unit-iii - inorganic chemistry - solutions(final)

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Section A : Straight Objective Type

1. Answer (4)

All these contain 3C – 2– bond.

2. Answer (3)

Na2B4O7⋅10H2O + H2O ⎯⎯→⎯+H 2B(OH)3 + 2B(OH)4

3. Answer (4)

Hydrolith – CaH2, oxidane – H2O, Azane – NH3.

4. Answer (2)

HClO4 does not contain peroxide linkage while others contain. Therefore HClO4 does not give H2O2 onhydrolysis.

5. Answer (1)

Factual.

6. Answer (1)

Tritium (1H3) is a heavy isotope of hydrogen which is obtained by nuclear reaction.

7. Answer (1)

Factual.

8. Answer (3)

No. d-orbital is involved because hybridisation of oxygen in H2O2 is sp3.

9. Answer (2)

)A(Na2 + 2H2O →

)B(2H +

)C(NaOH2 and

)D(Zn +

)C(NaOH2 → Na2ZnO2 +

)B(2H .

10. Answer (2)

Dielectric constant of H2O is 78.39 while that of D2O is 78.06. Hence, solubility of ionic compound is smallerin heavy water.

11. Answer (4)

Ortho hydrogen is more stable than para form and the latter always tends to revert to the stable ortho form.The magnetic moment of para hydrogen is zero since the nuclear spins neutralise each other.

12. Answer (2)

Alkali metals and alkaline earth metals such as Ca, Sr, Ba generally form such type of hydrides. They liberatehydrogen at the anode and confirm that they are ionic compounds and contain H– ion.

13. Answer (3)

The hydrogen molecules undergo thermal dissociation into atoms at high temperatures (around 2270 K) andlow pressures. H2 2H – xJ.

Inorganic Chemistry UNIT 3


Success Magnet (Solutions) Inorganic Chemistry

14. Answer (2)

2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2

2KI + H2SO4 + H2O2 → K2SO4 + I2 + 2H2O

2KMnO4 + 3H2O2 → )s(MnO2brown

2 + 2KOH + 3O2 + 2H2O

)black(PbS + 4H2O2 →

)white(4PbSO + 4H2O.

15. Answer (3)

44 344 21

gaswater

2NiK1270

steam2

coke)g(H)g(CO)g(OH)s(C +⎯⎯⎯ →⎯+

CO(g) + H2(g) + H2O(g) K770

OCr/OFe 3232 ⎯⎯⎯⎯⎯ →⎯ CO2(g) + H2(g).

16. Answer (2)

K2Cr2O7 + H2SO4 + 4H2O2 ⎯⎯ →⎯ether K2SO4 + blue

5CrO2 + 5H2O.

17. Answer (2)

Volume strength of H2O2 = 11.2 × M

= 11.2 × 3.57~ 40

18. Answer (2)

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2.

19. Answer (4)

4H2O2 + PbS → PbSO4 + 4H2O

4 mole 1 mole

0.8 mole 0.2 mole

volume strength of H2O2 = 11.2 × M

∴2.11

40M

22OH =

number of mole = M × VL

volume in litre =

2.1140

8.0

volume in ml = ml224100040

2.118.0 =××.

20. Answer (1)

Volume strength of H2O2 = 5.6 × N

3000N6.550020

6.550015

6.550010 ×=×+×+×

339.1)mixture(N22OH =

∴ Volume strength of H2O2 = 1.339 × 5.6 = 7.5.


Inorganic Chemistry Success Magnet (Solutions)

21. Answer (2)

OH8CrONa2OH3])OH(Cr[Na2NaOH2 2)B(

42224)A(

+→++

OHSONaOCrNaSOHCrONa2 242)C(

72242)B(

42 ++→+

22. Answer (1)

CaO + Ca(HCO3)2 → 2CaCO3 + H2O

For one litre water,

Meq. of CaO = Meq. of Ca(HCO3)2

or

2162

100062.1

2561000w ×=×

wCaO = 0.56 g

Thus CaO required for 103 litre H2O = 0.56 × 103 = 560 g.

23. Answer (4)

Ag2O + H2O2 → blackAg2 + H2O + O2.

24. Answer (2)

CaH + 2H O2 2 Ca(OH) + 2H2 2

CaCO3

(milky)

CO2

25. Answer (2)

Al is an amphoteric metal and it reacts with alkali like NaOH.

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

26. Answer (1)

More the polarisation power of cation , more is the stability of the complex formed by it. Be2+ has higher chargedensity and hence, higher polarisation power.

27. Answer (3)

Air contains O2, N2 and CO2. Li and Mg combines with O2 and N2 both. NaOH is the absorbent of CO2.

28. Answer (3)

Anhydrous MgCl2 can not be prepared by simply heating MgCl2.6H2O because it gets hydrolysed by its ownwater of cyrstallization.

MgCl2.6H2O → MgO + 2HCl + 5H2O

29. Answer (1)

Beryllium and magnesium atoms are smaller in size and their electrons are strongly bond to the nucleus. Theyneed large amounts of energy for excitation of electrons to higher energy levels which is not available in theBunsen flame.



30. Answer (1)

(NH ) C O4 2 2 4

NaOH

ΔNH3

(A) (B)

CaCl solution2

CaC O (C)2 4

NH + HCl3 NH Cl4

(B) (white fumes) .

31. Answer (3)

106 g of Na2CO3 = 100 g CaCO3

∴ 0.0106 g Na2CO3 = 0.01 g CaCO3

103 ml H2O = 103 g of H2O

∴ 103 g of H2O contains 0.01 g of CaCO3

∴ 106 g of H2O contains 6

310

1001.0 × g

= 10 ppm.

32. Answer (1)

Fire extinguisher contains NaHCO3 and compound X can be an acid (like CH3COOH) that can decomposeNaHCO3 to give CO2.

NaHCO3 + H+ → Na+ + H2O + CO2

NaHCO3 and acid (X) are kept in a sealed flask (but placed inside separately). In case of emergency (whenthere is fire), seal is broken and X and NaHCO3 react instantly to give CO2 which can put off fire.

33. Answer (2)

Compounds of Li+, Na+, and K+ have solubility in the order Li+ < Na+ < K+.

34. Answer (3)

BeCl2 has more covalent character due to polarisation and hence more hydrolysis.

35. Answer (1)

LiF has lowest solubility among the group-I metal halides due to high lattice energy.

36. Answer (1)

Among silver halides only AgF is soluble in water. On moving down in group II, the solubility of fluoridesdecreases.

37. Answer (3)

Al4C3 reacts with water to give CH4.

CaC2 reacts with water to give C2H2.

38. Answer (3)

MgSO + NH OH + Na HPO4 4 2 4 Mg(NH ) PO + 2NaCl + H O4 4 2

white ppt..



39. Answer (3)

Like alkali metal all alkaline earth metals dissolve in liquid ammonia. The dilute solutions are bright blue incolor due to solvated electrons but concentrated solutions are bronze coloured due to the formation of metalclusters.

M + (x + 2y) NH3 → M2+ (NH3)x + 2e– (NH3)y

40. Answer (4)

Water, nitrogen and CO2 react with alkali metals.

41. Answer (3)

On moving down in group-II, the solubility of sulphates decreases.

42. Answer (2)

Ca(OH) + Ca(HCO )2 3 2 2CaCO + 2H O3 2

Ca(OH) + CO2 2 CaCO + H O3 2

(A)

(milkiness)

43. Answer (1)

BeSO4 is soluble in water.

Be(OH)2 is insoluble in water.

Be is an amphoteric metal so can react with NaOH.

44. Answer (1)

More the polarisation power of cation, more is the covalent character in its compounds.

45. Answer (1)

2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O]

[H2O2 + [O] → H2O + O2] × 5

2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2

2 mol KMnO4 ≡ 5 mol H2O2

∴ 1 mol KMnO4 ≡ 25

mole H2O2

46. Answer (1)

On moving down in group II, the thermal stability of carbonates, sulphates increases. Be is an amphotericmetal, so its oxide is amphoteric. Mg is typical metal, so its oxide is basic.

47. Answer (4)

NaOH is the absorbent of CO2. It absorbs moisture from air.

48. Answer (1)

Factual

49. Answer (1)

Factual

50. Answer (3)

Factual

51. Answer (3)

Factual

52. Answer (1)

Factual



53. Answer (1)

Factual

54. Answer (2)

Factual

55. Answer (3)

Factual

56. Answer (3)

Factual

57. Answer (1)

Factual

58. Answer (4)

All metal nitrates are water soluble.

58(a). Answer (3) IIT-JEE-2008

Since nitrates are more soluble in water.

∴ nitrates are less abundant in earth’s crust.

59. Answer (2)

Br2 + .concand.hotKOH → KBr + KBrO3 + H2O

60. Answer (3)

CuO + NH3 ⎯→⎯Δ Cu + N2 + H2O

61. Answer (1)

AgCl + Na2S2O3 → Na3[Ag(s2O3)2] + NaCl

62. Answer (2)

2KMnO4 → K2O + 2MnO + 5[O]

K2O + 2HCl → 2KCl + H2O

MnO + 2HCl → MnCl2 + H2O] × 2

2HCl + [O] → H2O + Cl2 × 5

2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

63. Answer (2)

H3PO2 is a monobasic acid.

64. Answer (2)

Zn reduces dilute HNO3 to N2O.

65. Answer (1)

H3BO3 + H2O → [B(OH)4]– + H3O

+

66. Answer (1)

H3BO3 + OT231 → [B(OH)3O1T

3]– + +OT331



67. Answer (2)

Ga exists as a co-ordinated complex with both +1 and +3 oxidation states.

68. Answer (3)

CO2 is linear and HgCl2 is also linear.

69. Answer (2)

HF is highly corrosive in nature

SiO2 + 4HF → SiF4 + 2H2O

70. Answer (1)

CO, NO and N2O are neutral oxides.

71. Answer (1)

CH4 has lower molecular weight and hence, more volatility.

72. Answer (3)

H – O – S – O – O – S – O – H

O

O

O

O

73. Answer (1)

Among HClO, HClO2, HClO3 and HClO4, HClO is the weakest acid, hence its conjugate base ClO– is thestrongest base.

74. Answer (1)

P

PP

P

six P – P single bonds.

75. Answer (3)

Solid PCl5 exist as [PCl4]+[PCl6]

–.

76. Answer (3)

Due to small size of Li+, Li cannot form alums.

77. Answer (1)

Solder is a major constituent of Sn.

78. Answer (3)

Measuring the solid state magnetic moment.

79. Answer (3)

Ammonia can be dried by CaO.

80. Answer (2)

Because the reaction is endothermic. So on increasing temperature, increase the amount of NO2 gas whichis brown coloured.

