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Section A : Straight Objective Type1. Answer (4)
All these contain 3C 2 bond.2. Answer (3)
Na2B4O710H2O + H2O +H
2B(OH)3 + 2B(OH)43. 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
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Success Magnet (Solutions) Inorganic Chemistry
14. Answer (2)2KMnO4 + 3H2SO4 + 5H2O2 K2SO4 + 2MnSO4 + 8H2O +
5O22KI + 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
gaswater2Ni
K1270
steam2
coke)g(H)g(CO)g(OH)s(C + +
CO(g) + H2(g) + H2O(g) K770OCr/OFe 3232 CO2(g) + H2(g).
16. Answer (2)K2Cr2O7 + H2SO4 + 4H2O2
ether K2SO4 +
blue5CrO2 + 5H2O.
17. Answer (2)Volume strength of H2O2 = 11.2 M
= 11.2 3.57~
4018. Answer (2)
2Al + 3H2SO4 Al2(SO4)3 + 3H22Al + 2NaOH + 2H2O 2NaAlO2 +
3H2.
19. Answer (4)4H2O2 + PbS PbSO4 + 4H2O4 mole 1 mole0.8 mole 0.2
molevolume strength of H2O2 = 11.2 M
2.1140M 22OH =
number of mole = M VL
volume in litre =
2.1140
8.0
volume in ml = ml2241000402.118.0
=
.
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.
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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 + H2OFor 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 + 3H226. 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.
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Success Magnet (Solutions) Inorganic Chemistry
30. Answer (1)
(NH ) C O4 2 2 4 NaOH NH3(A) (B)
CaCl solution2
CaC O (C)2 4NH + HCl3 NH Cl4(B) (white fumes) .
31. Answer (3)106 g of Na2CO3 = 100 g CaCO3 0.0106 g Na2CO3 =
0.01 g CaCO3103 ml H2O = 103 g of H2O
103 g of H2O contains 0.01 g of CaCO3
106 g of H2O contains 6
3 101001.0 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 + CO2NaHCO3 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 2white
ppt.
.
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Inorganic Chemistry Success Magnet (Solutions)
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)y40. 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 + 5O22 mol KMnO4 5
mol H2O2
1 mol KMnO4 25
mole H2O246. 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
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Success Magnet (Solutions) Inorganic Chemistry
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-2008Since nitrates are more soluble in
water. nitrates are less abundant in earths crust.
59. Answer (2)Br2 + .concand.hot
KOH KBr + KBrO3 + H2O
60. Answer (3)
CuO + NH3
Cu + N2 + H2O61. Answer (1)
AgCl + Na2S2O3 Na3[Ag(s2O3)2] + NaCl62. Answer (2)
2KMnO4 K2O + 2MnO + 5[O]K2O + 2HCl 2KCl + H2O
MnO + 2HCl MnCl2 + H2O] 22HCl + [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)3O1T3] + +OT331
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Inorganic Chemistry Success Magnet (Solutions)
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 natureSiO2 + 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
O73. 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.
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Success Magnet (Solutions) Inorganic Chemistry
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 Na2S2O35H2O
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) = 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 forma 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).
Hg2SnClClHgSnClClHgSnClHgCl2SnCl
4222
22422
++
++
95. Answer (3)Both Cu+2 and V+4 have one unpaired e so possess
same colour.
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Inorganic Chemistry Success Magnet (Solutions)
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 transitionbut 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(n=
+=
99. Answer (2)When SO2 gas is passed into acidified 272OCr , SO2
is oxidised to 24SO , 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 24724246
+++ ++
+
+
It is a disproportionation reaction. In disproportionation
reaction, equivalent weight of the species
undergoingdisproportionation = Eq. wt. (in oxidation) + Eq. wt. (in
reduction)
.M23
1M
2MMnOof.wt.Eq
1M
.)oxidn(.wt.Eq,OMnOMn2M
.)redn(.wt.Eq,OMnOMn
4
47246
24246
=+=
=
=
+
+
+
+
103. Answer (2)In acidic medium, +
+ 247
MnOMn
1 M = 5 N
H2O2 100 mL of 5 N KMnO4
In basic medium, 2447 OMnOMn +
+
1 M = 3 N
H2O2 V mL of 3 N KMnO4 3V 500
V = 3500
mL
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Success Magnet (Solutions) Inorganic Chemistry
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)
MnO2106. 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
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Inorganic Chemistry Success Magnet (Solutions)
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.
