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CH – 1 SOME BASIC CONCEPTS OF CHEMISTRY

1. (b)

Gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO2 is absorbed. So the remaining gas is

CO.

So, weight of remaining gaseous product CO is g8.22820

2

So, the correct option is (b).

2. (a)

(1) Mass of water = 18 x 1 = 18 g

Molecules of water = mole x NA = A

N18

18

(2) Moles of water = 4

104.22

00224.0

Molecules of water = mole x NA = 10-4

NA

(3) Molecules of water = mole x NA = A

N18

18.0

=10—2

NA

(4) Molecules of water = mole x NA = 10-3

NA

3. (a)

Temperature dependent unit is molarity.

4. (c)

Mole ratio of (i) carbon = 1.712

7.85 (ii) Hydrogen = 3.14

1

3.14

Simplest ratio of carbon and hydrogen = 1 : 2

So, empirical formula is CH2

Molecular formula = molar mass 14

84

massformulaempirical

massmolar= 6

So, molecular formula = 126 HC

Note : molecular mass is obtained from following data 42 mg of the compound contains 3.01 x 1020

molecules.

5. (c)

16.9 g AgNO3 is present in 100 mL solution.

∴ 8.45 g AgNO3 is present in 50 mL solution

5.8 g NaCl is present in 100 mL solution

∴ 2.9 g NaCl is present in 50 mL solution.

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30 YEAR’S BOOK CHEMISTRY PAGE: 78

mass of AgCl precipitated

= 0.049 × 143.5 g

= 7g AgCl

6. (b)

∵ mass of 1 mol (6.022 × 1023

atoms) of carbon = 12 g

If Avogadro Number (NA) is changed

then mass of 1 mol (6.022 × 1023

atoms) of carbon

23

20

106.022

106.02212

g1012

3

7. (d)

∵ 1 mole water = 6.02 × 1023

molecules

∴ 18 mole water = 18 × 6.02 × 1023

molecules

So, 18 mole water has maximum number of molecules.

8. (d)

MgCO3(s) → MgO(s) + CO2(g)

moles of MgCO3 = 84

20= 0.238 mol

From above equation

1 mole MgCO3 gives 1 mole MgO

∴ 0.238 mole MgCO3 will give 0.238 mole MgO

= 0.238 × 40 g = 9.523 g MgO

Practical yield of MgO = 8 g MgO

∴ % purity = 84%1009.523

8

9. (b)

10. (a)

MgOO2

1Mg 2

32

56.0

24

0.1

4

07.0

12

5.0

2

x

4

0.07x -

12

0.5

Oxygen is limiting reagent so 02

x

4

0.07

2

0.07x

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Excess Mg = mole4

0.07 -

12

0.5

Mass of Mg is = 1 0.7 12 = 0.16 gram.

11. (a)

H2 + CI2 2HCI

22.4 It 11.2 It

= 1 mole = 0.5 mole

Limiting reagent is CI2. So, 1 mole HCI is formed.

12. (c)

Millimoles of solution of chloride = 0.05 × 10 = 0.5

Millimoles of AgNO3 solution = 10 × 0.1 = 1

So, the millimoles of AgNO3 are double than the chloride solution.

∴ XCl2 + 2AgNO3 → 2AgCl + X(NO3)2.

13. (a)

M 0.01106.02

106.02

206.02100

1000106.02M

23

21

23

20

14. (c)

No. of molecules

Moles of CO2 = 44

44 = 1 NA

Moles of O3 = 48

48 = 1 NA

Moles of H2 = 2

8 = 4 4NA

Moles of SO2 = 64

64 = 1 NA

15. (b)

The number of atoms in 0.1 moles of a triatomic gas

= 0.1 × 3 × 6.023 × 1023

= 1.806 × 1023

16. (d)

No. of milli equivalent of HCl = 20 × 0.05 = 1.0

No. of milli equivalent of Br(OH)2 = 30 × 0.1 × 2 = 6.0

After neutralization, no. of milli equivalents in 50 ml. of solution = (6 – 1) = 5

Total volume of the solution = 20 + 30 = 50 ml

∴ No. of milli equivalent of OH– is 5 in 50 ml

M 0.11050

10005][OH

3

17. (b)

OHO2

1H

2

64g

210g

2

(2mol)(5mol)

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In this reaction oxygen is the limiting agent. Hence amount of H2O produced depends on the amount of O2 taken

∵ 0.5 mole of O2 gives H2O = 1 mol

∴ 2 mole of O2 gives H2O = 4 mol

18. (d)

Writing the equation for the reaction, we get

OHPbCl2HClPbO

2

278g71207

2

73g36.52

223g16207

From this equation we find 223 g of PbO reacts with 73 g of HCl to form 278 g of PbCl2. If we carry out the reaction

between 3.2 g HCl and 6.5 g PbO.

Amount of PbO that reacts with 3.32 g HCl

= g 9.773.273

223

Since amount of PbO present is only 6.5 g so PbO is the limiting reagent.

Amount of PbCl2 formed by 6.5 g of PbO

g 6.5223

278

Number of moles of PbCl2 formed

278

6.5

223

278 moles = 0.029 moles

19. (a)

Element % At. wt. Relative number Ratio

C 38.71 12 3.2312

38.71 1

3.23

3.23

H 9.67 1 9.671

9.67 3

3.23

9.67

O 100 – (38.71 + 9.67) = 51.62 16 3.2316

51.62 1

3.23

3.23

Thus, empirical formula is CH3O.

20. (c)

Writing the equation of combustion of propane (C3H8), we get

O4H3CO5OHC 22

L 5 vol5

2

L 1 vol1

83

From the above equation we find that we need 5 L of oxygen at NTP to completely burn 1 L of propane at N. T.P.

If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure. The same

results are obtained. i.e. 5 L is the correct answer.

21. (a)

C8H10CO2Mn 22

2

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Fe2+

ions will also oxidise to Fe3+

.

From above equation 2 moles MnO4 required to oxide 5 moles of oxalate.

Thus number of moles of MnO4 required to oxidise one mole of oxalate = 2/5 = 0.4.

And number of moles of MnO4 required to oxidise one mole of Fe

2+ =1/5 = 0.2

Total = 0.6

22. (c)

Volume

MassDensity

3

3

cm

gram 1cm gram 1

Density

MassVolume

3

3cm 1

cm gram 1

gram 1

∴ Volume occupied by 1 gram water = 1 cm3 or volume occupied by

18

106.02323

molecules of water = 1 cm

3

[∴ 1 g water = 18

1 moles of water]

Thus volume occupied by 1 molecules of water

23

106.023

181

cm

3 = 3.0 × 10

–23 cm

3.

i. e. the correct answer is option (c).

23. (b)

The balance chemical equation is:

34 5SO6H2MnO → O3H5SO2Mn 24

From the equation it is clear that

Moles of

4MnO require to oxidize 5 moles of

3SO are 2/5.

24. (d)

Average isotopic mass of

2890

2202819990200X

100

404189218000

100

19996

= 199.96 amu

25. (c)

Molarity of H2SO4 solution.

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18.41.8410098

100098

Suppose V ml of this H2SO4 is used to prepare 1 lit. of 0.1 M H2SO4

∴ V × 18.02 = 1000 × 0.1

Or 18.02

0.11000V

= 5.55 ml.

26. (a)

4

6

2

OH

4

7

nOMKnOMK

Change in oxidation number of Mn in basic medium is 1. Hence mole of KI is equal to mole of KMnO 4.

27. (c)

2Al2O3 + 3C → Al + 3CO2

Gram equivalent of Al2O3 ≡ gm equivalent of C

Now equivalent weight of Al = 93

27

Equivalent weight of C = )OCC(gm34

122

40

No. of gram equivalent of Al = 9

102703

= 30 × 10

3

Hence,

No. of gram equivalent of C = 30 × 103

Again,

No. of gram equivalent of C = weight equivalent gram

gram in mass

⇒ 30 × 103 =

3

mass

⇒ mass = 90 × 103 g = 90 kg

28. (a)

No. of molecules in different cases

(a) ∵ 22.4 litre at STP contains = 6.023 × 1023

molecule of H2

∴ 15 litre at STP contains = 23

106.02322.4

15

(b) ∵ 22.4 litre at STP contains = 6.023 × 1023

molecule of N2

∵ 5 litre at STP contains = 23

106.02322.4

5

(c) ∵ 2 gm of H2 = 6.023 × 1023

molecules of H2

∵ 0.5 gm of H2 = 23

106.0232

0.5

(d) Similarly 10 g of O2 gas 23

106.02332

10 molecules

Thus, (a) will have maximum number of molecules.

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29. (c)

litre 20 vol.2

3

litre 30 vol.3

2

litre 10 vol.1

2 2NH3HN

It is given that only 50% of the expected product is formed hence only 10 litre of NH3 is formed.

N2 used = 5 litres, left = 30 – 5 = 25 litres

H2 used = 15 litres, left = 30 – 15 = 15 litres

30. (b)

2 g of H2 means one mole of H2, hence contains 6.023 × 1023

molecules. Others have less than one mole, so have less

no. of molecules.

31. (d)

Suppose the mol. wt. of enzyme = x

Given 100 g of enzyme wt of Se = 0.5 gm

∴ In x g of enzyme wt. of Se = x100

0.5

Hence, 100

0.578.4

x

∴ x = 15680 = 1.568 × 104

32. (d)

Specific volume (volume of 14 gm) of cylindrical virus particle = 6.02 × 10–2

cc/gm

Radius of virus (r) = 7 Å = 7 × 10–8

cm

Length of virus = 10 × 10–8

cm

Volume of virus

8282 1010)10(7

7

22r lπ

= 154 × 10–23

cc

Wt. of one virus particle = volumesecific

volume

∴ Mol. wt. of virus = Wt. of NA particle

23

2

23

106.02106.02

10154

= 15400 g/mol = 15.4 kg/mole

33. (c)

gm 197

3BaCO → BaO + CO2

∵ 197 gm of BaCO3 released carbon dioxide = 22.4 litre at STP

∴ 1 gm of BaCO3 released carbon dioxide = 197

22.4 litre

∴ 9.85 gm of BaCO3 released carbon dioxide = 9.85197

22.4 = 1.12 litre

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34. (c)

Element % At. wt. Relative number Ratio

C 40 12 12

403.33

3.33

3.33 1

H 6.66 1 1

6.66 6.66

3.33

6.66 2

O 53.34 16 16

53.34 3.33

3.33

3.33 1

Empirical formula of organic compound = CH2O.

35. (c)

According to Stoichiometry they should react as follow

mole 1moles 5

2

moles 0.8moles 4

3 5O4NH →

moles 1.2moles 6

2

mole 0.8moles 4

O6H4NO

Thus for 1 mole of O2 only 0.8 mole of NH3 is consumed. Hence O2 is consumed completely.

36. (c)

There is 0.334 g of Fe per 100 g of Hb.

∴ 100 g Hb ⟶ 0.334 g Fe.

∴ 67200 g Hb ⟶ 100

672000.334 = 224.448 g

56 g Fe ⟶ 1 Fe atom

224.448 ⟶

56

1224.4 4 atoms of He

37. (d)

We know that all non-zero digits are significant and the zeros at the beginning of a number are not significant.

Therefore number 161 cm, 0.161 cm and 0.0161 cm have 3, 3 and 3 significant figures respectively.

38. (c)

As the sum of the percentage of C, H & N is 100. Thus it does not contains O atom.

Table for empirical formula

Element % At. wt. Rel. Number Ratio

C 40.00 12 12

403.66

3.33

3.66 1.09 ~ 1

H 13.33 1 1

13.33 13.33

3.33

13.33 4

N 46.67 14 14

46.67 3.33

3.33

3.33 1

Hence, empirical formula = CH4N

39. (d)

15(32)

22(78)

66 (g)15OH2C → 12CO2(g) + 6H2O(g)

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∴ 156 gm of benzene required oxygen = 15 × 22.4 litre

∴ 1 gm of benzene required oxygen = 156

22.415 litre

∴ 39 gm of Benzene required oxygen = 84.0156

3922.415

litre

40. (d)

Molecular weight of ZnSO4.7H2O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287.

∴ Percentage mass of zinc (Zn) = % 22.65100287

65

41. (b)

Molecular weight of C60H122 = (12 × 60) + 122 = 842.

Therefore weight of one molecule number sAvagadro'

HC of weightMolecular 12260

g 101.4106.023

842 21

23

42. (c)

On calculation we find

6

5

101.171.37

)10(1.79 20.2)(29.2

As the least precise number contains 3 significant figures therefore answers should also contains 3 significant figures.

43. (a)

No of moles of nitride ion

0.314

4.2 mol = 0.3 × NA = 0.3 NA nitride ions

Valence electrons = 8 × 0.3 NA = 2.4 NA

(5 + 3 due to charge) One N3–

ion contains 8 valence electrons.

44. (a)

5 M H2SO4 = 10 N H2SO4, (∵ Basicity of H2SO4 = 2)

N1V1 = N2V2,

10 × 1 = N2 × 10 or N2 = 1 N

45. (c)

According to Avogadro’s law ‚equal volumes of all gases contain equal numbers of molecules under similar conditions

of temperature and pressure‛. Thus if 1 L of one gas contains N molecules, 2 L of any gas under the same conditions

will contain 2N molecules.

46. (a)

Average atomic mass = 100

mass] Atomic2 of Al [Relativemass] Atomic1 of Abundance [Relative 2

100

11811019 = 10.81

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47. (a)

4.4 g CO2 = 0.144

4.4 mol CO2 (mol. wt. of CO2 = 44)

= 6 × 1022

molecules

= 2 × 6 × 1022

atoms of O.

48. (b)

6.02 × 1023

molecules of CO = 1 mole of CO

6.02 × 1024

CO molecules = 10 moles CO

Which is equal to 10 gm atoms of O and 5 gm molecules of O2

49. (c)

C2H4 + 3 O2 → 2CO2 + 2H2O

28kg 96 kg

∵ 28 kg of C2H4 undergo complete combustion by = 96 kg of O2

∴ 2.8 kg of C2H4 undergo complete combustion by = 9.6 g of O2.

50. (a)

Cp / Cv = 1.4 shows that the gas is diatomic.

22.4 litre at NTP ≡ 6.02 × 1023

molecules

11.2 L at NTP = 3.01 × 1023

molecules

= 3.01 × 1023

× 2 atoms = 6.02 × 1023

atoms

51. (d)

The reaction may given as

Z2O3 + 3H2 → 2Z + 3H2O

0.1596 g of Z2O3 react with H2 = 6 mg = 0.006 g

∴ 1 g of H2 react with 0.006

0.1596 = 26.6 g of Z2O3

∴ Eq. wt. of Z2O3 = 26.6 (from the definition of eq. wt.)

Eq. wt. of Z + Eq. wt. of O = E + 8 = 26.6

⇒ Eq. wt. of Z = 26.6 – 8 = 18.6

Valency of metal in Z2O3 = 3

valency

wt. Atomicmetal of wt. Eq.

∴ At. wt. of Z = 18.6 × 6 = 55.8

52. (d)

At NTP 22400 cc of N2O = 6.02 × 1023

molecule

∴ 1 cc N2O = 22400

106.0223

molecules

22400

106.02323

atoms

22

10224

1.8 atoms

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No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22

Hence no. of electrons 2222400

106.0223

electrons = 22400

101.3223

53. (a)

Mg++

+ Na2CO3 → MgCO3 + 2Na+

1g eq. 1g eq.

Mili eq. of Mg2+

= 112

12

wt. atomic

10wt(g)3

∴ 12 mg Mg++

= 1 milli eq. Na2CO3

54. (a)

Mol wt. of CCl4 = 154 g = 22.4 L at STP

∴ Density = 22.4

154gL

–1 = 6.875 gL

–1

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1. (b)

According to Hund’s Rule of maximum multiplicity, the correct electronic configuration of N -atom is

Option (b) violates Hund’s Rule.

2. (c)

In H–like atom energy of 2s = 2p. orbital.

3. (d)

42 = 16

4. (d)

Ti(22) = 1s22s

22p

63s

23p

64s

23d

2

5. (a)

6. (d)

18

9

834

104.41045

103106.63

λ

hcE

7. (a)

It is 3P orbital with magnetic Q.N = 0

So, it should be 3PZ

8. (d)

AN

hcE

λ

λ

231027106.02103106.62

λ

8101.19

Hence, none of the given option is correct.

9. (c)

Energy of an electron at infinite distance from the nucleus is zero. As an electron approaches the nucleus, the electron

attraction increases and hence the energy of electron decreases and thus becomes negative. Thus as the value of n

decreases, i.e., lower the orbit is, more negative is the energy of the electron in it.

10. (c)

n = 3 → 3rd

shell

l = 1 → p sub shell.

m = –1 is possible for two electrons present in an orbital.

11. (b)

C = v

nm 50106

103

v

C15

17

λ

CH – 2 STRUCTURE OF ATOM

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12. (a)

Orbital angular momentum

1)(2

h

π

For p orbital = 1

So, ππ 2

h2

2

h

13. (c)

Electronic configuration of Rb = [Kr] 5s–1

Set of quantum number, n = 5

= 0, ∴ s orbital

m = 0, s = + 1/2

14. (a)

(n = 4, l = 3) ⇒ 4f subshell

Since, maximum no. of electrons in a subshell = 2(2l + 1)

So, total no. of electron in 4f subshell

= 2 (2 × 3 + 1) = 14 electrons.

15. (c)

MOT configuration of O2 and

2O :

2O : (σ1s)2 (σ*1s)

2 (σ2s)

2 (σ*2s)

2 (σ2pz)

2

)2p2p( 2

y

2

x ππ )2p2p( 1

yx *π*π 1

Number of unpaired electrons = 2, so paramagnetic.

O2: (σ1s)2 (σ*1s)

2 (σ2s)

2 (σ*2s)

2 (σ2pz)

2

)2p2p( 2

y

2

x ππ )2p2p( 0

yx *π*π 1

Number of unpaired electrons = 1, so paramagnetic.

16. (c)

Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.

2

2

2

1

2

n

1

n

113.6ZE Δ

17. (a)

ns → (n – 2)f → (n – 1)d → np [n = 6]

18. (b)

Given E1 = 25eV E2 = 50 eV

1

1

hcE

λ

2

2

hcE

λ

1

2

2

1

E

E

λ

λ

2

1

50

25

1

2 λ

λ ∴ 1 = 22

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19. (a)

O2 16 KK(σ2s)2 (σ*2s)

2 (σ2pz)

2 (π2px)

2 (π2py)

2 2

2O 15 Remove one electron from π*2py from O2 2.5

2O 17 KK(σ2s)2 (σ*2s)

2 (σ2pz)

2 (π2px)

2 (π2py)

2 1.5

2

2O 18 KK(σ2s)2 (σ*2s)

2 (σ2pz)

2 (π2px)

2 (π2py)

2 1

Since, the bond length decreases as the bond order increases, hence,

2O have least bond length.

20. (b)

Total no. of atomic orbital in a shell = n2.

Given n = 4; Hence number of atomic orbitals in 4th

shell will be 16.

21. (d)

m1011000.66

106.6

mv

h 3534

λ

22. (b)

Bond order of Be2 = 0, hence Be2 cannot exis t.

23. (a)

Molecular orbital configuration of

1

z

1

y2

x2

z

2

y22222

22p*

2p* σ2p

2p

2p2s*σ σ2s1s*σ σ1sN

π

π

π

π

Bond order = 22

610

0

z

1

y2

x2

z

2

y2222

22p*

2p* 2p

2p

2p2s* 2s1s* 1sN

π

πσ

π

πσσσσ

Bond order = 2.52

510

2p 2p

2p2s* 2s1s* 1sN 2

x2

z

2

y2222

2 σπ

πσσσσ

Bond order = 32

410

∴ The correct order is = 22

2

2 NNN

24. (b)

m = 2l + 1, thus for l = 2, m = 5, hence values of m will be –2, –1, 0, +1, +2.

Therefore, for l = 2, m cannot have the value – 3.

25. (d)

The number of sub shell is (2 + 1). The maximum number of electrons in the sub shell is 2 (2 + 1) = (4 + 2)

26. (d)

K.E per atom

2

104.0104.41919

20

19

102.02

100.4

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27. (a)

∆P = m∆v

Substituting the given values of ∆x and m, we get

1 × 10–18

gcms–1

= 9 × 10–28

g × ∆v

or 28

18

109

101v

Δ

= 1.1 × 109 cms

–1 ≃ 1 × 10

9 cm s

–1

i.e. option (a) is correct.