81. Answer (3)

The electronegativity of N is 3.0 and that of carbon is 2.5. On moving from left to right in a period,electronegativity increases.



82. Answer (2)

HClO4 and H2SO4 both are strong acids, so the mixture of these will be stronger.

83. Answer (2)

S – S bond energy is high.

84. Answer (4)

XeF2 – sp3d – linear (3 lone pairs)

XeF4 – sp3d2 – square planar (2 lone pairs)

XeF6 – sp3d3 – distorted octahedral (one lone pair)

85. Answer (3)

Factual, hypo is Na2S2O3·5H2O

86. Answer (4)

F2 is most reactive non metal.

87. Answer (2)

The 3 : 1 mixture of concentrate HCl and concentrate HNO3 is called aqua regia.

88. Answer (2)

ZnO22–

89. Answer (3)

μ = BM)2n(n +

90. Answer (3)

All have hybridisation sp3 and so tetrahedral.

91. Answer (1)

The sum of the 3rd and 4th ionization energies of nickel is much higher than the sum of 1st and 2nd ionizationenergies. Hence, Ni (II) is more common. In case of platinum, the difference between IE3 + IE4 and IE1 + IE2is much less. Hence, Pt (IV) is more common.

92. Answer (2)

Lowest oxide of Cr is CrO which is basic. The highest oxide is CrO3 which is acidic. In between, Cr2O3 isamphoteric. Higher the oxidation state of the metal, more easily it can accept electrons and hence, greateris the acidic character.

93. Answer (2)

The electronic configuration of Hg(I), i.e., Hg+ is [Xe] 4f 145d106s1 and thus has one electron in the valence6s-orbital. If this were so, all Hg(I) compounds should be paramagnetic but actually they are diamagnetic. Thisbehaviour can be explained if we assume that the singly filled 6s-orbitals of the two Hg+ ions overlap to form

a Hg-Hg covalent bond. Thus, Hg+ ions exist as dimeric species, i.e., +22Hg .

94. Answer (1)

SnCl2 is a strong reducing agent and hence, reduces HgCl2 first to Hg2Cl2 (white) and then to Hg (black).

Hg2SnClClHgSnCl

ClHgSnClHgCl2SnCl

4222

22422

+⎯→⎯+

+⎯→⎯+

95. Answer (3)

Both Cu+2 and V+4 have one unpaired e– so possess same colour.



96. Answer (2)

Ti3+ has one electron in the d-orbital (3d 1) which can absorb energy corresponding to yellow wavelength andjump from t2g to eg set of d-orbitals.

97. Answer (3)

Mn in −4MnO is in +7 state with d° configuration. The deep purple colour of KMnO4 is not due to d-d transition

but due to charge transfer (from O to Mn) reducing the oxidation state of Mn from +7 to +6 momentarily.

98. Answer (3)

Using the formula,

3n

.M.B)2n(nµ

=+=

99. Answer (2)

When SO2 gas is passed into acidified −272OCr , SO2 is oxidised to −2

4SO , while −272OCr is reduced to Cr3+.

100. Answer (4)

])CN(Au[Na4OH2ONaCN8Au4 222 →+++ + 4NaOH

101. Answer (3)

solidyellowLemon

CrFFCrO 6atm25

C17023 ⎯⎯⎯ →⎯+ °

102. Answer (4)

OH2OMn2OMnH2OMn3 247

24

24

6++⎯→⎯+ −

++

+−

+

It is a disproportionation reaction. In disproportionation reaction, equivalent weight of the species undergoingdisproportionation = Eq. wt. (in oxidation) + Eq. wt. (in reduction)

.M23

1M

2M

MnOof.wt.Eq

1M

.)oxidn(.wt.Eq,OMnOMn

2M

.)redn(.wt.Eq,OMnOMn

4

47

24

6

24

24

6

=+=∴

=⎯→⎯

=⎯→⎯

−

−

+

−

+

+

−

+

103. Answer (2)

In acidic medium, +−

+⎯→⎯ 2

47

MnOMn

1 M = 5 N

∴ H2O2 ≡ 100 mL of 5 N KMnO4

In basic medium, 24

47

OMnOMn+

−

+⎯→⎯

1 M = 3 N

∴ H2O2 ≡ V mL of 3 N KMnO4

∴ 3V ≡ 500

V = 3

500mL



104. Answer (3)

In −e],)CN(Fe[FeK 6IIIII transition is possible between FeII and FeIII

But in K2FeII [Fe(CN)6] no such transition is possible.

105. Answer (2)

MnO2

106. Answer (3)

−<−>− ⎯⎯ →⎯⎯⎯ →⎯ 272

mediumacidic

7pH24

mediumbasic

7pH272 OCrCrOOCr

107. Answer (1)

ClNH4Cl)NH(HgHg)aq(NH2ClHg

ClNHCl)NH(Hg)aq(NH2HgCl

42)Black(

3)Y(

22

4.pptwhite

23)X(

2

++⎯→⎯+

+⎯→⎯+

108. Answer (1)

KMnO4 in presence of dilute H2SO4 acts as oxidising agent. KMnO4 in presence of dilute HCl also acts asoxidising agent but oxygen produced is used up partly for oxidation of HCl.

109. Answer (2)

Cyanogen2

2 )CN(Cu2CN2Cu2 +⎯→⎯+ +−+

↓⎯→⎯+ −+ CuCNCNCu

).complexlelubSo()I(cupratetetracyano.Pot

43 ])CN(Cu[KKCN3CuCN ⎯→⎯+

110. Answer (1)

K3[Cu(CN)4] is more stable and hence does not ionize to give Cu2+ ions. In contrast, K2[Cd(CN)4] is less stableand hence, ionises to give Cd2+ ions. Thus, when H2S is passed through the solution of these two complexes,Cu2+ in form of the complex, K3[Fe(CN)4], remains in solution but Cd2+ gets precipitated as yellow ppt. of CdS.

111. Answer (1)

Absence of unpaired e–.

112. Answer (4)

AgCl + NH4OH → [Ag(NH3)2]+ soluble complex.

113. Answer (3)

Ni (CO)4 → sp3 unpaired electrons = 0

[Ni(CN)4]–2 dsp2 unpaired electrons = 0

[NiCl4]–2 sp3 unpaired electrons = 2.

113(a). Answer (2) (IIT-JEE-2008)

[Ni(CO)4] --------- sp3 hybridization

[Ni(CN)4]–2 --------- dsp2 hybridization



114. Answer (1)

In organo metallic compound there is direct bond between carbon and metal.

115. Answer (1)

SNa])CN(Ag[NaNaCNSAg 222 +⎯→⎯+

116. Answer (3)

Py is a neutral ligand and Cl– an anionic ligand carries one unit negative charge.

117. Answer (2)

[Ni(CN)4]2– is a square planar complex and trans [Pt(NH3)2Cl2] both have zero dipole moment.

118. Answer (4)

Coordination number is number of mono dentate ligands present in coordination sphere.

119. Answer (4)

Factual

120. Answer (3)

Factual

121. Answer (3)

Factual

122. Answer (3)

Factual

123. Answer (3)

Do yourself

124. Answer (3)

Based on facts

125. Answer (1)

Based on facts

126. Answer (4)

Fe+3 + SCN– → colourred

3)CNS(Fe

Co+2 + SCN– → colourblue

24 ])SCN(Co[ −

127. Answer (3)

Hg2Cl2 + NH4OH → Hg + HgNH2Cl.

128. Answer (1)

Phosphate ion gives yellow ppt., it is interfering for basic radical.

129. Answer (3)

Red ppt of [Ni(DMG)2] complex.

130. Answer (2)

Zn+2 and Cu+2 form soluble complex.



131. Answer (2)

SnS + (NH4)2Sn → lelubso2SnS .

132. Answer (1)

Both Bi+3 and Sn+4 → give ppt with H2S in acidic medium.

133. Answer (3)

Microcosmic salt gives bead test.

134. Answer (3)

Cu+2 give ppt in acidic medium where as Cr+3 in alkaline medium.

135. Answer (4)

AgNO3 + Na2S2O3 → white

322 OSAg → black

2SAg

136. Answer (2)

NH3 + HCl → NH4Cl

∴ equilibrium shifts towards left.

137. Answer (1)

)orange(722

mediumacidic

CO

)yellow(42 OCrNaCrONa 2⎯⎯ →⎯

138. Answer (3)

By decreasing (H+) ion concentration pH increases.

139. Answer (4)

HgCl2

140. Answer (2)

Green colour of methyl borate.

141. Answer (3)

CuSO4 + 2KI → white

22ICu + K2SO4

KI + HgCl2 → KCl + HgI2 orange

42KI HgIK⎯→⎯

KI + Pb(NO3)2 → 3

.pptyellow2 KNO2PbI +

142. Answer (3)

NaHCO3 ⎯→⎯Δ Na2CO3

Na2CO3 + H2O + CO2 → 2NaHCO3

143. Answer (3)

Ba+2 + CrO4–2 →

.pptyellow4BaCrO



144. Answer (1)

Cr(OH)3 + NaOH + Na2O2 → .pptyellow

24CrO −

145. Answer (4)

Nessler’s reagent

K2(HgI4)–2 2K+ + (HgI4)

–2

146. Answer (2)

Corrosive sublimate,

HgCl2 and SnCl2 doesnot give test.

147. Answer (3)

BaCO3 + H2SO4 → BaSO4↓

148. Answer (3)

AgNO3

H SO conc.2 4

Cu

HCl

NO2

Cu(NO ) + Ag3 2

AgClwhite ppt.

Section - B : Multiple Choice Questions

1. Answer (1, 2, 3)

As the temperature is raised, the proportion of ortho hydrogen increases upto a limiting mixture containing 75%ortho hydrogen. The nuclear spins of the two atoms in the hydrogen molecule are either in the same direction(ortho form) or in opposite direction (para form) and give rise to spin isomerism.

2. Answer (1, 2, 3, 4)

It has been found that occlusion of H2 is greatest when Pd is in finely divided form. Colloidal Pd which adsorbs2050 times its own volume of H2 is still stronger reducing agent.

3. Answer (1, 2, 3, 4)

2MnO4– + H+ + 5H2O2 medium

acidic⎯⎯⎯ →⎯ 2Mn2+ + 8H2O + 5O2

2MnO4– + 3H2O2 medium

basic⎯⎯ →⎯ 2Mn+4O2 + 2OH– + 3O2 + 2H2O

I2 + H2O2 + 2OH– ⎯→⎯ 2I– + 2H2O + O2

2I– + H2O2 + 2H+ ⎯→⎯ I2 + 2H2O.