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Success Magnet (Solutions) Inorganic Chemistry
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
white322 OSAg
black2SAg
136. Answer (2)NH3 + HCl NH4Cl equilibrium shifts towards
left.
137. Answer (1)
)orange(722
mediumacidicCO
)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.pptyellow
2 KNO2PbI +
142. Answer (3)
NaHCO3
Na2CO3Na2CO3 + H2O + CO2 2NaHCO3
143. Answer (3)
Ba+2 + CrO42 .pptyellow
4BaCrO
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Inorganic Chemistry Success Magnet (Solutions)
144. Answer (1)
Cr(OH)3 + NaOH + Na2O2 .pptyellow
24CrO
145. Answer (4)Nesslers 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)
AgNO3H SO conc.2 4
CuHCl
NO2Cu(NO ) + Ag3 2AgCl
white ppt.
Section - B : Multiple Choice Questions1. 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 mediumacidic
2Mn2+ + 8H2O + 5O2
2MnO4 + 3H2O2 mediumbasic
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.
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Success Magnet (Solutions) Inorganic Chemistry
6. Answer (1, 3, 4)Acidic nature (ka for H2O2 = 1.55 1012 &
kw for H2O = 1014 at 25CE.g., NaOH + H2O2 NaHO2 + H2O
Sodium 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 o Cu/Cuo
H/H 22== ++
8. Answer (1, 4)For diatomic gases (H2 & CO) has
V
PCC
=
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 CaSO4.2H2OIn 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 120C, it
loses
th43
of its water of crystalization and form plaster of Paris.
ParisofPlaster24
C120
Gypsum24 OH2
1.CaSOO.2HCaSO
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Inorganic Chemistry Success Magnet (Solutions)
While on heating at 200C 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
compoundsRMgCl + BeCl2 R2Be + 2MgCl2For Be2C2BeO + 2C Be2C +
CO2
20. Answer (1, 2, 3, 4)Chile saltpetre NaNO3Trona
Na2CO3.2NaHCO3.3H2OIndian saltpetre KNO3Sylvine KClAll 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)Betts 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 0G and T for a particular metal oxide. The metal shown
byline can reduce all other metal which lies above it. At point of
intersection of 0G = so process will be inequilibrium. Just by
heating, we can make veG0 += 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.
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Success Magnet (Solutions) Inorganic Chemistry
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
O34. Answer (1, 2, 3, 4)
612
12
122
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) -Al2O3
called rhombic lattice or corundum(ii) -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.
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Inorganic Chemistry Success Magnet (Solutions)
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 ReRe quadruple bond and i.e., whyReRe
bond length is abnormally short. If ReRe bond points along z-axis,
then square planar ReX4 unit willuse s, px, py and 22 yxd orbitals
for the formation of four ReX -bonds. ReRe bond is comprised of one
,2 and one -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 notdue to
d-d transition but due to charge transfer. Charge transfer in 4MnO
and 24CrO lies in visible region while
in 4ORe and 4TcO , it lies in UV region. Hence, 4MnO and 24CrO
are coloured and 4ORe and 4TcO arecolourless.
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
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
Mn
CuFe
Radius
21 30
Se
Ti
v1 Cr
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.
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Success Magnet (Solutions) Inorganic Chemistry
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 electronCN 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 xx3d5 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.
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Inorganic Chemistry Success Magnet (Solutions)
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 or 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)
Factual79. Answer (1, 2, 3)
32324 SOSOOFeFeSO2 ++
80. Answer (2, 3)Factual
Section - C : Linked ComprehensionC1. 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.
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Success Magnet (Solutions) Inorganic Chemistry
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 1012 at 293 K which is only slightly higher than
that of water (1.0 1014).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., LiCl2H2O.
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.
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Inorganic Chemistry Success Magnet (Solutions)
C6. 1. Answer (1)Hhydration of LiCl = 499 382 = 881 kJ mol1
Hsolution = Hhyd. HUHsolution(LiCl) = 881 (840) = 41 kJ
mol1.
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
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Success Magnet (Solutions) Inorganic Chemistry
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 2332272 ++++ ++++
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
.conc424 +++
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)
Mohrs salt is FeSO4(NH4)2SO46H2O and is used as a reducing
agent.