28. (a)

We know ∆p.∆x π4

h

Or m.∆v.∆x = π4

h [∴ ∆p = m∆v+

Since ∆p = ∆x (given)

∴ ∆p.∆p = π4

h or m∆v m ∆v =

π4

h

Or (∆v)2 =

2 4 mπ

h

Or ∆v =

h

mmπ

h

2

1

42

Thus, option (a) is the correct option.

29. (b)

(ii) is not possible for any value of because varies from 0 to (n – 1) thus for n = 2, can be only 0, 1, 2.

(iv) is not possible because for = 0, m = 0.

(v) is not possible because for = 2, m varies from –2 to +2.

30. (b)

Magnetic quantum no. represents the orientation of atomic orbitals in an atom. For example p x, py & pz have

orientation along X-axis, Y-axis & Z-axis.

31. (d)

We know that ∆x . ∆p ≥ π4

h

∆x.m∆v > π4

h

∆v > xm

h

4

3110

34

109.11100.14

106.626v

πΔ

= 7

1094

66

π

= 5.79 × 10

6 m/sec

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32. (b)

We know that

21

nn

zEE

Given 1

2

12

mol kJ 3282

EE

11

2

24

mol kJ 82mol kJ4

328

2

EE

33. (c)

2

2

2

1n

1

n

1IE

h

1v

115

34

18

s 103.0816

1

1

1

106.625

102.18

34. (c)

Amongst isoelectric species, ionic radii of anion is more than that of cations. Further size of anion increase with

increase in-ve change and size of cation decrease with increase in +ve charge. Hence, ionic radii decreases from O– to

Al+++

.

35. (d)

v

cor

hchvE λ

λ

m 103.75108

103 8

15

8

λ

In nanometer = 3.75 × 10

Which is closest to 4 × 101.

36. (a)

Suppose the nucleus of hydrogen atom have charge of one proton i.e., The electron revolves in radius of r around it.

Therefore the centripetal force is supplied by electrostatic force of attraction i.e.

2

22

r

ze

r

mv

Or r

ze

r

mv 22

Or K.Er

ze

2

1mv

2

1 22

now total energy (En) = K.E + P.E in first excited state

r

zemv

2

1E

22

r

ze

r

ze

2

122

–3.4 eV = r

ze

2

1 2

∴ K.E = r

ze

2

12

= + 3.4 eV

37. (c)

Both CN– and CO have 14 electrons.

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38. (a)

Quantum number n = 3, l = 2, m = +2 represent an orbital with

2

1s 22

yxxy 3dor 3d

Which is possible only for one e–.

39. (c)

The molecule which contains same number of electrons are called isoelectronic. Eg.,

N2 = CO = CO– = 2

2O = 14e

40. (c)

The energy of photon,

19

103.03hv

E

λ

Or 19

834

103.03

103106.626

λ

Or 7

103.03

19.878 λ

= 6.56 × 10

–7 m = 656 nm

41. (c)

Energy of electron in 2nd orbit of Li+2

2

2

n

z13.6

eV 30.6(2)

(3)13.62

2

Energy required = 0 – (–30.6) = 30.6 eV

42. (d)

By Heisenberg uncertainty Principle 4π

hΔpΔx (which is constant)

As ∆x for electron and helium atom is same thus momentum of electron and helium will also be same therefore the

momentum of helium atom is equal to 5 × 10–26

kg. m.s–1

.

43. (d)

Given: Radius of hydrogen atom = 0.530Å, Number of excited state (n) = 2 and atomic number of hydrogen atom (Z) =

1. We know that the Bohr radius.

Z

n(r)

2

Radius of atom 0.5301

(2)2

= 4 × 0.530 = 2.12 Å

44. (a)

We know that ions which have the same number of electrons are called isoelectronic. We also know that both CO and

CN– have 14 electrons, therefore these are isoelectronic.

RJ VISION PVT. LTD.


45. (c)

According to deBroglie

17

34

10

106.626hmv

mv

h

λ

λ

⇒ p = 6.626 × 10–17

kg m/s

46. (d)

The orbitals which have same energy are called degenerate orbitals eg. p x, py, pz.

47. (a)

Given mass of an electron(m) = 9.1 × 10–28

g;

Velocity of electron (v) = 3 × 104 cm/s;

Accuracy in velocity = 0.001% = ;100

0.001

Actual velocity of the electron

(∆v) = 3 × 104 ×

100

0.001= 0.3 cm/s.

Planck’s constant (h) = 6.626 × 10–27

erg-sec.

∴ Uncertainty in the position of the electron

(∆x) = vm

h

4 0.3)10(9.1224

7106.62628

27

= 1.93 cm

48. (a)

State of hydrogen atom (n) = 1 (due to ground state)

Radius of hydrogen atom (r) = 0.53 Å.

Atomic number of Li(Z) = 3.

Radius of Li2+

ion 0.173

(1)0.53

Z

nr

22

1

49. (d)

as m = (2 + 1), hence m = –1 means,

–1 = 2 + 1. = 1

i.e., least value of = 1. So it cannot be present in s-orbital. Because for s orbital = 0.

50. (c)

K.E. of emitted electrons = hʋ - hʋ (i.e. smaller than hʋ)

51. (b)

P.E. = work done = r

zedr

r

ze 2r

2

2

52. (b)

Wave nature of electron was shown by Davisson and Germer. Davisson and germer demonstrated the physical reality

of the wave nature of electrons by showing that a beam of electrons could also diffracted by crystals just like light of x-

rays.

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53. (b)

Cathode rays are never electromagnetic waves.

54. (b)

The sub-shell are 3d, 4d, 4p and 4s, 4d has highest energy as n + value is maximum for this.

55. (a)

The ionization energy of any hydrogen like species (having one electron only) is given by the equation

2

422

h

e mZ 2I.E

π or I.E ∝ Z

2

Since the atomic number of H is 1 and that of He is 2, therefore, the I.E. of He+ is four times (2

2) the I.E. of H i.e., 13.6 ×

4 = 54.4 eV

56. (c)

Outer electronic configuration of Cl = 1

z

2

y

2

x

2 3p3p3p3s

Outer electronic configuration of Cl– =

2

z

2

y

2

x

2 3p3p3p3s

Hence Cl– contain four lone pairs of electron

57. (c)

Energy of an electron in Bohr’s orbit is given by the relationship. eVn

13.6E

2n

58. (b)

The sub-shell with lowest value of (n + ) is filled up first. When two or more sub-shells have same (n + ) value of

subshell with lowest value of ‘n’ is filled up first therefore the correct order is

Orbital 4s 3d 4p 5s 4d

n + 4 + 0 3 + 2 4 + 5 + 4 + 2

Value = 4 5 5 5 6

59. (d)

3 means f-subhsell. Maximum no. of electrons = 4 + 2 = 4 × 3 + 2 = 14

60. (d)

This is as per the definition of Pauli’s Exclusion principle.

61. (a)

Cu+ = 29 – 1 = 28e

–

Thus the electronic configuration of Cu+ is Cu

+(28) = 1s

2 2s

2 2p

6

18e

1062 3d 3p 3s

62. (c)

n = 2, 1 = 1 means 2 p-orbital. Electrons that can be accommodated = 6 as p sub-shell has 3 orbital and each orbital

contains 2 electrons.

63. (b)

No. of orbitals in a sub-shell = 12

⇒ No. of electrons = 241)2(2

RJ VISION PVT. LTD.


64. (c)

N(7) = 1s22s

22p

2

N2+(s)

= 1s2, 2s

2 1

x2p

Unpaired electrons = 1.

65. (d)

Is uncertainty principle which was given by Hiesenberg and not Bohr’s postulate

66. (a)

Radius of nth

orbit = r1 n2. (for H-atom)

67. (a)

No. of radial nodes in 3p-orbital

= (n – – 1)

[for p orbital = 1]

= 3 – 1 – 1 = 1

68. (b)

Both He and Li+ contain 2 electrons each therefore their spectrum will be similar.

RJ VISION PVT. LTD.


1. (d)

Elements B Ga Al In Tl

Atomic radii (pm) 85 135 143 167 170

2. (d)

Element (X) electronic configuration 1s22s

22p

3

So, valency of X will be 3.

Valency of Mg is 2.

Formula of compound formed by Mg and X will be Mg3X2.

3. (a)

Z = 114 [Rn]86

7s2 5f

14 6d

10 7p

2 14

th gp. (carbon family)

4. (a)

5. (c)

6. (b)

Be2 = 1s

2 = Li

+

7. Bonus

F = 133 pm

O2 = 140 pm

Na+ = 102 pm

There is no correct option.

8. (a)

As the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increases and

hence the electron gain enthalpy becomes more negative, hence the correct order is:

Ca < Al < C < O < F

9. (a)

As the positive charge increases on metal cation, radius decreases. This is due to the fact that nuclear charge in the

case of a cation is acting on lesser number of electrons and pulls them closer.

10. (a)

IE1 of Na = – Electron gain enthalpy of Na+

= – 5.1 volt

11. (c)

12Mg 15P 17Cl 20Ca

160p 110 99 197 (pm)

Cl < P < Mg < Ca

12. (b)

O < S < F < CI

Electron gain enthalpy – 141 – 200 – 333 – 349 kJ mol–1

13. (c)

Among the isoelectronic species, size increases with the increase in negative charge. Thus S2–

has the highest negative

charge and hence largest in size followed by Cl–, K

+ and Ca.

CH – 3 CLASSIFICATION OF ELEMENTS AND PERIODICITY

IN PROPERTIES

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14. (d)

The smaller the atomic size, larger is the value of ionisation potential. Further the atoms having half filled or fully fille d

orbitals are comparatively more stable, hence more energy is required to remove the electron from such ato ms.

15. (d)

The stability of + 1 oxidation state increases from aluminium to thallium i.e.,

Al < Ga < In < Tl

16. (a)

Proton affinity decreases in moving across the period from left to right due to increase in charge, within a group the

proton affinities decreases from to p to bottom.

Nitrogen family > Oxygen family > Halogens

17. (b)

18. (a)

For isoelectronic species, size of anion increase as negative charge increases whereas size of cation decreases with

increase in positive charge. Further ionic radii of anions is more than that of cations. Thus the correct order is

Ca++

< K+ < Ar < Cl

– < S

– –

19. (d)

The amount of energy released when an electron is added to an isolated gaseous atom produce a monovalent anion is

called electron gain enthalpy.

Electron affinity value generally increase on moving from left to right in a period however there are exceptions of this

rule in the case of those atoms which have stable configuration. These atoms resist the addition of extra electron,

therefore the low value of electron affinity

3.83.62.01.48ClFSO

On the other hand Cl because of its compariti very bigger size than F, allow the addition of an extra electron more

easily.

20. (d)

Due to odd number of electrons in ClO2.

21. (a)

Ionic radii are inversely proportional to effective nuclear charge.

Ionic radii in the nth

orbit is given as

Z

1ror

Z

anr n

0

2

n

When n = principal quantum number

Z-effective nuclear change.

22. (a)

Among the given options, only Fe shows variable oxidation states so it can form two chlorides, viz. FeCl2 and FeCl3.

RJ VISION PVT. LTD.


23. (b)

This is because of inter-electronic repulsions between lone pairs.

B.E.: F – F Cl – Cl

(kJ mol–1

) 158.8 242.6

24. (a)

The electronic configuration clearly suggest that it is a d-block element (having configuration (n – 1)d1 – 10

ns 0 – 2

) which

starts from III B and goes till II B. Hence with d3 configuration it would be classified in the group.

25. (b)

Along the period, I.P. generally increases but not regularly. Be and B are exceptions. First I.P. increases in moving from

left to right in a period, but I.P. of B is lower than Be.

26. (a)

Be – 1s22s

2; B – 1s

22s

22p

1; C – 1s

22s

22p

2; N – 1s

22s

22p

3; O – 1s

22s

22p

4. IP increases along the period. But IP of Be > B.

Further IP of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation

energy.

27. (a)

Mg = 1s22s

22p

63s

2

After removing of 2 electron, the magnesium acquired noble gas configuration hence removing of 3rd electron will

require large amount of energy.

28. (d)

First ionisation potential of Be of greater than boron due to following configuration

4Be = 1s2, 2s

2 5B = 1s

2, 2s

22p

1

Order of attraction of electrons towards nucleus 2s > 2p, so more amount of energy is required to remove the electron

with 2s-orbital in comparison to 2p orbital.

29. (a)

In a period on moving from left to right ionic radii decreases.

(a) So order of cationic radii is

Cr2+

> Mn2+

> Fe2+

> Ni2+

and

(b) Sc > Ti > Cr > Mn (correct order of atomic radii)

(c) For unpaired electrons

Mn2+

(Five) > Ni2+

(Two) < Co2+

(Three) < Fe2+

(Four)

(d) For unpaired electrons

Fe2+

(Four) > Co2+

(Three) < Ni2+

(Two) < Cu2+

(One)

30. (b)

Greater is the positive charge on atom, large is effective nuclear charge. Hence , smaller is the size.

31. (b)

Electronic configuration of element with atomic number 118 will be [Rn] 5f14

6d10

7s27p

6. Since its electronic

configuration in the outer most orbit (ns2np

6) resemble with that of inert or noble gases, therefore it will be noble gas

element.

RJ VISION PVT. LTD.


32. (a)

Atomic number of the given element is 15 and it belongs to 5th group. Therefore atomic number of the element below

the above element = 15 + 18 = 33.

33. (d)

In general non-metals are more electronegative then metals.

34. (d)

Abnormally high difference between 2nd and 3rd ionization energy means that the element has two valence electrons,

i.e., configuration (d).

35. (d)

Atomic volume is the volume occupied by one gram of an element. Within a period from left to right, atomic volume

first decreases and then increases due to increases of nuclear charge and increase in the number of electrons in the

valence shell.

36. (c)

Element with Z = 33

(1s2 2s

2p

63s

2p

6d

104s2p

3) lies in fifth (or 15th) group.

37. (c)

Amongst isoelectronic ions, the size of the cation decreases as the magnitude of the charge increases.

38. (c)

Proton (H+) being very small in size would have very large hydration energy.

39. (c)

N, O and F (p-block elements) are highly electronegative non metals and will have the strongest tendency to form

anions by gaining electrons from metal atoms.

40. (c)

Electron affinity values are high in case of halogen because halogens have seven electrons (ns2np

5) in the valence shell,

they have a strong tendency to acquire the nearest inert gas configuration by gaining an electron from the metallic

atom and form halide ions easily.

41. (a)

Metallic character decreases in a period and increases in a group.

42. (c)

Elements (a), (b) and (d) belong to the same group since each one of them has two electrons in the s sub shell. In

contrast, element (c) has seven electrons in the valence shell and hence does not lie in the same group in which

elements (a), (b) and (c) lie.

43. (a)

Pauling scale of electronegativity was helpful in predicting

(i) Nature of bond between two atoms

(ii) Stability of bond

By calculating the difference in electronegativities polarity of bond can be calculated.

RJ VISION PVT. LTD.


1. (c)

The structure of CiF3 is

The number of lone pair of electrons on central Cl is 2.

2. (c)

NO : (1s)2, (

*1s)

2, (2s)

2,(

*2s)

2,( 2pz)

2,(2px)

2 = (2py)

2,(

*2px)

1=(

*2py)

0

BO = 5.22

510

CN- : (1s)

2,(

*1s)

2,(2s)

2,(

*2s)

2,(2px)

2=(2py)

2,(2pz)

2

32

4-10 BO

CN : (1s)2, (

*1s)

2,(2s)

2,(

*2s)

2,(2px)

2=(2py)

2,(2pz)

1

5.22

4-9 BO

CN+ : (1s)

2, (

*1s)

2, (2s)

2,(

*2s)

2, (2px)

2=(2py)

2

22

4-8 BO

Hence, option (c) should be the right answer.

3. (b)

Total no. of electrons in CN. is 14.

Total no. of electrons in CO is also 14.

Hence B.O. of both CN. & CO is 3.

4. (c)

5. (d)

243 ])([ NHCu - dsp

2 hybidisaiton

6. (a)

Neutral structure is most stable than charged.

7. (d)

(i) Structure of SiCl4 is tetrahedral

Structure of

4PCl is tetrahedral

CH – 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE

RJ VISION PVT. LTD.


(ii) Diamond & SiC both are isostructural because both have tetrahedral arrangement and central atom is sp3

hybridised.

(iii) Hybridiation of NH3 [σ = 3, lp = 1]

sp3 geometry : tetrahedral

(iv) Structure of XeF4 is square planar.

Structure of XeO4 is tetrahedral

So, XeF4 and XeO4 are not isostructural.

8. (a)

Given species: O2, 1

2O , 1

2O , -2

2O

Total number of electrons

O2 → 16e–

1

2O → 17e–

1

2O → 15e–

2

2O → 18e–

1

2O O2 1

2O 2

2O

Bond order 2.5 2 1.5 1

* Stability order [ 1

2O > O2 > 1

2O > 2

2O ]

9. (d)

Species 2

2O

2O

2O

Bond order 3 2.5 1.5

10. (c)

In

2NO , the bond angle is 180.

11. (b)

Species

2O 2O

2O

Bond order 2.5 2 1.5

RJ VISION PVT. LTD.


12. (d)

Species 2

3SO

3ClO

Shape Pyramidal Pyramidal

13. (b)

Species 4XeO

Shape Tetrahedral

14. (b)

Sp2 (triangular planer)

15. (c)

CO2 CH4 NH3 NF3

= 0 = 0 = 1.47D = 0.23D

16. (d)

17. (d)

(a) N2 →

2N

B.O 3 2.5

Bond energy decreases

Magnetic behaviour changes from diamagnetic to paramagnetic.

(b) O2 →

2O

B.O. 2 2.5

Bond energy increases

Magnetic behaviour does not change

(c) C2 →

2C

B.O. 2 2.5

Bond energy decreases

Magnetic behaviour changes from diamagnetic to paramagnetic.

(d) NO → NO+

B.O. 2 2.5

Bond energy increases

Magnetic behaviour changes from paramagnetic to diamagnetic.

18. (b)

No. of electrons in CO = 6 + 8 = 14

No. of electrons in NO+ = 7 + 8 – 1 = 14

∴ Co and NO+ are isoelectronic species.

Isoelectronic species have identical bond order.

RJ VISION PVT. LTD.


19. (c)

Applying VSEPR theory, both NF3 and H2O are sp3 hybridized.

20. (a)

21. (a)

Molecular orbital configuration of

2O is

2O (17) = σ1s2, σ*1s

2, σ2s

2, σ*2s

2,

1

y

2

x

2

y

2

x

2

z 2p*2p*,2p2p,σ2p ππππ

22. (a)

23. (b)

(BH3)2 or (B2H6)

It contains two 3 centre – 2 electron bonds and present above and below the plane of molecules compounds which do

not have sufficient number of electrons to form normal covalent bonds are called electron deficient molecules.

24. (d)

Bond order = 2

NN ab

2He = σ(1s)2 σ*(1s) B.O. = 0.5

2O = KK σ(2s)2 σ*(2s)

2 σ(2pz)

2 π(2px)

2 π(2py)

2 π*(2px)

2 π*(2py)

1 B.O. = 1.5

NO = KK σ(2s)2 σ*(2s)

2 π(2px)

2 π(2py)

2 σ*(2pz)

2 π*(2px)

1 B.O. = 2.5

2

2C = KK σ(2s)2 σ*(2s)

2 π(2px)

2 π(2py)

2 σ(2pz)

2 B.O. = 3.0

25. (a)

O2 = KK (σ2s)2(σ*2s)

2(σ2pz)

2(π2px)

2(π2py)

2(π*2px)

1(π*2py)

1

2O = KK (σ2s)2(σ*2s)

2(σ2pz)

2(π2px)

2(π2py)

2(π*2px)

2(π*2py)

1

26. (a)

Both 2

2O and B2 has bond order equal to 1

] π πσσσσ 1

z

1

y

2222

2 2p2p2s* 2s 1s* 1s[(10)B

Bond order = 12

2

2

46

2

NN ab

B2 is known in the gas phase

2222

2 2s*σ σ2s 1s*σ σ1sO 2y

2

x

2

y

2

x

2

z 2p*2p*2p2pσ2p ππππ

Bond order = 16)(82

1

RJ VISION PVT. LTD.