4. Answer (2, 4)

All are s-block hydride but BeH2 & MgH2 has tendency to form polymeric hydride only because both arecovalent in nature.

5. Answer (1, 2, 4)

All three are the properties of water.



6. Answer (1, 3, 4)

Acidic nature (ka for H2O2 = 1.55 × 10–12 & kw for H2O = 10–14 at 25ºC

E.g., NaOH + H2O2 → NaHO2 + H2OSodium hydro peroxide (Acidic salt)

H2O2 acts as both oxidising and reducing agent. E.g., It oxidises iodises to iodine and reduces halogen tohalogen acids.

7. Answer (2, 3, 4)

Reduction potential )(EoCell of copper is greater than hydrogen.

V0.34E0,E oCu/Cu

oH/H 2

2== ++

8. Answer (1, 4)

For diatomic gases (H2 & CO) has V

P

CC=γ

∴ 40.1CC

V

P =

9. Answer (1, 2, 4)

The metals of group 7, 8, 9 do not form hydrides. The region of the periodic table from groups 7 - 9 which do notform hydrides is referred to as the hydride gap. In group 6, Cr alone forms the hydride. Mn is in group 7, so it doesnot form interstitial hydride.

10. Answer (1, 2, 3)

The metals of group 7, 8 and 9 donot form interstitial hydrides.

11. Answer (3, 4)

Among alkali metals only K, Rb and Cs form superoxides while Na and Ba can form peroxide.

12. Answer (1, 2, 3, 4)

Ca, Sr, Ba are highly electropositive metals.

13. Answer (1, 2, 3, 4)

Be predominantly forms covalent compounds. Beryllium halides are covalent polymers.

14. Answer (1, 2, 3)

All alkali metals do not have ccp structures.

15. Answer (1, 2, 3, 4)

Gypsum is CaSO4.2H2O

In the electrolysis of fused CaH2, H2 is liberated at anode. Among alkaline earth metals, Be and Mg do notgive flame test.

16. Answer (1, 3)

Explanation → Hydration energy depends upon size of the ion, smaller the size greater the charge density,more will hydration energy and on moving down the group hydration energy decreases due to increase in size.

17. Answer (1, 2)

Factual

18. Answer (1, 2, 3, 4)

Gypsum on heating first changes from monoclinic to orthorhombic form without loss of water at 120°C, it loses

th43

of its water of crystalization and form plaster of Paris.

ParisofPlaster

24C120

Gypsum24 OH

21.CaSOO.2HCaSO ⎯⎯⎯ →⎯ °



While on heating at 200°C it changes to dead plaster of burnt plaster. On strongly heating it decomposes tocalcium oxide.

19. Answer (1, 2, 3)

Due to smaller size it forms many complexes such as (BeF3)–, (BeF4)

2, [Be(H2O)4]2+ etc.

Organometallic compounds

RMgCl + BeCl2 → R2Be + 2MgCl2For Be2C

2BeO + 2C → Be2C + CO2

20. Answer (1, 2, 3, 4)

Chile saltpetre → NaNO3

Trona → Na2CO3.2NaHCO3.3H2O

Indian saltpetre → KNO3

Sylvine → KCl

All above are the ores of alkali metals.

21. Answer (1, 2, 3, 4)

Be shows diagonal relationship with Al.

Both Be and Al are amphoteric metals.

22. Answer (1, 2, 3)

Factual

23. Answer (1, 2, 4)

Bett’s process is not used for the purification of bauxite.

24. Answer (1, 2, 3)

Anglesite is not a phosphatic ore.

25. Answer (2, 4)

Mg and Al are more electropositive than hydrogen.

26. Answer (1, 2)

Ti or Zr are heated in evacuated vessel with I2.TiI4 or Zr I4 is formed and voltalizes. On heating it decomposeto metal and I2.

27. Answer (1, 2, 4)

Leaching of Al2O3 is carried by NaOH while leaching of Au & Ag is carried by NaCN.

28. Answer (1, 2, 3, 4)

Elligham diagram is the graph plotted between 0GΔ and T for a particular metal oxide. The metal shown by

line can reduce all other metal which lies above it. At point of intersection of 0G =Δ so process will be in

equilibrium. Just by heating, we can make veG0 +=Δ for reaction like 2M(s) + O2(g) ⎯→⎯ 2MO2(s).

29. Answer (1, 2, 3)

Down the group solubility of carbonate of alkaline earth metal decrease and down the group thermal stabilityincreases. Carbonate of alkaline earth metal are generally hydrated.

30. Answer (1, 2, 3)

These unreactive metal oxide like Hg, Pb, Cu are reduced by air/anion of ore. Here no external reducing agentis added.



31. Answer (1, 3, 4)

Based on facts

32. Answer (1, 2, 3, 4)

SO2 acts as a bleaching agent in the presence of moisture.

33. Answer (2, 4)

H – O – S – O – O – H

O

O

34. Answer (1, 2, 3, 4)

6

1

2

1

2

1

22

12PtFO,FO,FO

+−+−+

35. Answer (3, 4)

KI3 → K+ + I3–

36. Answer (1, 2)

He and Ne do not form clathrates.

37. Answer (3)

Due to its small size

38. Answer (2, 4)

According to N.Bartett (pt F6) is a powerful oxidising agent. The ionisation energy of ‘Xenon and Kr’ is almostidentical with of dioxygen.

39. Answer (1, 2)

Al2O3 exist in two most stable form

(i) α-Al2O3 called rhombic lattice or corundum

(ii) γ-Al2O3 called cubic lattice

40. Answer (1, 2, 3, 4)

H N

B

BN

H

B H

N

H

H

H

(Borazole)

(Benzene)

Since nitrogen is more electronegative than boron, therefore ‘–ve’ charge is retained by nitrogen. Isoteric meanssame number of atoms and electrons.

41. Answer (1, 2)

Alumina(Al2O3) reacts with both acid and base.

42. Answer (3, 4)

Cr3+ and V3+ have unpaired electron.



43. Answer (1, 2)

The metals of 4d-series and 5d-series have nearly the same size.

44. Answer (1, 2)

In [La · (EDTA)(H2O)4] · 3H2O, co-ordination number of La is 10 while in La2(SO4)3, 9H2O, the co-ordinationnumber of La is 12.

45. Answer (1, 2, 3)

Lanthanides [From Ce (58) to Lu (71)]

Actinides [From Th (90) to Lr (103)].

46. Answer (1, 2, 4)

It is unusual compound and made up of two ReX4 units link together with Re–Re quadruple bond and i.e., whyRe–Re bond length is abnormally short. If Re–Re bond points along z-axis, then square planar ReX4 unit willuse s, px, py and 22 yx

d−

orbitals for the formation of four Re–X σ-bonds. Re–Re bond is comprised of one σ,

2π and one δ-bond.

47. Answer (1, 2)

In all the four CrO,TcO,ORe,MnO 24444

−−−− all the four metal ions have d° configuration and here colour is not

due to d-d transition but due to charge transfer. Charge transfer in −4MnO and −2

4CrO lies in visible region while

in −4ORe and −

4TcO , it lies in UV region. Hence, −4MnO and −2

4CrO are coloured and −4ORe and −

4TcO are

colourless.

48. Answer (1, 2, 3)

Na2Cr2O7 is hygroscopic and cannot be used as primary standard in volumetric estimations.

49. Answer (1, 2, 4)

After removal of one electron from K, the resulting structure becomes like that of noble gas Ar. Hence, the2nd IE of K is much higher.

50. Answer (1, 2)

Reduction potential depend on ionisation energy, enthalpy of sublimation and hydration energy. Enthalpy ofsublimation depend on the packing.

51. Answer (1, 2, 3)

Acidic or basic character of transition metal depend on the oxidation state of metal. In higher oxidation statethese are acidic, and in lower oxidation state, these are basic.

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

Radius

21 30

Se

Ti

v1Cr

Mn

CuFe

eF O is non-stoichiometric compound because of variable oxidation state.

52. Answer (1, 3, 4)

OH4NOCrOCr)NH( 22327224 ++⎯→⎯ (green coloured solid)

Once the reaction started, enough heat produced to continue on its own.



53. Answer (1, 2)

This is a complex in which partial spin pairing takes place with rise in temperature & hence magnetic momentis temperature dependent.

54. Answer (1, 2, 4)

[ ] 452I SONO)OH(Fe + complex No is 3 e– donor in this complex.

55. Answer (1, 3)

In [Fe(CN)6]4–, all the six electrons of d-orbital get paired up while in [Fe(CN)6]

3–, there is one unpaired electron

CN– is a stronger ligand than H2O.

56. Answer (1, 2, 3)

K2[HgI4] (aq)

2K+ + [HgI4]–2

57. Answer (1, 2, 4)

Correct name of [Mn(CN)5]2– is penta cyano manganate (III) ion.

58. Answer (1, 3, 4)

[Pt(NH3)2Cl2] is square planer complex.

59. Answer (3, 4)

[Cu(H2O)6]+2 → is paramagnetic and hybridization is sp3d2.

60. Answer (2, 3)

In octahedral crystal field, the de– on a metal occupy t2g set of orbitals before they occupy the eg set of orbitalst2g orbitals are dxy, dyz, dxy.

61. Answer (3, 4)

Mn+2 = [Ar] 3d54s0

Mn+2 in [Mn(H2O)6]+2 xx xx xx xx xx xx

3d5

4s 4p 4d

62. Answer (2, 4)

CN– is strong field ligand.

63. Answer (3, 4)

64. Answer (1, 2, 3)

Only [Cu(NH3)4]SO4 has 35 (EAN)

EAN = [Atomic number – Oxd. state] + 2 × C.N.

65. Answer (1, 2)

CN– is a strong field ligand so pairing occur and inner orbital complex is formed having d2sp3 hybridisation.

66. Answer (1, 2, 3)

All the statements are correct except (4).

67. Answer (1, 2, 3, 4)

Orbital splitting depends upon all the four factors.



68. Answer (1, 3)

Fe+3 present in coordination sphere.

69. Answer (1, 2, 3)

Experimentaly found (fact)

70. Answer (1, 2)

Metal ion which have three or less than three e– always form inner orbital complex.

71. Answer (1, 2, 3)

Ligands which are capable of accepting e– from d-orbital of metal ion/atom to its vacant π or π∗−orbital.

72. Answer (1, 3, 4)

Pt have a tendency to form square planar complexes even with weak ligand, Ni+2 form square planar complexwith strong ligand.

73. Answer (1, 2, 3, 4)

As size of metal decrease, stability increase, stronger will be ligand stronger will be complex.

74. Answer (1, 3, 4)

Al+3, Sn+2, Pb+2 form soluble complex with NaOH.

75. Answer (2, 3, 4)

Zn+2, Co2+, Ni+2, Mn+2 give ppt with H2S in ammonical medium.

