the reaction

24. Define molarity of a solution. Give its SI units.

Sol. Molarity is defined as the number of moles of solute per litre of the solution. Its SI unit is mol L–1.

25. Acid contains 13% by mass H2SO4. What is the molality of the solution?

Sol. 13% by mass H2SO4 means 100 g solution contain 13 g H2SO4

g 13w42SOH

98

13n

42SOH = 0.133 mol

OH2w = 87 g

Molality = 100087

133.01000

w

n

A

B

m = 1.53 molal 26. Calculate the molarity of solution obtained by dissolving 0.212 g of Na2CO3 in 250 cm3 of solution. Sol. Molar mass of Na2CO3 = 106 g mol–1

Number of moles of Na2CO3 = 106

212.0

= 2.0 × 10–3 mol

Volume of solution = 250 cm3 = 1/4 L

M = L4/1

mol 100.2

V

n 3

Molarity = 8 × 10–3 = 0.008 M



27. Zinc and hydrochloric acid react according to the reaction.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

if 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are

produced?

Sol. )g(H)aq(ZnCl2HCl(aq) )s(Zn 22mol 0.52mol 30.0

0.30 mol Zn requires 2 × 0.30 = 0.60 mol HCl, but only 0.52 mol of HCl are given.

Number of mole of H2 produced = HCln2

1

= mol. 26.052.02

1

28. 3.00 of H2 react with 29.0 g of O2 to yield H2O.

(i) Which is the limiting reactant?

(ii) Calculate the maximum amount of H2O that can be formed

(iii) Calculate the amount of the reactant which remains unreacted.

Sol. 3.00 g H = 50.12

00.3 mol H and

29.00 g O2 = 32

00.29 = 0.90 mol O2

The balanced equation for the reaction is 2H2 + O2 2H2O

1.50 mol H2 require 2

50.1 = 0.75 mol O2 to give 1.50 mol H2O. Since more oxygen is present than

required, it follows that

(a) H2 is the limiting reactant

(b) the maximum amount of H2O that can be formed is 1.50 mol H2O (27 g H2O)

(c) 0.90 – 0.75 = 0.15 mol O2 remains unreacted in the reaction mixture

29. Commercially available concentrated hydrochloric acid contains 38% HCl by mass.

(i) What is the molarity of this solution? The density of this solution is 1.19 g/mL.

(ii) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCl?

Sol. (i) Given: 38% HCl by mass

100 g solution contains 38 g HCl

Molar mass of HCl = 36.5 g/mol

nHCl = g/mol 36.5

g 38 = 1.04 mol

Mass of solution = 100 g

Density of solution = 1.19 g mL–1

Volume of solution = 1.19 g mL–1 × 100 g = 119 mL = 0.119 L

M = M 74.8L 0.119

mol 04.1

V

n

(ii) Suppose the volume of 8.74M HCl required is V L. This is a dilution problem no. of mol of HCl

remains same.

n(conc) = ndilute

Mconc.Vconc. = Mdil Vdil

8.74 × V = 0.10 × 1

V = 0.011 L = 11.0 mL



STRUCTURE OF ATOM

30. Which experiment led to the discovery of neutrons?

Sol. Neutrons were discovered by Chadwick in 1932. Neutrons are produced when we bombard a thin foil

of beryllium with fast moving -particles. These neutral particles were found to have mass 1.675 × 10-

27 kg.

n CHeBe 10

126

42

94

31. An atom contains five protons and six neutrons. What is the nuclide symbol for the nucleus? What is

its atomic number and mass number?

Sol. Atomic number = number of protons = 5

Mass number = number of protons + number of neutrons = 5 + 6 = 11

The symbol is B115

32. Write the complete symbol for (a) the nucleus with atomic number 56 and mass number 138 (b) Nucleus with four protons and five neutrons

Sol. (a) Ba13856

(b) Atomic number = 4 Mass number = 9

The symbol is Be94

33. What is the frequency of violet light with a wavelength of 408 nm?

Sol. 1-9

-18

m 10408

ms 1000.3c

= 7.35 × 1014 s–1 34. The red spectral line of lithium occurs at 671 nm. Calculate the energy of one photon of this light. Sol. The frequency of this light is

1-147

-18s 1047.4

m 1071.6

ms 1000.3c

Hence, the energy of one photon is

E = h = 6.63 × 10-34 Js × 4.47 × 1014 s–1

= 2.96 × 10–19 J 35. What was the shortcoming of the Thomson model of atom? Sol. The Thomson model of atom could account for the electrical neutrality of atom, but it could not

explain the results of gold foil scattering experiment carried out by Rutherford. 36. What do you understand by black body and black body radiations? Sol. The ideal body which emits and absorbs all frequencies, is called a black body and the radiation

emitted by this body is called black body radiation. The exact frequency distribution of the emitted radiation (i.e. intensity versus frequency curves of the

radiation) from a black body depends upon its temperature. 37. What is photoelectric effect? Sol. The phenomenon involving ejection of electron from some metal surfaces when irradiated with light of

appropriate frequency.

38. State at least four ways in which positive canal rays differ from cathode rays.

Sol.

Positive rays Cathode rays

(i) These are positively charged rays.

(ii) The charge-to-mass ratio depends on

ions & is variable.

(iii) Mass depends on ions and is variable

(iv) These have a charge of mostly +1, but

sometimes have +2, +3 and so on.

(i) These are negatively charged rays.

(ii) The charge-to-mass ratio has a

definite value.

(iii) These have a definite mass.

(iv) These always have a charge of -1.



39. The wavelength of a beam of light is 42.0 mm. What is (i) its wavelength in cm (ii) its frequency (iii) its wave number (iv) the energy of one of its photons.

Sol. = 42.0 mm (i) 1 mm = 0.1 cm

= 42.0 mm = 4.2 cm

(ii) = m 1042

ms 1000.3c

3

-18

= 7.14 × 109 s–1

(iii) m42

11310

= 23 m–1

(iv) E = h = 6.63 × 1034 Js × 7.14 × 109 s-1 4.74 × 10–24 J

40. The following are representative wavelengths in the infrared, ultraviolet and X-ray regions of the electromagnetic spectrum, respectively: 1.0 × 10-6 m, 1.0 × 10-8 m, and 1.0 × 10-10 m. (i) What is the energy of a photon of each radiation? (ii) Which has the greater amount of energy per photon and which has the least?

Sol. (i)

c

hEIR

= m 100.1

ms 103.00Js 1063.6

8

-1834

= 1.99 × 10–19 J

m 100.1

ms 103.00Js 1063.6chE

8

-1834

uv

= 1.99 × 10-19 J Ex-ray = 1.99 × 10-15 J

(ii) X-rays has the greatest amount of energy per photon and infrared has the lowest amount of energy.

41. A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 mm. Calculate the rate of emission of quanta per second.

Sol. Power of the bulb = 25 Js–1 Energy of one photon = E

= 6

834

1057.0

100.31063.6ch

= 3.49 × 10-19 J. Rate of emission of quanta per second

= 1-99

s 1016.71049.3

25

42. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of

wavelength 6800 Å. Calculate the threshold frequency (0) and work function of the metal.

Sol. = 6800 Å = 6.80 × 10–7 m

as velocity = zero KE also equals to zero. Hence, Energy of photon = Work function

W = E = h7

834

1080.6

100.310626.6c

W = 2.92 × 10–19 J

W = h0

Threshold frequency 0 = h

W

11434

19s1041.4

10626.6

1092.2



43. A photon of wavelength 4 × 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon in eV (ii) the kinetic energy of the emitted electron in joules. (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10-19 J)

Sol. = 4 × 10–7 m (i) Energy of the incident photon

= E = h

c

E = m104

ms 100.3Js10626.6

7

-1834

= 4.9695 × 10-19 J

E = eV 102.3eV106020.1

109695.4

19

19

(ii) Kinetic energy (KE) = Energy of photon – Work function = 3.102 – 2.13 = 0.972 eV = 0.972 × 1.6020 × 10 –19 J (iii) Velocity of the photoelectron =

= em

KE2

31

19

101.9

10557.12

= 5.85 × 105 m s–1 44. The first line of the lyman series of hydrogen atom emission spectrum results from a transition from n

= 2 level to n = 1 level. What is the wavelength of the emitted photon?

Sol. 12H

2L

7 mn

1

n

110097.1

1

122

7 m2

1

1

110097.1

17 m4

310097.1

m10097.13

4

7

= 1.22 × 10-7 m = 122 nm 45. What is the maximum number of emission lines when the excited electron of a H-atom in n = 6 drops

to the ground state?

Sol. No. of spectral line = 2

)1n(n

n = 6

No. of spectral lines = 2

)16(6

= 15 lines 46. Calculate the mass of a photon with wave length 3.6 Å.

