Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (223) Section A : Straight Objective Type 1. Answer (4) All these contain 3C – 2 – bond. 2. Answer (3) Na 2 B 4 O 7 ⋅10H 2 O + H 2 O ⎯ ⎯→ ⎯ + H 2B(OH) 3 + 2B(OH) 4 3. Answer (4) Hydrolith – CaH 2 , oxidane – H 2 O, Azane – NH 3 . 4. Answer (2) HClO 4 does not contain peroxide linkage while others contain. Therefore HClO 4 does not give H 2 O 2 on hydrolysis. 5. Answer (1) Factual. 6. Answer (1) Tritium ( 1 H 3 ) 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 H 2 O 2 is sp 3 . 9. Answer (2) ) A ( Na 2 + 2H 2 O → ) B ( 2 H + ) C ( NaOH 2 and ) D ( Zn + ) C ( NaOH 2 → Na 2 ZnO 2 + ) B ( 2 H . 10. Answer (2) Dielectric constant of H 2 O is 78.39 while that of D 2 O is 78.06. Hence, solubility of ionic compound is smaller in 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 liberate hydrogen 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) and low pressures. H 2 2H – xJ. Inorganic Chemistry UNIT 3
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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.
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.
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
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.
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.
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.
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)
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.
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.
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.
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.
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.
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.
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.
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.
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.
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)
Both statements are correct and statement-2 is the correct explanation.
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., Δl = ± 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 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.
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 turns
black 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 + K2SO4
Kl + H2SO4 Hl + K2SO4
Hl + 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 + 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.
(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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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
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