76. Answer (1, 2, 3, 4)

77. Answer (2, 3)

Both Fe4 [Fe(CN)6]3 & Cu2 [Fe(CN)6] are coloured.

78. Answer (1, 2, 4)

Factual

79. Answer (1, 2, 3)

32324 SOSOOFeFeSO2 ++⎯→⎯Δ

80. Answer (2, 3)

Factual

Section - C : Linked Comprehension

C1. 1. Answer (4)

Z = 43, [Kr]4d55s2

The element with atomic number 43 will be in group 7 and elements of groups 7, 8 and 9 do not formhydrides.

2. Answer (4)

BeH2 is a covalent hydride, therefore, it doesnot conduct electricity at all. CaH2 conducts electricity in thefused state while ZrH2 is an interstitial hydride and conducts electricity at room temperature.

3. Answer (3)

If red hot Pd is cooled in H2, it adsorbs or occludes about 935 times its own volume of H2 gas.



C2. 1. Answer (3)

At room temperature, ordinary hydrogen contains 75% of ortho hydrogen and 25% of para hydrogen. Asthe temperature is lowered, the percentage of ortho hydrogen in the mixture decreases while that of parahydrogen increases and at about 20 K, it is pure para hydrogen. In contrast, when a sample of ordinarydihydrogen is heated say to 400 K or above, the ratio of ortho and para hydrogen remains to be the same(3 : 1). Thus, it is possible to obtain pure para hydrogen but it is not possible to obtain pure orthohydrogen.

2. Answer (3)

Ortho and para hydrogen are nuclear spin isomers.

3. Answer (1)

The melting point of O – H2 is 0.15 K lower than that of hydrogen containing 75% O – H2.

C3. 1. Answer (2)

In SrO2, the oxidation state of oxygen is –1 so it is a peroxide and gives H2O2 on treatment with a diluteacid.

SrO2 + .dil

42SOH → SrSO4 + H2O2

2. Answer (2)

A stronger acid displaces a weaker acid from its salts

acidkerwea223

acidstronger222 OHBaCOCOOHBaO +→++

Pure H2O2 turns blue litmus red but its dilute solution is neutral to litmus. It thus, behaves as a weak acid. Itsdissociation constant is 1.55 × 10–12 at 293 K which is only slightly higher than that of water (1.0 × 10–14).Thus, hydrogen peroxide is only a slightly stronger acid than water.

3. Answer (2)

TiO2 and PbO2 are not peroxides. Anhydrous BaO2 is not used for the preparation of H2O2 due to formationof a protective layer on it

2242cold.dil

4222 OHSONaSOHONa +→+ .

C4. 1. Answer (1)

Due to high polarisation power, Li forms hydrated salt, e.g., LiCl·2H2O.

2. Answer (1)

LiCl is covalent while others are ionic compounds.

3. Answer (1)

Li has lowest reduction potential.

C5. 1. Answer (2)

Factual

2. Answer (1)

Be shows the co-ordination 4. It does not have vacant d-orbital in its outermost orbit.

3. Answer (2)

Anhydron is Mg(ClO4)2.



C6. 1. Answer (1)

ΔHhydration of LiCl = –499 – 382 = –881 kJ mol–1

ΔHsolution = ΔHhyd. – ΔHU

ΔHsolution(LiCl) = –881 – (–840) = –41 kJ mol–1.

2. Answer (1)

The ionic compounds having higher hydration energy than lattice energy are soluble in water.

3. Answer (2)

More the charge density of ion, more is the hydration energy.

C7. 1. Answer (2)

Boron nitride, (BN)x, is called inorganic graphite.

2. Answer (3)

742C160

2C100

33 OBHHBOBOH ⎯⎯⎯ →⎯⎯⎯⎯ →⎯ °°

3. Answer (4)

K2CO3 + BN → KCNO + KBO2.

C8. 1. Answer (2)

Hydrolysis of SiCl4 follows SN2 mechanism.

2. Answer (3)

Hybridisation is sp3d.

3. Answer (2)

SN2 mechanism, so inverted configuration.

C9. 1. Answer (3)

Birkland-Eyde process for the manufacture of HNO3 is now obsolete.

2. Answer (1)

In Ostwald process, NH3 is catalytically oxidised to NO.

3. Answer (4)

In the laboratory, Nitric acid can be prepared by heating a KNO3 or NaNO3 with conc. H2SO4.

C10. 1. Answer (3)

[X] is S and [Y] is Na2S3

2. Answer (4)

NaI2OSNaIOSNa2)B(

6422)A(

322 +→+

3. Answer (2)

Na2S2O3 + AgBr → Na3[Ag(S2O3)2] + NaBr



C11. 1. Answer (3)

−+⎯⎯ →⎯+ 6K278

6 PtFXePtFXe

2. Answer (3)

)s(XeF)g(F2)g(Xe 4bar7

K873

)ratio5:1(2 ⎯⎯ →⎯+

3. Answer (4)

CrO42– is tetrahedral while XeF4 is square planar.

C12. 1. Answer (4)

In nitrogen family on moving down the group both thermal and electrical conductivity increases due toincrease in delocalisation of electron from nitrogen to bismuth.

2. Answer (4)

Greater electronegativity and higher oxidation state in responsible for greater acidic character.

3. Answer (4)

Due to presence of empty d-orbital in Sb. It can expand covalency upto 6.

C13. 1. Answer (1)

K2Cr2O7 is a powerful oxidising agent. In the presence of dil.H2SO4, one molecule of K2Cr2O7 gives 3 atomsof available oxygen.

2. Answer (2)

yellow

24

7pH272 CrOOCr −>− ⎯⎯⎯ →⎯

3. Answer (2)

OH7Fe6Cr2H14Fe6OCr 23322

72 ++⎯→⎯++ ++++−

C14. 1. Answer (2)

Cu(I) has d0 configuration.

2. Answer (2)

4224 SOKCuIKI2CuSO +⎯→⎯+

22 ICuI2CuI2 +⎯→⎯

3. Answer (2)

Au3+ does not undergo disproportionation.

C15. 1. Answer (3)

OHKHSO2OMnSOH2KMnO2 2472.conc

424 ++⎯→⎯+

2. Answer (2)

The minimum reduction potentials required to oxidise water to dioxygen is E0 > 0.185 volt.

here both (1) and (3) have negative reduction potential.

3. Answer (2)

Mohr’s salt is FeSO4·(NH4)2SO4·6H2O and is used as a reducing agent.



C16. 1. Answer (3)

[Co(NH3)5NO2]Cl2

2. Answer (1)

[Co(en)2Cl2]+ → cis-trans.

isomerismopticalligandbidentateox

ligandbidentateen

⎭⎬⎫

→→

3. Answer (4)

(i) (Cu(NH3)4][PtCl4]

(ii) [Cu(NH3)3Cl][Pt(NH3)Cl3]

(iii) [Cu(NH3)2Cl2][Pt(NH3)2Cl2]

(iv) [Cu(NH3)Cl3][Pt(NH3)3Cl]

C17. 1. Answer (4)

[Cu(NH3)4]+2 is a square planar complex.

2. Answer (2)

One unpaired e– in [Cu(NH3)4]+2 but no unpair e– in [Fe(CN)4]

–2.

3. Answer (4)

[Cu(en)2]+2 is more stable due to chelation.

C18. 1. Answer (2)

A → CrCl3

B → Na2CrO4

C → Na2Cr2O7

D → (NH4)2Cr2O7

G → N2

I → Li3N

J → NH3

H → Cr2O3

2. Answer (3)

CrO2Cl2 is chromyl chloride

3. Answer (2)

OH2NNONH 2224 +⎯→⎯Δ

C19. 1. Answer (2)

)B(2

)C(4

.dil42

)A(SHZnSOSOHZnS +→+



K2Cr2O7 + H2SO4 + )B(

2SH → K2SO4 + Cr2(SO4)3 + H2O + S

)D(S + O2 →

)E(2SO

SO2 + H2S → H2O + )D(

S

2. Answer (1)

ZnSO4 + NaOH → Na2ZnO2 + H2SO4

3. Answer (4)

Based on facts

C20. 1. Answer (3)

)B(3

KOH

)A(4224 NHOC)NH(

Δ⎯⎯ →⎯

)fumeswhite(4

)B(3 ClNHHClNH ⎯→⎯+

ClNH2OCaCOC)NH( 4

.pptwhite

)C(42

CaCl

)A(4224

2 +⎯⎯⎯ →⎯

)colourless(

2KMnO

)C(42 MnOCaC 4 +⎯⎯⎯ →⎯

2. Answer (2)

ClNHNClClNH 43)excess(

2)B(3 +→+

3. Answer (2)

Based on facts

C21. 1. Answer (1)

(A) is Pb3O4 (red lead)

2. Answer (3)

HMnO4 is pink.

3. Answer (3)

BaCrO4 is obtained as yellow ppt.

Section - D : Assertion - Reason Type1. Answer (2)

Because of small size of Li+ more covalent character is present is LiBH4 and it is very reactive towards water.

2. Answer (1)

Both A and R are correct and R is the correct explanation of A.

3. Answer (3)

The dielectric constant of pure H2O2 is 93.7, (which also increases on dilution, 97 for 90% pure; 120 for 65% pure).The dielectric constant of water is 82.

4. Answer (3)

Glycerol, acetanilide and phosphoric acid act as negative catalyst for the decomposition of H2O2 and thus,decomposition of H2O2 is checked off.



5. Answer (4)

Chemical reaction of D2O are slower than H2O. Heavier isotope (deuterium) is less reactive and bond energyof O – H bond is lesser than O – D bond.

6. Answer (1)

Both A and R are correct and R is the correct explanation of A.

7. Answer (3)

The reaction between hydrazine and H2O2 is highly exothermic and takes place with a large increase involumes so that it can propel a rocket

NH2·NH2 + 2H2O2 ⎯⎯ →⎯ )II(Cu N2↑ + 4H2O↑

8. Answer (4)

The hydrides of N, O and F such as NH3, H2O and HF have unusually higher boiling point due to associationof its molecules by means of intermolecular H-bonding.

9. Answer (1)

Both (A) and (R) are correct and (R) is the correct explanation of A.

10. Answer (3)

The standard enthalpies of formation of alkali metal chlorides become more and more negative as we movedown the group, i.e., ΔfH° of KCl is more negative than that of NaCl. Therefore, the above reaction proceedsbetter with KF than with NaF.

11. Answer (4)

When aqueous alkalimetal salt solutions are electrolysed, H2 gas is liberated at cathode.

12. Answer (1)

Both (A) and (R) are correct and is the correct explanation.