27. (d)

PCl3

28. (b)

(O2) = σ1s2 σ*1s

2 σ2s

2 σ*2s

2 2

x

2

z 2pσ2p π1

y

1

x

2

y 2p*2p*2p π ππ

Bond order = 22

4

2

610

2

aNNb

ionO2

= σ1s2 σ*1s

2 σ2s

2 σ*2s

2

1

x

2

y

2

x

2

z 2p* 2p2p σ2p πππ

Bond order = 2

12

2

5

2

510

2

NN ab

2O = σ1s2 σ*1s

2 σ2s

2 σ*2s

2

1

y

2

x

2

y

2

x

2

z 2p*2p*2p2pσ2p π ππ π

Bond order = 2

11

2

3

2

710

2

NN ab

2

2O = σ1s2 σ*1s

2 σ2s

2 σ*2s

2

2

y

2

x

2

y

2

x

2

z 2p*2p* 2p2pσ2p πππ π

Bond order = 12

2

2

810

2

NN ab

29. (d)

4BF hybridisation sp3, tetrahedral structure.

4NH hybridisation sp3, tetrahedral structure.

30. (a)

Hybridisation 2

1 [No. of valence electrons of central atom + no. of monovalent atoms attached to it–

2NO 2

1H [5 + 0 + 1 – 0] = 3 = sp

2

3NO 2

1H [5 + 0 + 1 – 0] = 3 = sp

2

2NH

2

1H [5 + 2 + 1 + 0] = 4 = sp

3

4

NH 2

1H [5 + 4 + 0 – 1] = 4 = sp

3

SCN– = sp

i.e.,

2NO and

3NO have same hybridisation.

31. (b)

32. (c)

(a) :NH4

sp

3 hybridisation

(b) CH4 : sp3 hybridisation

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(c) SF4 : sp3d hybridisation

(d) : BF4

sp3 hybridisation

∴ Correct choice : (c)

33. (b)

In ,23NO nitrogen is in sp2

Hybridisation, thus planar in shape. In H3O+, oxygen is in sp

3 hybridisation, thus tetrahedral in shape.

∴ Correct choice : (b)

34. (c)

62

255 :SbCl

2

5

means sp3d

2

43 SF ,I and PCl5; all have sp

3d hybridisation.

35. (b)

32

33 : BF3

means sp

2

32

15 : NO 2

means sp

2

36. (c)

For neutral molecules:

No. of electron pairs = No. of atoms bonded to it 2

1 [Group number of central atom – valency of the central atom].

For ions:

No. of electron pairs = No. of atoms bonded to it 2

1 [Group no. of central atom- valency of the central atom ± no. of

electron]

On calculating no. of electron pairs in given molecules. We find that in the given molecule hybridisation is:

BF3 → sp2

2NO → sp2

2NH → sp3

H2O → sp3

37. (d)

Due to intermolecular hydrogen bonding in methanol, it exist at associated molecule.

38. (d)

Calculating the bond order of various species

:O2

kkσ2s

2 σ*2s

2

2

z2pσ

2

y

2

x

2

y

2

x 2p*2p*2p2p π π ππ

2

bondingnon in electrons ofNumber bonding in electrons ofNumber B.O.

1.5or

2

58

NO : 2

x

22 2p2s*σ σ2s kk π 1

x

2

z

2

y 2p*σ2p 2p ππ

RJ VISION PVT. LTD.


2.5or 2

38

2

NNB.O.

ab

2

x

222

2 2p 2s* 2s kk:C πσσ 2

z

2

y 2p 2p σπ

3or 2

38

2

NNB.O.

ab

122

2 1s*σ1sHe σ

0.5or 2

12

2

NNB.O.

ab

From these values we find the correct order of increasing bond order is

2

22

2

2 CNOOHe

39. (b)

From the structure of three species we can determine the number of lone pair electrons on central atom (i.e., N atom)

and thus the bond angle.

We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between

Nitrogen and oxygen atoms. Thus, smaller is bond angle.

The correct order of bond angle is:

222 ONNONO

i.e. option (b) is correct.

40. (d)

The shape of ozone molecule is

In it we find 2σ and 1π bond, i.e., option (d) is correct.

41. (c)

All these structures exhibits resonance and can be represented by the following resonating structures.

More is the single bond character. More will be the bond length. Hence, the correct order is:

CO < CO2 <

3CO

RJ VISION PVT. LTD.


42. (c)

Hybridisation of Br is

3BrO

Total valence electrons = 7 + 3 × 6 = 25

Charge = –1 hence

Total = 25 + 1 = 26

52(R)3(Q)8

26

Hybridisation = dsp3

Hybridization of Xe in XeO3

Total valence electrons

= 8 + 3 × 6 = 26 = 8

26 = 3(Q) + 2(R) = 5

Hybridization = dsp3

In both cases, the structure is trigonal pyramidal.

43. (d)

In SnCl4 there is sp3 hybridisation so the structure is tetrahedral. In

4

3

4

2

4 NH,PO,SO the structure is tetrahedral because

in all hybridisation is sp3. But in SCl4, sp

3d hybridization present so shape is different which is see saw.

44. (b)

2NO will have linear shape as it will have sp hybridisation.

45. (a)

In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite direction so

in the former case they are added up whereas in the latter case net result is reduction of dipole moment. It has been

shown in the following figure:

46. (d)

In BF3, AlF3 & NF3 all fluoride atoms are symmetrically oriented with respect to central metal atom but in ClF3 three

fluorine atoms are arranged as follows:

Here two bonds are in equatorial plane & one bond is in axial plane.

47. (d)

Electronic configuration of the molecule according to molecular orbital theory, is

σ 1s2 σ*1s

2 σ2s

2 σ*2s

2 σ2pz

2 (π2px

2 = σ2py

2)

(π*2px2 = π2py

2). σ*2pz

2 σ3s

2 σ*3s

2 σ3pz

2

(π3px2 = π3py

2) (π*3px

1 = 3py

1)

Last two electrons are unpaired. So no. of unpaired electron is 2.

RJ VISION PVT. LTD.


48. (b)

Statement (a), (c), (d) are correct. Statement (b) is incorrect statement.

AB5 may have two structures as follows:

49. (b)

SF4 has permanent dipole moment.

SF4 has sp3d hybridization and see saw shape (irregular geometry)

Whereas XeF4 shows square planar geometry SiF4 has tetrahedral shape and BF3 has Trigonal planar shape. All these

are symmetric molecule. Hence μ ≠ 0.

50. (a)

BF3 is sp2 hybridised. So, it is trigonal planner. NH3, PCl3 has sp

3 hybridisation hence has trigonal bipyramidal shape, IF3,

has sp3d hybridization and has linear shape.

51. (c)

As difference of electronegativity increases % ionic character increases and covalent character decreases i.e., negativity

difference decreases covalent character increase.

Further greater the charge on the cation more will be its covalent character. Be has maximum (+2) charge.

52. (b)

The compound, of which central atom is octetless known as electron deficient compound. Hence B2H6 is electron

deficient compound.

53. (c)

In BrF3, both bond pairs as well as lone pairs of electrons are present. Due to the presence of lone pairs of electrons

(1p) in the valence shell, the bond angle is contracted and the molecule takes the T-shape. This is due to greater

repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pa irs.

RJ VISION PVT. LTD.


54. (a)

Only those d orbitals whose lobes are directed along X, Y and Z directions hybridise with s and p orbital. In other three

d orbitals namely dxy, dyz and dxz, the lobes are at an angle of 45° from both axis, hence the extent of their overlap with

s and p orbitals is much lesser than 22 yxd

and 2z

d orbitals.

55. (d)

In a linear symmetrical molecule like BeF2, the bond angle between three atoms is 180°, hence the polarity due to one

bond is cancelled by the equal polarity due to other bond, while it is not so in angular molecules, like H2O.

56. (a)

Thus here bond angles between

X4 – M – X2 = 180°

X1 – M – X3 = 180°

X5 – M – X6 = 180°

57. (a)

Shape (structure) of a species can be ascertained on the basis of hybridisation of its central atom, which in turn can be

known by determining the number of hybrid bonds

(H). Thus A)CX(V2

1H

For SiF4 : 4;0)04(42

1H sp

3 (Tetrahedral structure)

For SF4: 5;0)04(62

1H sp

3d (See saw structure)

Hence the two compounds have different structures.

58. (c)

As sigma bond is stronger than the π(pi) bond, so it must be having higher bond energy than π(pi) bond.

59. (d)

To form ,NO3

nitrogen uses one p-electron for π-bond formation two p-electrons for σ-bond formation. 2s electrons

are used for coordinate bond formation. Thus there is no lone pair on nitrogen but four bond pairs.

RJ VISION PVT. LTD.


60. (b)

In 2

3SO

S is sp3 hybridised, so

16S = 1s2, 2s

2 2p

6,

ionhybridisat sp

1

z

1

y

1

x

2

3

3p3p3p3s

unhybride

1

xy3d

In ‘S’ unhybride d- orbit is present, which is involved in π bond formation

8O = 1s2, 2s

2

1

z

1

y

2

x 2p2p2p

In oxygen two unpaired p- orbital is present in these one is involved in σ bond formation while other is used in π bond

formation.

Thus in ,SO2

3

pπ and dπ orbitals are involved for pπ – dπ bonding.

61. (a)

In X — H - - - - Y, X and Y both are electronegative elements (i.e., attracts the electron pair) then electron density on X

will increase and on H will decrease.

62. (a)

For π-overlap the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

63. (c)

In XeF2 and

2IF . Both XeF2 and

2IF are sp3d hybridized and have planar shape.

64. (c)

We know that bond angles of NH3 = 107°,

4NH = 109.5°, PCl3 = 100°. Therefore bond angle of

4NH is maximum.

65. (c)

In P–O bond, π bond is formed by the sidewise overlapping of d-orbital of P and p-orbital of oxygen. Hence it is

formed by pπ and dπ overlapping.

66. (a)

We know that the electrostatic force that binds the oppositely charged ions which are formed by transfer of electron

from one atom to another is called ionic bond. We also know that cation and anion are oppositely charged particles

therefore they form ionic bond in crystal.

67. (c)

Dissociation energy of any molecules depends upon bond order. Bond order in N2 molecule is 3 while bond order in

2N is 2.5.

Further we know that more the Bond order, more is the stability and more is the BDE.

68. (a)

Boron in BCl3 has 6 electrons in outer most shell. Hence BCl3 is a electron pair deficient compound.

69. (d)

HF form linear polymer structure due to hydrogen bonding.

RJ VISION PVT. LTD.


70. (d)

PF3 has pyramidal shape

Phosphorus exist in sp3 hybridisation state hence it exist in tetrahedral shape. But due to presence of lone pair its

shape is pyramidal.

F–H----F—H----F—H----F

71. (b)

NO+ = σ1s

2 σ*1s

2 σ2s

2 σ*2s

2 σ2px

2

π2py2 = π2pz

2

Bond order of NO+ = )N(N

2

1ab

362

14)(10

2

1

Similarly, Bond order of NO 5)(102

1

2.5(5)2

1

Bond order of NO– = 2(4)

2

16)(10

2

1

Bond order of 7)(102

1O2

1.5(3)2

1

By above calculation, we get decreasing bond order

NO+ > NO > NO

– >

2O

72. (c)

As dipole moment = electric charge × bond length

D. M. of AB molecules

= 4.8 × 10–10

× 2.82 × 10–8

= 13.53 D

D. M. of CD molecules

= 4.8 × 10–10

× 2.67 × 10–10

= 12.81 D

Now % ionic character

compound ionic pure of moment Dipole

bond the of moment dipole Actual then % ionic character in

76.94%10013.53

10.41AB

% ionic character in

80.23%10012.81

10.27CD

73. (b)

XeF4 hybridisation is = 2

1(V + X – C + A) hence V = 8 (no. of valence v

–)

RJ VISION PVT. LTD.


X = 4 (no of monovalent atom) 60)04(82

1 sp

3d

2

C = 0 charge on cation

A = 0 (charge on anion). The shape is square planar shape.

74. (c)

Bond order between P – O 1.254

5

structures resonating of no. total

direction possible all in bonds of no.

or formal charge on oxygen = 0.754

3

(Formal charge = No. of electrons in valence shell

electrons free of No,electrons bonding of No.

2

1

75. (a)

Paramagnetic character is based upon presence of unpaired electron.

17Cl– = 1s

2, 2s

22p

6, 3s

2 2

z

2

y

2

x 3p3p3p

4Be = 1s2, 2s

1

10Ne2+

= 1s2, 2s

2

1

z

1

y

2

x 2p2p2p

33As+ = 1s

2, 2s

2 2p

6, 3s

2 3p

6 3d

10

4s2

0

z

1

y

1

x 4p4p4p

Hence, only Cl– do not have unpaired electrons.

76. (d)

Total no. of electrons in 2

2O = 16 + 2 = 18. Distribution of electrons in molecular orbital

σ1s2, σ*1s

2, σ2s

2, σ*2s

2, ,2p2

xσ2

y2pσ

2

y

2

x

2

z 2p*2p* ,2p πππ

Anti bonding electron = 8 (4 pairs)

77. (c)

The electronic configuration of As is

As = 1s2, 2s

2, 2p

6,

tiondhybridisa sp

11

z

1

y

1

x

1

3

3d3p3p3p3s

Thus, the hyrbidisation involved in the AsF5 molecule is trigonal bipyramidal. So, the hybrid orbitals used are s, p x, py,

pz, 2

zd

78. (b)

We know that in O2 bond, the order is 2 and in

2O bond, the order is 1.5. Therefore the wrong statement is (b).

RJ VISION PVT. LTD.


79. (d)

In alkynes the hybridisation is sp i.e. each carbon atom undergoes sp hybridisation to form two sp -hybrid orbitals. The

two 2p-orbitals remain unhybridised. Hybrid orbitals form one sigma and two unhybrisdised orbitals form π-bonds.

Hence two π bond and one sigma bond between C – C lead to cylindrical shape.

80. (a)

We know that due to polar nature, water molecules are held together by intermolecular hydrogen bonds. The structure

of ice is open with large number of vacant spaces, therefore the density of ice is less than water.

81. (a)

The bond represented by dots form the 3 -centred electron pair bond. The idea of three centred electron pair bond B–

H–B bridges is necessitated because diborane does not have sufficient electrons to form normal covalen t bonds. It has

only 12 electrons instead of 14 required to give simple ethane like structure for diborane.

82. (c)

The N – O bond length decreases in the order.

83. (d)

In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding

molecular orbital. So that Nb = 8 and Na = 2.

32)(82

1)N(N

2

1order Bond ab

84. (d)

The bond length of O – O in O2 is 1.21 Å, in H2O2 it is 1.48 Å and in O3 it is 1.28 Å.

∴ Correct order of bond length is H2O2 > O3 > O2.

85. (d)

As there is no lone pair on boron in BCl3 therefore no repulsion takes place. But there is a lone pair on nitrogen in NCl3.

Therefore repulsion takes place. Thus BCl3 is planar molecule but NCl3 is a pyramidal molecule.

86. (b)

Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons. In NO the presence of

unpaired electron is clear. Therefore it is paramagnetic.

87. (c)

The C – C bond distance decreases as the multiplicity of the bond increases. Thus, bond distance decreases in the

order: butane (1.54 Å) > benzene (1.39 Å) > ethene (1.36 Å) > ethyne (1.20 Å) . Thus in butane, C – C bond distance is

the largest.

RJ VISION PVT. LTD.


88. (c)

The b.p. of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-

bonding but in o-nitrophenol it is intermolecular H-bonding.

89. (c)

Chemical bonds.

90. (b)

The angle between the bonds formed by px and py orbitals is the minimum i.e., 90°

91. (b)

The stability of the ionic bond depends upon the lattice energy which is expected to be more between Mg and F due

to +2 charge on Mg atom.

92. (b)

BH3 has sp2 hybridization and hence does not have tetrahedral structure while all others have tetrahedral structures.

93. (d)

H-bond is the weakest

94. (b)

The removal of an electron from a diatomic molecule may increases the bond order as in the conservation O 2(2) →

2O (2.5) or decrease the bond order as the conversion, N2(3.0) →

2N (2.5). As a result, the bond energy may increase or

decrease, thus, statement (b) is incorrect.

95. (a)

According to Fajan rules, ionic character increases with increase in size of the cation and decrease in size of the anion.

Thus, CsF has higher ionic character than NaCl and hence bond in CsF is stronger than in NaCl.

96. (c)

The strength of the interactions follows the order Vander Waal’s < hydrogen – bonding < dipole-dipole < covalent.

97. (b)

For compounds containing ions of same charge, lattice energy increases as the size the ions decrease. Thus, NaF has

highest lattice energy. The size of cation is in the order Na+ < K

+ < Rb

+ < Cs

+

98. (b)

Sigma bond is stronger than π-bond. The electrons in the π bond are loosely held. The bond is easily broken and is

more reactive than σ-bond. Energy released during sigma bond formation is always more than π bond because of

greater extent of overlapping.

99. (c)

H – F shows strongest H-bonds.

100. (b)

CO2 has sp-hybridization and is linear. SO2 and 2

3CO are planar (sp2) while

2

4SO is tetrahedral (sp3).

101. (c)

The bond length decreases in the order sp3 > sp

2 > sp.

Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest.

102. (c)

Polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. Greater the

electronegativity difference more is the polarity of the bond.

RJ VISION PVT. LTD.


N – Cl O – F N – F N – N

3.0–3.0 3.5–4.0 3.0–4.0 3.0–3.0

As the electronegativity difference between N and F is maximum hence this bond is most polar.

103. (b)

CCl4 has sp3 hybridisation, tetrahedral geometry and all bond angles of 109° 28’.

104. (d)

The star marked carbon has a valency of 5 and hence this formula is not cor rect.

105. (a)

Linear combination of two hybridized orbitals leads to the formation of sigma bond.

106. (d)

With the increase of electronegativity and decrease in size of the atom to which hydrogen is covalently linked, the

strength of hydrogen bond increases. As F is most electronegative thus HF shows maximum strength of hydrogen

bond.

107. (a)

A σ-bond is stronger than a π-bond hence option (a) is not correct. Sigma (σ) bonds are formed by head on overlap of

unhybridized s-s, p-p or s-p orbitals and hybridised orbitals (sp, sp2, sp

3, sp

3d and sp

3d

2) hence σ bonds are strong

bonds. Where as Pi (π)-bonds are formed by side ways overlap of unhybridised p- and d orbitals-hence π bonds are

weak bonds.

108. (c)

As we move in period form Li → Be → B → C, the electronegativity (EN) increases and hence the EN difference between

the element and Cl decreases and accordingly the covalent character increases Thus option (c) i.e.,

LiCl < BeCl2 < BCl3 < CCl4 is correct.

109. (b)

BeF2 is linear and hence has zero dipole moment while H2O, being a bent molecule, has a finite or non-zero dipole

moment.

110. (b)

BF3 involves sp2-hybridization

111. (d)

In metallic bonds each ion is surrounded by equal no. of oppositely charged ions hence have electrostatic attraction on

all sides and hence do not have directional characteristics.

112. (a)

For linear arrangement of atoms the hybridisation should be sp (linear shape, 180° angle). Only H 2S has sp3-

hybridization and hence has angular shape while C2H2, BeH2 and CO2 all involve sp - hybridisation and hence have

linear arrangement of atoms.

113. (b)

Equilateral or triangular planar shape involves sp2 hybridization.

114. (a)

The overlap between s- and p-orbitals occurs along internuclear axis and hence and angle is 180°.

RJ VISION PVT. LTD.


1. (a)

Van der waal constant ‘a’, signifies intermolecular forces of attraction.

Higher is the value of ’a’, easier will be the liquefaction of gas.

2. (d)

In real gas equation, nRTnbVV

anP

)(

2

2

Van der Waal’s constant, ‘a’ signifies intermolecular forces of attraction.

3. (d)

)s(SrCO3 )g(CO)s(SrO 2

2COp PK

Maximum pressure of CO2 = 1.6 atm

P1V1 = P2V2

0.4 20 = 1.6 V2

V2 = 5L

4. (a)

Real gases show ideal gas behaviour at high temperatures and low pressures.