Sol.

m

h

8

3410

100.3m

10626.6106.3

m = kg106.3100.3

10626.6

108

34

m = 6.135 × 10-33 kg



47. Calculate the momentum of a particle which is associated with a de-Broglie wavelength of 0.1 nm. (h = 6.63 × 10-34 kg m2 s-1)

Sol. p

h

Here, h = 6.63 × 10–34 kg m2 s–1 and = 0.1 nm = 10–10 m

10

34

10

1063.6p

= 6.63 × 10-24 kg ms-1

48. Calculate the product of uncertainty in position and velocity of a proton of mass 1.67 × 10-27 kg. (h = 6.62 × 10-35 kg mass-1)

Sol. x p 4

h

x m4

h

x 27

34

1067.114.34

1062.6

x 3.156 × 10–8 m2 s–1 49. What is the wavelength and frequency of the limiting line of the Balmer series in the spectrum of

atomic hydrogen?

Sol. For the Balmer series nL = 2. As nH gets larger and approaches infinity (nH ), 2H

n

1 gets smaller

and smaller and approaches zero. So, for limiting line, nH =

227

1

2

110907.1

1

= 1.097 × 107 × 1m4

1

= m10097.1

4

7 = 3.64× 10-7 m

= 364 nm

1-147

18s 1024.8

m1064.3

ms1000.3c

50. Calculate the energy required to cause the ionization of one mole of hydrogen atom. Sol. Energy required to cause the ionization of H-atoms means the energy to be supplied to remove an

electron from nL = 1 to nH = , when the energy of the electron zero. For one mole of H atom

E=

2H

2L

An

1

n

1hcR N

= 6.023 × 1023 × 6.63 × 10-34 × 3 × 108 × 1.097 × 107

21

1J. mol

= 131.42 × 104 J/mol = 1314.2 × 103 J/mol = 1314.2 kJ/mol

51. How will you differentiate between an orbit and an orbital? Sol.

Orbit Orbital (i) It is definite path followed by an electron (i) It is the part of space around the nucleus in which

the probability of finding the electrons is maximum (ii) It can have more than two electrons. For example, second orbit has eight electrons and third has eight electrons and third has eighteen electrons.

(ii) It cannot have more than two electrons.

(iii) An orbit is circular in shape (iii) Orbitals have different shapes. For example, s-orbital is spherical, p-orbital is dumbbell shape.

(iv) Two orbits of an atom cannot have the same energy

(iv) Two orbitals can have the same energy. For example px, py and pz have same energy.



52. How much energy is required to ionize a H-atom if the electron occupies n = 5 orbits? Compare your

answer with the ionization energy of H-atom (energy required to remove the electron from n = 1

orbit.)

Sol. 1-2

5

n mol Jn

1012.13E

Energy required to remove the electron completely from n = 5 orbit in an H-atom is

mol/J5

1012.13

2

5

= 5.2480 × 104 J/mol

= 8.71 × 10-20 J/atom

I.E. of H-atom is 2.178 × 10-18 J/atom

Energy required to ionize H-atom when electron is in n = 5 orbit a 25

1times the energy required to

ionize the H-atom when the electron is in n = 1 orbit.

53. The mass of an electron is 9.1 × 10-31 kg. If its kinetic energy is 4.55 × 10-25 J. Calculate its wave

number.

Sol. KE =

2/12

m

KE 2 m

2

1

2/1

31

25

101.9

1055.42

= 103 ms–1

1

number w aveand m

h

1-234

-1331

s m kg10626.6

s m 10kg101.9

h

mv

v = 1.37 × 106 m–1

54. Calculate the de-Broglie wavelength of an electron traveling at 1% of speed of light.

Sol. me = 9.1 × 10-31 kg

h = 6.62 × 10-34 kg m2 s-1

= 1% of 3.0 × 108 = 3.0 × 106 ms-1

= equation) Brogliede(m

h

1631

-1234

ms100.3kg101.9

s m kg1062.6

= 2.42 × 10-10 m

= 242 pm

55. Two particles A and B are in motion. If the wavelength associated with particle A is 5.0 × 10-8 m,

calculate the wavelength associated with particle B if its momentum is half of A.

Sol. AB p2

1p

A = 5.0 × 10-8 m From de-Broglie equation, we can write

AB

Bp

h2

p

h

AAB 2h

h2

8B 100.52

B = 1.0 × 10–7 m



56. An electron is moving with a kinetic energy of 4.55 × 10-25 J. Calculate the wavelength of this moving electron. Given: me = 9.1 × 10-28g and h = 6.62 × 10-34 Js

Sol. K.E. = 2

m 2

em

2KE

= 631

2510

101.9

1055.42

= 103 ms–1 From de-Broglie equation

331

34

10101.9

1062.6

m

h

= 7.27 × 10–7 m

= 7.27 × 10-7 m = 727 nm 57. What are the drawbacks of the Bohr’s model? Sol. Drawbacks of Bohr’s model include the following: (i) The model could not explain the spectra of atom containing more than one electron.

(ii) It was observed that in the presence of a magnetic field, each spectral line gets split into closely spaced lines. This phenomenon, known as Zeeman effect, could not be explained by Bohr’s model.

(iii) de-Broglie suggested that electron has dual character. It has particle as well as wave nature. Bohr had treated electron only as a particle.

(iv) According to Heisenberg’s uncertainty principle, it is impossible to determine simultaneously the exact position and the momentum of a small moving particle like an electron. The postulate of Bohr that electrons move in well defined orbits around the nucleus is not valid.

(v) If there are only just a few stationary states in an atom, the spectra of elements should consist of

very few lines. But a detailed examination of spectra of elements containing many electrons,

reveals the presence of series of lines.

58. What is meant by dual nature of light? Calculate the wavelength (in angstroms) associated with an electron traveling at a speed of 2.19 × 106 m/s.

Sol. Light was shown by Einstein to consist of particles (photons), each of energy E = h, where h is Planck’s constant.

But light is electromagnetic radiation characterized by its wavelength and frequency , de-Broglie proposed that there are wave properties associated with particles. For a particle of mass, m and

speed , the wavelength is related to momentum (m) by the de-Broglie relation = m

h. The wave

character of electrons is confirmed by the diffraction patterns.

The wave properties of a particle are described by a wave function , from which we can get the probability of finding the particle in different regions of space.

The particle nature of light was explained by Einstein in the photoelectric effect.

m

h

1-631

1-234

s m102.19 kg1011.9

sm kg1062.6

= 3.318 × 10-10 m = 3.32 Å

59. What are the possible values of m for an electron with = 3?

Sol. The value of m can have values from - to + .

In this case, m can have values – 3, – 2, –1, 0, +1, +2, +3 60. How many nodal spheres does a 5s electron charge cloud have?

Sol. The number of nodal spheres for an ns-orbital is n – 1. The 5s electron charge cloud will have four

nodal spheres.

61. What is the probability of finding a 4d electron right at the nucleus?

Sol. There is no probability of finding a d electron right at the nucleus.



62. State Hund’s rule of maximum multiplicity.

Sol. Hund’s rule states that the lowest energy arrangement of electrons in a subshell is obtained by

putting electrons into separate orbitals of the subshell with parallel spin before pairing electron.

It also states that “in degenerate atomic orbitals, no pairing of electrons start until and unless each one is singly filled.”

63. Explain the meaning of and 2.

Sol. is a wave function which represents the amplitude of the electron wave, is obtained as a solution

to Schrodinger wave equation. However, the square of the wave function, 2 gives the probability of finding the electron in the region in space around the nucleus.

64. What is the orbital angular momentum of an electron in 2s orbitals?

Sol. Orbital angular momentum =

2

h)1(

For 2s electron, = 0

Therefore, orbital angular moment is =

2

h)10(0

= 0 (zero) 65. Write the four quantum numbers for the 19th electron in chromium atom (Z = 24). Sol. The electronic configuration of Cr (Z = 24) is [Ar] 3d5 4s1. The 19th electron is present in 4s atomic

orbital and the four quantum number are

n = 4, = 0, m = 0, ms = + 2

1

66. Draw the shapes (boundary surfaces) of the following orbitals: (i) 2py (ii) 3dz2 (iii) 3dx2 – y2

Sol.

PERIODIC PROPERTIES AND CHEMICAL BONDING

67. What is meant by Periodic classification of elements?

Sol. The periodic classification of elements means an arrangement in which similar atoms are grouped

together while dissimilar are separated from one another.

68. State the Modern Periodic Law.

Sol. The modern periodic law states that “when the elements are arranged in order of increasing atomic

number, then their physical and chemical properties are periodic function of their atomic number”.

69. What would be the IUPAC name and symbol for the element with atomic number 120? Sol. The roots for 1, 2 and 0 are un, bi and nil respectively. Hence the name is unbinilium and the symbol

is Ubn. 70. Write down the general outer electronic configuration of

(i) p-block elements (ii) d-block elements and (iii) f-block elements

Sol. (i) p-block elements: ns2 np1–6 (ii) d-block elements: (n – 1)d1-10ns1-2 (iii) f-block elements: (n – 2)f1 – 14 (n – 1)d0 – 1ns2 71. Arrange the following elements in the increasing order of metallic character. B, Al, Mg, K Sol. Metallic character increases down the group and decreases along the period. The correct order is B < Al < Mg < K

y

z

x

2py

y x

z

22 yxd3

z

y

2zd3



72. Arrange the following elements in the increasing order of non-metallic character. C, Si, N, F Sol. The correct order is Si < C < N < F 73. Which of the following species will have the largest and the smallest size?