13. Answer (4)

C2H5OH + Na → C2H5O–Na+ +

21

H2

14. Answer (1)

Both statements are correct and statement (2) is the correct explanation of statement (1).

15. Answer (4)

Phenolphthalein is not a good indicator for weak alkali titrations as 50% neutralisation of K2CO3 gives KHCO3during titration which is a weak base

K2CO3 + HCl → KHCO3 + KCl

16. Answer (1)

Both statements are correct and statement-2 is the correct explanation of statement-1.

17. Answer (2)

The polarisation power of the cations of alkaline earth metals is more.

18. Answer (1)

Both statements are correct and statement-2 is the correct explanation.

19. Answer (1)




20. Answer (2)

Both statements are correct.

21. Answer (3)

22. Answer (1)


23. Answer (1)


24. Answer (1)


25. Answer (1)


26. Answer (2)

Sulphur exists as S2 in vapoure phase and hence, paramagnetic like O2.

27. Answer (2)

In the reaction of H2S with SO2, SO2 behaves as an oxidising agent.

28. Answer (4)

H3PO4 is less acidic than H3PO3 and H3PO2.

29. Answer (2)


30. Answer (1)


31. Answer (3)

Electronegativity of elements decreases on moving down the group.

32. Answer (2)


33. Answer (3)

Phosphorus exhibits pentavalency.

34. Answer (2)

BF3 has least Lewis acidity due to pπ-pπ back bonding.

35. Answer (4)

H3PO3 can form only one acidic salt.

36. Answer (4)

−+ +⎯→⎯+ Cl6])OH(Al[2OH12ClAl 362262

37. Answer (2)

F2 is stronger oxidising agent due to its higher reduction potential.

38. Answer (4)

H3PO2 (Hypophosphorus acid) is a monobasic acid.



39. Answer (2)


40. Answer (1)


41. Answer (1)


42. Answer (3)

In V2O5, colour is due to charge transfer.

43. Answer (2)

Due to lanthanoid contraction, all the lanthanoids ions are of almost same size, so they have almost similarchemical and physical properties.

44. Answer (3)

The difference is due to occurrence of a wide range of oxidation states in actinoids. Also, their radioactivitycauses hindrance in their study.

45. Answer (4)

Orange

272

mediumbasic

mediumacidic

Yellow

24 OCrCrO −−

⎯⎯⎯⎯⎯ ⎯← ⎯⎯⎯⎯⎯ →⎯

46. Answer (1)


47. Answer (2)

In [FeF6]3–, Fe is in +3 state and has d5-configuration. F– is weak field ligand and so [FeF6]

3– is high spincomplex. d-d transitions in this arrangement is spin forbidden because spin will be reversed. Also all d-dtransitions are against Laporte selection rules (i.e., Δl = ± 1), i.e., why complex is colourless.

48. Answer (3)

FeCl3 is a salt of strong acid and weak base.

49. Answer (1)


KCl12])CN(Fe[Fe])CN(Fe[K3FeCl4)blueussian(Pr

364643 +⎯→⎯+

50. Answer (3)

In CuSO4· 5H2O, four H2O molecules are co-ordinated to the central Cu2+ ion and one H2O molecule has

H–bond with −24SO .

51. Answer (2)

In La (57), the last electron goes to 5d instead of 4f (violation of Aufbau principle).

52. Answer (4)

Actinoids show larger number of oxidation states. The energy gap between 4f- and 5d- subshell is small.

53. Answer (4)

[(PPh3)3RhCl] does not show geometrical as well as optical isomerism.

54. Answer (3)

µ of K4[Fe(CN)6] is zero and of K3[Fe(CN)6] is 3 B.M.



55. Answer (3)

It is bidentate σ-bonded complex.

56. Answer (2)

[FeF6]–4 has unpair e¯ but in [Fe(CN)6]

–4 complex all e¯ are paired.

57. Answer (4)

[Cu(NH3)4]+2 is dsp2 hybridized square planar complex.

58. Answer (4)

[Cu(NH3)2]+ BF4̄ is colourless complex.

59. Answer (2)

Mn+2 has 5 unpair e¯.

60. Answer (4)

Potassium ferricyanide is weakly paramagnetic due to presence of one unpair e¯.

61. Answer (3)

It is inner orbital octahedral complex with d2sp3 hybridization.

62. Answer (1)


63. Answer (2)


64. Answer (1)


65. Answer (3)

Phosphomolybdate ion is yellow coloured.

66. Answer (3)

Pb(OH)2 is insoluble in NH4OH but (white ppt) soluble in NaOH

Pb(OH)2 + NH4OH No Reaction

Pb(OH)2 + 2NaOH Na2[Pb(OH)4]

67. Answer (1)

Hg CO HgO + Hg + CO2 3 2→ ↑(ppt.)

black(gas)

68. Answer (3)

Due to formation of potassium dicyanoargentate (I)

AgCN + KCN K [Ag(CN)2]

(ppt.) (soluble)

69. Answer (3)

NaOH / H2O2 will convert Cr3+ to CrO4̄ which become soluble while Fe(OH)3 remain as ppt.

70. Answer (3)

NaOH + Al(OH)3 Na [Al(OH)4]

NaOH form soluble complex with aluminium hydroxide not with ferric hydroxide.



71. Answer (1)


72. Answer (1)


73. Answer (2)

Ba+2 give test of Sr+2 & Ca+2 ions.

74. Answer (3)

[Ag(S2O3)2]–3 is soluble complex which changes to white ppt. Sodium thiosulphate Ag2S2O3 and finally turns

black due to formation of Ag2S.

75. Answer (1)

[Fe(CN)6]4– + 2Cu2+ Cu2[Fe(CN)6]↓

76. Answer (1)


77. Answer (2)

KCl + H2SO4 HCl + K2SO4

Kl + H2SO4 Hl + K2SO4

Hl + H2SO4 Br2 + SO2 + H2O

78. Answer (1)


79. Answer (1)


80. Answer (2)

2KMnO4 K2MnO4 + MnO2 + O2

Black residue is due to potassium manganate and manganese dioxide

81. Answer (3)

IInd A group are insoluble in yellow ammonium sulphide where as IInd (B) group are soluble.

Section - E : Matrix-Match Type

1. Answer - A(p, q, r), B(p, q, s), C(q), D(r)

H2O2 behaves as an oxidising agent as well as reducing agent in both acidic solution and basic solution.

⎪⎩

⎪⎨⎧

→+→++

−

+

)mediumalkalineIn(OH2e2OH

)mediumacidicIn(OH2e2H2OHnaturegsinOxidi

22

222

H2O2 → 2H+ + O2 + 2e (In acidic medium)

H2O2 + 2OH– → 2H2O + O2 + 2e (In alkaline medium)

H2O2 acts as a bleaching agent for delicate articles like hair, silk, wool, ivory, etc. The bleaching action is dueto oxidation by nascent oxygen and hence is permanent.

H2O2 → H2O + [O]

The pure liquid has weak acidic nature. Ka = 1.55 × 10–12. H2O2 is less acidic than carbonic acid.



2. Answer - A(p, s), B(q, r), C(q, r), D(q, r)

(A) Dilute H2O2 is a valuable antiseptic and germicide for washing wounds, teeth and ears under the nameof perhydrol.

(B) Water gas is CO + H2

Preparation of H2 by water gas

22OFe

Cº45022 COH2OHHCO

32

+⎯⎯⎯ →⎯++

(C) The maximum quantity of commercial hydrogen is produces by “Bosch’s” process, water gas is firstproduced.

43421

gas water

2C100

2 HCOOHC +⎯⎯⎯ →⎯+ °

(D) By partial oxidation or cracking natural gas (CH4) gives water gas and H2.

2C820

CrNi

Steam24 HCOOHCH +⎯⎯⎯ →⎯+

°

−

22CrNi

2 HCOOHCO +⎯⎯⎯ →⎯+ −

3. Answer - A(p, q), B(p), C(p, q, r, s), D(q, r, s)

(A) Formula of heavy water is D2O and it is colourless compound.

(B) Heavy water is used as a moderator in nuclear reaction.

(C) D2O is prepared by the reaction of H2O with D2. D2 is colourless with low boiling point molecule and itis diatomic gas.

(D) H2 is diatomic, colourless and low boiling point molecule.

4. Answer - A(p, r), B(q, r), C(r, s), D(q, r)

Dihydrogen forms three types of hydrides, ionic, covalent and interstitial hydrides. Ionic or salt like (saline)hydrides are formed by alkali metals, alkaline earth metals with exception of Be and Mg and some highlyelectropositive members of lanthanide series.

Covalent or molecular hydrides are formed by all the true non metals (except zero group elements) and theelements such as Al, Ga, Sn, Pb, Sb, Bi, Po etc., which are normally metallic in nature.

Metallic or interstitial hydrides are formed by many d-block and f-block elements at elevated temperatures.These hydrides are often non-stoichiometric.

Complex metal hydrides such as LiAlH4 and NaBH4 are powerful reducing agents and are widely used inorganic reactions.

5. Answer - A(q, r, s), B(q, r), C(p, q), D(p, q)

CaH2 and LiH are ionic hydrides and act as reducing agents.

6. Answer - A(q, r), B(p, s), C(p, s), D(p, q, r, s)

Lithium shows diagonal relationship with Mg. Except LiF, all other halides of Li show somewhat covalent natureas the small Li+ cation brings more and more polarisation in the molecule as the size of the halide ion, X–,increases.

Sea water contains 2.0 – 2.9% sodium chloride. Mg is also found in sea water. It is an essential constituentof chlorophyll.

Ca is present in natural waters and causes hardness in water. It is an essential constituent of bones and teeth.Egg and sea shells contain CaCO3.



7. Answer - A(q, r, s), B(p, r), C(q, s, r), D(p, r)

Sorel's cement (or magnesia cement) is MgCl2, 5MgO.xH2O. Albite is NaAlSi3O8. Carnallite is KCl.MgCl2.6H2O. Glauber's salt is Na2SO4.10H2O.

8. Answer - A(p, q, r, s), B(p, q, r, s), C(p, q, s), D(s)

(A) Salts of Be undergo hydrolysis to form hydroxo complex, BeCl2 on reaction with H2O gives the fumesof HCl it acts as a lewis acid and soluble in water.

(B) Similar to above

(C) Due to low charge density salts of Mg are not acts as a Lewis acid and for rest answers same as above.

(D) In alkaline earth metals for the oxides solubility increases down the group.

9. Answer - A(p, r, s), B(s), C(p, q, r), D(p, q, r)

(A) Since sodium is very reactive, hence it reacts with air therefore it stores in kerosene or benzene.

(B) Formula of baking powder is NaHCO3.