5. (c)

H2

32

16

O2

32

1

CH4

32

2 mole ratio

6. (d)

Density = g/ml 3.415000.0821

285

RT

PM

7. (b)

This type of attractive force operates between the polar molecules having permanent dipole and the molecules lacking

permanent dipole.

HCl is polar (μ ≠ 0) and He is non polar (μ = 0), thus gives dipole-induced dipole interaction.

8. (c)

Higher the critical temperature more easily will be the gas liquify. Now since most easily liquefiable gas show larger

deviation, NH3 will show maximum deviation from ideal behavior.

9. (d)

‚Molar mass increase, ‘a’ increases

size of molecule increase, ‘b’ increase

b(L/mol) a(bar. L2/mol

2)

H2 → 0.02661 CH4 → 2.25

He → 0.0237 O2 → 1.36

O2 → 0.03183 H2 → 0.244

CO2 → 0.04267

CH – 5 STATES OF MATTER

RJ VISION PVT. LTD.


10. (b)

M

1r

1

2

2

1

M

M

r

r

4

M

r

3r 2

1

1

4

M9 2

M2 = 36 g/mole

11. (c)

⇒ AM

36

150

200 ⇒

AM

36

3

4

⇒

AM

36

9

16

⇒ 20.254

81MA

So, most approximate answer is 32.

12. (a)

Given

P1 = 1.5 bar T1 = 273 + 15 = 255 K V1 = V

P2 = 1.0 bar T1 = 273 + 25 = 298 K V2 = ?

2

22

1

11

T

VP

T

VP

298

V1

288

V1.5 2

V2 = 1.55 V i.e., volume of bubble will be almost 1.6 time to initial volume of bubble.

13. (c)

For free expansion of an ideal gas under adiabatic condition

q = 0, ∆T = 0, w = 0

14. (a)

Given nCO = 2Nn

PCO + 2NP = 1 atm

Partial pressure of a gas = mole fraction of gas × total pressure

1nn

nP

2

2

2

NCO

N

N

12n

n

2

2

N

N

atm 0.52

1

15. (b)

A

B

B

A

M

M

r

r

49

M

2

1

49

M

10

V20

V

BB

12.25494

1MB

A

B

M

M

B

B

t

V

A

A

t

V

RJ VISION PVT. LTD.


16. (d)

Average velocity = M

8RT

π

i.e., v ∝ T

1.41T

2T

V

V

1

2

17. (a)

Ideal gas during spontaneous expansion into vacuum does not do any external work.

18. (d)

Pa 416480.03

4028.314

16.02

6

V

nRTP

19. (c)

At any constant temperature the K.E. of gaseous molecules remains same. (K.E. ∝ T).

Thus, options (c) is correct answer.

20. (c)

Due to intermolecular H-bonding the surface tension of H2O is more than other liquid. One H2O molecule is joined

with 4 another H2O molecules through H–bond.

Hydrogen bonding is in order H2O > C2H5OH > CH3OH.

21. (a)

At higher temperature and low pressure real gas acts as an ideal gas .

22. (c)

Average molar kinetic energy = kT2

3

As temperature is same hence average kinetic energy of CO and N2 is same.

23. (d)

(Given)1

8

O of Moles

H of Moles

2

2

1

8

O of weight

H of weight .

H of M.W.

O of M.W.

2

2

2

2

2

1

132

28

O of weight

H of weight

2

2

24. (a)

Applying Boyle’s law P1V1 = P2V2 for both gases

3

200P3P400

1000

500

3

400P'3P'

1000

666.6600

⇒ PT = P + P'

= tor 2003

600

3

400

3

200

RJ VISION PVT. LTD.


25. (a)

Rate of diffusion depend upon molecular weight

2

1

2

1

M

M

r

r

⇒ r1 \ r2 if M1 = M2

Hence, compounds are N2O and CO2 as both have same molar mass i.e. 22

26. (b)

At low pressure and high temperature real gas nearly behave like ideal gas. Hence deviation is minimum from ideal

behaviour.

27. (d)

By Ideal gas equation

P1 V = n1 RT

n1 ∝ P1 and n2 ∝ P2

2

1

2

1

P

P

n

n ⇒ 0.30

570

170

n

n

2

1

28. (a)

Given initial volume (V1) = 600 C, C: Initial pressure (P1) = 750 mm and final volume (V2) = 500 c.c. according to Boyle’s

law.

P1V1 = P2V2

Or 750 × 600 = P2 × 500

Or 900mm500

600750P2

Therefore increases in pressure = (900 – 750) = 150 mm.

29. (b)

Given initial volume (V1) = 500 ml; Initial temperature (T1) = 27°C = 300 K and final temperature (T2) = –5°C = 268 K.

From Charle’s law:

2

2

1

1

T

V

T

V or

268

V

300

500 2

Where V2 = New volume of gas

ml 446.66268300

500V2

30. (d)

A substance exists as a liquid above its melting point and below its Boiling point.

31. (a)

By definition of Nernst distribution law.

When a solute is shaken with two immiscible liquids, having solubility in both, the solute distributes itself between the

two liquids in such a way that the ratio of its concentrations in two liquids is constant at a given temperature, provided

the molecular state of the solute remains the same in both the liquids.

32. (d)

u α T or u1/u2 = 21 TT

273927

27327

=

1200

300 =

2

1

u2 = 2u1

33. (b)

At low temperature and high pressure.

RJ VISION PVT. LTD.


34. (d)

Most probable velocity (α) = M

2RT

Mean velocity M

8RT)v(

π

Root mean square velocity (u) = M

3RT

∴ α : v : u = M

2RT :

M

8RT

π :

M

3RT

= 3:8

:2π

35. (a)

PV = 3

1 mnu

2 =

3

1Mu

2

= 2

1

3

2. Mu

2 =

3

2E per unit vol.

36. (c)

PV = nRT or P = V

n RT = CRT.

Hence 1 = 1 × 0.082 × T is T = 0.082

1 = 12 K

37. (a)

Average kinetic energy depends only on temperature

KT

2

3K.E.

38. (c)

In the equation PV = nRT, n moles of gas have volume V.

39. (d)

In the ideal gas, the intermolecular forces of attraction are negligible and hence it cannot be liquefied.

40. (c)

R = 0.082 litre atm K–1

mole–1

.

41. (c)

Molecules in an ideal gas move with different speeds.

42. (b)

Velocity and hence the K.E is maximum in the gaseous state.

43. (b)

PV = 3

1mnu

2 =

3

1Mu

2 or u = 3PV/M (∵ n = NA)

At STP u ∝ M

1

i.e., higher will be the molar mass lower will be the value of urms.

Molecular masses of H2, N2, O2 and HBr are 2, 28, 32 and 81. Hence the correct order is HBr < O2 < N2 < H2.

44. (d)

According to Boyle’s law at constant temperature, V ∝ P

1 or PV = constant

RJ VISION PVT. LTD.


45. (c)

Here volume is constant. Again the mass of H2 is fixed so the number of moles of the gas do not change. As

temperature increases the pressure also increases. The rate of collis ion among the gas molecules and their energy also

increase.

46. (d)

At low pressure and high temperature: At low pressure volume correction for 1 mole of a gas in negligible, i.e., b = 0

thus the gas equation becomes

2V

aP V =RT

or RTV

a1

RT

PVZ

m

m

At higher pressure, the pressure correct for 1 mole of gas in negligible i.e., 0V

a2

Or (P + 0) (V – b) = RT

Or P(Vm – b) = RT

Or PVm = RT + Pb

Or RT

Pb1

RT

PVZ m

47. (c)

Vander Waal’s equation for 1 mole:

2V

aP (V – b) = RT

Here,

2V

aP represent the intermolecular forces. (V – b) is the corrected volume.

48. (a)

Absolute zero is the temperature at which kinetic energy of gas molecules becomes zero i.e., all molecular motio n

ceases.

49. (c)

According to kinetic gas equation 3

1PV mNu

2rms, u = root mean square velocity

⇒ u2 =

mN

3PV or u ∝

m

1 i.e. u ∝ 2

1

m

50. (b)

T

PVconstant or

2

22

1

11

T

VP

T

VP

2

1

22

11

T

T

VP

VP

51. (a)

Charle’s Law – The volume of the given mass of a gas increases or decreases by 273

1 of its volume at 0°C for each

degree rise or fall of temperature at constant pressure.

Vt = V0

273

t1 at const. pressure

52. (d)

PV = nRT = M

mRT

Or PM = V

mRT = dRT ⇒ d =

RT

PM

RJ VISION PVT. LTD.


1. (b)

The reaction for ∆fHo(XY)

)()(2

1)(X

2

122

gXYgYg

Bond energies of X2, Y2 and XY are X, XX

,2

respectively

∆H= 20042

X

XX

On solving, we get

20042

XX

X = 800 kJ/mole

2. (b)

Work done in irreversible process

W = –Pext V

= – 2.5 [4.5 – 2.5] = – 5 L atm

= – 5 × 101.3J = – 505J

Since system is well insulated q = 0

By FLOT E = q + W

E = W = – 505 J.

3. (a)

G = H – TS

For equilibrium G = 0

H = TS

KS

HT

eq425

6.83

10005.35.

Since, the reaction is endothermic it will be spontaneous at T > 425 K.

4. (c)

Entropy of a molecule is directly proportional to number of bond.

5. (b)

Non-expansion work done by the system.

6. (a)

Formation of CO2 from carbon and dioxygen gas can be represented as

C(s) + O2(g) → CO2(g); ∆fH = – 393.5 kJ mol–1

(1 mole = 44 g)

Heat released on formation of 44 g CO2

= – 393.5 kJ mol–1

g 35.244g

mol kJ 393.51

= – 315 kJ

7. (c)

CH – 6 THERMODYNAMICS

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8. (b)

H = U + ng RT

3.32.11.21000

300221.2

G = H Ts

calK 7.263.31000

203003.3

9. (b)

Bonus According to AIPMT Answer Key 2014.

G = 2.303 RT log Ksp

63.3 103 = 2.303 8.31 298 log Ksp

11.09 = log Ksp

8 1012

= Ksp

10. (c)

Applying Hess’s law, equation (i) can be obtained by adding equations (ii) and (iii).

∴ x = y + z

11. (c)

O2HCO2OCH 222x

2x

4

O4H3CO5OHC 22x)5(52

x)(583

2x + 5(5 – x) = 16

⇒ x= 3L

∴ Heat released

317222022.4

2890

22.4

3

12. (c)

∆G = –nFE°

For the reaction 4323

2n

960 × 103 = –4 × 96500 × E°

E° = –2.5 volt

13. (d)

1.673

5

R2

3

R2

5

C

C

V

P

14. (a)

H2O )( ⇌ H2O(g) + Q

∆E = 37558 J/mol

∆E = 37.56 kJ mol–1

15. (c)

273

101.435

T

HS

3

ΔΔ

= 5.260 cal/mol – K

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16. (a)

Since, in the first reaction gaseous products are forming from solid carbon hence en tropy will increase i.e. ∆s = + ve.

C(gr.) + 2

1O2(g) → CO(g); ∆S° = + ve

Since, ∆G° = ∆H° – T∆S hence the value of ∆G decrease on increasing temperature.

17. (b)

Given ∆H

A2

1 ⟶ B + 150 …(1)

3B ⟶ 2C + D –125 …(2)

E + A ⟶ 2D +350 …(3)

To calculate ∆H operate

2 × eq. (1) + eq. (2) – eq. (3)

∆H = 300 – 125 – 350 = – 175

18. (c)

4H(g) → 2H2(g) ; ∆H = –869.6 kJ

Or 2H2(g) → 4H(g) ; ∆H = 869.6 kJ

H2(g) → 2H(g) ; ∆H = 2

869.6 = 434.8 kJ

19. (d)

Given ∆H = 30 kJ mol–1

T = 273 + 27 = 300 K

1

4

T

T mol J300

103

T

ΔHΔS

= 100 J mol–1

K–1

20. (a)

Fe2O3 + CO(g) → 2FeO(s) + CO2(g)

∆H = –26.8 + 33.0 = +6.2 kJ

21. (d)

(1) Kp > 0 (iv) Spontaneous

(2) ∆G < R T In Q (i) Non spontaneous

(3) Kp = 0 (ii) Equilibrium

(4) ΔS

ΔHT (iii) Spontaneous and endothermic

22. (d)

H2O )( H2O(g)

∆H = 40630 J mol–1

∆S = 108.8 JK–1

mol–1

∆G = ∆H – T ∆ S When ∆G = 0, ∆G – T ∆ S = 0

K 373.4mol J108.8

mol J40630

S

HT

1

1

Δ

Δ

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23. (a)

∆S for the reaction 22 Y

2

3X

2

1 ⇌ XY3

∆S = 50 – (30 + 60) = –40 J

For equilibrium ∆G = 0 = 0 = ∆H – T ∆ S

K 7540

30000

S

HT 0

Δ

Δ

24. (b)

Enthalpy of reaction

= B. E(Reactant) – B.E(Product)

= [B.E(C=C) + 4 B.E(C–H) + B.E(H–H)] – [B.E(C–C) + 6B.E.(C–H)]

= [606.1 + (4 × 410.5) + 431.37)] – [336.49 + (6 × 410.5)]

= –120.0 kJ mol–1

25. (b)

∆G = ∆H – T ∆ S

At equilibrium, ∆G = 0

⇒ 0 = (170 × 103J) – T (170 JK

–1)

⇒ T = 1000 K

For spontaneity, ∆G is –ve, which is possible only if T > 1000 K.

26. (c)

The reaction for formation of HCl can be written as

H2 + Cl2 → 2HCl

H – H + Cl – Cl → 2(H – Cl)

Substituting the given values, we get enthalpy of formation of

2HCl = –(862 – 676) = –186 kJ.

∴ Enthalpy of formation of

93kJ.kJ2

186HCl

27. (b)

For the reaction

PCl5(g) ⇌ PCl3(g) + Cl2(g)

The reaction given is an example of decomposition reaction and we know that decomposition reactions are

endothermic in nature, i.e., ∆H > 0.

Further

∆n = (1 + 1) – 1 = +1

Hence, more number of molecules are present in products which shows more randomness i.e.

∆S > 0 (∆S is positive)

28. (d)

We know that q (heat) and work (w) are not state functions but (q + w) is a state functions. H – TS (i.e. G) is also a state

functions. Thus II and III are not state functions so the correct answer is option (d).

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29. (b)

HClCl2

1H

2

122

∆HHCl = reactant of B.E. – products of B.E.

2402

1430

2

190 B.E. of HCl

∴ B.E. of HCl = 215 + 120 + 90 = 425 kJ mol–1

30. (d)

This reaction show the formation of H2O, and the X2 represents the enthalpy of formation of H2O because as the

definition suggests that the enthalpy of formation is the heat evolved or absorbed when one mole of substance is

formed from its constituent atoms.

31. (d)

We know that

∆H = ∆E + P∆V

In the reactions, H2 + Br2 → 2HBr this is no change in volume or ∆V = 0

So, ∆H = ∆E for this reaction

32. (a)

If ∆Gsystem = 0 the system has attained equilibrium is right choice.

In it alternative (d) is most confusing as when ∆G > 0, the process may be spontaneous when it is coupled with a

reaction which has ∆G < 0 and total ∆G is negative, so right answer is (a) .

33. (d)

∆G = ∆H – T∆S

When the reaction is in equilibrium, ∆G = 0

0 = ∆H – T∆S ⇒ T = ΔS

ΔH

K 285.7105

100030T

34. (a)

= – 358.5 kJ

The resonance energy provides extra stability to the benzene molecule so it has to be overcome for hydrogenation to

take place.

So ∆H = –358.5 – (–150.4) = –208.1 kJ

35. (a)

Measure of disorder of a system is nothing but Entropy. For a spontaneous reaction,

∆G < 0. As per Gibbs Helmnoltz equation, ∆G = ∆H – T∆S

Thus ∆G is –ve only

When ∆H = –ve (exothermic)

and ∆S = +ve increasing disorder)

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36. (c)

For a spontaneous reaction

∆G(–ve), which is possible if ∆S = +ve, ∆H = +ve

and T∆S > ∆H *As ∆G = ∆H – T∆S+

37. (c)

As MgO is a oxide of weak base hence some energy is lost to break MgO(s). Hence enthalpy is less than –57.33 kJmol–1

38. (b)

W = –p∆V

= –3(6 – 4) = –6 litre atmosphere

= –6 × 101.32 = –608 J

39. (a)

For a spontaneous process, ∆Stotal is always positive

40. (b)

∆G = ∆H – T∆S

= –382.64 – (298 × (-145.6) × 10–3

)

= –339.3 kJ mol–1

41. (d)

H2(g) + Br2(g) → 2HBr(g)

∆H°= (B E)reactant – (B E)product

= (433 + 192) – (2 × 364)

= 625 – 728 = –103 kJ

42. (c)

∆H = ∆E + nRT

∆n = 3 – (1 + 5)= 3 – 6 = –3

∆H – ∆E = (–3RT)

43. (d)

XeF4 sublimes at room temp., while N2O3 does not exist at room temp. It decomposes even at –30°C.

44. (b)

∆G = –P∆V = Work done

∆V is the change in molar volume in the conversion of graphite to diamond.

L102.25

12

3.31

12ΔV 3

= –1.91 × 10

–3 L

Work done = – (–1.91 × 10–3

) × P × 101.3 J

9794atm101.3101.91

mol J1895P

3

1

1 atm = 105 × 1.013 Pa

⇒ P = 9.92 × 108 Pa

45. (a)

T

HS

ΔΔ

∆S (per mole) = 273

6000

T

moleper H

Δ= 21.98 JK

–1 mol

–1

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46. (c)

Given Cp = 75 JK–1

mol–1

18

100n mole. Q = 1000 J ∆T = ?

Q = nCp∆T

K 2.475100

181000T

Δ

47. (a)

T

qΔS

q → required heat per mole

T → constant absolute temperature

Unit of entropy is JK–1

mol–1

48. (a)

C(s) + O2(g) → CO2(g) ∆H = –94 kCal/mole …(i)

H2(g) + 2

1O2 → H2O(g) ∆H = –68 kCal/mole …(ii)

CH4 + 2O2 → CO2 + 2H2O, ∆H = –213 kCal/mole …(iii)

C(s) + 2H2 → CH4(g), ∆H = ? …(iv)

Eqn. (iv) can be obtained by eq. (i) + eq. (ii) × 2 – eq. (iii)

C(s) + O2 → CO2(g)

2H2 + O2 → 2H2O(g)

CO2 + 2H2O → CH4(g) + 2O2

C(s) + 2H2(g) → CH4(g)

So, 4CHΔH = –94 + 2 (–68) – (–213)

= –94 –136 + 213 = –17 k Cal/mole

49. (d)

For isothermal reversible expansion

w = q = nRT × 2.303 log

1

2

v

v

= 2RT × 2.303 log2

20

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, cal. 9.2T

9.2T

T

qΔS

50. (a)

Internal energy is dependent upon temperature and according to firs t law of thermodynamics total energy of an

isolated system remains same, i.e., in a system of constant mass, energy can neither be created nor destroyed by any

physical or chemical change but can be transformed from one form to another

∆E = q + w

For closed insulated container, q = 0, so, ∆E = +w, as work is done by the system

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51. (a)

2H2O2(l) → 2H2O(l) + O2(g) ∆H = ?

∆H = *2 × (∆Hf of H2O(l)) + (∆Hf of O2) – (2 × ∆Hf of H2O2(l))]

= [(2 × –286) + (0) – (2 × –188)]

= [–572 + 376] = –196 kJ/mole

52. (b)

As volume is constant hence work done in this process is zero hence heat supplied is equal to change in internal

energy.

53. (b)

CH4(g) + 2

1O2(g) → CH3OH(l) ∆H = ?

∴ ∆H = *(∆H of combustion of CH3OH) – (∆H of combustion of CH4)]

= [(–y) – (–x)] = –[–y + x) = x – y

Given ∆H = –ve

∴ x – y < 0

Hence x < y.