Mg, Mg2+, Al, Al3+

Sol. Largest size: Mg Smallest size : Al3+ 74. Arrange the following ions in order of increasing size:

Be2+, Cl–1, S2–, Na+, Mg2+

Sol. The correct order of increasing size is Be2+ < Mg2+ < Na2+ < Cl– < S2–

75. Which of the following species has a largest radius? Mg, Na, Na+, Mg2+, Al?

Sol. Na. The size decreases as one goes to the right in the Periodic Table, also decreases when the charge on the ion increases.

76. Arrange the following in order of decreasing size. Li +, Na+, K+ and Mg2+.

Sol. The correct order is K+ > Na+ > Li+ > Mg2+ Li+ has the smallest size because of its highest effective nuclear charge. 77. Out of Na and Mg, which has a higher second ionization enthalpy and why? Sol. Na has a higher second ionization enthalpy because the electron is to be removed from the Na+

which has a stable noble gas configuration of Ne. 78. State which member of each of the following pairs has the greater electron gain enthalpy.

(i) Cl, S (ii) O, S (iii) Si, P (iv) N, C

Sol. (i) Cl (ii) S (iii) Si (iv) C

79. What are isoelctronic species? The following species are isoelectronic with the noble gas. Arrange

these in order of increasing size: K+, S2–, Cl–, Ca2+

Sol. The species having the same number of electrons are called isoelectronic species. Since, all these

ion have equal number of electrons, greater the nuclear charge, smaller the size, the correct order is

Ca2+ < K+ < Cl– < S2–

80. Why are the electron gain enthalpy of Be and Mg positive?

Sol. The ns subshell is completely filled in Be and Mg. The gain of electron is strongly endothermic. Hence, these elements have positive electron gain enthalpy.

81. Which of the following will have the most negative electron gain enthalpy and which the least negative?

P, S, Cl, F Explain your answer. Sol. Electron gain enthalpy generally becomes more negative across a period as we move from left to

right. Within a group, electron gain enthalpy becomes less negative down a group. However, addition of an electron to the 2p orbital leads to greater repulsion than adding an electron to the larger 3p orbital. Hence, the element with most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorus

82. Among the elements B, Al, C and Si (i) which has the highest first ionization enthalpy? (ii) which has the most negative electron gain enthalpy? (iii) which has the longest atomic radius? (iv) which has the most metallic character? (v) Explain diagonal relationship with an example of group 1 elements

Sol. (i) The highest first ionization enthalpy Carbon

(ii) Most negative electron gain enthalpy Carbon

(iii) Largest atomic radius Aluminium

(iv) Most metallic element Aluminium 83. Define electrovalency. Sol. The electrovalency is defined as the number of electron lost or gained by an atom. It is equal to the

number of unit charges on the ion.



84. Give two examples of compounds in which the central atom has incomplete octet.

Sol. BeH2 H Be H

(Be has only 4 electrons)

BCl3 Cl

Cl

BCl

(B has only 6 electrons)

85. Write the Lewis structure of sodium sulphate, Na2SO4. Sol. The total number of valence electron is 2(Na) + S + 4 (O) = 1 × 2 + 6 + 4 × 6 = 32 So, 16 electron pairs must be accommodated. The Lewis structure of sodium sulphate, Na2SO4 is

2

O

O

O

S ONa2

The (2–) charge belongs to the entire anion and is not localized on one atom. 86. Define lattice enthalpy and how is it related to the stability of the ionic compound. Sol. The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one

mole of solid ionic compound into gaseous constituent ions. 87. How are bond order and bond length related to each other? Sol. Bond order and the bond length are inversely related.

Bond order length Bond

1

88. What do you understand by the term resonance hybrid? Sol. A resonance hybrid is the composite structure that results from resonance.

The energy of the O3 resonance hybrid (III) is lower than either of the two canonical forms (I) and (II). Resonance hybrid is more close to actual structure.

89. Write the resonance structure for the nitrite ion, 2

NO .

Sol. The resonance structures for the nitrite ion are

]ONO[]ONO[

90. (i) Arrange the following elements in order of increasing electronegativity: N, Al, C, B

(ii) Arrange the following bonds in order of increasing bond polarity: P – H, H – O, C – Cl Sol. (i) The increasing order of electronegativity is Al < B < C < N (ii) The increasing order of bond polarity is P – H < C – Cl < H – O 91. Select the element that has the greatest electronegativity in each set

(i) Cl, P or S (ii) F, O or S (iii) Br, Cl or O Sol. (i) Chlorine (ii) Fluorine (iii) Oxygen 92. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. Sol. The correct order of increasing ionic character is N2 < ClF3 < SO2 < LiF < K2O

93. Write formal charges of the atoms in the carbonate ion, 23

CO .

Sol.

O

C

O

O (a)

(c) (b)

2-

O

O

O O

O

O

O

O

O

(I) (II) (III)



Formal charge on O atom (a)

= 6 – 4 - 2

1 × 4 = 0

Formal charge on O atom (b) and (c) each is

= 6 – 6 - 2

1 × 2 = - 1

Formal charge on C atom is 4 – 0 - 2

1 × 8 = 0

94. Out of the following four resonance structures for the CO2 molecule, which are important for describing the bonding in the molecule and why?

)i(

OCO

)II(

OCO

)III(

OCO

)IV(

2

OCO

Sol. Structures (I), (II) and (III) are all important in describing the bonding in CO2 as each and every atom has a complete octet. While in structure (IV), C has incomplete octet and can not contribute towards the resonance hybrid.

95. ClF3 is a T-shape molecule while NF3 has a trigonal pyramidal shape. Explain. Sol. In ClF3, Cl, the central atom has five electron pair (three are bond pair and two are lone pair). It is a

AX3L2 type molecule. NF3 is AX3L, type molecule and has a trigonal pyramidal shape. 96. Describe the hybrid orbitals used by each carbon atom in the following molecules:

(i) HO

H

H

CCHCC

H

H

|

|2

\

/ (ii) N C – C N

Sol. (i) (ii)

97. How many sigma () and pi () bonds are present in each of the following molecules? (i) CaC2 (ii) C3O2 (iii) Toluene (iv) 4-Methylphenol

Sol. (i) 1, 2 (ii) 2, 2 (iii) 15, 3 (iv) 16, 3 98. Which out of the two molecules OCS and CS2 has a higher dipole moment and why? Sol. OCS; as dipole are not cancelled

||

SCO

99. Predict the dipole moment of a molecule of the type (i) AX4 having a square planar geometry. (ii) AX5 having a square pyramidal shape (iii) AX6 having an octahedral geometry.

Sol. (i) zero (ii) non-zero (iii) zero due to regular structure

HO

H

H

CCHCC

H

H

|

|2

\

/

sp2 sp

3

N C – C N

sp

H – C – H

H

Toluene

O – H

H

H – C – H

4-Methylphenol



100. Explain the formation and difference between a sigma () bond and a pi() bond.

Sol.

-bond -bond

(a) -bond is formed by the axial overlapping of

half-filled atomic orbitals.

(a) -bond is formed by the lateral or side-

ways overlapping of atomic orbitals

(b) It is formed when two s-orbitals overlap or

when one orbital with directional character (p-

orbital) overlaps along its axis.

(b) It is formed by the overlapping of two

parallel p orbitals which are available

after -bonds have been formed

(c) A -bond has a cylindrical shape about the

bond axis

(c) A -bond has an electron distribution

above and below the bond axis.

(d) A bond is a strong bond (d) A -bond is a weak bond

101. Why does He2 molecule not exist?

Sol. Helium atom has two electrons in its 1s orbitals. In this case the repulsive forces dominant over the

attractive forces during the approach of two He atoms. As a result, the energy of the system

increases which leads to instability. Since the energy of the individual He atoms is smaller than that of

the system when they are close to each other, they stay separate and do not form He2 molecule. The

MO configuration of He2 is 1s2 *1s2. As Nb = Na, bond order is equal to zero. Hence, He2 molecule

does not exist. According to MOT in He2 molecule Number of bonding & antibonding electron are

same so its bond order is zero.

102. Which of the following has a greater dipole moment and why? NH3 or NF3

Sol. NH3 and NF3 molecules both have trigonal pyramidal geometry. In both molecules, N is the central

atom with a lone pair of electrons. But in NH3, the bond dipoles (moments) are pointing towards N-

atom while in NF3, bond moment points away from the N-atom. Therefore, the dipole moment of NH3

is greater than that of NF3

103. The dipole moment of hydrogen halides decreases from HF to HI. Explain this trend.

Sol. In H – X, X atom is more electronegative than H. Therefore, the H – X bond is polar and the direction

of dipole is from H to X. As the electronegativity of X decreases from F to I, the charge separation

decreases. Therefore, the dipole moment decreases.

104. Calculate the ionic character of HCl. Its measured dipole moment is 3.436 × 10-30 Cm. The HCl bond

length is 1.29 Å.