(C) Since it can eject electrons easily therefore used in photoelectric cell and for rest answers same as aand b.

(D) It is photoelectric material.

10. Answer - A(p, q, s), B(p, r), C(q, s), D(p, q, r, s)

(A) Davy and Lunge mechanism (for H2SO4)

NO + NO2 → N2O3

2SO2 + N2O3 + O2 + H2O → ate)(Intermedi

4.NO2HSO

2HSO4.NO + H2O → 2H2SO4 + NO + NO2

∴ all three (p, q, s) are produced in the reaction

(B) Iron pyrite (FeS2) produces. Sulphur on distillation

SSFeFeS 43ondistillati

2 +⎯⎯⎯⎯ →⎯

FeS2 is used in the preparation of H2SO4 in lead chamber process.

(C) Copper when reacts with HNO3 to give NO and NO2 in molar ratio of 2 : 1

7Cu + 20HNO3 → 7Cu(NO3)2 + 2NO2 + 10H2O + 4NO

(D) SO2 + NO2 → SO3 + NO(g)

SO3 + H2O → H2SO4(l)

SO2 is reduced to ‘s’ in presence of moist SO2 + H2S → S + H2O

In part ‘A’ (above) NO and NO2 both produced when intermediate reacts with H2O.

11. Answer - A(q), B(r), C(p), D(q, r)

On hydrolysis of Bi3+ gives its oxides on acidification [AlO2]+ gives its hydroxide.

12. Answer - A(q, r), B(r), C(p, s), D(s)

The hybridisation of Xe in XeF2 is sp3d but it has three lone pair of electrons which occupy all the threeequatorial positions and hence, its shape becomes linear.

The hybridisation of Xe in XeF4 is sp3d2 but shape is square planar because it has two lone pair of electrons.

The hybridisation of Xe in XeOF4 is sp3d2 but it has one lone pair and hence, its shape is square pyramidal.



13. Answer - A(p, q), B(p), C(r, s), D(r)

The hybridisation of Xe in XeF6 is sp3d3 but it has one lone pair of electrons and so, its shape becomesdistorted octahedral.

The hybridisation of the central atom in IF7 is sp3d3 and its shape is pentagonal bipyramidal.

14. Answer - A(p, s), B(p, s), C(q, r), D(q)

In BCl3, hybridisation of the central atom is sp2 and its geometry is trigonal planar. It does not form bridgestructure due to larger size of Cl-atom. In B2H6, B is sp3 hybridised and it has bridge structure.

F—B N—H|

|

|

|F

F H

H

Both B and N are sp3 hybridised.

15. Answer - A(p, q, s), B(p, q), C(s), D(q, r)

BeCl2 in vapour phase is monomeric and dimeric but in solid state it is polymeric.

Boron nitride (BN) is similar to the structure of graphite

NB

N

BN

B

B

N

N

B

Structure of Boron nitride.

BeCl2, AlCl3, SiCl4 are Lewis acids.

16. Answer - A(p, q, s), B(p, q, s), C(q, s), D(r)

Iron pyrites and Fool's gold is FeS2. Sulphide ores are concentrated by froth floatation process.

Galena is PbS. Haematite is Fe2O3. It is usually red in colour. Lead is mainly extracted from galena ore.

17. Answer - A(p, q, s), B(p, q), C(q, r), D(p, q)

Stainless steel contains Fe, Cr and Ni

In Permalloy, there is 21% Fe, 78% Ni and carbon. In German silver there is 56% Cu, 24% Zn and 20% Ni.In Alnico, there is 60% Fe, 12% Al, 20% Ni and 8% Co.

18. Answer - A(q), B(p, q), C(r, s), D(q, r)

Liquation process is based on the difference in fusibility of the metal and impurities. When the impurities areless fusible than the metal itself, this process is employed. This method is used to purify the metals like Bi,Sn, Pb, Hg, etc.

Distillation process is used for those metals which are easily volatile. This is used for the purification of Zn,Cd, Hg etc.

Many of the metals such as copper, silver, gold, aluminium, lead etc., are purified by electrolytic refining of metals.

19. Answer - A(p, q), B(q, r, s), C(q, r, s), D(p, q, s)

(A) Extraction of Zn is carried by Parke Process

(B) Cyanide of Au and Ag is soluble complexes

(C) + 1 oxidation state of Au and Cu show disproportionation.



20. Answer - A(q, s), B(q), C(p), D(r)

The highest oxidation state +8 is shown by ruthenium and osmium in oxides, fluorides, oxo-anions and fluorocomplexes.

The two elements osmium and iridium have highest densities 22.67 g mL–1 and 22.61 g mL–1 respectively.

Cr(24) — [Ar] 3d54s1

Cr has 6 unpaired electrons.

Technetium (TC) was the first synthetic element and is radioactive.

21. Answer - A(p, r), B(q, r), C(p, q, s), D(p, q, s)

In K2MnO4, the oxidation state of Mn is +6.

OH2MnOK2OKOH4MnO2 2422)orepyrolusite(

2 +→++

2K2MnO4 + Cl2 → 2KMnO4 + 2KCl

KMnO4 acts as an oxidising agent in alkaline, neutral or acidic solutions.

23242232oreChromite

32 CO8OFe2CrONa8O7CONa8OCr.FeO4 ++→++

2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

22. Answer - A(p, q), B(p, q), C(s), D(r)

Fe, Pt do not form amalgams.

23. Answer - A(p, q, r), B(q, r), C(q), D(q, s)

FeCl3 and FeF3 are almost black coloured solid FeF3 is sparingly soluble in water but gives Na3[FeF6] insolution which does not gives +ve test for Fe+3.

FeCl3.6H2O is used as both oxidising and as a mordant in dyeing. Ferrates are only stable in strongly alkalinesolution.

24. Answer - A(p, q, s), B(p, r, s), C(p, s), D(s)

(A) Na2Cr2O7 compound is hygroscopic, oxidising agent and coloured.

(B) CrO3 (anhydride of chromic acid) is good oxidising agent, corrosive and coloured.

(C) CrO2Cl2 is a oxidising agent and coloured due to charge transfer.

(D) Cr2O3 is a green coloured solid.

25. Answer - A(p, q, r), B(p, q, r), C(q, r, s), D(s)

FeSO4 + 6KCN → K4[Fe(CN)6] + K2SO4

KCl12])CN(Fe[Fe])CN(Fe[K3FeClblueussianPr

364643 +→+

42.pptbrownddishRe

62644 SOK2])CN(Fe[Cu])CN(Fe[KCuSO2 +→+

Extraction of metals involving aqueous solution is known as hydrometallurgy. Ag, Au, Cu, etc., are extractedby this process.

26. Answer - A(r), B(p, q), C(q, s), D(p, q, s)

SCN–, NO2– ambidentate ligands.

[Cr(H2O)6]Cl3 has 3 hydrate isomers.



27. Answer - A(p, s), B(p, q, r), C(q, s), D(p, r, s)

OX (oxalato) is a bidentate ligand and so it is a chelating ligand.

en (ethylene diammine) is a bidentate chelating ligand.

28. Answer - A(p, r), B(p, s), C(q, r), D(q, s)

Cu+ show sp3, Ni form tetrahedral complex when R – C6H5 in [Ni X2 (PR3)2] and when R is cyclo-hexyl thencomplex will square planar.

29. Answer - A(p, q, r, s), B(p, q, r), C(r), D(p, q, r)

Stability of a complex depend on the nature of metal ion, electronic configuration, nature of ligand and canalso be increased by chelation.

Crystal field splitting energy depends on the size of metal ion, electronic configuration and type of ligand.Isomerism depend on the type of ligand and tendency to form outer orbital complex or inner orbital complexdepend on nature of metal in electronic configuration and type of ligand.

30. Answer - A(p, q, s), B(p, r, s), C(p, q), D(p, q)

Co and CN– are σ donar and accept the e– from d-orbital of metals CO, CN–, CH2 = CH2, also fromsynergic bonding.

31. Answer - A(p, q), B(p, q), C(r, s), D(r)

CN–, NO2–, CNS– etc., are ambident ligands. All ambidentate ligands are monodentate ligands. Polydentate

ligands are chelating ligands. EDTA is ethylene diammine tetra acetate and is hexa dentate ligand.

32. Answer - A(r), B(p, q), C(p, q), D(s)

Zinc oxide (ZnO) is also called zinc white or chinese white or philosopher's wool. It is a white powder. Itbecomes yellow on heating and again turns white on cooling.

PbO is known in two forms :

(i) a yellow powder commonly known as 'massicot' and

(ii) a buff coloured crystalline form known as litharge.

33. Answer - A(p), B(p, q, r), C(p, q), D(p, q, r, s)

Al3+ is in group III and its group reagent is NH4OH + NH4Cl.

Cu2+, Bi3+ are in group II and group reagent is H2S + dil. HCl.

Zn2+ is in group IV and its group reagent is H2S + NH4OH.

34. Answer - A(p, s), B(p, s), C(r), D(q)

Zn2+ and Co2+ are in group IV.

Sr2+ is in group V and its group reagent is (NH4)2CO3 + NH4Cl + NH4OH.

35. Answer - A(q, r), B(p, s), C(q, r), D(q, s)

Ag NO3 is called Lunar caustic. All metal nitrates are water soluble.

SnCl2 is a white crystalline solid. It is soluble in water, alcohol and ether.

36. Answer - A(p, r), B(p), C(s), D(q, r)

H2S and SO2 gas reduce acidified K2Cr2O7 to Cr3+, SO2 and CO2 both turn lime water milky.

2242.dil

4232 COOHSONaSOHCONa ++→+

OHCaCOCO)OH(Ca 2)milky(32

waterLime2 +→+

2242.dil

4232 SOOHSONaSOHSONa ++→+



OHCaSOSO)OH(Ca 2)milky(32

waterLime2 +→+

OHSO3Cr2H2SO3OCr 22

4)green(

32

272 ++→++ −++−

242.dil

422 HNO2SONaSOHNaNO2 +→+

2KI + H2SO4 + 2HNO2 → K2SO4 + 2H2O + 2NO + I2I2 + starch → blue colour.

422.dil

422 SONaSHSOHSNa +→+

COOHCH2PbSPb)COOCH(SH 3)Black(

232 +→+

37. Answer - A(p, q, r), B(p, r), C(p, r, s), D(p, r)

(A) OH2ClNHCl)NH(HgHgOHNH2ClHg 242422 +++⎯→⎯+

(B) 4222 SnClHg2SnClClHg +⎯→⎯+

(C)

4222

42222

SnClHg2SnClClHg

SnClClHgSnClHgCl2

+⎯→⎯+

+⎯→⎯+

(D)pptgrey

HgCuHgCl2 ⎯→⎯+

38. Answer - A(q, s), B(r), C(p), D(p)

↑⎯⎯⎯⎯ →⎯−2

SOHdil23 COCO 42

↑+⎯→⎯++−223423 NO)NO(CuCuSOHNO

]NOS)CN(Fe[Na 52 Purple coloured compound.