54. (c)

Heat of reaction

∆H = Eforward – Ebackward

= 19 – 9 = 10 kJ/mol

55. (c)

322 SOO

2

1SO

o

)f(SO

o

)f(SO 23HHH ΔΔΔ

= –98.2 + 298.2 = 200 kJ/mole

56. (b)

For the reaction

2 ZnS → 2 Zn + S2; ∆G1° = 293 kJ …(1)

2 Zn + O2 → ZnO; ∆G2° = –480 kJ …(2)

S2 + 2O2 → 2SO2; ∆G3° = –544 kJ …(3)

∆G° for the reaction

2 ZnS + 3 O2 → 2 ZnO + 2 SO2

Can be obtained by adding eqn. (1), (2) and (3)

⇒ ∆G° = 293 – 480 – 544 = –731 kJ

57. (a)

T

H

point Melting

fusion of heat LatentS fΔ

Δ

11mol K J

300

2930

= 9.77 J K–1

mol–1

58. (c)

As all reactant and product are liquid

∆n(g) = 0

∆H = ∆E - ∆nRT ∆H = ∆E (∵ ∆n = 0)

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59. (a)

∆E = ∆Q – W

For adiabatic expansion, ∆Q = 0

⇒ ∆E = –W

The negative sign shows decrease in Internal energy, which is equal to the work done on the system by the

surroundings.

60. (d)

For a cyclic process

∆E = 0, ∆H = 0 & ∆G = 0. As all depend upon final state and initial state, w doesn’t depend initial and final state.

61. (b)

For an isothermal process ∆E = 0

62. (a)

We know from the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero

temperature is taken to be zero.

63. (b)

Given C + O2 = CO2, ∆H° = – x kJ …(i)

2CO2 = 2CO + O2, ∆H° = +y kJ …(ii)

Or CO2 = CO + 1/2O2, ∆H° = +y/2 kJ …(iii)

From eq. no (1) and (3)

CO,O2

1C 2 ∆H° = y/2 – x =

2

2xy kJ

64. (b)

Hydration energy of Cl+ is very less than H

+ hence it doesn’t from Cl

+ (aq) ions.

65. (d)

Entropy change

∆S = ∆Sproduct – ∆Sreactant

= 2 (186.7) – (223 + 130.6)

= 373.4 – 353.6

= 19.8 JK–1

mol–1

66. (b)

Entropy states the randomness or disorderness of the system. At absolune zero, the movement of molecules of the

system or randomness of the system is zero, hence entropy is also zero.

67. (a)

The relation between free energy change and equilibrium constant is given by Nernst equation

Q lnnF

RTEE o

At equilibrium, E = 0 and Q = KC

K lnnF

RTEo …(i)

Again ∆G = –nFE° …(ii)

Put in (i)

;K lnnF

RT

nF

ΔGo

∆G° = –RT ln K

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68. (b)

Enthalpy of formation of C2H4, CO2 and H2O are 52, –394 and –286 kJ/mol respective. (Given)

The reaction is

C2H4 + 3O2 → 2CO2 + 2H2O

Change in enthalpy,

(∆H) = ∆Hproducts – ∆Hreactants

= 2 × (–394) + 2 × (–286) – (52 + 0)

= –1412 kJ/mol

69. (a)

If more trans-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the K

constant, concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.

70. (b)

This is combustion reaction, which is always exothermic hence

∆H = –ve

As the no. of gaseous molecules are increasing hence entropy increase

Now ∆G = ∆H – T∆S

For a spontaneous reaction

∆G = –ve

Which is possible in this case as ∆H = –ve and ∆S = +ve.

71. (d)

∆G is negative for a spontaneous process.

72. (c)

During isothermal expansion of ideal gas,

∆T = 0. Now H = E + PV

∵ ∆H = ∆E + ∆(PV)

∴ ∆H = ∆E + ∆(nRT);

Thus if ∆T = 0., ∆H = ∆E

i.e., remain unaffected

73. (b)

∆ng = 2 – 4 = –2, ∆H = ∆E – 2RT.

74. (b)

1 M H2SO4 = 2g eq. of H2SO4

Hence y = 2x or y2

1x

75. (d)

As ∆H = ∆E + ∆ngRT

If np < nr; ∆ng = np – nr = –ve

Hence ∆H < ∆E

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1. (d)

Meq of HCl = 1515

175

Meq of NAOH = 515

125

Meq of HCl in resulting solution = 10

Molarity of [H+] in resulting mixture = 10

1

100

10

pH = -log[H+] = -log 0.110

1

2. (a)

Solubility of BaSO4, )L (mol233

1042.2 1-3

s )L (mol1004.1

-15

)(4 sBaSO ⇌ )()(2

4

2aqSOaqBa

ss

22

4

2]][[ sSOBaKsp

= (1.04 x 10

—5)2

= 1.08 x 10-10

mol2 L

-2

3. (a)

)()( 22 gBgA ⇌ xkJ);(2 HgX

On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction .

On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.

So, high pressure and low temperature favours maximum formation of product.

4. (a)

22 H3N K NH2 13 → (1)

22 ON K NO2 2 → (2)

22 O2

1H

K O H 32 → (3)

For reaction 23 O2

5NH2 OH3NO2 2 (4)

Equation (4)

= equation (2) + 3 equation (3) – equation (1)

1

332

K

K.KK , Option (a).

5. (a)

Equilibrium constant is not affected by presence of catalyst hence statement (a) is incorrect.

CH – 7 EQUILIBRIUM

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6. (c)

422 OCAg Ag2 + 242OC

2.2 10–4

M 1.1 10–4

M

]OC[]Ag[K2

422

sp

]101.1.[]102.2[ 424

12sp 103.5K

7. (d)

n

pc RTKK )( = 825 1056.3)29808314.0(108.5

8. (a)

Catalyst do not affect the value of equilibrium.

9. (b)

n

pc RTKK )(

nc

P

RTK

K

)(

1

10. (a)

N1V1 – N2V2 = N.V.

0.1 × 1 – 0.01 × 1 = N × 2

[OH–] = N 0.045

2

0.09NR

pOH = –log (0.045) = 1.35

∴ pH = 14 – pOH = 14 – 1.35 = 12.65

11. (a)

HClO4 and NaClO4 cannot act as an acidic buffer.

12. (a)

N2(g) + O2(g) ⇌ 2NO(g); K

(g)O2

1(g)N

2

122 ⇌ NO(g); K'

when a reaction is multiplied by 1/2 then

K' = (K)1/2

13. (d)

14. (c)

15. (a)

21

0

1

2

T

1

T

1

2.303R

ΔH

K

Klog

T2> T1 So Kp<Kp’ (exothermic reaction)

(Assuming T2> T1, although it is not mentioned, which temperature is higher )

If T1> T2 then Kp>K’p then answer should be (2).

16. (d)

According to Le-Chatalier principle.

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17. (c)

;0.10

K;0.037

c

Kα aa

Ka = (0.037)2 × 0.10 = 1.37 × 10

–4

18. (c)

pH = pKa + log [Acid]

[Salt]

5 = 4 + log [Acid]

[Salt]

∵ pKa = – log Ka

Given Ka = 1 × 10–4

∴ pKa = – log (1 × 10–4

)

= 4

Now, from Handerson equation

pH = pKa + log[Acid]

[Salt]

putting the values

5 = 4 + log [Acid]

[Salt]

log[Acid]

[Salt] = 5 – 4 = 1

taking antilog

[Salt]/[Acid] = 10 = 1

10

19. (a)

Kw at 25°C = 1 × 10–14

At 25°C

Kw = [H+] [OH

–] = 10

–14

At 100°C (given)

Kw = [H+] [OH

–] = 55 × 10

–14

∵ for a neutral solution

[H+] = [OH

–]

∴ [H+]2 = 55 × 10

–14

or [H+] = (55 × 10

–14)1/2

∵ pH = – log [H+]

On taking log on both side

– log [H+] = – log (55 × 10

–14)1/2

pH = 2

1 log 55 + 7 log 10

pH = – 0.87 + 7

= 6.13

RJ VISION PVT. LTD.


20. (a)

CaCO3 → x

2

3x

2 COCa

CaC2O4 → y

2

42y

2 OCCa

∴ [Ca2+

] = x + y

Now, Ksp (CaCO3) = [Ca2+

] ][CO2

3

Or 4.7 × 10–9

= (x + y)x

Similarly, Ksp (CaC2O4) = [Ca2+

] ]O[C 2

42

Or 1.3 × 10–9

= (x + y)y

On solving, we get

[Ca2+

] = 7.746 × 10–5

M

21. (b)

BF3 is Lewis acid (e– pair acceptor)

22. (c)

Solubility of alkali metal is maximum among the following. Among ZnS (1.7 × 10–5

) & Cus (8 × 10–37

), ZnS has higher

value of Ksp.

23. (c)

A2 + B2 ⇌ 2AB ]][B[A

[AB]K

22

2

c

33

23

c104.2103

)10(2.8K

0.624.23

(2.8) 2

24. (c)

2SO2 + O2 ⇌ 2SO3 K = 278 (given)

SO3 ⇌ SO2 + 2

1O2

K

1K' 24 1061035.97

278

1

25. (a)

Lets take an example of an acidic buffer CH3COOH and CH3COONa.

CH3COOH ⇌ CH3COO– + H

+; CH3COONa ⇌ CH3COO

– + Na

+

When few drops of HCl are added to this buffer, the H+ of HCl immediately combine with CH3COO

– ions to form

undissociated acetic acid molecules. Thus there will be no with CH3COO– ions to form undissociated acetic acid

molecules. Thus there will be no appreciable change in its pH value. Likewise if few drops of NaOH are added, the OH–

ions will combine with H+ ions to form unionized water molecule. Thus pH of solution will remain constant.

26. (a)

(AlCl3, LiCl & BeCl2) all these solutions are acidic due to cationic hydrolysis, where as BaCl2, is salt of strong base and

strong acid, hence its solution will almost neutral i.e., pH ≈ 7.

27. (b)

Given pH = 12

Or [H+] = 10

–12

RJ VISION PVT. LTD.


Since, [H+][OH

–] = 10

–14

∴ 2

12

14

1010

10][OH

Ba(OH)2 ⇌ 2ss

2 2OHBa

[OH–] = 10

–2

2s = 10

–2

2

10s

2

Ksp = 4s3

32

2

104

= 5 × 10–7

28. (c)

Ksp = [Ag+] [Cl

–]

1.8 × 10–10

= [Ag+] [0.1]

[Ag+] = 1.8 × 10

–9 M

Ksp = [Pb+2

] [Cl–]2

1.7 × 10–5

= [Pb+2

] [0.1]2

[Pb+2

] = 1.7 × 10–3

M

29. (c)

BF3 behaves as lewis acid

30. (b)

Given [NH3] = 0.3 M, [NH4+] = 0.2 M, Kb = 1.8 × 10

–5.

[base]

[salt]logpKpOH b [pKb = –log Kb; pKb = –log 1.8 × 10

–5]

∴ pKb = 4.74

4.560.47710.30104.740.3

0.2log4.74

pH = 14 – 4.56 = 9.436

31. (c)

2A(g) + B(g) ⇌ 3C(g) + D(g)

Mole ratio 2 1 3 1

Molar concentration at t = 0 1 1 0 0

Equilibrium molar concentration 0.50 0.75 0.75 0.25

(0.75)(0.50)

(0.25)(0.75)

[B][A]

[D][C]K

2

3

2

3

c

32. (c)

Boron in B2H6 is electron deficient.

33. (d)

Kb = 10–10

; Ka = 10–4

or pKa = 4

RJ VISION PVT. LTD.


For the buffer solution containing equal concentration of B– and HB

pH = pKa + log ][

][

Acid

Salt= pKa + log

x

x= pKa + log 1 = pKa

pH = pKa = 4

The octahedral complex ion [CoCl2(NH3)4]+ i.e., tetra amminedichloro cobalt (III) ion exis ts as cis and trans isomers.

34. (d)

2C(s) + O2(g) ⇌ 2CO2(g)

∆n = 2 – 1 = 1

∴ Kc and Kp are not equal.

35. (d)

Acid

Salt logpKpH a

log [H+] = log Ka – log

Acid

Salt

log [H+] = log Ka + log

Salt

Acid

Salt

AcidK][H a

= 1.8 × 10–5

× 0.2

0.1 = 9 × 10

–6 M

36. (d)

Ba(OH)2(s) → (aq)

2

(aq) 2OHBa

pH = 12 or pOH = 2

[OH–] = 10

–2 M

Ba(OH)2 → 2

2

100.5Ba

+ 210

2OH

[∴ Concentration of Ba2+

is half of OH–]

Ksp = [Ba2+

] [OH–]2

= [0.5 × 10–2

] [1 × 10–2

]2

= 0.5 × 10–6

= 5 × 10–7

M3

37. (d)

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant K h can be calculated as

10-

5

14

b

wh 105.65

101.77

101

K

KK

38. (d)

(CH3)3 B is electron deficient, thus behave as a lewis acid.

39. (c)

Given, CH3COOH ⇌ CH3COO– + H

+;

5

a 101.5K1

…(i)

HCN ⇌ H+ + CN

–;

01

a 104.5K2

or H+ + CN

– ⇌ HCN;

10

a

'

a104.5

1

K

1K

2

2 …(ii)

RJ VISION PVT. LTD.


∴ From (i) and (ii), we find the equilibrium constant (Ka) for the reaction,

CN– + CH3COOH ⇌ CH3COO

– + HCN, is

'

aaa 21KKK

45

10105.6510

3

1

104.5

101.5

5

40. (b)

Given: Equilibrium constant (K1) for the reaction

HI(g) 8;K(g);I2

1(g)H

2

1122 …(i)

To find equilibrium constant for the following reaction

H2(g) + I2(g) ⇌ 2HI(g); K2 = ? …(ii)

For this multiply (i) by 2, we get

2HI(g) ⇌ H2(g) + I2(g); K1 = 82 = 64 …(iii)

[Note: When the equation for an equilibrium is multiplied by a factor, the equilibrium constant must be raised to the

power equal to the factor]

Now reverse equation (iii), we get

H2(g) + I2(g) ⇌ 2HI(g); 64

1K …(iv)

[Note: For reversible reaction, the equilibrium constant of the backward reaction is inverse of the equilibrium constant

for the forward reaction.]

Equation (iv) is the same as the required equation (ii), thus K2 for equation (ii) is 64

1

i.e., Option (b) is correct.

41. (c)

Given reaction are

X ⇌ Y + Z …(i)

And A ⇌ 2B …(ii)

Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then

1

9

KP

K

2

P1 (given)

After dissociation,

X ⇌ Y + Z

At equilibrium (1 – α) α α

[Let mole of X dissociation with α as degree of dissociation]

Total number of moles = 1 – α + α + α

= (1 + α)

Thus 1Z1Y1X .P1

P;P1

P;.P1

1P

α

α

α

α

α

α

111P .P1

1/.P

)(1.P

1K

1

α

α

α

α

α

α …(i)

Similarly for A ⇌ 2B

At equilibrium (1 – α) 2α

1K

RJ VISION PVT. LTD.


We have,

2

2

2P P

1

1/

1

P2K

2

α

α

α

α …(ii)

Dividing (i) by (ii), we get

2

1

P2

P1

2

2

1

2

P2

P1

P

P.

4

1

K

Kor

.P4

.P

K

K

α

α

Or

2

1

P

P.

4

19

1

9

K

K

2P

P1

1

36

P

P or

2

1 or P1 : P2 = 36 : 1

42. (b)

For the reaction

At equilibrium x)2(12 (g)2AB

⇌ x2

2x

(g)B2AB(g)

2

2

2

2

c][AB

][B[AB]K or

33

4

4x

x

2

2

cx)}{2(1

x(2x)K

= x3[(1 – x) can be neglected in denominator (1 – x) = 1]

The partial pressure at equilibrium is calculated on the basis of total number of moles at equilibrium.

Total number of moles

= 2 (1 – x) + 2x + x = (2 + x)

∴ Px)(2

x)2(1P

2AB

where P is the total pressure.

,Px)(2

2xPAB

P

x)(2

xP

2B

Since x is very small so can be neglected in denominator

Thus, we get

Px)(1P2AB

PxPAB

P2

xP

2B

Now,

2

2

AB

B

2

AB

PP

PPK

22

22

Px)(12

x.PP(x)

2

33

P12

.Px

[∴ 1 – x ≃ 1]

3

1

Pp33

P

2K xor

P

2.K xor

2

.Px

RJ VISION PVT. LTD.


43. (b)

The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals

increase as we move from Mg to Ba and thus the solution of BaCl2 in water will be most basic and so it will have

highest pH.

44. (c)

For this reaction Keq. Is given by

3

33

Fe(OH)

OHFeK

= (Fe3+

) (OH–)3 [∴ [solid] = 1].

If (OH–) is decreased by

4

1 times then for reaction equilibrium constant to remain constant, we have to increase the

concentration of [Fe3+

] by a factor of 43 i.e., 4 × 4 × 4 = 64. Thus option (c) is correct answer.

45. (b)

[H3O]+ for a solution having pH = 3 is given by

[H3O]+ = 1 × 10

–3 moles/litre

[∴ [H3O]+ = 10

–pH]

Similarly for solution having pH = 4,

[H3O]+ = 1 × 10

–4 moles/litre and for pH = 5

[H3O]+ = 1 × 10

–5 moles/litre

Let the volume of each solution in mixture be IL, then total volume of mixture solution L = (1 + 1 + 1 ) L = 3L

Total [H3O]+ ion present in mixture solution = (10

–3 + 10

–4 + 10

–5) moles

Then [H3O]+ ion concentration of mixture solution

M3

0.0011M

3

101010 543

= 0.00037 M = 3.7 × 10–4

M.

46. (a)

Given [H3O+] = 1 × 10

–10 M

at 25° [H3O+] [OH

–] = 10

–14

4

10

14

1010

10][OH

Now, [OH–] =

OHp10 = 10

–4 =

OHp10

∴ pOH = 4.

47. (d)

Given,

N2 + 3H2 ⇌ 2NH3; K1 …(i)

N2 + O2 ⇌ 2NO; K2 …(ii)

H2 + 2O2

1 ⇌ H2O; K3 …(iii)

We have to calculate

4NH3 + 5O2 → 4NO + 6H2O; K = ?

RJ VISION PVT. LTD.


or 2NH3 + 2O

2

5 → 2NO + 3H2O

for this equation, 5/2

2

2

3

3

2

2

][O][NH

O][H[NO]K

but ,3

22

2

3

1]][H[N

][NHK

]][O[N

[NO]K

22

2

2

& 2

1

][O][H

O][HK

2

3

2

2

3 or 3/2

2

3

2

3

2

3][O][H

O][HK

Now operate,

1

3

32

K

.KK

2

3

3

22

3/2

2

3

2

3

2

22

2

][NH

]][H[N

][O][H

O][H

]][O[N

[NO]

K][O][NH

O][H[NO]5/2

2

2

3

3

2

2

1

3

32

K

KKK

48. (a)

Given Ka = 1.00 × 10–5

, C = 0.100 mol

For a weak electrolyte,

Degree of dissocation

1%100.100

101

C

Kaα) ( 2

5

49. (d)

HNO2 is a weak acid and NaNO2 is salt of that weak acid and strong base (NaOH).

50. (a)

For a solution of 10–8

M HCl [H+] = 10

–8

[H+] of water = 10

–7

Total [H+] = 10

–7 + 10

–8 = 10 × 10

–8 + 10

–8

10–8

(10 + 1) = 11 × 10–8

51. (a)

First option is incorrect as the value of Kp given is wrong. It should have been

2

OCH

CO

P][PP

PK

24

2

52. (d)

Given Kb = 1.0 × 10–12

[BOH] = 0.01 M [OH] = ?

x)c(1eq ct

BOH0t

⇌

cxcx00

OHB

x)(1

cx

x)c(1

xcK

222

b

RJ VISION PVT. LTD.


x)0.01(1

0.01x101.0

212

On calculation, we get, x = 1.0 × 10–5

Now [OH

–] = cx

= 0.01 × 10–5

= 1 × 10–7

mol L–1

53. (a)

The solution formed from isomolar solutions of sodium oxide, sodium sulphide, sodium selenide H2O, H2S, H2Se &

H2Te respectively. As the acidic strength increases from H2O to H2Te thus pH decreases and hence the correct of pHs is

pH1 > pH2 > pH3 > pH4

54. (a)

IVth

group needs higher S2–

ion concentration. In presence of HCl, the dissociation of H2S decreases hence produces

less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H2S which is sufficient to

precipitate IInd

group radicals.