Sol. Dipole moment corresponding to 100% ionic character of HCl

= 1.602 × 10-19 C × 1.29 × 10-10 m

(1Å = 10-10m)

= 20.67 × 10-30 Cm

Actual dipole moment of HCl

= 3.436 × 10-30 Cm

Percentage ionic character

= %171001067.20

Cm10436.3Cm 30

30

N

H

H

H

= 4.9 × 10-30

cm

N

F F

F

= 0.80 × 10-30

cm



105. Considering x-axis as the internuclear axis which out of the following form a sigma bond: (i) 1s and 1s (ii) 1s and 2px (iii) 2py and 2py (iv) 2px and 2py (v) 1s and 2s

Sol. (i) 1s and 1s (ii) 1s and 2px

(iii) 2py – 2py (-bond) (iv) 2px – 2py zero overlap (v) 1s and 2s (-bond)

106. Explain hybridization in ethyne. Why two type of bond length exist in PCl5.

Sol. Ethyne is sp-hybridised. The two unhybridised p-orbitals form -bond.

Two types of bond length exist in PCl5 are due to equatorial & axial bonds. The axial bonds are longer than equatorial bonds to minimise the repulsion.

107. Why does antibonding molecular orbital have higher energy than either of the atomic orbitals? Sol. In antibonding MO, the attraction between the electrons and the nuclei is less than the mutual

repulsion of electrons and this results in net increase in energy. 108. Write the correct order of Boiling point.

i) HF > HI > HBr > H – Cl ii) (H2O > H2Te > H2Se > H2S

iii) H2O > H – F (Heat of vapourisation) iv) ether

33alcohol

23 CHOCH OHCHCH

v) H2SO4 > HNO3 > HCl Sol. i) Because of H-bonding. Highest molecular mass

iii) When we boiling H – F it forms H2F2 & still H-bonding is there as size is small. But H – O bond is bigger & until its all bonds breaks it does not go in gaseous state.

iv) In alcohol, intermolecular H-bonding occurs through OH & molecules get associated so its boiling point is high.

v)

+

1s 1s

+ x

sigma bond

- + + x

1s 1s 2px sigma bond

1s 2s

x

sigma bond

1s 2s 2py 2py

pi bond

2py2py

HO

HO

S

O

O

HO – N

O

O-

(No H-Bond)

H-Bonding along with OH groups

H-Bonding along with one OH groups

H

1s

2pz

sp

2py

2py

2pz

sp H

BOND

- BOND

sp sp

1s

Excited State of C

sp

Two sp hybrid orbitals

Two Unhybridized

p-orbital for bond

2s 2p

Cl

Cl

P

Cl

Cl

Cl



109. o-Nitrophenol is less soluble in water than p-nitrophenol. Explain. Sol. 110. KHF2 exists KHCl2 does not. Explain.

Sol. 2HFK :

salt Fermi

]FH.....F[K

In the above structure molecule is H-bonded to F– ion. KHCl2 – does not exist because Cl shows no H-bonding. 111. Dipole moment of BF3 is zero but NF3 has net dipole moment.

Sol. BF3 is planar but NF3 is pyramidal due to lone pair so dipole will not be cancelled

112. Which of the following is less stable and why? 2

H or 2

H

Sol. Both 2

H and 2

H have a bond order of 2

1but

2H has one electron in antibonding *1s orbital.

Therefore H 2

is less stable than 2

H

113. Explain why the bond order of N2 is greater than N2+, but the bond order of O2 is less than that of O

2.

Sol. N2 N 2

+ e–; O2 O 2

+ e–

In N2, removal of one electron takes place from lower energy bonding MO. This results in the decrease in stability and the bond order.

In N2, the bond order is 3 but in 2

N , the bond order is 2.5.

In O2, removal of one electron takes place from higher energy antibonding MO. This results in the increase in stability and the bond order.

In O2, the bond order is 2 but in O 2

, the bond order is 2.5

114. N2O is isoelectronic with CO2. Draw its two resonating structures, that satisfy the octet rule.

Sol.

ONN ONN

O H O

N O

OH

N = O

p-Nitrophenol O

o-Nitrophenol

due to intramolecular H-bonding –OH group is not available to form H-Bond with water



STATES OF MATTER 115. Which of the following two graphs will be a straight line at constant temperature? P versus V or P

versus V

1(at constant T)

Sol. P versus V

1 graph will be a straight line.

116. What is P – V isotherm? Sol. The graph plotted between P and V for a gas at constant temperature is known as P – V isotherm. 117. Why is it not possible to cool a gas at 0 K? Sol. It is not possible to cool gas at 0K because all gases condense to liquids or solids before this

temperature is reached. 118. What is the increase in volume, when the temperature of 600 ml of air increases from 27°C to 47°C

under constant pressure? Sol. Charles law is applicable as the pressure and amount remains constant.

2

121

2

2

1

1

T

TV Vor

T

V

T

V

ml 640 ml 600K 300

K 320V1

The increase in volume of air = 640 ml – 600 ml = 40 ml 119. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C? Sol. As temperature and mass of gas constant, use Boyle’s law P1V1 = P2V2

P2 = P1

3

3

2

1

dm 200

dm 500bar 1

V

V = 2.5 bar

120. Calculate the number of nitrogen molecules present in 2.8 g of nitrogen gas?

Sol. Number of moles of nitrogen = mol 1.0mol g28

8.21-

Number of nitrogen molecules = 0.1 mol × 6.02 × 1023 mol-1 = 6.02 × 1022 121. What is the density of oxygen. O2 in grams per litre at 25°C and 0.850 atm?

Sol. RT

PMd

K 298mol atm L 0821.0

mol g 0.32atm 850.0d

1-

-1

= 1.11 g L–1

122. A sample of N2 gas occupies a volume of 1.0 L at a pressure of 0.5 atm at 40°C. Calculate the pressure if the gas is compressed to 0.225 cm3 at 6°C.

Sol. V1 = 1.0 L = 103 cm3 V2 = 0.225 cm3 P1 = 0.5 atm T1 = 40 + 273 = 313 K T2 = - 6 + 273 = 267 K

2

22

1

11

T

VP

T

VP

1

2

2

112

T

T

V

VPP

= K313cm225.0

cm10267Katm 5.03

33

P2 = 1896 atm

P

V

1



123. The drain cleaner, Drainex contains small bits of aluminium which reacts with caustic soda to produce

hydrogen. What volume of hydrogen at 20°C and one bar will be released when 0.15 g of aluminium

reacts? (R = 0.083 L bar mol-1 K-1)

Sol. 2Al + 2NaOH + 2H2O 2Na AlO2 + 3H2

2 mol Al = 3 mol H2

2

3

mol 27g

0.15g released n

1-H2

= 8.33 × 10-3 mol

P

nV

RTHH

2

2 =

bar 1

K 293K mol bar 0.083Lmol 1033.8 -1-13

= 0.203 L

203V2H ml

124. Chloroform, CHCl3 is a volatile (easily vapourised) liquid solvent. Calculate the density of chloroform

vapour at 99°C and 745 mmHg. Give the answer in gram per litre.

Sol. M = 119.5 g mol-1

T = 99 + 273 = 372 K

P = atm 98.0760

745

d = RT

PM =

K 372K mol atm L 0821.0

mol 119.5g atm 98.01-1-

-1

d = 3.83 g L–1

125. What will be the pressure exerted by a mixture of 3.2 g of a methane and 4.4 g of carbondioxide

contained in a 9 dm3 flask at 27°C?

Sol. mol 2.0mol g 16

g2.3n

1-CH4

mol 1.0mol 44g

g4.4n

1-CO2

RTV

np 4

4

CHCH

RTV

2.0p and RT

V

1.0p

42 CHCO

V

RT1.0

V

RT2.0ppp

24 COCHmixture

9

3000821.03.0pmixture

atm 821.0pmixture

126. What is the effect of temperature on density of a liquid?

Sol. Increase in temperature of a liquid increases its volume marginally because of increase in its average

kinetic energy and therefore density decreases. Decrease in temperature of a liquid decreases the

volume and therefore increases its density.

127. Explain why 15 g of steam at 100°C melts more ice than 15 g of liquid water at 100°C?

Sol. More heat is required for vapourisation than for melting. Melting needs only enough energy for the

molecules to escape from their sites in the solid. For vapourisation, enough energy must be supplied

to break most of the intermolecular attractions. That’s why 15 g of steam melts more ice than 15 g of

water at 100°C.



128. Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a

balloon to a certain volume at 100°C if 3.5 g He is required to inflate the balloon to half the volume at

25°C.

Sol. Volume of 3.5 g He at 25°C and pressure P is

P 4

R 2985.3

P

RT

4

5.3

P

nRTV

(1)

To fill the balloon with H2 to double this volume means,

V2V2H and T = 373 K. Hence,

P

373R

2

w

P

nRTV2

(2)

Dividing (2) by (1) gives

2985.3

4

2

373w2

or w = 2.796 g

129. The pressure exerted by 12 g of an ideal gas at temperature t°C in a vessel of volume V L is one

atom. When the temperature is increased by 10 degrees at the same volume, the pressure increases

by 10%, calculate the temperature t and volume V (molecular weight of the gas = 120).

