Download solution-chemistry

MODERN VIDYA NIKETAN SR. SEC. SCHOOL
CLASS : XII
SUBJECT : CHEMISTRY
SOLUTION
1.
In Frenkel defect, a cation is missing from its lattice site and occupies interstitial side. Eg. Nacl. Electrical
conductance increases due to this defect.
2.
1
x
 KP n
m
log
x
1
 logK  logP
m
n
So the slope indicates
1
where n is order of reaction.
n
3.
CH3
|
CH3  C  Cl will react fastest as it will form most stable carbocation.
|
CH3
4.
At temperature above 983K, coke is better reducing agent. However below this temperature, carbon monoxide
is a better reducing agent than coke.
5.
Ethylcyclohexane carboxylate
6.
CH3CH2NH2 will give carbylamine reaction and  CH3CH2 2 NH will not
CH3CH2  NH2  CHCl3  KOH 
CH3  CH2  N  C  3KCl  H2O
7.
Essential amino acids are those which cannot be synthesized in body e.g. Lysine, Leucine, valine etc
8.
In R3P = O there is p  d bonding which is not possible in R3N  O
9.
Henry’s law
PA  k Hx A
Its application is in dissolution of CO2 in cold drinks.
10.
KA
e

Ea
RT
slope 
l n K = l n A  Ea
nK
RT
A slope of
11.
Ea
R
l nK Vs 1 graph will be equal to  Ea .
T
nA
R
420  425K
Al 2O3  2H2O  2NaOH 
 2NaAl O2  3H2O
Sodium meta
aluminate
(water soluble)
Page 1
NaAl O2  2H2O 
 Al  OH3  NaOH
2Al  OH3 
 Al 2O3  3H2O
12.
(i)
Because sulphur cannot undergo sp3d2 hydridisation in presence of H atoms.
(ii)
As we move down the group in hydrides of group 15, bond angle decreases due to decrease in bond
pair-bond pair repulsion.
13.
(i)
O
OH
||
|
C6H5  CH  CH  C  H
|
CH3
ONa
OH
(ii)
14.
(i)
A low concentration of H+ ion pH  4  5 helps in removal of H2O, but further increase in H+
concentration may protonate the nucleophile.
(ii)
Presence of anhd. AlCl3, NH2 acts as a Lewis base.
NO2
15.
conc.HNO3
conc.H2SO4
(i)
NH2
Sn, HCl
N C
CHCl3, KOH

(ii)
16.
C6H5MgBr
H2 O,H
CH3  C  N 
 CH3  C  NMgBr 
 CH3  C  O
|
|
C6H5
C6H5
H OH
C
CHOH
CHOH
O
CHOH
CH
CH2OH
Glucose does not respond to 2, 4-DNP test
Page 2
17.
(i)
Globular protein
(ii)
Fibrous protein
(1)
Sherical in shape
(1)
Linear in shape
(2)
Part of blood and egg yalk
(2)
Part of dead tissues like hair and nails
Nucleoside – has nitrogenous base and pentose sugar
Nucleotide – has nitrogenous base, pentose sugar and phosphate group
OR
18.
(1)
As source of energy
(2)
As structural material
(3)
Reserve food
(4)
Maintenance and body growth
Calcium bicarbonate makes water hard. Soap (RCOONa) will react with the salt to form corresponding calcium
salt which will be precipitated and wasted
3RCOONa  Ca HCO3 2 
 RCOO 2 Ca  2NaHCO3
The synthetic detergents do not react with Ca HCO3 2 and therefore can be used without being precipitated
19.
Atomic mass, M = 60g/mil, density,
Z
l = 6.23 g/cm3 edge length a = 400pm = 4×10—8cm
l  a3  NA
M
6.23   4  108   6.022  1023
3

60
= 4 four atoms per unit cell are present in FCC lattice
So, r 
20.
(a)
a
2 2

400
 141.44pm
2  1.414
Oxidation state of Pt
x  4  1  2
x  2
El . Confign of Pt (II)
Xe
5d
6s
6P
In strong field of CN—, it rearranges as
Xe
dsp2
So shape of Pt  CN 4 
2
hybridisation
is square planar.
Page 3
FeSO4 . NH4 2 SO4  6H2O (Mohis salt) gives free Fe2 ions in solution but CuSO4 and NH3 react as
(b)
CuSO4  4NH3 
 Cu NH3 4  SO4
As there is no free Cu2 ion
21.
22.
23.
3
3
WBRT 1.26  10  10  g  0.0821  300
MB 

