Rao IIT Academy/ ISC - Board _Chemistry_Solutions 1 1 SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR | PART I Question 1: (a) (i) positive , lesser (ii) efficient , 74 (iii) decreases , common-ion effect (iv) Cannizzaro , no hydrogen (v) paramagnetic , dimagnetic (b) (i) (2) –0.372°C (ii) (3) 1155 (iii) (2) Secondary Alcohol (iv) (4) Octahedral and sp 3 d 3 (v) (1) +3 and + 4 (c) (i) As temperature increases, concentration of H + and OH – increases and hence pH value decreases. At 298 K. 14 H OH 10 for neutral solution pH = 7 and at high temperature pH will be less than 7. (ii) 3 2 Fe e Fe for 1 mole of 3 , Fe 1 mole of e is required for 3 moles of 3 , Fe 3 moles of e are required 3 3 3 96500 e F C Q I t 3 96500 2 t 3 96500 2 3600 t 40.2 t hrs XII - ISC BOARD Date: 20.03.2015 CHEMISTR Y - SOLUTIONS
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(ii) NaCl show metal excess defect due to anionic vacancies. When crystals of NaCl are heated in an atmosphere
of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse to thesurface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electrons by sodiumatoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites. As a resultcrystal now has excess Na+ ions. The anionic sites occupied by unpaired electrons are called F–centres. Theyimpart yellow colour to the crystals of NaCl. The colour results by excitation of these electrons when theyabsorb energy from visible light falling on the crystal.
(iii) 2( ) 2 33 2g g gN H NH
K (equilibrium constant) = 6.0 × 10–2 1mol L
Temperature (T) = 715 K
12 0.25H mol L
13 0.06NH mol L
2 ?N
23
32 2
NHK
N H
23
2 32
NHN
K H
2
32
0.06
6.0 10 0.25
20.0036
6.0 10 0.015625
12 3.84N mol L
(iv) 2 16.2 10H mol L 1410H OH
Now putting the value of H
14
210
6.2 10OH
13 11.61 10OH mol L (v) Le–Chatelier’s principle : The principle states that if an external stress is applied to a reacting system at
equilibrium, the system will adjust itself in such a way that the effect of the stress is reduced.
(ii) Each Na+ ion is surrounded by six Cl ions and each Cl ion is surrounded by six Na ions.(iii) Number of Cl ions = 4
Number of Na ions 4
(iv) In the NaCl crystal, Cl ions occupy all the FCC positions and Na+ ions occupy the octahedral voids.(c)(i) Increase in temperature – forward reaction
Decrease in temperature – backward reaction(ii) Increase in pressure – forward reaction
Decrease in pressure – backward reaction(iii) Increase in conc. of [N2O4] – forward reaction
Decrease in conc. of [N2O4] – backward reaction(iv) Removal of NO2 causes decrease in [NO2] – forward reactionQuestion 4:
(b)(i) Due to small size of transition metal atoms or their cations and high effective nuclear charge, they have
a high positive charge density on them. This high positive charge density makes the atoms or cations toattract the lone pairs of e– from the ligands.
The transition metal cation or atoms have vacant d-orbital in which they can accomodate the lone pairs
of e– donated by ligands and thus can form L M coordinate bonds.(ii) Paramagnetic character increases with the increase in the number of unpaired electrons. The larger is the
number of unpaired e– in central atom or ion greater is the paramagnetic character.
Hence from Sc to Mn number of unpaired e– increases and then from Fe to Zn decreases. Hence paramagneticcharacter also increases from Sc to Mn and then decreases from Fe to Zn.
SECTION CQuestion 8:(a)(i) Glycerol to formic acid
HO OHOH
3H– C – OHFormic
acid
Pd(I) CatalystH O2 2
Glycerol
O
(ii) Chlorobenzene to phenolCl OH
+ OH–
Chlorobenzene Phenol
NaOHpressure, H+
(iii) Diethyl ether to ethanol
3 2 2 3 2 5 2 5Diethyl ether Ethanol
CH CH O CH CH HI C H OH C H I
(iv) Phenol to anilineOH NH2
+ NH3
Phenol Aniline
ZnCl /2
High pressure + H2O
(b)
(i) 3 2 2 3 2Ethanol Iodoform
CH CH OH 4I 6 NaOH CHI HCOONa 5 NaI 5H O
(ii) Chlorobenzene is heated with sodium metal in presence of dry ether gives diphenyl compound.
Ethylamine reacts with Nitrous acid in cold condition to form unstable diazonium salt and which decomposesto give alcohol with liberation of N2 gas.
2 5 2 2 2 5 2 2C H NH + NaNO + 2HCl C H N Cl NaCl + 2H O +
2 5 2 2 5 2C H N Cl H O C H OH + HCl N diethylamine reacting with Nitrous acid give yellow oily N–Nitrosoamines
22 5 2 5 2 5 2
N nitroso diethylamineC H NH C H HO N = O C H N N = ONaNO
HCl
(ii) Acetaldehyde and benzaldehyde
Acetaldehyde reacts as reducing agent which reduces Fehling solution. The aldehyde gets oxidized tocarboxylate ion and Cu2+ is reduced to Cuprous oxide
23 3 2
Carboxylateion
Acetaldehyde
O|
CH C H + 2Cu 5OH CH C = O Cu O||O
Benzaldehyde
CHO(aromatic aldehyde) are not oxidised by Fehling solution.
(c)(i) Compounds in ascending order of their basic strength:
diethylether < Aniline < Methylamine < ethylamineIn aliphatic amines, due to presence of electron donating alkyl group, electron density on nitrogen increases.Hence lone pair is easily available for protonation.In aniline the lone pair of nitrogen is in conjugation with benzene and less available for protonation henceaniline is less basic than aliphatic amine and ethers are inert in nature due to stable C O C bond .
(ii) (a) Polyester :Monomer used :Ethylene Glycol 2 2OH CH CH OH
Dimethyl terephthalate H C – O – C3 C – O – CH3
O OType of Polymerization : Condensation polymerization.
(b) Bakelite :Monomer used :Phenol :
OH
Formaldehyde – HCOOH
Type of Polymerization : Condensation polymerization.
CH COOH + HO – CH – CH3 2 3 CH – COO – C H + H O3 2 5 2Con. H SO2 4
dil. H SO2 4
(E)Ethyl acetate
CH – CHO – NH3 3 CH – CH – OH3 CH – CH = NH3
NH2
Acetaldamine(C)
(b)
(i)Cold
3 2 2 2 3 2 2 2Ethylamine
CH CH NH HNO CH CH OH N H O
72 2
2 43 2 3 2
K Cr Odil. H SO ,CH CH OH (O) CH COOH H O
CH – COOH + NH3 3 CH – C – O NH3 – +
4
O
CH – C – O NH Br + 2KOH 3 2 – +
4 +
O CH – NH + 2KBr + 2H O + CO3 2 2 2
Methylamine
(ii) When globular proteins undergo coagulation or precipitation to give fibrous proteins, due to coagulation thenative shape of protein is destroyed and biological activity is lost. That is why coagulated proteins so formedare called denaturated proteins.During denaturation, secondary and tertiary structures are destroyed but primary structure remains intact.