Sol. 1st Case 1 × V = )t273(R120

12 (1)

2nd Case 1.1 × V = )t283(R120

12 (2)

Dividing (1) by (2)

gives)t283(

)t273(

1.1

1

t = -173°C Substituting t in eq. (1) gives V = 0.82 L 130. The compressibility factor for one mole of a Vander Waal gas at 0°C and 100 atm pressure is found

to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the Vander Waals constant a.

Sol. 273082.01

V1005.0

nRT

PVZ

V = 0.1119 L

Vander Waals equation for 1 mol of a gas RT)bV( V

aP

2

b << V

RT (V) V

aP

2

2VPV

RTa

V

RT

V

aP

2

PV

RT

V

a2

a = RTV-PV2

VRT

PV1V

RT

PVV

RT

a

a = (1 – 0.5) × 0.1119 × 0.082 × 273

= 1.252 atm L2 mol–2

131. Draw diagram for liquification of Andrew’s isotherm of CO2 at various temperatures.



132. What would be the S.I units for the quantity PV2T2/n?

Sol. 1-242232

mol Km Nmol

)K()m)(Nm(

133. Pay load is defined as the difference between the mass of displaced air and the mass of balloon. Calculate the pay load when a balloon of radius 10m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 & R = 0.083 bar dm3 K-1 mole-1)

Sol. Radius of the balloon = 10 m

Volume of the balloon = 333 cm 5.4190m) 10(7

22

3

4r

3

4

Volume of He filled at 1.66 bar and 27°C = 41905 m3 Calculation of mass of He

PV =nRT = RTM

w

or K) (300 )mol K dm bar 083.0(

)dm 10(4190.5 bar) (1.66 )mol kg 104(

RT

MPVw

1-1-3

33-13

= 1117.5 kg

Total mass of the balloon alongwith He = 100 + 1117.5 = 1217.5 kg Maximum mass of the air that can be displaced by balloon to go up = Volume × Density = 4190.5 m3 × 1.2 kg m–3 = 5028.6 kg

Pay load = 5028.6 – 1217.5 kg = 3811.1 kg

134. What would be the pressure exerted by a mixture of 3.2gm of methane 24.49 gm of carbon dioxide placed in a 9 dm3 flask at 27°C?

Sol. mol 2.0mol g16

g2.3n

1-CH4

mol 1.0mol g 44

g 4.4n

1-CO2

RTV

np 4

4

CHCH

RTV

1.0p

2CO and RTV

2.0p

4CH

V

RT1.0

V

RT 2.0ppp

24 COCHmixture

9

3000821.03.0pmixture

Pmixture = 0.821 atm 135. What is Vander Waal’s equation for Real gas & give units for constant ‘a’ and ‘b’? Sol. The van der Waals equation for n mole of gas is

nRTnb)-(V V

anP

2

2

Where a and b are van der Waals constant and their value depend upon the nature of the gas. They are characteristics of the gas. Unit a = atm L2 mol-1 and b = L mol–1



EQUILIBRIUM

136. Explain Lechatalier Principle. 137. Explain what do you understand by saturated solution? Sol. A solution in which no more solute is dissolved is a called saturated solution 138. In what type of system could an equilibrium be attained between a liquid and its vapour? Sol. Closed system

139. For the reaction: (g)(g)

2)g(2 2HI I H

Kc = 54.8 at 700K. Calculate the value of Kp? Sol. ng = 0 so Kp = Kc = 54.8

140. (g)

3(g)

2)g(2 2NH 3H N H° = -92.3 kJ/mol.

Predict whether the increase or decrease in temperature favour the high yield of ammonia? Sol. As the reaction is exothermic, so decrease in temperature shift the equilibrium to right and low

temperature is favourable for high yield of ammonia. 141. What is the effect of temperature change on a reaction? Sol. Temperature effects the reaction in two ways: (a) increases rate of reaction

(b) effect equilibrium constant 142. What is the effect of an increase in pressure on the equilibrium composition of the reaction

(a) (g)

2)g(

42 2NO ON (b) (g)(g)

2)g(2 2HI I H

Sol. (a) Reverse direction (b) no effect Increase in pressure will shift equilibrium to the side having lesser number of moles. 143. Write the conjugate acid and conjugate base of H2O Sol. Conjugate acid: H3O

+ Conjugate base: OH–

144. What is the pH of the solution whose [OH–] = 6.2 × 10-9 M? Sol. [OH–] = 6.2 × 10–9

[H3O+] =

9

14

102.6

10

= 1.61 × 10–6

pH = - log (1.61 × 10–6) = 5.79 145. What is the pH of a neutral solution at 37°C, when Kw = 2.5 × 10-14 Sol. For the neutral solution, [H3O

+] = [OH–] Kw = [H3O

+]2 = 2.5 × 10–14 [H3O

+] = 1.58 × 10-7 pH = -log (1.58 × 10-7) = 6.80 146. The acetic acid has a Ka = 1.74 × 10–5.

Calculate the H3O+ concentration in 0.01 M acetic acid solution?

Sol. CH3COOH + H2O CH3COO– + H3O+

[H3O+] = C = C CK

C

Ka

a

[H3O+] = 01.01074.1 5

= 4.17 × 10–4 M 147. Calculate the pH of 0.01 M acetic acid solution. Given Ka (CH3COOH) = 1.8 × 10–5 M Sol. CH3COOH H+ + CH3COO– Initial 0.01 0 0 0.01 – 0.01 0.01 0.01

01.0

]01.0[K

2

a

= 1.8 × 10–5 H+ = C 1 – 1

2 = 1.8 × 10-3 or H+ = CKa

= 4.24 × 10–2 [H+] = 0.01 = 0.01 × 4.24 × 10–2 = 4.24 × 10–4 pH = -log [H+] = -log [4.24 × 10–4] pH = 4 – log 4.24 = 3.37



148. Calculate the degree of hydrolysis of 0.01 M sodium acetate solution.[Ka = 1.8 × 10–5]

Sol. CH3COONa + H2O CH3COOH + NaOH

h1

Ch

K

KK

2

a

wh

; h = degree of hydrolysis

and CK

Kh

a

w

=

01.0108.1

101

5

14

= 2.36 × 10–4

149. Calculate the pH of a buffer solution containing 20 ml of 0.1 M CH3COOH and 20 ml of 0.1M

CH3COONa. [pKa (CH3COOH) = 4.74]

Sol. pH = pKa + log ]acid[

]salt[ ; [salt] = [acid]

= pH = pKa

150. Calculate the Kh and degree of hydrolysis of the 0.1 M NH4CN solution at 25°C

Given: Ka(HCN) = 4.8 × 10-10 and

Kb(NH4OH) = 1.8 × 10–5

Sol. NH4CN is a salt of weak acid and weak base

Kh = 510

14

ba

w

108.1108.4

101

KK

K

= 1.16

and 16.1KKK

K

h1

hh

ba

w

= 1.08

h = 1.08 – 1.08h h = 08.2

08.1 = 0.52

151. A 0.02 M solution of pyridine hydrogen chloride has pH = 3.44. Calculate the ionization constant of

pyridine?

Sol. Pyridinium hydrochloride is a salt of strong acid & weak base

pH = 7 – 2

1(log C + pKb)

Kb = dissociation constant of pyridine

3.44 = 7 – 2

1(log 0.02 + pKb)

pKb = 8.82

Kb = antilog (-8.82) = 1.513 × 10–9

152. Calculate the pH of a solution obtained by dissolving 4.92 g of sodium acetate in water to a volume of

400 m.

Ka(CH3COOH) = 1.8 × 10–5

Sol. [Ac–] = 400

1000

82

92.4 = 0.15 mol L–1

Ka = 1.8 × 10–5 pKa = 4.74

CH3COONa is a salt of strong base and weak acid

pH = 7 + 2

1[pka + log C]



= ]15.0log74.4[2

17 = 8.96

153. The pka of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the hydrolysis constant of ammonium acetate at 298 K and also degree of hydrolysis and pH of its (a) 0.01 M (b) 0.04 M solution

Sol. Ka = 1.74 × 10–5, Kb = 1.78 × 10–5

55

14

ba

wh

1078.11074.1

101

KK

KK

= 3.23 × 10–5

h = 5h 1023.3K = 5.68 × 10–3

pH of a solution of salt of weak acid and weak base is

pH = 7 + 2

1(pka – pkb)

= 7 + 2

1(4.76 – 4.75)

= 7 + 2

1 × 0.01 = 7 + 0.005

= 7.005

154. (a) Write the conjugate bases for the bronsted acids: HF, H2SO4 and HCO 3

Sol. Conjugate base will be F–, 4

HSO , 23

CO

(b) Conjugate acid for the following bronsted bases: 2

NH , NH3 and HCOO

Sol. NH3, 4

NH , HCOOH

155. Assume complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002M HBr (d) 0.002M KOH

Sol. HCl + aq H+ + Cl–

[H+] = [HCl] = 3 × 10–3

pH = - log [3 × 10–3] = 2.52

(b) NaOH + aq Na+ + OH–

[OH–] = 5 × 10–3 M or pOH = -log(5 × 10–3)

[H+] = M 102105

10 123

14

= - 0.6990 + 3 = 2.3

pH = -log[2× 10-12] = 11.70 pH = 14 – 2.3 = 11.7

(c) HBr BrHaq

[H+] = 2 × 10–3

pH = -log [2 × 10–3] = 2.70

(d) KOH aq

K+ + OH–

[OH–] = 2 ×10–3

[H+] = 123

14105

102

10

pH = -log[5× 10–12] = 11.30

156. If 0.561gm KOH is dissolved in water to give 200 mL of solution at 298K, calculate the concentration

of potassium, hydroxide ion & hydrogen ions. What is its pH?