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Inorganic Chemistry Success Magnet (Solutions)
C16. 1. Answer (3)[Co(NH3)5NO2]Cl2
2. Answer (1)[Co(en)2Cl2]+ cis-trans.
isomerismopticalligandbidentateoxligandbidentateen
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 CrCl3B Na2CrO4C Na2Cr2O7D (NH4)2Cr2O7G N2I
Li3N
J NH3H Cr2O3
2. Answer (3)CrO2Cl2 is chromyl chloride
3. Answer (2)
OH2NNONH 2224 +
C19. 1. Answer (2)
)B(2
)C(4
.dil42)A(
SHZnSOSOHZnS ++
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Success Magnet (Solutions) Inorganic Chemistry
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(
)fumeswhite(4
)B(3 ClNHHClNH +
ClNH2OCaCOC)NH( 4.pptwhite
)C( 42CaCl
)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.
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Inorganic Chemistry Success Magnet (Solutions)
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
NH2NH2 + 2H2O2 )II(Cu N2 + 4H2O8. 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., fH 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 C2H5ONa+ + 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 + KCl16. 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)
Both statements are correct and statement-2 is the correct
explanation.
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Success Magnet (Solutions) Inorganic Chemistry
20. Answer (2)Both statements are correct.
21. Answer (3)22. Answer (1)
Both statements are correct and statement-2 is the correct
explanation.23. Answer (1)
Both statements are correct and statement-2 is the correct
explanation.24. Answer (1)
Both statements are correct and statement-2 is the correct
explanation.25. Answer (1)
Both statements are correct and statement-2 is the correct
explanation.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)
Both statements are correct.
30. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
31. Answer (3)Electronegativity of elements decreases on moving
down the group.
32. Answer (2)Both statements are correct.
33. Answer (3)Phosphorus exhibits pentavalency.
34. Answer (2)BF3 has least Lewis acidity due to p-p 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.
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Inorganic Chemistry Success Magnet (Solutions)
39. Answer (2)Both statements are correct.
40. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
41. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
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)Both statements are correct and statement-2 is the
correct explanation.
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., l = 1), i.e., why complex is
colourless.
48. Answer (3)FeCl3 is a salt of strong acid and weak base.
49. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
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 hasHbond 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) of K4[Fe(CN)6] is zero and of K3[Fe(CN)6] is 3
B.M.
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55. Answer (3)It is bidentate -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]+ BF4 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)Both statements are correct and statement-2 is the
correct explanation.
63. Answer (2)Both statements are correct.
64. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
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 ReactionPb(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 CrO4 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.
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71. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
72. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
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 turnsblack due
to formation of Ag2S.
75. Answer (1)[Fe(CN)6]4 + 2Cu2+ Cu2[Fe(CN)6]
76. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
77. Answer (2)KCl + H2SO4 HCl + K2SO4Kl + H2SO4 Hl + K2SO4Hl +
H2SO4 Br2 + SO2 + H2O
78. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
79. Answer (1)Both statements are correct and statement-2 is the
correct explanation.
80. Answer (2)2KMnO4 K2MnO4 + MnO2 + O2Black 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 Type1. 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(OH2e2H2OH
naturegsinOxidi22
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
1012. H2O2 is less acidic than carbonic acid.
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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 name
of perhydrol.
(B) Water gas is CO + H2Preparation of H2 by water gas
22OFeC450
22 COH2OHHCO32
+ ++
(C) The maximum quantity of commercial hydrogen is produces by
Boschs process, water gas is firstproduced.
43421
gas water2
C1002 HCOOHC + +
(D) By partial oxidation or cracking natural gas (CH4) gives
water gas and H2.
2C820CrNi
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 it
is 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.
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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 fumes
of 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 a
and 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 43ondistillati2 +
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 + H2OIn 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.
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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.
FB NH||
||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 NiIn 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.
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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 mL1 and 22.61 g mL1
respectively.Cr(24) [Ar] 3d54s1Cr 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 + 2KClKMnO4 acts as an oxidising agent in
alkaline, neutral or acidic solutions.
23242232oreChromite
32 CO8OFe2CrONa8O7CONa8OCr.FeO4 ++++
2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2ONa2Cr2O7 + 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.
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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 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. Polydentateligands 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 +++
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Inorganic Chemistry Success Magnet (Solutions)
OHCaSOSO)OH(Ca 2)milky(32
waterLime2 ++
OHSO3Cr2H2SO3OCr 224)green(3
22
72 ++++++
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
SnClHg2SnClClHgSnClClHgSnClHgCl2
++
++
(D)pptgrey
HgCuHgCl2 +
38. Answer - A(q, s), B(r), C(p), D(p) 2SOHdil23 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 256FeSO2266 4 + +
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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]] 32Cr(OH)3 + 4NaOH + 3[O] 2Na2CrO4 + 4H2O
-------------------------------------------------------------------------------------------------------
2Cr(OH)3 + 4NaOH + 3H2O2 2Na2CrO4 + 8H2OH2O2 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 + O24. A + B C
(gives fumes)
C + NaOH A Gas A is NH3
B is HCland C is NH4ClNH3 gives brown precipitate with Nesslers
reagent. D is Nesslers reagent (K2HgI4)
E formed is OHg
NH I2Hg
oxydimercuric ammonium iodide
NH3 + 3NaOH + K2HgI4 OHg
NH I2Hg
+ 4KI + 3NaI + 2H2O.