Section - F : Subjective Type1. (i) Chromium hydroxide is converted into soluble yellow sodium chromate.

3]]O[OHOH[ 222 ×+⎯→⎯

OH5CrONa2]O[3NaOH4)OH(Cr2 2423 +⎯→⎯++

OH8CrONa2OH3NaOH4)OH(Cr2 242223 +⎯→⎯++

(ii) Zinc dissolves in caustic potash solution evolving hydrogen

↑+⎯→⎯+ 222 HZnOKKOH2Zn

(iii) Deuteromethane is evolved.

43234 CD3)OD(Al4OD12CAl +⎯→⎯+

(iv) OHOHHCOHHC 256FeSO

22664 +⎯⎯⎯ →⎯+



2. It is clear from point (iv) that (X) is H2O2.

42

)yellow(acidictanPerti42

)X(22224 SOH2TiOHOHOH2)SO(Ti +⎯→⎯++

(i)

2242)X(

224

2222

2424

OOH2MnOK2OHKOH2KMnO2

OOH]O[OH

]O[OHMnOK2KOH2KMnO2

++⎯→⎯++

+⎯→⎯+

++⎯→⎯+

KOH4OMnO2OH2MnOK2 2.)pptBrown(

2242 ++⎯→⎯+

(ii) 22)X(

22 OHCl2ClOH +⎯→⎯+

(iii) ↑++⎯→⎯++ 22424222 IOH2SOKSOHKI2OH

3. H2O2 as oxidant.

Chromium hydroxide is oxidised by H2O2 in presence of NaOH into sodium chromate.

[H2O2 ⎯→⎯ H2O + [O]] × 3

2Cr(OH)3 + 4NaOH + 3[O] ⎯→⎯ 2Na2CrO4 + 4H2O

-------------------------------------------------------------------------------------------------------

2Cr(OH)3 + 4NaOH + 3H2O2 ⎯→⎯ 2Na2CrO4 + 8H2O

H2O2 as reductant.

Potassium ferricyanide is reduced to ferrocyanide in presence of KOH by H2O2.

2K3[Fe(CN)6] + 2KOH ⎯→⎯ 2K4 [Fe(CN)6] + H2O + [O]

H2O2 + [O] ⎯→⎯ H2O + O2

--------------------------------------------------------------------------------------------------------------------

2K3[Fe(CN)6] + 2KOH + H2O2 ⎯→⎯ 2K4[Fe(CN)6] + 2H2O + O2

4. A + B → C (gives fumes)

C + NaOH → A

∴ Gas A is NH3

B is HCl

and C is NH4Cl

NH3 gives brown precipitate with Nessler’s reagent.

∴ D is Nessler’s reagent (K2HgI4)

∴ E formed is OHg

NH I2Hg

oxydimercuric ammonium iodide

NH3 + 3NaOH + K2HgI4 → OHg

NH I2Hg

+ 4KI + 3NaI + 2H2O.



5. Gas B is chlorine (Cl2)

Ethyl alcohol gives anaesthetic CHCl3

∴ E = CHCl3

The milky precipitate C is CaCO3

∴ A is a compound of Ca and Cl

CaCO3 + H2O + CO2 → Ca(HCO3)2

D = Ca(HCO3)2

The inert gas at room temperature by oxidation of NH3 is N2

∴ F = N2

∴ A also has O

A is a compound of Ca, O, Cl

∴ A is CaOCl2 i.e. bleaching powder

CaOCl2 + H2O → Ca(OH)2 + Cl2

3CaOCl2 + 2NH3 → 3CaCl2 + 3H2O + N2

Cl2 + H2O → 2HCl + O; O + CH3CH2OH → CH3CHO + H2O

CH3CHO + 3Cl2 → CCl3CHO + 3HCl; 2CCl3CHO + Ca(OH)2 → 2CHCl3 + Ca(HCOO)2

6. 6NaOH + 4S + 5H2O → Na2S2O3·5H2O + 2Na2S + 3H2O

∴ B is Na2S2O3·5H2O

Na2S + )excess(

S4 → Na2S5

∴ C is (Na2S5)

A + S → Na2S2O3

∴ A is Na2SO3

D is Na2S2O3

Na2S2O3 + H2SO4 → Na2SO4 + SO2 + )yellow(

S + H2O

∴ From description F is SO2

E is sulphur

2Na2S2O3 + I2 → NaI + Na2S4O6

∴ G is Na2S4O6 (sodium tetrathionate).

7. C + K2CrO4 → D (yellow ppt.)

∴ D is BaCrO4

C is a carbonate

∴ C is BaCO3

∴ A is Ba(NO3)2

B is BaCl2



BaCO3 + H2O → Ba(OH)2 + CO2

∴ E is Ba(OH)2

F is CO2

6Ba(OH)2 + 6Cl2 → .ppt

23 )ClO(Ba + 5BaCl2 + 6H2O

∴ G is Ba(ClO3)2.

8. D = AgI (from property)

B = KI (from 3rd reaction)

C = CuI2

∴ A = CuSO4.

9. The first part is bead test using microcosmic salt

∴ C is NaPO3

and A is Na(NH4)HPO4

Na(NH4)HPO4 ⎯→⎯Δ NaPO3 + H2O + NH3

Na2HPO4 + NH4Cl → Na(NH4)HPO4 + NaCl

∴ B is Na2HPO4

E is Na2SiO3 (present in glass)

So, D is a oxide of Si

D is SiO2

Na2SiO3 + 6HF → Na2SiF6 + 3H2O

∴ F is Na2SiF6.

10. Oxide B is amphoteric as it dissolves in NaOH. BeO is the only amphoteric oxide

∴ B = BeO

A = Be

BeO + 2NaOH → Na2BeO2 + H2O

∴ C = Na2BeO2

BeO + C + Cl2 → BeCl2 + CO

∴ D = BeCl2

E = CO

BeCl2 + )airof(

2OH2 → Be(OH)2 + )fumes(

HCl2

F = HCl

BeO + 4NH4F → (NH4)2BeF4 + 2NH3 + H2O

(NH4)2BeF4 ⎯→⎯Δ BeF2 + 2NH4F



11. Red ppt. is of HgI2∴ D is HgI2

A is iodide salt

Only KI dissolves free I2∴ KI + I2 → KI3∴ A = KI

B = KI3

OH8I5MnSO2SOK6SOH8KMnO2KI10 2)brown(

244242)pink(

4 +++→++

C = I22KI + HgI2 → K2HgI4E = K2HgI4 or Nessler’s reagent.

12. Ellingham diagram can be used to predict which metal/non-metal can be used to extract any other metal fromits oxide by using smelting process.

Any metal having its metal-metal oxide line above any other metal can be extracted by other metals havingtheir metal-metal oxide line below in Ellingham diagram. Ex. Al can be used to extract Cr, Fe etc. from theiroxide.

13. (i) 4FeCr2O4 + 16NaOH + 7O2 → )A(

42CrONa8 + 2Fe2O3 + 8H2O

(ii))B(

42)A(

42 SOHCrONa2 + → Na2Cr2O7 + Na2SO4 + H2O

(iii) Na2Cr2O7 + 3C + H2O ⎯→⎯Δ 2NaOH + Cr2O3 + 3CO

(iv) Cr2O3 + Al → 2Cr + Al2O3

(v) Na2Cr2O7 + 4NaCl + 6H2SO4 → 6NaHSO4 + 2CrO2Cl2 + 3H2O

14. In Ellingham diagram the line of Al – Al2O3 is below metal-metal oxide line of other metals. Thus coupledreaction of its oxide with other metal will have positive Gibb’s free energy change value. Which doesn’t favourits reduction using smelting.

15. (i) AlCl3, 6H2O exists as [Al(H2O)6]Cl3. Upon heating AlCl3 undergoes hydrolysis and forms Al2O3.

(ii) AlCl3 lacks back-bonding as in BCl3 because of large size of Aluminium.

Aluminium metal forms complete octet by coordinate bridges by chlorine atoms between two Al atoms.

(iii) On heating, borax first swells up due to elimination of water molecules. On further heating, it melts to aliquid which then solidifies to a transparent glassy mass.

444 3444 21

beadGlassy

322)strong(

742OH10

2742 OBNaBO2OBNaOH10OBNa2

+⎯→⎯⎯→⎯⋅ Δ

−

Δ

When the hot bead is touched with a coloured salt, B2O3 displaces the volatile oxides and combines withbasic oxides to form metaborates.

e.g. CuSO + B O CuO · B O + SO4 2 3 2 3 3

Cu(BO ) 2 2Blue

Colour of metaborates :Cu

BlueFe

GreenCo

BlueCr

GreenNi

Brown



16. (i)

OH3NONH)NO(Sn4)dil(HNO10Sn4

OH3NONHH8HNO2

4]H2)NO(SnHNO2Sn[

234233

2343

233

++⎯→⎯+

+⎯→⎯+

×+⎯→⎯+

(ii)

CO6SO)NH(3FeSOSOK2OH6SOH6])CN(Fe[K

6]OHCOHCOOH[

3]SO)NH(SOHNH2[

6]NHHCOOHOH2HCN[

HCN6FeSOSOK2SOH3])CN(Fe[K

42444224264

2

424423

32

4424264

+++⎯→⎯++

×+⎯→⎯

×⎯→⎯+

×+⎯→⎯+

++⎯→⎯+

(iii) 223C375275

Cu3 SiCl)CH(SiClCH2

°−⎯⎯→⎯+

(iv)

OH2SiFCaSO2SiOSOH2CaF2

OH2SiFHF4SiO

2]HF2CaSOSOHF[

2442422

242

4422

++⎯→⎯++

+⎯→⎯+

×+⎯→⎯+

(v)

OH2PbO)NO(Pb2)dil(HNO4OPb

2]OH)NO(PbHNO2PbO[

PbOPbO2OPb

2223343

2233

243

++⎯→⎯+

×+⎯→⎯+

+⎯→⎯

17. (i) N2O3 > P2O3 > As2O3

Acidic character decreases down the group.

(ii) NF3 > NCl3 > NBr3

Stability decreases due to increasing size of halogen atom.