Sol. KOH = 200

1000

56

561.0 = 0.050 M

As KOH K+ + OH– [K+] = [OH–] = 0.05 M

[H+] = 05.0

10

]OH[

K 14w

= 2.0 × 10–13



pH = -log [H+] = -log[2 × 10–13] = 13 – 0.3010 = 12.699 157. The solubility of Sr(OH)2 at 298K is 19.23 g/L of solution. Calculate the concentration of Sr+2 and OH–

ions and the pH of the solution [Atomic mass of Sr = 87.6] Sol. Molar mass of Sr(OH)2 = 87.6 + 34 = 121.6g mol–1

Solubility of Sr(OH)2 = 1581.06.121

23.19 mole/lit.

Assume complete dissociation,

Sr(OH)2 Sr+2 + 2OH–

[Sr+2] = 0.1581, [OH–] = 2 × 0.1581 = 0.3162 pOH = -log (0.3162) = 0.5 pH = 14 – 0.5 = 13.5

158. Predict the acidic, basic or neutral nature of the solution of following salt: NaCl, KBr, NaCl, NH4NO3, NaNO2, KF

Sol. NaCN, NaNO2, KF solution are basic, as they are salts of strong base, weak acid. NaCl, KBr solutions are neutral, as they are the salt of strong acid, strong base NH4NO3 solution is acidic.

THERMODYNAMICS

159. What are state functions? Give example. Sol. State functions are thermodynamic function which are independent of how the change is

accomplished. They depend only up on initial & final state of system. eg. Internal energy and enthalpy.

160. How can we change the internal energy of a system? Sol. The internal energy of a system can be changed into ways: (a) Either by allowing heat to flow into system or out of the system

(b) Work done on the system or by the system (c) By adding matter to the system 161. Is internal energy a state function? What is the change in internal energy for a cyclic process? Sol. Yes, the internal energy is a state function as it depends upon initial and final state of the system i.e.

H = u2 – u2

For a cyclic process, change in internal energy u is zero because initial and final state are same 162. What is the change in internal energy of the system enclosed in a bomb calorimeter? Sol. In bomb calorimeter, the volume is constant so the amount of heat exchanged between system and

surrounding is equal to the change in the internal energy.

qv = m × s × T 163. In a certain process, 500 J of work is done on a system which gives off 200 J of heat. What is the

change in internal energy for the process? Sol. Work is done on a system, so W = 500 J Heat is released by the system q = - 200J

u = q + w = - 200 + 500 = 300 J 164. State the first law of thermodynamics for

(a) adiabatic process (b) a system undergoes a change in which internal energy remains constant

165. Define heat capacity? Is it an extensive property Sol. The heat capacity (C) of a substance is the amount of heat required to raise the temperature by one

degree Celsius. Yes, heat capacity is an extensive property as it is directly proportional to the amount of a system.

166. Drive the relationship: H = E + ngRT 167. The enthalpy change (H) for the reaction:

)g(3

(g)2

)g(2 NH23H N

is -92.38 kJ at 298K. What is E at 298K? Sol. H = E + ng RT



E = H - ng RT E = -92.38 – (-2) × 8..314 × 10–3 × 298 = - 92.38 + 4.96 = -87.42 KJ 168. What do you understand by standard state of a substance?

Sol. The standard state of a substance is the state at 1 bar pressure and at specified temperature.

169. What would be the heat released when

(a) 0.25 mol of hydrochloric acid is neutralized by 0.25 mol of NaOH

(b) 0.02 mol of sulphuric acid solution is mixed with 0.2 mol of potassium hydroxide solution

Sol. (a) 0.25 mole of HCl + 0.25 mol of NaOH

The net equation is:

H+ (0.25 moles) + OH– (0.25 moles)

H2O (0.25 moles)

Therefore heat released = 57.1 × 0.25 kJ

= 14.3 kJ

(b) 1 mol of H2SO4 contains 2 mole of H+

0.02 mole of H2SO4 contain 0.04 moles of H+ ions

The net equation is

H+ (0.04 moles) + OH– (0.2 moles) H2O (0.04 moles)

Therefore heat released would be = 0.04 × 57.1 = 2.284 kJ

170. Calculate H° for the reaction

3C(gr) + 4H2(g) C3H8(g)

Given:

(a) )e(

2)g(

2)g(2

)g(83 OH4CO3O5HC H = -2220 kJ

(b) ;COOC)g(2

)g(2

)gr( H = - 394 kJ

(c) )e(

2)g(2

)g(2 OHO

2

1H ; H° = - 286 kJ

Sol. The target equation is

)g(83

)g(2

)gr(HCH4C3

Step I: We need 3 moles of C(gr)_ on LHS so multiplying equation (ii) by 3, we get

)g(

22)gr(

CO3O3C3 …(1)

H° = 3 × (-394) = -1182 kJ

Step II: Multiplying the equation (iii) by 4

)e(

2)g(2

)g(2 OH4O2H4

H° = 4 × (-286) = - 1144 kJ …(2)

Step III: Reverse equation (i)

)g(2

)g(83

(e)2

)g(2 O5HCO4H CO3

H° = 2220 kJ …(3)

Now Adding (1), (2) and (3)

)g(83

)g(2

)gr(HCH4C3



H° = -1182 – 1144 + 2220

= - 106 kJ

171. Calculate the heat of combustion of glucose from the following data:

(a) )g(2

)g(2

)gr(COOC ; H = - 394 kJ

(b) )g(

2

)g(

2)g(2 OHO

2

1H ; H = - 286 kJ

(c) 6126)g(2

)g(2

)grap(OHCO3H6C6 ; H° = -1170 kJ

Sol. Our aim is to calculate the H° of glucose for the reaction

)g(

2)g(

2)g(2

)s(6126 OH6CO6O6OHC H° = ?

Given:

)g(2

)g(2

)g(COOC ; H° = -394 kJ …(1)

)e(

2

)g(

2)g(2 OHO

2

1H H° = -286 kJ …(2)

6126)g(2

(g)2

)g(OHCO36H C6 ; H° = -1170 kJ …(3)

(1) Multiply (a) and equation (b) by 6 (2) Reverse equation (c)

(3) Add (1), (2) and (3)

H° = -2910 kJ 172. Calculate the bond energy of H – Cl

Given:

BE (H – H) = 436 kJ mole–1

BE (Cl – Cl) = 242 kJ mol–1 )HCl(H0

f = -91 kJ mol–1

Sol. )g(

(g)

2

)g(

2 HClCl2

1 H

2

1 kJ/mole 91)HCl(HH 0

f0f

- 91 = )ClH(BE)ClCl(BE2

1)HH(BE

2

1

-91 = )ClH(BE2422

1436

2

1

BE(H – Cl) = 218 + 121 + 91 = 430.0 kJ/mole

173. The following reaction is non-spontaneous at 25°C

)g(

2)s()s(

2 O2

1 Cu2OCu ; G° = 141 kJ

If S° = 75.8 JK–1, above what temperature the reaction become spontaneous?