55. (d)

For an acid-base indicator

HIn ⇌ H+ + In

–

[HIn]

]][In[HK ln

or [H+] =

][In

[HIn]K In

or log H+ = log KIn + log

][In

[HIn]

Taking negative on both sides

–log[H+] = –log KIn – log

][In

[HIn]

or log [HIn]

][In

= pH – pKIn

56. (c)

For AX2; Ksp = 4s3

∴ 3.2 × 10–11

= 4s3

or 4

103.13s

11 = 2 × 10

–4

57. (c)

Ksp for AgI = 1 × 10–16

In solution of KI, I– would be due to the both AgI and KI, 10

–4 solution KI would provide = 10

–4 I

–

AgI would provide, say = x I– (x is solubility of AgI)

Total I– = (10

–4 + x), Ksp of AgI = (10

–4 + x)x

⇒ Ksp = 10–4

x + x2

as x is very small ∴ x2 can be ignored

∴ 10–4

x = 10–16

or )(mol1010

10x 112

4

16

y)(solubilit

RJ VISION PVT. LTD.


58. (b)

B(OH)3 does not provide H+ ions in water instead it accepts OH

– ion and hence it is Lewis acid.

B(OH)3 + H2O ⇌ [B(OH)4]– + H

+

59. (c)

The higher is the tendency to donate proton, stronger is the acid. Thus the correct order is

R – COOH > HOH > R – OH > CH ≡ CH depending upon the rate of donation of proton.

60. (d)

For reaction to proceed from right to left rate forward

crate backward

KQ i.e. the reaction will be fast in backward direction.

i.e. rb > rf.

61. (b)

pOH = pKb + log [Base]

[Salt]

or pKb = pOH – log [Base]

[Salt]

but pOH + pH = 14 or pOH = 14 – pH

∴ 14 – pH – log bpK

0.1

0.1

62. (b)

Na2CO3 is a salt of weak acid H2CO3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH

more than 7.

Na2CO3 + 2H2O → acid weak

32base strong

CO2H2NaOH

63. (d)

Given s = 0.5 × 10–4

moles/lit

[MX2 ⇌ M2+

+ 2X–]

∵ For MX2, Ksp = s × (2s)2 = 4s

3

Ksp = 4 × (0.5 × 10–4

)3 = 4 × 0.125 × 10

–12

= 0.5 × 10–12

= 5 × 10–13

64. (c)

For the reaction

BaO2(s) ⇌ BaO(s) + O2(g); ∆H = +ve

At equilibrium Kp = 2OP

[For solid and liquids concentration term is taken as unity]

Hence, the value of equilibrium constant depends only upon partial pressure O 2. Further on increasing temperature

formation of O2 increases as this is an endothermic reaction.

65. (c)

Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting

tendency of proton is

I– < HS

– < NH3 < RNH2

66. (c)

CH3COOH ⇌ CH3COO– + H

+

RJ VISION PVT. LTD.


COOH][CH

]][HCOO[CHK

3

3a

Given that,

[CH3COO–] = [H

+] = 3.4 × 10

–4 M

Ka for CH3COOH = 1.7 × 10–5

CH3COOH is weak acid, so in it [CH3COOH] is equal to initial concentration. Hence

COOH][CH

)10(3.4 )10(3.4101.7

3

445

5

44

3101.7

103.4103.4COOH][CH

= 6.8 × 10–3

M

67. (a)

M2S ⇌

x2xS2M

Solubility product = (2x)2(x) = 4x

3

= 4 (3.5 × 10–6

)3 = 1.7 × 10

–16

68. (b)

CH3COOH is weak acid while NaOH is strong base, so one equivalent of NaOH cannot be neutralized with one

equivalent of CH3COOH. Hence the solution of one equivalent of each does not have pH value as 7. Its pH will be

towards basic side as NaOH is a strong base hence conc. of OH– will be more than the conc. of H

+.

69. (b)

In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the Ka values for successive

loss of protons from these acids differ by at least a factor of 10–3

i.e., 21 aa KK

H2X ⇌ H+ + HX

– (

1aK )

HX– ⇌ H

+ + X

2– (

2aK )

70. (b)

Because NH3 after losing a proton (H+) gives

2NH

NH3 + H2O ⇌ OHNH 32

(Conjugate acid-base pair differ only by a proton)

71. (a)

MgCO3(s) → MgO(s) + CO2(g)

MgO & MgCO3 are solid and they donot exert any pressure and hence only pressure exerted is by CO 2.

Therefore Kp = 2COP

72. (a)

Given: Hydroxyl ion concentration [OH–] = 0.05 mol L

–1.

We know that the [H+][OH

–] = 1 × 10

–14

or [H+] =

1314

1020.05

101

mol L–1

RJ VISION PVT. LTD.


We also know that

Ph = –log[H+] = –log[2 × 10

–13]

= –log2 – log10–13

= –log2 –(–13) log10

= –0.3010 + 13.0000 = 12.6990

Since the value of pH > 7, therefore the solution is basic.

73. (d)

2xx

2 2ABBA

Solubility product = [x] [2x]2 = 4x

3

4 × 10–12

= 4x3 or 3

12

4

104x

∴ x = 10–4

74. (d)

No. of moles of NaOH = 0.140

4

[Molecular weight of NaOH = 40]

No. of moles of OH– = 0.1

Concentration of OH– =

litre 1

0.1 = 0.1 Mole/L

As we know that, [H+] [OH

–] = 10

–14

∴ [H+] = 10

–13 (∵ OH

– = 10

–1)

75. (c)

Lewis acid is that compound which have electron deficiency. e.g. BF3, SnCl2.

76. (d)

For the reaction

XeF6(g) + H2O(g) ⇌ XeOF4(g) + 2HF(g)

O]][H[XeF

][HF][XeOFK

26

2

41 …(1)

and for the reaction

XeO4(g) + XeF6(g) ⇌ XeOF4(g) + XeO3F2(g)

]][XeF[XeO

]F][XeO[XeOFK

64

2342 …(2)

For reaction:

XeO4(g) + 2HF(g) → XeO3F2(g) + H2O(g)

2

4

223

][HF][XeO

O]][HF[XeOK

∴ From eq. no. (1) and (2)

K = K2 / K1

77. (d)

A2(g) + B2(g) ⇌ 3C(g) ⇌ D(g)

step-1 step-2

RJ VISION PVT. LTD.


78. (d)

pH = pKa + log10 [Acid]

[Salt]

For small concentration of buffering agent and for maximum buffer capacity [Acid]

[Salt]

≅ 1

∴ pH = pKa

79. (d)

CuS ⇌

2

S

2

SSuC

Ksp = S2 or S =

spK

For Binary salts like CuS & HgS, solubility, S = spK

⇒ SCuS = ,1031

SHgS =45

10

For Ag2S → s

2

2s

S2Ag

Ksp = 4s3 or 3

44

3 sp

SAg4

10

4

KS

2

∴ The order is CuS > Ag2S > HgS

80. (b)

A2 ⇌ 2A Equilibrium constant is given by

Kc = reactants of terms conc.

product of terms conc.

][A

[A]

2

2

Since the value given is very small, hence conc. of products is less. It means the reaction is slow.

81. (b)

According to equation

2HI ⇌ H2 + I2

At t = 0 (2 moles) 0 0

At equilibrium (2 - 2α) moles α mole α mole

Total moles at equilibrium = 2 - 2𝛂 + α + α = 2 mole

82. (d)

The buffer system present in serum is H2CO3 + NaHCO3 and as we know that a buffer solution resist the change in pH

therefore pH value of blood does not change by a small addition of an acid or a base.

83. (d)

Rate constant of forward reaction (Kf) = 1.1 × 10–2

and rate constant of backward reaction (Kb) = 1.5 × 10–3

per minute.

Equilibrium constant (Kc) = 7.33101.5

101.1

K

K3

2

b

f

84. (a)

Molarity (M) = 10 M. HCl is a strong acid and it is completely dissociated in aqueous solutions as:

HCl ⟶ H+ + Cl

–

10 M 10 M

So, for every moles of HCl, there is one H+.

Therefore [H+] = HCl or [H

+] = 10.

pH = –log[H+] = –log[10] = –1

RJ VISION PVT. LTD.


85. (b)

For Bi2S3. Ksp– = (2s)2 . (3s)

3

= 4s2 . 27s

3 = 108s

5

or 5

70sp

108

101

108

K5s

For MnS. Ksp = s2

or 16

sp 107Ks

for CuS 73

sp 108Ks

for Ag2S Ksp = 2s2.s = 4s

3

or 3 sp

4

Ks

Thus MnS has maximum solubility.

86. (c)

In presence of the given salts, the solubility of AgCl decreases due to common ion effect.

87. (d)

Solid ⇌ Liquid

It is an endothermic process. So when temperature is raised, more liquid is formed. Hence adding heat will shift the

equilibrium in the forward direction.

88. (b)

AgBr has the highest solubility in 10–3

M NH4OH. All other solvents will dissolve AgBr poorly. Moreover bromides of

Ag+, 2

2Hg and 2

2Cu are water insoluble.

89. (d)

According to Le-chatelier’s principle whenever a constraint is applied to a system tends to readjust so as to nullify the

effect of the constraint.

90. (a)

Sodium borate is a salt of strong base (NaOH) and weak acid (H3BO3). Hence its aqueous solution will be basic.

91. (b)

An aqueous solution of acetic acid dissociates as

CH3COOH + H2O ⇌ CH3COO– + H3O

+

92. (a)

For reaction (1)

]][O[N

[NO]K

22

2

1

and for reaction (2)

[NO]

][O][NK

21

21

222

Therefore 2

2

1K

1K

93. (c)

NaCl is a salt of strong acid and strong base hence its aqueous solution will be neutral ie pH = 7. NaHCO 3 is an acidic

salt hence pH < 7. Na2CO3 is a salt of weak acid and strong base. Hence its aqueous solution will be strongly basic i.e.,

pH > 7.

RJ VISION PVT. LTD.


1. (d)

V 59.1E ,o

HBrO/Br

01

2

rBrOBH

,3

51

rOBrOBH V 5.1Eo

/BrO-3

HBrO

o

cellE for the disproportionation of HBrO,

o

cellE = o

/BrO

o

HBrO/Br -32

EEHBrO

= 1.59 -1.5

=0.095 V = + ve

Hence, option (d) is correct answer.

2. (c)

n-factor of 54

MnO

n-factor of 22

42

OC

Ratio of n-factor of

4MnO and 2

42OC is 5 : 2

So, molar ratio in balanced reaction is 2 : 5

The balanced equation is

OHCOMnHOCMnO 22

22

424 81021652

3. (a)

4. (c)

FeCI2 and SnCI2 (both are reducing agent and have lower oxidation no.)

5. (a)

Higher the value of reduction potential higher will be the oxidizing power whereas the lower the value of reduction

potential higher will be the reducing power.

6. (c)

OHCOKClSOKSOHOCHKClO 22

16

42

6

42422

5

3

i.e. maximum change in oxidation number is observed in Cl (+5 to –1)

7. (b)

On reaction with hot and concentrated alkali a mixture of chloride and chlorate is formed

3Cl2 + 3NaOH(excess) Hot O3HNaClO5NaCl 23

51

8. (d)

3

4PO = x + 4 (–2) = –3; x – 8 = –3; x = +5

2

4SO = x + 4 (–2) = –2; x – 8 = –2; x = +6

2

72OCr = 2x + 7 (–2) = –2; 2x – 14 = –2; 2x = 12; x = +6

CH – 8 REDOX REACTIONS

RJ VISION PVT. LTD.


9. (b)

2

3SO → S is in +4 oxidation state

2

42OS → S is in +3 oxidation state

2

62OS → S is in +5 oxidation state

10. (d)

Zinc gives H2 gas with dil. H2SO4/HCl but not with HNO3 because in HNO3,

3NO ion is reduced and give NH4NO3,

N2O, NO and NO2 (based upon the concentration of HNO3)

[Zn + 6%)(nearly

32HNO → Zn(NO3)2 + 2H] × 4

HNO3 + 8H → NH3 + 3H2O

NH3 + HNO3 → NH4NO3

4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H2O

Zn is on the top position of hydrogen in electrochemical series. So Zn displaces H2 from dilute H2SO4 and HCl with

liberation of H2.

Zn + H2SO4 → ZnSO4 + H2

11. (b)

Oxidation number of a compound must be 0. Using the values for A, B and C in the four options we find that A3(BC4)2

is the answer.

12. (d)

Zn → Zn+2

+ 2e– …(1)

8e– + 10H+ + O3HNHNO 243 …(2)

operate eq. (1) × 4 + eq. (2) × 1

4Zn + 10H+ +

3NO → 4Zn+2

+

4NH + 3H2O

13. (d)

Pyrophosphoric acid H4P2O7

Let oxidation state of phosphorus is x

(4 × 1 + (–2) × 7 + 2x) = 0

∴ 2x = 10 or x = 5

14. (c)

2NaOH + H2SO4 → ation)(neutraliz

242 O2HSONa (not redox)

3

0Light

2

0

O2O3 (not redox reaction)

22

Light

02

02 ON2ON

(redox reaction)

Here oxidation of N2 & reduction of O2 is taking place

H2O(l) ΔH2O(g) (not redox reaction)

15. (c)

Carbon has the maximum oxidation state of +4, therefore carbon dioxide (CO2) cannot act as a reducing agent.

RJ VISION PVT. LTD.


16. (a)

Losing of electron is called oxidation.

17. (a)

Let x = oxidation no. of Cr in K2Cr2O7.

∴ (2 × 1) + (2 × x) + 7 (–2) = 0

Or 2 + 2x – 14 = 0 or x = +6.

18. (c)

2

4

61

2

1

2

2

OSHOBa

→ 1

2

1

2

2

4

62

OHOSBa

In this reaction, none of the elements undergoes a change in oxidation number of valency.

19. (a)

O.N. of P in H3PO3 (phosphorous acid)

3 × 1 + x + 3 × (–2) = 0 or x = +3

In orthophosphoric acid (H3PO4) O.N. of P is +5, in hypophosphorous acid (H3PO2) it is +1 while in metaphosphoric

acid (HPO3), it is +5.

20. (b)

The element undergo oxidation itself and reduces other is known as reducing agent. In this reaction O N of Ni Changes

from 0 to +2 and hence Ni acts as a reducing agent.

RJ VISION PVT. LTD.


1. (c)

O3 is reduced into O2 ion and

Ag2O is reduced to Ag so

H2O2 is reducing agent in both (a) and (b).

2. (b)

∴ Correct choice: (b)

3. (b)

H2O2 + [O] OxidationH2O + O2↑

4. (a)

O2H2Na 2A'' →

B''2

'C'H2NaOH

B''222

C''D''HZnONa2NaOHZn

B''2442

D''HZnSOSOdil.HZn

Na produces golden yellow colour with smokeless flame of Bunsen burner.

5. (c)

Fe + dil.H2SO4 → FeSO4 + H2↑

3Fe + Steam

2O4H → Fe3O4 + 4H2↑

Cu + dil. HCl → No reaction

Copper does not evolve H2 from acid as it is below hydrogen in electro chemical series.

2Na +C2H5OH → 2C2H5ONa + H2↑

6. (b)

2 acid

22 base1 acid

21 base

(g)H(aq)OHO(l)H(aq)H

In this reaction H– acts as bronsted base as it accepts one proton (H

+) from H2O and for H2.

7. (c)

Volume strength = 5.6 × Normality

= 5.6 × 1.5 = 8.4 L

8. (b)

Temporary hardness is due to presence of bicarbonates of calcium and magnesium and permanent hardness is due

to the sulphates or chlorides of bond of calcium and magnesium.

9. (d)

O – O – H bond angle in H2O2 is 97°.

10. (b)

Electrostatic forces of attraction are reduced to 1/80th

in water.

CH – 9 HYDROGEN

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11. (b)

In the structure of ice each molecule of H2O is surrounded by three H2O molecules in hexagonal honey comb

manner. On the other in water, each molecule is surrounded by four neighbouring molecules randomly which results

an open cage like s tructure. As a result there are a number of ‘hole’ or open species. In such a structure lesser

number of molecules are packed per ml. When ice melts a large no. of hydrogen bonds are broken. The molecule s

therefore move into the holes or open spaces and come closer to each other than they were in solid state. this result

sharp increase in the density. Therefore ice has lower density than water.

12. (c)

H(g) → H+(g) + e

–.

13. (a)

The complex salt of metaphosphoric acid sodium hexametaphosphate (NaPO3)6, is known as calgon. It is represented

as Na2[Na4(PO3)6]

14. (b)

is the structure of H2O2.

15. (d)

In this reaction H2S is oxidized to sulphur and H2O2 is reduced to H2O, hence this reaction show oxidation-reduction

both i.e., redox reaction.

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1. (c)

So, the most acidic should be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic.

2. (a)

For 2nd

group hydrides, on moving down the group metallic character of metals increases so ionic character of metal

hydride increases. Hence the option (a) should be correct option.

3. (c)

ion hy drated of size

1 mobility Ionic

Smaller size hydrated ion in aq. soln – Rb

+(aq)

Larger size hydrated ion in aq. soln – Li

+(aq)

Lowest ionic mobility in aq. soln ⟶ Li

+(aq) due to high hydration.

4. (c)

All the above compounds absorb CO2, but only superoxide ( 2KO ) releases oxygen.

5. (c)

Thermal stability order

K2CO3 > Na2CO3 > CaCO3 > MgCO3

Therefore MgCO3 release CO2 most easily

MgCO3 Δ MgO + CO2

6. (a)

7. (c)

8. (c)

Na2CO3 is basic due to hydrolysis of 2

3CO ion

OHCO 2

23

⇌ OHHCO3

9. (d)

At the cathode, since the discharge potential of Na+ ions is lower than that of H

+ ions on the mercury, cathode, so

Na+ ions are discharged while H

+ ions remain in the solution.

Na+ + e

– → Na

2Na + xHg → Amalgam Sodium

2HgxNa

10. (c)

Tertiary halide can show ionic reaction with MF so, MF should be moist ionic for reaction to proceed forward. Hence

‘M’ should be ‘Rb’.

11. (b)

All alkali metal salts are ionic (except Lithium) and soluble in water due to the fact that cations get hydrated by water

molecules. The degree of hydration depends upon the size of the cation. Smaller the size o f a cation, greater is its

hydration energy.

CH – 10 THE s-BLOCK ELEMENTS

RJ VISION PVT. LTD.


Relative ionic radii:

Cs+ > Rb

+ > K

+ > Na

+ > Li

+

Relative ionic radii in water or relative degree of hydration

Li+ > Na

+ > K

+ > Rb

+ > Cs

+

12. (c)

All the a lkali metals when heated with oxygen form different types of oxides for example lithium forms lithium oxide

(Li2O), sodium forms sodium peroxide (Na2O2), while K, Rb and Cs form their respective superoxide.

OLiO2

12Li 22

13. (b)

(A) Plaster of paris = OH2

1CaSO 24 (B) Epsomite = MgSO4.7H2O

(C) Kieserite = MgSO4.H2O (D) Gypsum = CaSO4.2H2O

14. (c)

2Al(s) + 2NaOH(aq) + 2H2O(l) → aluminate meta sod.

22 3H2NaAlO

15. (b)

Active ingredient in bleaching powder for bleaching action is Ca(OCl)2

16. (c)

Melting point of halides decreases as the size of the halogen increases. The correct order is

CaF2 > CaCl2 > CaBr2 > CaI2.

17. (b)

(s)colourless

2(g)3(s) CaOCOCaCOA

Δ

2(aq)B

2(s) OH)(Ca2COCaO

C2(aq)322 )Ca(HCO2COCa(OH)

O(g)HCOCaCO)Ca(HCO 22(g)3(s)2(s)3A

∴ Correct choice: (b)

18. (a)

Lattice energy decreases more rapidly than hydration energy for alkaline earth metal hydroxides.

19. (b)

Be2+

is very small, hence its hydration enthalpy is greater that its lattice enthalpy.

(At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)

20. (a)

NaOH is a strong alkali. It combines with acidic and amphoteric oxides to form salts. Since CaO is a basic oxide hence

does not reacts with NaOH.

21. (d)

The stability of alkali metal hydrides decreases from Li to Cs. It is due to the fact that M–H bonds becomes weaker

with increase in size of alkali metals as we move down the group from Li and Cs. Thus the order of stability of

hydrides is

LiH > NaH > KH > RbH > CsH

i.e., option (d) is correct answer.

RJ VISION PVT. LTD.