V
2.57  103  0.2
K
2.303
 Pi 
log 

t
 2Pi  Pt 
K360 
2.303
35
log
 2.17  103 s 1
360
 70  54 
K720 
2.303
35
log
 2.24  103 S1
720
 70  63
(i)
The positively charged sol particles of ferric hydroxide get their charge neutralized by interacting with
 

negative chloride ions Cl . As a result, they get coagulated.
(ii)
Smoke and dust in factories have dispersions of electrically charged particles in air. When these are
passed between two electrodes at a potential difference of 50,000V. these particles are discharged and
precipitated.
24.
(i)
Hofmann bromamide reaction
O
||
Heat
CH3  CH2  C  NH2  Br2  4KOH 
CH3CH2NH2  2KBr  K2CO3  2H2O
Pr opanamide
(ii)
Hydroboration oxidation
H2O2 / OH
3CH3  CH  CH2  BH3 
  CH3CH2CH2  B 
 3CH3CH2CH2OH  H3BO3
H2O
Propane

(iii)
Sandmeyer reaction
N NCl
Cl
CuCl/HCl
N2
Benzene diazonium
Chloride
Page 4
Cl
Cl OH
Cl OH
Cl OH
N
N
N
slow
step
OH
NO2
25.
O
(a)
Cl OH
O
O
O
O
OH
fast
step
Cl
N
O
O
N
O
O
O
C N NHCONH2
H
(b)
Br
Br2 UV light
(c)
26.
H2C  C  CH  CH3 

| |
A  

CH3 CH3


Cl H
| |
B  CH3  C  C  CH3
| |
CH3 CH3
C  Cl  CH2  CH  CH  CH3
|
|
CH3 CH3
D  CH3  C  C  CH3
| |
CH3 CH3
O
||
E  CH3  C  CH3
F  O  C  CH  CH3
| |
CH3 CH3
O
||
G  H C H
27.
(a)
H2O
H2N  CH2  COOH  H2N   CH2 5  COOH 
 NH2  CH2  CO  NH   CH2 5  COOH
NH  CH  CO  NH  CH 
2
2 5
 CO 
n
Page 5
(b)
Vulcanization is addition of a small amount of sulphur to molten rubber at temperature ranging
between 373 to 415 K. Vulcanized rubber is more elastic, less water absorbing and more resistant to
oxidation and towards organic solvents.
28.
(a)
Fuel cells convert the energy of combustion of certain fuels into electrical energy without the use of a
heat engine e.g. H2  O2 fuel cell.
Reactions
(b)
at anode
:
2H2  g   4OH  aq. 
 4H2O  l   4e
at cathode
:
O2  g   2H2O  l   4e 
 40H  aq.
overall
:
2H2  g   O2  g  
2H2O  l 
0
0
0
Ecell
 Ecathode
 Eanode
= 0.80  0.34
= 0.46V
The reaction will be
Cu S   2Ag aq. 
Cu2 aq.  2Ag  S 
Ecell  E
0
cell
Cu2 
0.0591

log10
2
n
 Ag  
= 0.46 
0.0591
4
log10
2
 0.12
 0.46 
.0591
 2.6020
2
 0.353V
OR
(a)
Discharging
Pb  S   SO24  aq. 
PbSO4  S   2e  oxidation
PbO2  S   SO24  aq.  4H aq.  2e 
PbSO4  S   2H2O  l Reducation
Pb  S   PbO2  S   4H  aq.  2SO24 aq. 
2PbSO4  S   2H2O  l 
Page 6
During Charging
PbSO4  S   2e 
Pb  S   SO24 aqReduction
PbSO4  S   2H2O  l  
Pb  S   PbO2  S   4H aq.   2SO24 aq. 
(b)
1
I  aq.  Fe3  aq. 
Fe2  aq.   I2  g 
2
0
Ecell
 0.77  0.54
 0.23V
29.
(a)
2KMnO4  3H2SO4  5H2S 
K2SO4  2MnSO4  8H2O  5S
(b)
Ce has electronic configuration Xe 4f1 5d1 6s2. In Ce (IV) it acquires noble gas configuration.
(c)
In Fe3+, the electronic configuration is
 Ar3d5
which is more stable than Fe2+ with electronic
configuration  Ar  3d
6
In Mn3 , the electronic configuration is  Ar  3d which is less stable than Mn2 with configuration  Ar  3d
4
30.
(d)
Cu2 has very high solvation energy which easily compensates 2nd ionization enthalpy of copper.
(e)
It is due to lanthanoid contraction
(a)
5
4Zn  10HNO3  dil 
 4Zn NO3 2  H2O  N2O
nitrousoxide
F
F Br F
F
F
(b)
square pyramidal
(c)
MI  MBr  MCl  MF
It is due to increase in electronegativity difference
(d)
It is due to highest E0value of F2 among halogens.
(e)
Chlorine reacts with water to evolve nascent oxygen which brings about both oxidation and bleaching.
As oxidizing agent
Cl 2  H2O 
2HCl  O
SO2  H2O  O 
 2HCl  O
Cl 2  SO2  2H2O 
 2HCl  H2SO4
Page 7
As bleaching agent
Cl 2  2H2O 
2HCl  O
 colourless product
Coloured substance   O  
Page 8