Sol. G° = H° - TS°

H° = G° + TS° = 141 + (298) (75.8 × 10–3 kJ K–1) = +163.6 kJ The temperature above which the reaction is spontaneous is given by

T = 1-3

JK 8.75

J106.163

S

H



= 2158 K Above 2158K, the reaction will become spontaneous 174. The standard free energy charge for the reaction:

(g)3

(g)

2

)g(

2 NH H2

3 N

2

1 is

G° = 26.9 kJ/mole at 700 K. Calculate the equilibrium constant at this temperature. Sol. G° = 26.9 kJ/mole = 26.9 × 103 J/mole R = 8.314 J mole–1 K–1 T = 700 K G° = - 2.303 RT log k 26.9 × 103 = -2.303 × 8.314 × 700 × log K log K = -2.007 K = 9.84 × 10–3-

175. Hydrogen peroxide (H2O2) decomposes according to the equation:

)g(

2)(

2)(

22 O2

1 OHOH

H° = -98.2 kJ and S° = +70.1J/K

(A) is this reaction spontaneous at 25°C (B) from the given data calculate the value of kp for this reaction at 25°C? Sol. (A) G° = H° - TS° G° = -98.2 kJ – (298 K) × (70.1 × 10–3 KJ/K) = - 119.09 kJ The reaction is spontaneous at 25°C is G° < 0

(B) Log Kp = RT303.2

G

= 29810314.8303.2

)09.119(

3

= 20.872 Kp = antilog (20.872) = 7.44 × 1020

176. Calculate the average S –F bond energy in SF6. The given data is:

kJ/mole 1100]SF[H 60f

kJ/mole 275]S[H )g(0f

kJ/mole 80]F[H )g(0f

Sol. (g))g()g(

6 6F SSF

Hdiss = 6BE(S – F)

= [(H 0f

S(g) + 6H 0f

F(g)] – [H 0f

(SF6)]

6BE(S – F) = [(275 + 6 × 80J] – (-1100) = 1855 BE(S – F) = 309.16 kJ/mole 177. Calculate the number of kJ necessary to raise the temperature of 60.0g of Al from 35 to 55°C. Molar

heat capacity of Al is 24J mol-1 K-1 Sol. = n × C × T

No. of kJ required = K) 3555(mole JK2427

601-1

= 1066.7 J = 1.07 kJ 178. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110, -393, 81 and 9.7 kJ mole-1

respectively. Find the value of H for the reaction

)g(2

)g(2

)g()g(42 CO3ONCO3ON

Sol. H = fH(Products) - fH(Reactant)

= [fH(N2O) + 3fH(CO2)]-[fH(N2O4) + 3f(CO)]



= [81 +3(–393)] – [9.7 + 3(–110)] = -777.7 kJ 179. The standard molar enthalpies of formation of cyclohexane and benzene at 25°C are -156 and +49 kJ

mol–1 respectively. The standard enthalpy of hydrogenation of cyclohexene () at 25°C is -119 kJ mol–1.

Use these data to establish the magnitude of resonance energy of benzene? Sol. Standard enthalpy of hydrogenation of cyclohexene = -119 kJ mole-1. This is observed at enthalpy of

hydrogenation of one double bond. If benzene is considered as cyclohexatriene, the observed enthalpy of the reaction

H = 3 × (-119) = - 357 kJ/mole–1

enthalpy of above reaction is

H = H°(C6H14) – [H 0f

(C6H6) + 3 H 0f

(H2)]

= - 156 – [49 + 0] = - 205 kJ/mole Resonance energy

= observed rH° - calculated rH° = (-357) – (-205) = -152 kJ/mole 180. Calculate the entropy change for rusting of iron according to the equation:-

(s)32

(g)2

)S(O2Fe 3O Fe4 H = -1648 kJ mole–1

Given that the standard entropies of Fe, O2 and Fe2O3 are 27.3, 205.0 and 87.4 JK–1 mol–1 respectively. Will the reaction be spontaneous at room temperature (25°C)? Justify your answer?

Sol. rS° = S°(Products) - S°(Reactant) = 2S°(Fe2O3) – [4S°(Fe) + 3S° (O2)] = 2 × 874 – [4 × 27.3 + 3 × 205.0] JK–1 mole–1 = -549.4 JK–1 mole–1

This is the entropy change of the reaction i.e. system (Ssystem). Now,

G° = H° - TS° = - 1648000J – 298 × (-549.4) = -1484279 JK-1 mole-1 181. Give reasons for the following:

(a) Neither q nor is a state function but (q + ) is a state function (b) The dissolution of NH4Cl in water is endothermic still it dissolves in water (c) A Real crystal has more entropy than ideal crystal

Sol. (a) (q + w = u. As u is a state function hence (q + w) is a state function

(b) On dissolution entropy increases i.e. S = +ve.

Though H = +ve but if TS > H, then according to equation, G = H - TS, G will be –ve. Hence the process is spontaneous.

(c) A real crystal has more disorder due to presence of defects where as ideal crystal has no disorder. Hence a real crystal has more entropy than ideal crystal.

182. Explain why the entropy of a pure crystalline substance is zero at 0 K? State the law on which it is based? Give the application of this law?

Sol. At 0 K, there is a perfect arrangement of the constituent particles of pure crystalline substance and there is no disorder at all. Hence its entropy is taken as zero. This statement is based on third law of thermodynamics.

+ 3H2



REDOX REACTION 183. In the following reactions, label the oxidizing agent and the reducing agent.

(i) P4(s) + 5O2(g) P4O10(s)

(ii) CO(g) + Cl2(g) COCl2(g)

Sol. (i) )s(OP)g(O5)s(P 104

agentgsinoxidi

2

agentreducing

4 (ii) )g(COCl)g(Cl)g(CO 2

agentgsinoxidi

2

agentreducing

184. Identify the oxidants and reductants in the following reactions:

(i) CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g)

(ii) 2H+(aq) + MnO2(s) + C2H2O4 Mn2+ + 2CO2(g) + 2H2O(l)

(iii) I2 (aq) + 232OS2 (aq) 2I–(aq) + S4O 2

6(aq)

(iv) Cl2(g) + 2Br –(aq) 2Cl–(aq) + Br+2 (aq)

Sol.

Reaction Oxidant Reductant

(i)

(ii)

(iii)

(iv)

Cl2(g)

C2H2O4

I2

Cl2

CH4(g)

MnO2

S2O 23

Br

185. Find the oxidation number of Cl in HCl, HClO, 4

ClO and Ca (OCl) Cl.

Sol. (i) HCl; O.N. of Cl is – 1

(ii) HClO; +1 + O.N. (Cl) – 2 = 0

O.N. (Cl) = +1

(iii) 4ClO ; O.N. (Cl) + 4 (-2) = - 1

O.N. (Cl) = + 7

(iv) Ca(OCl) Cl; + 2 – 2 + 2 × O.N. (Cl) = 0 ;

)1(

)1(l

OCl

C

/\

Ca

O.N. (Cl) = 0

186. Balance the following equation using oxidation number method

HNO3(aq) + Cu2O(s) Cu (NO3)2 (aq) + NO(g) + H2O() + 2OH–

Sol. HNO3(aq) + Cu2 O(s) Cu (NO3)2 (aq) + NO(g) + H2O() + 2OH–

(i) The oxidation number of N changes from +5 in HNO3 to +2 in NO.

(ii) The oxidation number of Cu changes from +1 in Cu2O to +2 in Cu (NO3)

(iii) Electron balance diagrams can written as follows

N+5 + 3e– N+2 (1)

Cu+1 Cu+2 + e– (2) (iv) In order that the number of electrons lost shall be equal the number gained, multiply diagram (1)

by 2 and (2) by 6

2N+5 + 6e– 2N+2

6Cu+1 6Cu+2 + 6e– Hence, the coefficients of HNO3 and of NO are 2 and those of Cu2O is 3 and Cu (NO3)2 is 6. Part

of the skeleton equation can now be written as

2HNO3 + 3Cu2O 2NO + 6 Cu(NO3)2 The two atoms of H form one H2O on the right.

2HNO3 + 3Cu2O 2NO + 6Cu (NO3)2 + H2O



Although the N atoms that change oxidation number have been balanced, other N atoms are present

as 3

NO ions in 6Cu (NO3)2. For this reason, we add 12HNO3 to the 2HNO3 already on the left side

and then balance H2O molecule. The complete balanced equation is

14HNO3(aq) + 3Cu2O(s) 6Cu (NO3)2 (aq) + 2NO(s) + 7H2O(l) 187. Balance the following equation using oxidation number method.

Zn(s) + HNO3(aq) NH4NO3(aq) + Zn(NO3)2 (aq) + H2O(l)

Sol. Zn(s) + HNO3(aq) NH4NO3(aq) + Zn(NO3)2 (aq) + H2o(l)

Step (1) Only the Zn and N atoms change oxidation number (noting that N atoms in 3

NO ions on the

right have not changed oxidation number)

3

3423

25

3

0

NONH)NO(ZnNOH Zn

Step (2) Now we determine the change in oxidation number in the reduction portions of the equation.

e8

3

3

423

2

e2

5

3

0

ONHN)NO(ZnHNO Zn

We multiply the coefficient of the Zn species by 4 so that the change in oxidation number is equal in absolute value.

e8

3

3

423

2

e8

5

3

0

ONHN)NO(Zn4HNO Zn4

Step (3) We now balance the remaining N atoms. Add 9HNO3 to what we already have on the left.

4Zn + 10HNO3 4Zn (NO3)2 + NH4NO3 Finally, we balance the O atom by adding H2O to one side or the other. So, we add 3H2O to the right. The H atom should also balance.

4Zn(s) + 10HNO3(aq) 4Zn(NO3) (aq) + NH4NO3 (aq) + 3H2O(l) 188. Explain following with example

(1) Intramolecular Redox

(2) Intermolecular redox reaction

(3) Disproportionation

(4) Conproportionation

(5) Metal displacement Redox Reaction

(6) Non-metal displacement Reaction Sol. (1) Intramolecular redox reaction: A redox reaction within the same molecule

2

116

3

51

O2

3ClKOClK

(2) Intermolecular redox reaction

2

4

20

4

20

SOZn Cu CuSO Zn

A redox reaction between two different reactant.

(3) Disproportionation

In this reaction, an element in one oxidation state is simultaneously oxdised & reduced.