22. (b)

Smaller the ion more is its ionic mobility in aqueous solution. Ionic radii of the given a lkali metals in the order Na+ <

K+ < Rb

+ < Cs

+ and thus expected ionic mobility will be in the order Cs

+ < Rb

+ < K

+ < Na

+. However due to high

degree of solvation (or hydration) because of lower size or high charge density, the hydrate ion size follows the order

Cs+ < Rb

+ < K

+ < Na

+ and thus conductivity order is Cs

+ > Rb

+ > K

+ > Na

+ i.e., option (b) is correct answer.

23. (a)

The solubility of sulphates of alkaline earth metals decreases as we move down the group from Be to Ba due to the

reason that ionic size increases down the group. The lattice energy remains constant because sulphate ion is so large,

so that small change in cationic sizes do not make any difference. Thus the order:

BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4

24. (a)

As the basicity of metal hydroxides increases down the group from Be to Ba, the thermal stability of their carbonates

also increase in the same order. Further group 1 compounds are more thermally stable than group 2 because their

hydroxide are much basic than group 2 hydroxides therefore, the order of thermal stability is:

BeCO3 < MgCO3 < CaCO3 < K2CO3

25. (c)

Ionic radii of alkali metals in water follows the order Li+ > Na

+ > K

+ > Rb

+ > Cs

+

Thus in aqueous solution due to larger ionic radius Li+ has lowest mobility and hence the correct order of ionic

mobility is

Li+ < Na

+ < K

+ < Rb

+

26. (b)

The given properties coincide with CaCO3

27. (a)

In Castner process, for production of (Na) Sodium metal, Sodium hydroxide (NaOH) is electrolysed at temperature

330°C.

28. (c)

As Cs+ ion has larger size than Li

+ and I

– has larger size than F

–, therefore maximum distance between centres of

cations and anions is in CsI.

29. (b)

Calcium is obtained by electrolysis of a fused mass consisting six parts calcium chloride and one part ca lcium fluoride

at about 700°C in an electrolytic cell.

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30. (c)

MgSO4 is the only alkaline earth metal sulphate which is soluble in water and for solubility hydration energy should

be greater than lattice energy i.e. hydration energy > lattice energy

31. (c)

2Na2CO3 + NO + 3NO2 → 4NaNO2 + Co

32. (d)

Gypsum is CaSO4.2H2O and plaster of Paris is (CaSO4)2.H2O. Therefore, gypsum contain a lower percentage of calcium

than plaster of Paris.

33. (d)

Sodium is obtained by electrolytic reduction of its chloride. Melting point of chloride of sodium is high (803°C) so in

order to lower its melting point (600°C), calcium chloride is added to it.

34. (d)

Be(OH)2 is amphoteric, but the hydroxides of other alkaline earth metals are basic. The basic strength increases

gradually.

35. (d)

Ca2+

ions is an essential element for the contraction of muscles.

36. (b)

MgO, N2O5 is strongly acidic, ZnO and Al2O3 are amphoteric, therefore, MgO is most basic.

37. (c)

Ca and CaH2 both react with H2O to form H2 gas,

Ca + 2H2O → Ca(OH)2 + H2

CaH2 + 2H2O → Ca(OH)2 + 2H2

Whereas K gives H2 while KO2 gives O2 and H2O2

2K + 2H2O ⟶ 2KOH + H2

2KO2 + 2H2O ⟶ 2KOH + O2 + H2O2

Similarly, Na gives H2 while Na2 O2 gives H2O2

2Na + 2H2O → 2NaOH + H2

Na2O2 + 2H2O → 2NaOH + H2O2

Likewise Ba gives H2 while BaO2 gives H2O2

Ba + 2H2O → Ba(OH)2 + H2

BaO2 + 2H2O → Ba(OH)2 + H2O2

38. (c)

Mixture of K2CO3 and Na2CO3 is called fusion mixture.

39. (c)

A cation is always much smaller than the corresponding atom, whereas an anion is always larger than the

corresponding atom, hence the size decreases in the order

Na– > Na > Na

+

40. (c)

Atomic size of K+ > Ca

2+ > Mg

2+ and that of Cl

– > F

–. therefore, Mg

2+/Cl

– ratio has the minimum value.

RJ VISION PVT. LTD.


41. (b)

20Ca = 1s22s

22p

63d

23p

6 = [Ar]4s

2

42. (b)

Ca(20) = 1s22s

23p

63s

23p

64s

2 = [Ar]4s

2

43. (b)

Washing soda is Na2 CO3. 10H2O

44. (d)

Because of larger size and smaller nuclear charge, alkali metals have low ionization potential relative to alkaline ear th

metals.

45. (c)

Within a period, the atomic size decreases from left to right. Fur ther atomic size increases down the group. Hence the

correct order is i.e., Na > Mg > Li > Be.

46. (a)

Within a group, ionic radius increases with increase in atomic number. the melting points decrease down the group

due to weakening of metallic bond. The electronegativity and the 1st ionization energy also decreases down the group.

RJ VISION PVT. LTD.


1. (b)

‘B’ has no vacant d-orbitals in its valence shell, so it can’t extend its covalency beyond 4, i.e. ‘B’ cannot form the ion

like MF63(-)

i.e. BF63(-)

Hence, the correct option is (b).

2. (b)

Due to inert pair effect.

3. (a)

Strong reducing behaviour of H3PO2

All oxy-acid of phosphorus which contain P.H bond act as reductant.

Presence of one –OH group and two P–H bonds.

4. (b)

Stability of +1 oxidation state due to inert pair effect Tl > In > Ga > Al

5. (b)

Due to strong H-bonding in HF molecule, boiling point is highest for HF

HF > HI > HBr > HI

6. (d)

(i) No. of electron in ONF = 24

No. of electron in

2NO = 24

both are isoelectronic

(ii) OF2 is a fluoride of oxygen not oxide of fluorine because EN of fluorine is more than oxygen

OF2 = oxygen difluoride

(iii) Cl2O7 is an anhydride of perchloric acid

2HClO4 OH

Δ

2 Cl2O7

(iv) O3 molecule is bent bent shape

7. (d)

8. (a)

Feldspars are aluminosilicates.

9. (a)

4

4SiO is basic structural unit of silicates.

10. (b)

Since Me3SiCl contains only one Cl, therefore it can’t form high molecular mass silicon polymer. It can fo rm only

dimer.

CH – 11 THE p-BLOCK ELEMENTS

RJ VISION PVT. LTD.


11. (d)

Boron nitride (BN) is known as inorganic graphite. The most stable form is hexagonal one. It has layered structure

similar to graphite.

12. (c)

Fused alumina (Al2O3) is a bad conductor of electricity. Therefore, cryolite (Na3AlF6) and fluorspar (CaF2) are added to

purified alumina which not only make alumina a good conductor of electricity but also reduce the melting point of

the mixture to around 1140 K.

13. (c)

Pyrosilicate [Si2O7]6–

14. (b)

p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in

boron of BF3 is neutralized partially to the maximum extent by back donation.

Hence, BF3 is least acidic.

15. (c)

Hydrolysis of substituted chlorosilanes yield corresponding silanols which undergo polymerization.

Polymerisation of dealkyl silandiol yields linear thermoplastic polymer.

16. (b)

Al2O3 can be converted to anhydrous AlCl3 by heating a mixture of Al2O3 and carbon in dry Cl2.

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17. (c)

Zeolite are microporous crystalline solid with well defined structure. They contain silicon, aluminium and oxygen in

their framework and cations, water & other molecules within their pores. The general formula of zeolite is

Mx/y[(AlO2)x(SiO2)y]. mH2O.

18. (b)

Na2B4O7. 10H2O O210H

Δ

anhydride Boric

32metaborate sod.

2anhydrous

74a2 OB2NaBOOBN Δ

CuO + B2O3 → beeda) (Blue metaborate cupric

22)Cu(BO

19. (b)

6HF + SiO2 → H2SiF6 + 2H2O

20. (d)

In graphite, each carbon is sp2-hybridized and the single occupied unhybridized p-orbitals of C-atoms over lap side

wise to give π-electron cloud which is delocalized and thus the electrons are spread out between the structures.

21. (a)

H3BO3 is a weak monobasic acid.

22. (c)

Aluminium can be extracted by electrolysis of pure alumina (Al2O3). Alumina ionizes as

32OAl ⇌ anode

33cathode

3 AlOAl

Al3+

+ 3e → Al (at cathode)

3

34AlO → 2Al2O3 + 3O2 + 12e (at anode)

Aluminium of 99.8% purity is obtained from this process.

23. (a)

Water gas is made by blowing steam through the layer of incandescent coal.

28kcalCOHCOHgasWater

2hot RedSteam2

24. (b)

In diamond each carbon atom is sp3 hybridized and thus forms covalent bonds with four other carbon atoms lying at

the corners of a regular tetrahedron.

25. (c)

Glass is a super cooled liquid.

26. (a)

SiCl4 gets hydrolysed in moist air and gives white fumes which are used as a smoke screen in warfare.

27. (d)

Potash Alum, K2SO4 . Al2(SO4). 24H2O is a double salt.

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1. (b)

-NO2 group exhibit –l effect and it decreases with increase in distance. In option (b) positive charge present on C-atom

at maximum distance so –l effect reaching to it is minimum and stability is maximum.

2. (c)

-l effect increases on increasing electronegativity of atom. So, correct order of –l effects is –NH2 < -OR < - F.

*Most appropriate Answer is option (a); however option (c) may also be correct answer.

3. (c)

The ortho and para isomers can be separated by steam distillation o-Nitrophenol is steam volatile due to

intramolecular hydrogen bonding while p-nitro phenol is less volatile due to intermolecular hydrogen bonding which

cause association of molecule.

4. (c)

Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electron

from a nucleophile.

5. (c)

6. (a)

Reason: Nucleophiles are electron rich species so act as Lewis base.

7. (a)

8. (b) 9. (a)

10. (a) 11. (c)

12. (c)

13. (a)

Volume of 1M H2SO4 = 10 m mol

Volume of NH3 consumed = 20 m mol

Weight of 0.280gg1000

2014N

37.33%1000.75

0.280%N

CH – 12 ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES

AND TECHNIQUES

RJ VISION PVT. LTD.


14. (b)

In Ethyne (CHCH) both carbon atoms are sp

hybrid as the hybridization of combustion

15. (c)

The blue colour is of Fe4[Fe(CN)6]3

16. (b)

In the carbonium ion the carbon atom carrying the positive charge is sp2 hybridized.

17. (b)

Stability depends on number of hyperconjugative structure.

18. (c)

Conformers are form of stereoisomers in which isomers can be interconverted by rotations about single bonds. I and

II are staggered and eclipsed conformers respectively.

19. (b)

Stability depends on number of hyperconjugative structures.

20. (d)

21. (b)

–NO2 is a powerful electron withdrawing group. Its presence on ring makes the ring less active

22. (c)

Decreasing order of deactivating effect of the given m-directing group is

> NO2 > –CN > –SO3H > –COOH

—NO2 group is most deactivating group due to strong – E, – I and – M effects.

23. (a)

IUPAC name of the structure is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-yonic acid.

24. (a)

It shows geometrical isomerism but does not show optical isomerism.

25. (a)

The correct name is 3-Bromoprop-1-ene

26. (c)

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Electrophilic rate order

Substitution hence presence of electron releasing group like CH3 in the nucleus facilitates nitration.

27. (a)

Conc. HNO3 decomposes NaCN and Na2S to avoid their interference.

NaCN + HNO3 → NaNO3 + HCN↑

Na2S + 2HNO3 → 2NaNO3 + H2S↑

28. (b)

Given wt of compound taken (w) = 0.35 g

Volume of nitrogen collected (V) = 55 ml

Room temperature (t K) = 300 K

Atmospheric pressure (P) = 715 mm

Aq. Tension (ρ) = 15 mm

Calculation-

Volume of N2 at NTP

ml760273

tV)(P

ρ

ml 46.098260273

3005515)(715

% of nitrogen

compound organic of wt22400100NTP at N of vol.28 2

16.46%0.3522400

10046.09828

29. (a)

Correct IUPAC name of above compound is trans-2-chloro-3-iodo-2-pentene

30. (a)

In SN Ar reactions, a carbanion is formed as an intermediate, so any substituent that increases the stability of

carbanion and hence the transition state leading to its formation will enhance the SNAr reactions. To compare the

rates of substitution in chlorobenzene, chlorobenzene having electron-withdrawing group, and chlorobenzene having

electron releasing group, we compare the structures carbanion I (from chlorobenzene). II (from chlorobenzene

containing electron-withdrawing group) and III (from chlorobenzene containing electron-releasing group).

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G withdraws electrons, neutralizes (disperses) –ve charge of the ring, stabilizes carbanion, facilitates SN reaction

(activation effect)

G release electrons. Intensifies –ve charge, destabilizes carbanion, retards SN reaction (deactivation)

NO2 is activating group and CH3 and OCH3 are deactiving group.

Hence, the correct order of nucleophilic substitution reactions.

31. (c)

Because of high electronegativities of the halogen atom, the carbon halogen (C – X) is highly polarised covalent

bond. Thus, the carbon atom of the C – X bond becomes a good site for attack by nucleophiles (electron rich

species). Nucleophilic substitution reactions are the most common reactions of alkyl halides.

32. (a)

Bond length order is

A1.54A1.40A1.34A1.10CCOCCCHC

33. (a)

The given compound is

4 ethyl, 3 proxyl hex – 1 – eme

34. (b)

Due to +M effect of –OH group and hyperconjugation of –CH3 group

35. (d)

Has a lone pair of electrons on oxygen atom, thus not an electrophile. Also octet is complete.

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36. (a)

Due to hydrogen bonding between the two OH groups, gauche conformation of ethylene glycol (a) is the most stable

conformation.

37. (b)

The bulky methyl groups are maximum away from each other.

38. (d)

Among –OH, –CH2OH, –NHCOCH3 and –OCH3, methoxy group has the highest +M effect.

39. (d)

Cl– is the weakest base and hence better leaving group.

40. (a)

Tertiary alkyl halide is most reactive towards nucleophilic substitution because the corresponding carbocation (3°) is

most stable. Aryl halide is least reactive due to partial double bond character of the C – Cl bond.

Presence of —NO2 groups in ortho and para positions increases the reactivity of the –Cl towards nucleophiles.

(CH3)3 – C – X > (CH3)2 – CH – X >

Or I < II < IV < III

41. (a)

Explanation: Presence of three —NO2 groups in o–, p– positions to phenolic groups (in III) makes phenol strongly

acidic because its corresponding phenate ion (conjugate base) is highly stabilized due to resonance. Conjugate base

of CH3COOH, II (i.e. CH3COO–) is resonance hybrid of two equivalent structures. The conjugate base of phenol, IV is

stabilized due to resonance note that here all resonating structures are not equivalent). The conjugate base of

cyclohexanol, I does not exhibit resonance, hence not formed.

RJ VISION PVT. LTD.


42. (d)

Alkenes with double bonds cannot undergo free rotation and can have different geometrical shapes with two

different groups on each and of the double bond.

43. (d)

44. (b)

C – 1 is sp hybridized (C ≡ C)

C – 3 is sp3 hybridized (C – C)

C – 5 is sp2 hybridized (C = C)

Thus the correct sequence is sp, sp3, sp

2.

45. (d)

The stability of carbanions is affected due to resonance, inductive effect and s -character of orbitals. Greater the

number of groups having +I group (alkyl group) lesser stable would be the carbanion.

Further stability of carbanion decreases with decrease in s-character. Benzene carbanions are stabilized due to

resonance, hence the correct order is

The correct order of stability of given carbanion is in order a > c > b > d.

46. (c)

A strong base can abstract and α-hydrogen from a ketone.

47. (a)

In the molecule

the number of stereoisomers is given by sum of geometrical isomers (because of presence of C = C) and optical

isomers (because of presence of chiral carbon atom). Number of geometrical isomers = 2 (one C = C is present).

Number of optical isomers = 2 (one chiral carbon atom).

Total number of stereoisomers = 2 + 2 = 4

48. (d)

The amount of s-character in various hybrid orbitals is as follows.

sp = 50%, sp2 = 33% and sp

3 = 25%

Therefore character of the C – H bond in acetylene (sp) is greater than that of the C – H bond in alkene (sp2

hybridized) which in turn has greater s character of the C – H bond in alkanes. Thus owing to a high s character of the

C – H bond in alkynes, the electrons constituting this bond are such a carbon atom can be easily removed as proton.

The acidic nature of three types of C – H bonds follows the following order

RJ VISION PVT. LTD.


≡ C – H > = C – H > –C – H

Further, as we know that conjugate base of a strong acid is a weak base, hence the correct order of basicity is

3

)()(

2

)(CHCHCHCHCCH

49. (c)

Out of the given compounds the most reactive towards nucleophilic attac k is

Phenoxide ion is stable due to resonance. i.e., the correct answer is option (c).

50. (c)

Electrophiles have high affinity for electrons. They attack at the site where electron density is highest. Electron

donating groups increases the electron density. The electron donating tendency decreases in the order:

–OH > –CH3 > –H > –Cl

Therefore, the correct order of reactivity towards electrophile is:

C6H5OH > C6H5CH3 > C6H6 > C6H5Cl

51. (d)

Nucleophilicity increases down the periodic table. I– > Br

– > Cl

– > F

–

52. (b)

53. (a)

Compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called meso-

compounds. The absence of optical activity in these compound is due to the presence of a plane of symmetry in their

molecules. E.g. mesotartaric acid is optically inactive.

54. (c)

The degree of hydrolysis increases as the magnitude of positive charge on carbonyl group increases. Electron

withdrawing group increases the positive charge and electron releasing group decreases the positive charge. Among

these NO2 & CHO are electron withdrawing group from which NO2 has more –I effect than –CHO. On the other hand

CH3 is a electron releasing group therefore the order of reactivity towards hydrolysis is:

RJ VISION PVT. LTD.


55. (b)

*marked are chiral carbons.

56. (a)

Amond the three given hybrid orbitals, sp hybrid orbital is most electronegative. Contribution of s in sp hybrid orbital

is maximum (50%) so this orbitals is closer to nucleus. Naturally it will have greater tendency to pull electron towards

it. Hence it becomes more electronegative and sp3 becomes least electronegative as it have only 25% S character.

57. (c)

It is 2, 3 dimethyl pentanoyl chloride.

58. (c)

General molecular formula of alkanols is C2H2n+2O ∙ *CnH2n+1OH)

59. (b)

Among the given compounds naphthalene is volatile but benzoic acid is non-volatile (it forms a dimer). So, the best

method for their separation is sublimation, which is applicable to compounds which can be converted directly into

the vapour phase from its solid state on heating and back to the solid state on cooling. Hence it is the most

appropriate method.

60. (a)

Correct IUPAC name of

is 4-Ethyl-3-methyl heptanes

61. (c)

Phenols are more acidic than alcohol as they are resonance stabilized whereas alcohols are not.

Further nitro is an electron withdrawing which increases acidic character and facilitates release of proton, whereas –

CH3 is an electron donating group. Which decreases acidic character, thus removal of H+ becomes very difficult.

[Chiral]

[Not chiral]

[chiral]

[chiral]

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62. (d)

More the number of alkyl groups, the greater the dispersal of positive charge and therefore more the stability of

carbocation hence.

63. (d)

1NS reaction is favoured by heavy groups on the carbon atom attached to halogen i.e. Benzyl > allyl > tertiary >

primary > secondary > primary > alkyl halides.

Obtained from SN1 path

This molecule is resonance stabilized.

64. (b)

Optical and geometrical isomerism pair up to exhibit stereoisomerism. This is because the isomers differ only in their

orientation in space.

65. (a)

Clockwise rotation.

Hence configuration is R.

If the eye travel in a clockwise direction, the configuration is specified as the order of priority is Br > Cl > CH3 > H.

66. (b)

In diphenylmethane monochlorination at following positions will produce structured isomers.

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67. (b)

H2C = CHCl is capable of showing resonance which develops a partial double bond, character to C–Cl bond, thereby

making it less reactive toward nucleophilic substitution.

68. (a)

Carbon atoms from C1 to C5 are chiral.

69. (b)

Compound which are mirror image of each other and are not super imposable are termed as enantiomers.

are anantiomers.

70. (b)

In the presence of UV rays or energy, by boiling chlorine, free radical is generated which attack the methyl carbon

atom of the toluene.

71. (a)

Amino group is ring activating while nitro group is deactivating. Hence, correct order is aniline > benzene >

nitrobenzene.