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5. Gas B is chlorine (Cl2)Ethyl alcohol gives anaesthetic CHCl3
E = CHCl3The milky precipitate C is CaCO3 A is a compound of Ca and
Cl
CaCO3 + H2O + CO2 Ca(HCO3)2D = 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 + Cl23CaOCl2 + 2NH3 3CaCl2 + 3H2O + N2Cl2 +
H2O 2HCl + O; O + CH3CH2OH CH3CHO + H2O
CH3CHO + 3Cl2 CCl3CHO + 3HCl; 2CCl3CHO + Ca(OH)2 2CHCl3 +
Ca(HCOO)26. 6NaOH + 4S + 5H2O Na2S2O35H2O + 2Na2S + 3H2O
B is Na2S2O35H2O
Na2S + )excess( S4 Na2S5
C is (Na2S5)A + S Na2S2O3
A is Na2SO3D is Na2S2O3
Na2S2O3 + H2SO4 Na2SO4 + SO2 + )yellow( S + H2O
From description F is SO2E 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
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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 NaPO3and A is Na(NH4)HPO4
Na(NH4)HPO4 NaPO3 + H2O + NH3Na2HPO4 + NH4Cl Na(NH4)HPO4 +
NaCl
B is Na2HPO4E is Na2SiO3 (present in glass)
So, D is a oxide of Si
D is SiO2Na2SiO3 + 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 = Na2BeO2BeO + C + Cl2 BeCl2 + CO
D = BeCl2E = CO
BeCl2 + )airof( 2OH2 Be(OH)2 + )fumes( HCl2
F = HCl
BeO + 4NH4F (NH4)2BeF4 + 2NH3 + H2O
(NH4)2BeF4 BeF2 + 2NH4F
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11. Red ppt. is of HgI2 D is HgI2
A is iodide saltOnly KI dissolves free I2 KI + I2 KI3 A = KI
B = KI3OH8I5MnSO2SOK6SOH8KMnO2KI10 2
)brown(244242
)pink(4 +++++
C = I22KI + HgI2 K2HgI4E = K2HgI4 or Nesslers 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 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 Gibbs free energy change
value. Which doesnt 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 a
liquid which then solidifies to a transparent glassy mass.
444 3444 21
beadGlassy322)strong(742OH10
2742 OBNaBO2OBNaOH10OBNa2
+
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
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16. (i)
OH3NONH)NO(Sn4)dil(HNO10Sn4OH3NONHH8HNO2
4]H2)NO(SnHNO2Sn[
234233
2343
233
+++
++
++
(ii)
CO6SO)NH(3FeSOSOK2OH6SOH6])CN(Fe[K6]OHCOHCOOH[
3]SO)NH(SOHNH2[6]NHHCOOHOH2HCN[
HCN6FeSOSOK2SOH3])CN(Fe[K
42444224264
2
424423
32
4424264
+++++
+
+
++
+++
(iii) 223C375275Cu
3 SiCl)CH(SiClCH2
+
(iv)
OH2SiFCaSO2SiOSOH2CaF2
OH2SiFHF4SiO2]HF2CaSOSOHF[
2442422
242
4422
++++
++
++
(v)
OH2PbO)NO(Pb2)dil(HNO4OPb2]OH)NO(PbHNO2PbO[
PbOPbO2OPb
2223343
2233
243
+++
++
+
17. (i) N2O3 > P2O3 > As2O3Acidic character decreases down
the group.