)g(O)(OH2)aq(OH2 2

21

2

1

2

1

2

)aq(POH3)g(PH)(OH3.)aq(OH3)s(P 2

1

2

3

3204

(4) Conproportionation reaction:



In this reaction an element from two different oxidation state gets changed into same oxidation

state in product.

OH2NNOHN 2

0

224

(5) Metal displacement Reaction:

A metal in a compound can be displaced by another metal in the uncombined state.

)aq(SOZn)s(Cu )s(Zn (aq)SOCu2- 4

4

2002- 4

4

2

(s)CaO 5 )s(V25Ca(s) )s(OV2- 202

5

5

2

(6) Non-metal displacement Reaction

The non-metal displacement redox reactions include H2 displacement.

)g(H (aq)HOaN2)(O2H )s(Na 20

2

12121

2

0

)g(H .)aq(ClFe .)aq(Cl H2 )s(Fe0

2

1

2

2-110

F2 + 2NaCl 2NaF + Cl2 189. Balance the following equations using the half-reaction method in the acidic medium.

(i) Zn + 3

NO Zn2+ + 4

NH

(ii) Zn + 3

NO Zn2+ + N2

Sol. (i)

oxidation

reduction

4

322

3

50

HNZnON Zn

oxidation half-reaction: [Zn Zn+2 + 2e–] × 4

reduction half-reaction: OH3NHe8H10NO 243

Net balanced redox-reaction: 4Zn + 3

NO + 10H+ 4Zn+2 + NH 4 + 3H2O

(ii)

oxidation

reduction

2

522

3

50

NZnON Zn

oxidation half-reaction: {Zn Zn+2 + 2e–} × 5

reduction half-reaction: 2NO 3

+ 12H+ + 10e– N2 + 6H2O

Net balanced redox reaction: 5Zn + 2NO 3

+ 12H+ 5Zn+2 + N2 + 6H2O

190. Balance the following redox-reaction in basic medium using the half-reaction method:

(i) Mn2+ + ClO 3 MnO2 + ClO2 (ii) Al +

3NO Al(OH)

4 + NH3

Sol. (i)

reduction

2

4

2

4

oxidation

3

522 OCl OMnOClMn

oxidation half-reaction: Mn+2 + 4OH– MnO2 + 2e– + 2H2O



reduction half-reaction: 2}OH2ClOeOHClO{ 223

Net balanced redox reaction: 2ClO 3

+ Mn+2 2ClO2 + MnO2

(ii)

reduction

3

3-

4

3-3

5

oxidation

0

HN )OH(AlON Al

oxidation half-reaction: {Al + 4OH– Al(OH) 4

+ 3e–} × 8

reduction half-reaction: { 3

NO + 6H2O + 8e– NH3 + 9OH–} × 3

8Al + 3 3

NO + 5OH– + 18H2O 8Al(OH) 4

+ 8NH3

191. Balance the following equation in basic medium by both the oxidation number and ion-electron methods and identify the reductants and the oxidants:

N2H4(g) + )aq(ClO3 NO(g) + Cl–(aq)

Sol. (i) N2H4(g) + ClO 3

(aq) NO(g) + Cl–(g)

Oxidation number method

e8

e6

-1-2

-3

5

4

2

2 lC NO2OCl HN

Net reaction is 6N2H4 + 8ClO 3

12NO + 8Cl– + 12H2O

Ion electron method

Oxidation half-reaction: N2H4 + 8OH– 2NO + 8e– + 6H2O] × 6

Reduction half-reaction: ClO 3

+ 6e– + 3H2O Cl– + 6OH–] × 8

Net reaction is 6N2H4 + 8ClO 3

12NO + 8Cl– + 12H2O

Reductant: N2H4 ; Oxidant: ClO 3

192. Find oxidation number of underlined element:

34 O NH N ; 722 OCrK ; 52 O SH ; 23OC ; 243 )OP(Ca

Sol. 34 O NH N : NH4+ NO

3

x + 4 – 1 = 0 x – 6 + 1 = 0

x + 3 = 0 x – 5 = 0

x = - 3 x = 5

722 OCrK : 2 + 2x – 14 = 0

2x – 12 = 0

x = 6

52 O SH It has two peroxy oxygen

+2 + x – 2 ×1 – 3 × 2 = 0 x = +6

23OC 3x – 4 = 0

x = 3

4

243 )OP(Ca : 2

Ca 34PO



In phosphate ion 34PO

x – 2 × 4 = - 3

x = + 5

193. What is the direction of the flow of electrons is the Daniel cell?

Sol. The electrons flow in the external circuit from the anode (zinc plate) to the cathode (copper plate)

194. Identify the strongest and weakest reducing agents from the following metals:

Zn, Cu, Ag, Na, Sn

Given: ; V 76.0E0Zn/Zn2

; V 34.0E0Cu/Cu2

; V 80.0E0Ag/Ag

; V 71.2E0Na/Na

; V 16.0E0Sn/Sn2

Sol. Na metal is the strongest reducing agent and Ag metal is the strongest oxidizing agent.

195. Calculate the standard e.m.f. of the following cell at 298 K using the standard electrode potential.

Al(s) |Al3+(aq) || Fe2+(aq) | Fe(s)

V44.0E and V 66.1E 0Fe/Fe

0Al/Al 23

What is the cell reaction?

Sol. 0L

0R

0cell

EEE

= - 0.44 – (-1.66)

V22.1E0cell

The cell reaction is 2Al(s) + 3Fe2+(aq) 2Al3+(aq) + 3Fe(s)

196. Can we store 1 M AgNO3 in a copper vessel ?

Given: V34.0E0Cu/Cu2

V80.0E0Ag/Ag

Sol. No, because, 0Cu/Cu2

E

is less than 0Ag/Ag

E

As a result copper will displace Ag+ from AgNO3 solution.

197. Is it possible to store:

(i) copper sulphate solution in a zinc vessel? (ii) copper sulphate solution in a nickel vessel?

(iii) copper sulphate solution in a silver vessel? (iv) copper sulphate solution in a gold vessel?

(Take help from electrochemical series)

Sol. (i) No, zinc will displace copper from CuSO4 solution because

0Cu/Cu

0Zn/Zn 22

EE

(ii) No, copper sulphate solution can not be stored in nickel vessel

(iii) Yes, copper sulphate solution can be stored in silver vessel as

0Ag/Ag

0Cu/Cu

EE2

. Silver will not displace copper from CuSO4 solution.

(iv) Yes, gold vessel can easily store CuSO4 solution.

198. Write the cell reaction for the following galvanic cells:

(i) Zn(s) | Zn2+ (aq) || Fe3+(aq) | Fe2+(aq) |Pt

(ii) Mg(s) | Mg2+ (aq) || Al3+ (aq) | Al (s)

Sol. (i) Zn(s) Zn2+ (aq) + 2e– (anode half reaction)

Fe3+ (aq) + e– Fe2+ (aq) (cathode half reaction)

The net cell reaction is

Zn(s) + 2Fe3+(aq) Zn2+(aq) + 2Fe2+(aq)

(ii) Mg(s) Mg2+ (aq) + 2e– (anode half reaction)



Al3+ (aq) + 3e– Al(s) (cathode half reaction)

Net cell reaction is

3Mg(s) + 2Al3+ (aq) 3Mg2+ (aq) + 2Al(s)

199. (i) How will you distinguish between the potential difference and the electromotive force? (ii) Consider the following half-reaction:

Cd(s) Cd2+(aq) + 2e– E° = +0.40 V

Ag+(aq) + e– Ag(s) E° = +0.80 V Write the cell diagram and calculate the standard cell e.m.f.

Sol. (i)

Potential difference Electropositive force

(i) It is the difference of the electrode potentials of the two electrodes when the cell is sending current through the circuit. (ii) It is less than the maximum voltage available from the cell. (iii) It can be measured by voltmeter also. (iv) It is an extensive quantity

(i) It is the potential difference between the two electrodes when no current is flowing through the circuit. That is when the circuit is open (ii) It is the maximum voltage available from the cell. (iii) It cannot be measured by voltmeter and can be measured by potentiometer (iv) It is an intensive quantity

(ii). The cell diagram is Cathodeanode

2 )s(Ag|)aq(Ag||)aq(Cd|)s(Cd

The standard cell e.m.f. is

E° = 0L

0R

EE

= + 0.80 – (- 0.40) = + 1.20 V 200. Which of the following metal (or metals) will oxidize Fe to Fe2+

Fe Fe2+ + 2e–E° = + 0.44 V

Given: Cu2+ + 2e– Cu E° = +0.34 V

Pb Pb2+ + 2e– E° = + 0.13 V

Ag+ + e– Ag E° = +0.80 V Sol. Any metal having higher reduction potential than iron will oxidise iron to ferrous ion. All three metals have E° value

)V 80.0E V 13.0E V,34.0E( 0Ag/Ag

0Pb/Pb

0Cu/Cu 22

higher than iron

)V44.0E( 0Fe/Fe2

All three metals will oxidize Fe to Fe2+

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Dr. Sangeeta Khanna Ph