(iii) HNO3 > H3PO4 > H3AsO4 > H3SbO4

18. (i)

OH3CO6HClKHSOSOHOCH3KClO

OH3CO6]O[3OCH3

]O[3HClKHSOSOHKClO

224424223

22422

4423

+++⎯→⎯++

+⎯→⎯+

++⎯→⎯+

(ii)

224424222424

2222

2442424

22424222

O5OH8MnSO2SOKSONa5ONa5SOH8KMnO2

5]OOH]O[OH[

]O[5OH3MnSO2SOKSOH3KMnO2

5]OHSONaSOHONa[

++++⎯→⎯++

×+⎯→⎯+

+++⎯→⎯+

×+⎯→⎯+

(iii)

OH3S3SONaSH2NaHSO2

S3OH2SH2SO

SOOHSONaNaHSO2

23223

222

22323

++⎯→⎯+

+⎯→⎯+

++⎯→⎯



19. (i)

NaI2OSNaIOSNa2

IKCl2ClKI2

OHClCa)COOCH(COOHCH2CaOCl

6422322

22

222332

+⎯→⎯+

+⎯→⎯+

++⎯→⎯+

A weighed quantity of Bleaching powder is suspended in water and treated with excess of acetic acidand KI. The liberated iodine is estimated by titrating it with a standard solution of hypo using starch asindicator.

(ii) Chlorides when heated with K2Cr2O7 and the H2SO4 evolve chromyl chloride (orange vapours) which whenpassed through lead acetate gives yellow ppt. of PbCrO4

COOHCH2PbCrO)COOCH(PbCrOH

HCl2CrOHOH2ClCrO

OH3ClCrO2KHSO6KCl4SOH6OCrK

2]OHClCrOHCl2CrO[

4]HClKHSOSOHKCl[

OHCrO2KHSO2SOH2OCrK

3)Yellow(42342

42222

vapourOrange222442722

2223

442

23442722

+⎯→⎯+

+⎯→⎯+

++⎯→⎯++

×+⎯→⎯+

×+⎯→⎯+

++⎯→⎯+

(iii) (fluoride salt)

62

)rodglassonlayerWaxy()acidsilicic(

223224

24222

224422

SiFH2)OH·SiO·(SiOHOH3SiF3

OH2SiFFH2SiO

FHCaSOSOHCaF

+⎯→⎯+

+⎯→⎯+

+⎯→⎯+

20. XeO2F2 :

Xe

F

F

O

Osee sawshape

sp d3 XeOF4 : F

O

F F

F

Xe

square pyramidal

XeO3F2 :

Xe

F

F

O

O

sp d3

O

trigonal bipyramidal

XeO3 :

O O O

sp3

Pyramidal

XeO4 :

O

O

O

sp3

Tetrahedral

O



21. It disproportionates in solution giving purple colored permanganate and brown colored MnO2.

4MnO42– + 4H+ → 3MnO4

– + MnO2 + 2H2O

acidic medion is obtained due to dissociation of H2CO3 formed.

22. [VO ]43– pH12 [VO ]3

2–·OH pH10 [V O ]2 6

3–·OH pH·9 [V O ]3 9

3–

colorless colorless colorless orange

[V O ]5 143–pH6.5V O2 5·(H O)2 n

pH2.2[V O ]10 286–pH1[VO ]2

+

pH·7

brown ppt. red

23. (i) Ni + 22+

CH C = NOH3

CH C = NOH3

+ 2NH OH4

CH C = N3

CH C = N3

Ni2+

N = C – CH3

N = C – CH3

OH ·············O

O ············HO ·

bis(dimethyl glyoximato) nickel (II)

(ii) Ni is in +2 oxidation state and the complex is square due to dsp2 hybridization.

(iii) The complex is diamagnetic due to absence of unpaired electron.

24. (i) SnCl2 reduces HgCl2 to Hg2Cl2 first and then to Hg

SnCl2 + 2HgCl2 → Hg2Cl2 + SnCl4

Hg2Cl2 + SnCl2 → 2Hg + SnCl4(ii) In the solution, the following equilibria exists :

Cr2O72– + H2O 2CrO4

2– + 2H+

In acidic medium (pH < 7), it exists as Cr2O72– ions and has orange color while in basic medium (pH > 7), it

exists as CrO42– ions and has yellow color.

25. Ksp of CuS is less than Ksp of ZnS. On passing H2S in acidic medium, the dissociation of H2S is suppresseddue to common ion effect and its provides a limited conc. of S2– ions enough of exceed Ksp of CuS but notZnS. Thus only CuS gets precipitated.

26. (i) 82 (vi) 34

(ii) 35 (vii) 33

(iii) 30 (viii) 86

(iv) 38 (ix) 54

(v) 53 (x) 86

27. (i) Tris (ethylenediammine) cobalt (III) chloride

(ii) Tris-oxalatocobaltate (III) ion

(iii) Decaammine-µ-peroxodicobalt (III) ion

(iv) Hexaammine chromium (III) hexa cyano cobaltate (III)

(v) Bis (dimethylglyoximato) nickel (II)

27(a). Answer (3) IIT-JEE-2008

IUPAC name is tetraamminenickel(II)-tetrachloronickelate(II).



28. (i) [NH4]2[FeF5(H2O)] (vi) [Ni(NH3)4] (ClO4)2

(ii) [Co(H2O)2(en)2]2 (SO4)3 (vii) [Ni(NH3)6]3 [Co(NO2)6]2

(iii) [Zn(NCS)4]2– (viii) [Ni(CO)2(PPh3)2]

(iv) Na3[Co(NO2)6] (ix) (en) Co Co(en)2 2

NH2

OH(SO )4 2

(v) [Fe(en)3] [Fe(CN)4] (x) (en) Co Co(en)2 2

NH2

OH(SO )4 2

29. (i) d 2sp3

(ii) sp 3d 2

(iii) dsp2

(iv) sp3

30. Au 53+ 8

d

5d 6s 6p

[AuCl ]¯4

5d 6s 6p

dsp2 hybridisation

Ga3+

3d10

4s 4p

[GaCl ]¯4

3d 4s 4p

sp3 hybridisation

31. A = [Cr(H2O)6]Cl3 ⎯⎯⎯ →⎯ 42SOH No reaction

(∴ All H2O molecule are present in co-ordination sphere)

B = ⎯⎯⎯ →⎯⋅ 42SOH2252 OHCl])OH(CrCl[ one mole of H2O is lost

Molecular weight of complex = 266.5

% loss = %75.61005.266

18 =×

C = ⎯⎯⎯ →⎯⋅ 42SOH22422 OH2Cl])OH(CrCl[ 2 moles of H2O is removed

∴ % loss in weight = %50.131005.266

182 =××



32. [Pt(NH3)4] [PtCl4] ; Tetraammineplatinum (II) tetrachloroplatinate (II)

[Pt(NH3)4] [NO3]2 ; Tetraammineplatinum (II) nitrate

Ag2[PtCl4]; Silvertetrachloroplatinate (II)

33. (A) = H2S2O8,

(B) = H2SO4,

(C) = H2O2 and

(D) = BaSO4

34. (i) Hg2(NO3)2 + 2KI Hg2I2 + 2KNO3

(ii) Hg2I2 + 2KI K2HgI4 + Hg

(iii) NiSO + 2H C — C == NOH 4 3

H C — C == NOH3

H C — C == N3

OH

H C — C == N3

O

NiN == C — CH3

HO

O

N == C — CH3

+ (NH ) SO + 2H O4 2 4 2+ 2NH OH4

35. A = Hg2(NO3)2,

B = Hg2Cl2,

C = HgCl2,

D = K2HgI4

E = Hg2,

F = FeSO4 · NO

36. A = PH4I,

B = PH3,

C = KI,

D = P2O5

E = Cu2I2

37. A = K2MnO4,

B = KMnO4,

C = KIO3,

D = Mn2O7,

E = MnO2

38. (i) Turns lime water milky, thus (A) is either CO2 or SO2 gas

(ii) (Y) gives alkaline solution and its solution forms white ppt (Z) with BaCl2 and (Z) on heating with acid giveseffervescences of CO2, so (Z) is BaCO3 and (Y) is metal carbonate

(iii) Since (Y) and (A) are formed from (X) and thus, (X) is metal bicarbonate and (A) is CO2



(iv)g8.16

X ⎯→⎯Δ

g4.4A +

g1.8B + Y

(v) The above data reveal that

2MHCO3 → CO2 + H2O + M2CO3

4.4 g CO2 is obtained by 16.8 g MHCO3

44 g CO2 is obtained by 168 g MHCO3

Molecular weight of MHCO3 = 2

168 = 84

∴ At. wt. of metal = 23

Hence metal is Na, i.e.,

)X(3NaHCO2 →

)A(2CO +

)B(2OH +

)Y(32CONa

39. White phosphorus ⎯⎯ →⎯Air waxy crystalline solid having garlic smell (A).

(A) + Hot water →

3PH thus smell,fish rotten having

(B)Gas + Acid (C)

CuSO4 solution + gas (B) → Black ppt of cupric phosphide (D)

Reactions involved :-

P4 + 3O2 → )A(

32OP2

2P2O3 + 6H2O → (B)

3PH + )C(

33POH3

3CuSO4 + 2PH3 →

)D(Black

23PCu + 3H2SO4.

40. Let us summarise the reaction

solutionyellow)B(SOHconc.NaCl(A) NaOH

gasOrange

Δ

42

CrystalOrange

⎯⎯⎯ →⎯⎯→⎯++

+

⎯→⎯+3

KI2

Cr

GreenI

H+

AgNO3

Red pptOrangesolutionCr

6+

Hence (A) = K2Cr2O7

(B) = CrO2Cl2 (C) = Na2CrO4

K2Cr2O7 + 4NaCl + 6 H2SO4 ⎯→⎯ 4NaHSO4 + 2KHSO4 + 2CrO2Cl2 ↑+ 3H2O

Orange crystal (B) (Orange)

(A)

CrO2Cl2 + 4NaOH ⎯→⎯ Na2CrO4 + 2NaCl + 2H2O



Na2CrO4 + 2AgNO3 → Ag2CrO4↓ + 2NaNO3

Red ppt

2Na2CrO4 + H2SO4 ⎯→⎯ Na2Cr2O7 + Na2SO4 + H2O

Cr2O72– + 14H+ + 6 I– ⎯→⎯ 2Cr3+ + 7H2O + 3I2

Quantitative analysis

CrO2Cl2 ≡ Na2CrO4 ≡ 2AgNO3

155 g 2 mol

155 g of CrO2Cl2 requires 2 moles of AgNO3

⇒ 0.155 g of CrO2Cl2 requires 0.002 mole (2 m.mole)

Similarly. 2 CrO2Cl2 ≡ 2 CrO42– ≡ Cr2O7

2– ≡ 3I2

⇒ 2 × 155 g

0.155 g will liberate 31023 −× moles = 1.5 milli mole

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unit-iii - inorganic chemistry - solutions(final)

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