I > II > III

–NO2 is an electron attracting group hence decrease the electron density on ring, whereas –NH2 group is electron

releasing group hence increases electron density on ring. Benzene is also rich due to delocalization of electrons.

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72. (b)

73. (c)

Geometrical isomers differ in spatial arrangement of atoms.

74. (a)

Both are resonating structures.

75. (b)

The IUPAC name of HCCHCHCHCHC65

2

4

2

32

2

1 is Hex-1-en-5-yne or 1-hexene-5-yne

The lowest number is given to the C = C double bond.

76. (a)

77. (b)

The principle of steam distillation is based on Dalton’s law of partial pressure. Suppose p 1 and p2 be the vapour

pressure of water vapour and the toluene at the distillation temperature. Toluene boils when total pressure is equal

to atmospheric pressure p

i.e., p = p1 + p2

or p2 = p – p1

as a result when toluene boil in the presence of steam. Its partial pressure P2 is less than atmospheric pressure.

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78. (a)

Electron releasing group (–CH3) decreases acidity while electron withdrawing group (–NO2) increases acidity.

79. (a)

A compound is said to exhibit optical isomerism if it atleast contains one chiral carbon atom, which is an atom

bonded to 4 different atoms or groups.

80. (a)

CH3 and NH2 are electron repelling groups, whereas NO2 and OH are electron attracting groups and they leave the

benzene ring deactivated. Due to stronger electron attraction (–I effect) effect of NO2 group C6H5NO2 shows least

reactivity towards electrophilic substitution.

81. (b)

Stability of an alkene depends upon the heat of hydrogenation of an alkene. The lower the heat of hydrogenation of

an alkene higher will be stability.

Order of stability Heat of hydrogenation (kJ/mol)

trans-2-butene 115.5

cis-2-butene 119.6 and

1-butene 126.8 respectively.

82. (a)

Electron withdrawing group (–NO2) increases acidity while electron releasing (–CH3, –H) decreases acidity. Also effect

will be more if functional group present at para position then ortho and meta position.

83. (c)

Chiral molecules are those molecules which have atleast one symmetric carbon atom (a carbon atom attached to 4

different groups). This is true in case of 3-methyl pentanoic acid.

84. (c)

223 sp

2

spspsp

3 CHCCHCH

85. (a)

When similar atoms are on opposite side the compound is in trans-form

86. (c)

The atom or group which has more power to attract electrons in comparison to hydrogen is said to have –I effect.

Thus higher the electronegativity of atom stronger will be the –I effect. As electronegativity of N, O and F follow the

order N < O < F hence based upon electronegative character order of –I effect is –NR2 < –OR < –F.

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87. (c)

The compound is diethyl ether (CH3CH2)2O which is resistant to nucleophilic attack by hydroxyl ion due to absent of

double or triple bond, whereas all other compounds given are unsaturated.

88. (a)

Due to absence of a symmetric (chiral) C-atom.

D—CH2—CH2—CH2Cl

Molecule is not a chiral molecule.

89. (d)

—Cl atom shows o/p-directive influence but deactivate the benzene ring, while [—OH/—CH3] groups show o/p

influence but activate the benzene ring but in these —OH group is more activating than —CH3. Hence order of

electrophilic substitution.

90. (c)

As in this compound the common groups i.e., highly electronegative halogen atoms are on opposite side, hence it is

a trans isomer.

Thus, its name is trans-2-chloro-3-iodo-2-pentene.

91. (d)

–OCH3 activates the benzene ring. –NO2 deactivates the ring. Hence the reaction of the given compounds with

electrophiles is in the order, I > II > III.

92. (c)

Order of stability : staggred anti > gauche > skew boat > eclipsed.

Newmann projection of n butane is given as

The maximum staggered conformation is most stable in which methyl groups are far apart as for as possible, due to

minimum repulsion between methyl groups and is also called anti conformation.

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93. (c)

In arenes, π electrons are delocalized, hence arens do not undergo addition reactions easily. Aromatic compounds

(Arenes) are highly stable and show resonance. e.g., Benzene is the simplest example.

94. (d)

When (–)2 bromo octane is allowed to react with sodium hydroxide under given conditions, where seco nd order

kinetics are followed, the product obtained is (+)2 octanol.

In this reaction Walden Inverson takes place so it is an example of SN 2-reaction.

95. (d)

Tautomerism is exhibited by the oscillation of hydrogen atom between two polyvalent atoms present in the molecule.

As option (d) has α-hydrogen atom. Therefore it shows tautomerism whereas other structures do not.

96. (c)

IR spectroscopy is used for the purification of cyclohexanone from a mixture of benzoic acid, isoamyl alcohol,

cyclohexane and cyclohexanone. Because in this method, each functional group upper at a certain peak. So,

cyclohexanone can be identified by carbonyl peak.

97. (c)

More is the electron - deficiency of the carbonyl carbon, greater will be the reactivity of the carbonyl compounds

towards nucleophilic addition.

98. (b)

Due to presence of four different groups on carbon, (C*) it is chiral.

99. (c)

Due to presence two similar methyl group at same carbon atom, above compound doesn’t show geometrical

isomerism.

100. (d)

Huckel’s rule states that for aromaticity there must be (4n + 2)π electron present in compound where n is an integer.

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101. (b)

When many substituents are present, the numbering is done from the end where the sum of locants is the lowest

(lowest sum rule)

2, 3, 6 TriMethyl Heptane (correct)

102. (b)

The possible isomers of the compound with molecular formula C7H8O is 5. These are (Anisol)

356 ,OCHHCalcohol benzyl

256 OHCHHC

and

103. (c)

The vital force theory suffered first death blow in 1828 when Wohler synthesized the Ist organic compound urea in

the laboratory from inorganic compounds reported below:

NH4CNO leading entrearrangemcharge isomeric to

Urea

22CONHNH

Later on a further blow to vital force theory was given by Kolbe (1845) who prepa red acetic acid, the first organic

compound, in laboratory from its elements.

104. (d)

The boiling point of o-nitrophenol is less than para-nitrophenol due to presence of intramolecular hydrogen

bonding. Since p-nitrophenol is less volatile in than o-nitrophenol due to presence of intermolecular hydrogen

bonding hence they can be separated by steam distillation.

105. (b)

Hydrazine (NH2NH2) does not contain carbon and hence on fusion with Na metal, it cannot form NaCN; consequently

hydrazine does not show Lassaigne’s test for nitrogen.

106. (b)

3-Ethyl-2-2methylpentane

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107. (c)

Chlorination of methane proceeds via free radical mechanism. Conversion of methyl chloride to methyl alcohol

proceeds via nucleophilic substitution. Formation of ethylene from e thyl alcohol proceeds via dehydration reaction.

Nitration of benzene is electrophilic substitution reaction.

108. (d)

Such question can be solved by considering the relative basic character of their conjugated bases which for H 2O,

C2H2, H2CO3 and C6H5OH are:

OHC,HCO,CHCOH,563

More the possibility for the dispersal of the negative charge, weaker will be the base. Thus the relative basic character

of the four bases is:

Thus the acidic character of the four corresponding acids will be

HC ≡ CH < H2O < C6H5OH < H2CO3

109. (a)

Resolution.

110. (c)

Diastereomers since they have different melting points, boiling points, solubilities etc.

111. (d)

Rotation around π bond is not possible. If any attempt is made to rotate one of the carbon atoms, the lo bes of π-

orbital will no longer remain coplanar i.e. no parallel overlap will be possible and thus π-bond will break. This is

known as concept of restricted rotation. In other words the presence of π-bonds makes the position of two carbon

atom.

112. (d)

Resonance structures are separated by a double headed arrow (⟷)

113. (d)

Due to +I-effect of the CH3 group, toluene has much higher electron density in the ring than benzene, nitrobenzene

and benzoic acid as nitro and carboxylic group show-I-effect and hence toluene is most reactive towards nitration.

114. (d)

Angle increases progressively

Sp3 (109°2.8'), sp

2 (120°), sp (180°)

115. (b)

116. (b)

2-Methyl-2-butene

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117. (d)

Carboxylic acids dissolve in NaHCO3 but phenols do not.

RCOOH 3NaHCO RCOONa + H2O + CO2

By evolving CO2 gas

118. (d)

Higher the possibility of delocalization of the positive charge, greater is stability of the species. Thus

22333256 HCCH3HC)(CHC)(CHHCHC

Benzyl carbocation is more stable than tert-butyl due to resonance in the former.

119. (d)

Organic compounds having same molecular formula but differ from each other in physical properties or chemical

properties or both are known as isomers.

120. (d)

spspspsp

2spspsp

3 CHCCHCHCHCHCH23223

121. (a)

sp3 orbital has 1/4 (25%) s-character. & 75% p character.

122. (d)

Shortest C – C distance (1.20 Å) is in acetylene.

As acetylene has sp hybridisation, the bond length increases in the order

)Asp(1.20

oCC <

)A(1.34spo2

CC < )Asp(1.54

oCC

123. (b)

Sodium cyanide (Na + C + N → NaCN). (Lassaigne’s test)

124. (a)

Hence it is homocyclic (as the ring system is made of one type of atoms, i.e. carbon) but not aromatic. As it does not

have (4n + 2)π electron required for aromaticity.

125. (a)

Kjeldal’s method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and

hydrogen. The method is not used in case of nitro, azo and azoxy compound. This method is basically used for

estimating nitrogen in food fertilizers and agricultural products.

126. (c)

Such isomers, which possess the same molecular and structure formula but differ in the arrangement of atoms

around the double bonded carbon atoms are known as geometrical isomers.

127. (d)

Nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne’s test. The organic compound is

fused with sodium metal as to convert these elements into ionisable inorganic substances,

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Na + C + N → NaCN

2Na + S → Na2S

2Na + X2 → 2NaX

The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

128. (d)

All the properties mentioned in the question suggest that it is a benzene molecule. Since in benzene all carbons are

sp2–hybridized, therefore, C – C – C angle is 120°

129. (c)

22 sp2

spsp2 CHCCH

130. (d)

Tetrachloroethene being an alkene has sp2-hybridizied C– atoms and hence the angle Cl–C–Cl is 120° while in

tetrachloromethane, carbon is sp3 hybridized, therefore the angle Cl–C–Cl is 109.5°.

131. (c)

The compound containing a chiral carbon atom i.e., (a carbon atom which is attached to four different substituents is

known as a chiral carbon atom) is optically active.

132. (b)

Five chain isomers are possible which are

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1. (c)

HCCCHCHspspspsp

22

2

Number of orbital require in hybridization = Number of -bonds around each carbon atom.

2. (d)

Hence the correct option is (d).

3. (a)

Correct order of acidic strength

CH CH > CH3 – C = CH > CH2 = CH2 > CH3 – CH3

acc. To EN and Inductive effect.

4. (c)

In conformation bond angle and bond length remain same.

5. (d)

Sigma bond can be considered to be cylindrically symmetrical.

6. (b)

7. (b)

8. (c)

CH – 13 HYDROCARBONS

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9. (b)

10. (c)

11. (a)

Huckel rule is not obeyed. It has only four electrons. Further it does not have continuous conjug ation.

12. (d)

13. (d)

Presence of 6p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance

with Huckel rule of aromaticity.

14. (a)

1-Butyne and 2-butyne are distinguish by NaNH2 because 1-Butyne react with NaNH2 due to presence of terminal

hydrogen.

CH3 — CH2 — C ≡ CH + NaNH2 → 1 – Butyne

CH3CH3 — C ≡ CNa + NH3

15. (a)

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16. (b)

In this case dehydration is governed by Saytezeff’s rule according to which hydrogen is preferentially eliminated from

the carbon atom with fewer number of hydrogen atoms i.e., poor becomes poorer. Thus, 2 methyl butane -2 is the

major product.

This reaction is governed by Markownikoff’s rule according to which when an unsymmetrical reagent e.g. HBr adds to

an unsymmetrical alkene, then the negative part of the reagent is added to that carbon atom of the double bond

which bears the least number of hydrogen atom. Thus, in above case. 2-methyl 2-bromo butane will be the major

product.

17. (b)

When both double and triple bonds are present, then double bond is considered as the principal group.

18. (d)

C6H5CH2Br + Mg ether

C6H5CH2MgBr (in ether as solution)

19. (b)

During cracking higher hydrocarbons (liquid) are converted to lower gaseous hydrocarbons.

20. (c)

If both the double and triple bonds are present the compound is regarded as derivative of alkene. Further if double

and triple bonds are at equidistance from either side, the preference is given to double bond.

21. (b)

We know that in case of an unsymmetrical alkene there is the possibility of forming two products. In such cases the

formation of major product is decided on the basis of Markownikoffs rule which is rationalized in terms of stability of

the intermediate carbocation. Also remember that 3° carbocation and 2° carbocation is more stable than 1°

carbocation.

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Of the two possibilities 2° carbocation is more stable so the product of the reaction expected was predominantly one

formed by 2° carbocation i.e.

i.e. 2-Bromo-3-Methylbutane

However some electrophilic addition reaction form products that are clearly not the result of the addition of

electrophile to the sp2 carbon bonded to the most hydrogens and the addition of a nucleophile to the other sp

2

carbon

In the above cases the addition of HBr to 3-methyl-1butene the two products formed are shown below.

In this case the major product is 2-Bromo-2-methylbutane i.e. option (b) is correct answer.

(Note: the unexpected product results form a rearrangement of carbocation intermediate . Please note that all

carbocation do not rearrange.

22. (d)

23. (c)

This reaction occurs according to Markownikoff’s rule which states that when an unsymmetrical alkene undergo

hydrohalogenation, the negative part goes to that C-atom which contain lesser no. of H-atom.

CH3 – CH2 – C ≡ CH + HCl →

24. (b)

CH3 – C ≡ C – CH2 – CH3 3O

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The glyoxal formed as an intermediate is oxidized by H2O2 to give the acids.

25. (a)

Heat of hydrogenation of alkene alkene of Stability

1

Hence, the alkene which will react fastest with H2 will be the least stable. Among the given options the compound

having least number of alkyl groups (R) will be the least stable.

Further the relative rates of hydrogenation decrease with increase of steric hindrance

26. (c)

C6H6 + CH2 = CH2 3AlCl C6H5CH2CH3

27. (d)

In presence of peroxide, HBr adds on alkenes in anti-markovanikov’s way thus

H3C CH = CH2 + HBr PeroxideH3C CH2CH2Br

Kharasch observed that the addition of HBr of unsymmetrical alkene in the presence of organic peroxide follows an

opposite course to that suggested by Markownikoff. This is termed anti-Markownikoff or peroxide effect.

28. (c)

29. (a)

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30. (a)

Temperature

R — CH2 — CH2OHC380350

O Aloo 32

R — CH = CH2 + H2O

While at 220°–250° C it forms ether

31. (c)

32. (d)

We know that

6CH3 – CH = CH2 Cether,0HB

o62

2(CH3CH2CH2)3 OH

2O2H

Propanol33223 BO2HOHCHCH6CH

33. (a)

C2H5MgBr + H2O →

34. (a)

Given

A 2ClB

alc./KOHC

O/HO 23 CH2O

Hydrocarbon

Since hydrocarbon C give only CH2O, on ozonolysis, C should be CH2 = CH2 hence going backward A should be

ethane. Thus the reactions are

(A)

33CHCH /hvCl2

(B)23 ClCHCH

KOH

alc.

(C)22 CHCH

Δ O/HO 23

(D)HCHO

35. (a)

Gasoline (petrol) is a mixture of alkenes, alkenes and aromatic hydrocarbons. The quality of a gasoline is determined

by the amount of branched chain hydrocarbons (2, 2, 4-trimethylpentane, commonly known as iso-octane) present in

it.

36. (a)

Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated non-carboxylic acids

yields higher alkane at anode.

2RCOOK OxidationicElectrolyt

CathodeAnode2K2RCOO

At anode 2RCOO– → 2RCOO + 2e

– → R — R + 2CO2

At cathode 2K+ + 2e

– → 2K

2K + H2O → 2KOH + H2↑

(Kolbe’s method)

37. (c)

We know that

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38. (a)

39. (a)

When 3, 3 dimethyl 2-butanol is heated with H2SO4 the major product obtained is 2, 3 dimethyl 2-butene.

40. (b)

6R – CH = CH2 CEther,0

HB o22

2(RCH2CH2)3

NaOHH 2O2 6RCH2CH2OH + 2H3BO3.

41. (a)

Cl2 ationChaininitihv 2Cl

42. (c)

CH3Cl + AlCl3 → leElectrophi

43 AlClCH

43. (d)

Of all the options listed CH3CH2CH2CH3 has the least number of C-atoms and hence has the lowest b.p.

44. (a)

Peroxide effect is observed only in case of HBr.

Therefore, addition of HCl to propene even in the presence of benzyoyl peroxide occurs according to Markonikov’s

rule‛

CH3 – CH = CH2 2O2CO)5H6(C HCl CH3 – CHCl – CH3

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45. (d)

The reactivity of H-atoms depends upon the stability of free radicals follows the order:

Tertiary > secondary > primary, therefore, reactivity of H-atoms follows the same order, i.e., tertiary > secondary >

primary.

46. (a)

Only C2H2 (acetylene) has acidic H-atoms and hence reacts with NaNH2 to form sodium salt, i.e.,

HC ≡ CH + NaNH2 → HC ≡ Can + NH3.

47. (c)

Reduction of alkynes with Na/liq. NH3 gives trans-alkenes. This reaction is called Birch reduction

48. (c)

On heating ethylene chloride (1, 1 dichloroethane) with alcoholic potash followed by sodamide alkyne is obtained

R – CH2 – CCl2 – R

alc.KOHR – CH = CCl – R

2NaNHR – C ≡ C – R

49. (c)

Benzene do not show addition reaction like other unsaturated hydrocarbons. However it show substitution reactions.

Due to resonance all the C – C bonds have the same nature, which is possible because of the cyclic delocalization of

π-electrons in benzene. Monosubstitution will give only a single product.

50. (d)

Br2 in CCl4(a) Br2 in CH3COOH(b) and alk. KMnO4(c) will react with all unsaturated compounds, i.e., 1, 3 and 4 while

ammonical AgNO3(d) reacts only with terminal alkynes, i.e., 3 and hence compound 3 can be distinguished from 1, 2

and 4 by ammonical AgNO3(d).

51. (a)

The acidity of acetylene or 1—alkynes can be explained on the basis of molecular orbital concept according to which

formation of C—H bond in acetylene involves sp-hybridised carbon atom. Now since a electrons are closer to the

nucleus than p electrons, the electrons present in a bond having more s character will be correspondingly more closer

to the nucleus. Thus owing to high s character of the C—H bond in alkynes (s = 50%), the electrons constituting this

bond are more strongly held by the carbon nucleus (i.e., the acetylenic carbon atom or the sp orbital acts as more

electronegative species than the sp2 and sp

3 with the result the hydrogen present on such a carbon atom (≡C—H) can

be easily removed as a proton.

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1. (a)

Fact

2. (d)

3. (d)

Acid rain is the rain water containing sulphuric acid and nitric acid which are formed from the oxides of sulphur and

nitrogen present in the air as pollutants and rain water a pH of 4-5.

4. (a)

Based on the features given below the gas must be SO2.

5. (d)

The oxidized hydrocarbons and ozone in presence of humidity cause photochemical smog.

Hydrocarbons + O2, NO2, NO, O, O3 → Peroxides, formaldehyde, peroxyacetylnitrate (PAN), acrolein etc.

It is oxidizing in nature and causes irritation to eyes, lungs, nose, asthamatic attack and damage plants.

6. (b)

The ideal value of D.O for growth of fishes is 8 mg/ . 7 mg is desirable range, below this value fishes get

susceptible to disease. A value of 2 mg/ or below is lethal for fishes.

7. (b)

Green chemistry may be defined as the programme of developing new chemical products and chemical processes or

making improvements in the already existing compounds and processes so as to make less harmful to human health

and environment. This means the same as to reduce the use and production of hazardous chemicals.

i.e. correct answer is option (b).

8. (c)

CO and oxides of Nitrogen are poisonous gases present in automobile exhaust gases.

9. (a)

Green house gases such as CO2, ozone, methane, the chlorofluoro carbon compounds and water vapour form a thick

cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in

temperature of atmosphere.

CH – 14 ENVIRONMENTAL CHEMISTRY

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