(ii) NF3 > NCl3 > NBr3Stability 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
S3OH2SH2SOSOOHSONaNaHSO2
23223
222
22323
+++
++
++
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Inorganic Chemistry Success Magnet (Solutions)
19. (i)
NaI2OSNaIOSNa2IKCl2ClKI2
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(PbCrOHHCl2CrOHOH2ClCrO
OH3ClCrO2KHSO6KCl4SOH6OCrK
2]OHClCrOHCl2CrO[4]HClKHSOSOHKCl[
OHCrO2KHSO2SOH2OCrK
3)Yellow(42342
42222
vapourOrange222442722
2223
442
23442722
++
++
++++
++
++
+++
(iii) (fluoride salt)
62
)rodglassonlayerWaxy()acidsilicic(
223224
24222
224422
SiFH2)OHSiO(SiOHOH3SiF3OH2SiFFH2SiO
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
Otrigonal bipyramidal
XeO3 :
O O O
sp3
Pyramidal
XeO4 :
O
O
O
sp3
Tetrahedral
O
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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 2OH pH10 [V O ]2 6 3OH pH9 [V O ]3 9
3
colorless colorless colorless orange
[V O ]5 14 3pH6.5V O2 5(H O)2 npH2.2[V O ]10 28 6pH1[VO ]2
+pH7
brown ppt. red
23. (i) Ni + 22+CH C = NOH3
CH C = NOH3+ 2NH OH4
CH C = N3
CH C = N3Ni2+
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 HgSnCl2
+ 2HgCl2 Hg2Cl2 + SnCl4Hg2Cl2 + SnCl2 2Hg + SnCl4
(ii) In the solution, the following equilibria exists :Cr2O72 +
H2O 2CrO42 + 2H+
In acidic medium (pH < 7), it exists as Cr2O72 ions and has
orange color while in basic medium (pH > 7), itexists 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--peroxodicobalt
(III) ion(iv) Hexaammine chromium (III) hexa cyano cobaltate
(III)(v) Bis (dimethylglyoximato) nickel (II)
27(a). Answer (3) IIT-JEE-2008IUPAC name is
tetraamminenickel(II)-tetrachloronickelate(II).
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Inorganic Chemistry Success Magnet (Solutions)
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 2NH2
OH(SO )4 2
(v) [Fe(en)3] [Fe(CN)4] (x) (en) Co Co(en)2 2NH2
OH(SO )4 2
29. (i) d 2sp3
(ii) sp 3d 2
(iii) dsp2
(iv) sp3
30. Au 53+ 8d
5d 6s 6p
[AuCl ]45d 6s 6p
dsp2 hybridisation
Ga3+3d10 4s 4p
[GaCl ]43d 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 lostMolecular
weight of complex = 266.5
% loss = %75.61005.26618
=
C = 42SOH22422 OH2Cl])OH(CrCl[ 2 moles of H2O is removed
% loss in weight = %50.131005.266182
=
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Success Magnet (Solutions) Inorganic Chemistry
32. [Pt(NH3)4] [PtCl4] ; Tetraammineplatinum (II)
tetrachloroplatinate (II)[Pt(NH3)4] [NO3]2 ; Tetraammineplatinum
(II) nitrateAg2[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 3H C C == NOH3
H C C == N3
OH
H C C == N3O
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 = K2HgI4E = Hg2,
F = FeSO4 NO
36. A = PH4I,
B = PH3,
C = KI,
D = P2O5E = Cu2I2
37. A = K2MnO4,
B = KMnO4,
C = KIO3,
D = Mn2O7,
E = MnO238. (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
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Inorganic Chemistry Success Magnet (Solutions)
(iv)g8.16
X
g4.4A +
g1.8B + Y
(v) The above data reveal that2MHCO3 CO2 + H2O + M2CO34.4 g CO2
is obtained by 16.8 g MHCO344 g CO2 is obtained by 168 g MHCO3
Molecular weight of MHCO3 = 2168
= 84
At. wt. of metal = 23Hence 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(
Black23PCu + 3H2SO4.
40. Let us summarise the reaction
solutionyellow)B(SOHconc.NaCl(A) NaOHgas
Orange
42
CrystalOrange
++
+
+3
KI2
CrGreenI
H+ AgNO3
Red pptOrangesolutionCr6+
Hence (A) = K2Cr2O7 (B) = CrO2Cl2 (C) = Na2CrO4
K2Cr2O7 + 4NaCl + 6 H2SO4 4NaHSO4 + 2KHSO4 + 2CrO2Cl2 +
3H2OOrange crystal (B) (Orange) (A)CrO2Cl2 + 4NaOH Na2CrO4 + 2NaCl
+ 2H2O
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Na2CrO4 + 2AgNO3 Ag2CrO4 + 2NaNO3 Red ppt
2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O
Cr2O72 + 14H+ + 6 I 2Cr3+ + 7H2O + 3I2Quantitative analysis
CrO2Cl2 Na2CrO4 2AgNO3155 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 Cr2O72
3I2 2 155 g
0.155 g will liberate 31023
moles = 1.5 milli mole