Centre for Advancement of Standards in Examinations

CENTRE FOR ADVANCEMENT OF STANDARDS IN EXAMINATIONS

(GEMS Asian Schools)

COMMON REHEARSAL EXAMINATION ─ JANUARY 2010

Grade : XII Subject: CHEMISTRY

Marking Scheme – SET II

Q.
No. / Value Points / Point
Wise
marks / Total
marks
1 / dipole-dipole forces / 1 / 1
2 / The aggregated particles of associated colloids formed above Kraft temperature and above CMC. / 1 / 1
3 / The H─X bond dissociation enthalpy decreases with increase in size of X. / 1 / 1
4 / Through the process of osmosis, the bacteria on salted meat lose water, shrivel and die. / 1 / 1
5 / Due to the presence of two weak P─H bonds. / 1 / 1
6 / The impurities are more soluble in the melt than in the solid state of the metal. / 1 / 1
7 / O=C=O H3O+
C6H5MgBr C6H5─C=O
Dry ether │
OMgBr
C6H5COOH / 1 / 1
8 / CH3
│
CH3─C─CH3
│
CH3─CH2─CH─CH─CH2─CH2─CH3
│
Cl / 1 / 1
9 / DTb = Kb x wsolute x 1000
Msolute x wsolvent
DTb and Kb are equal for both the solutions.
Hence wX x 1000 = DTb = wurea x1000
MX x wsolvent Kb Murea x wsolvent
1.5 = 0.75
MX 60
MX = 1.5 x 60
0.75
= 120 / ½
½
1 / 2
10 / Mond process : page 160 NCERT text
with equations
OR
DG for the reduction process should be negative.
Or when coupled with an oxidation process the DGtotal of the coupled redox reaction should be negative. / 1,1
1
1 / 2
11 / Nitrogen gas,
From Henry’s law
P = KH x Xgas
At same pressure XO2 is greater than XN2
Hence KH of Nitrogen gas is larger. / 1
½
½ / 2
12 / / ½
½
½
½ / 2
13 / a) azo-dye test for aniline-coloured dye formed with benzene diazonium chloride under ice-cold conditions.
b) carbyl amine test for methyl amine – offensive smell of carbyl amine or
Hinsberg test,description. / 1
1 / 2
14 / P = CH3CH2CHCH3
│
OCOCH3
Q = CH3CH=CHCH3
R = CH3CH2COCH3
S = CH3CH2CHCH3
│
I / ½ x 4 / 2
15 / a)Water soluble and fat soluble vitamins
Vitamin K
b) Ribose sugar, Uracil and phosphoric acid / ½
½
1 / 2
16 / a)The ─CO─NH─ bond that holds the amino acids together in the primary structure of protein. It is formed by loss of water between amino acids.
b)It is the sequence of amino acids in a polypeptide chain of protein. / 1
1 / 2
17 / rate(new) = 50/100 = k[A/2]2[nB]
rate(old) k[A]2[B]
½ = n/4
n = 2
Therefore concentration of B should be doubled. / 1
1 / 2
18 / a) N─H bond is less polar than O─H bond due to lower electronegativity of N.
b) Aniline being basic forms salt with AlCl3 which is a Lewis acid / 1
1 / 2
19 / k = ln(R0/R)
t
k after 135 min
k = ln(2.08/1.91) = 6.32 x 10─4
135
k after 347 min
k = ln(2.08/1.67) = 6.33 x 10─4
347
k after 683 min
k = ln(2.08/1.35) = 6.33 x 10─4
683
Value of k is almost constant for all experimental results hence it is a first order reaction
t1/2 = 0.693/k = 0.693/ 6.33 x 10─4
= 1.094 x 103 min / ½
½
1
1 /
3
20 / a) I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
b) (NH4)2Cr2O7 → N2 + 4H2O + Cr2O3
c) 2XeF2 + 2H2O → 2Xe + 4HF + O2 / 1
1
1 / 3
21 / a) [Al(H2O)2(OH)4]─
b)
NH3
O2N NH3
fac- isomer

O2N NH3
NO2
NH3
H3N NO2
mer- isomer


O2N NO2
NH3
c) Due to d8 configuration of Ni2+ ion d2sp3 hybridisation not possible. / 1
½
½
1 / 3
22 / a) colloidal solution with alcohol as dipersion medium.
b) The ability of a catalyst to direct a reaction to produce a particular product.
c) The movement of colloidal particles under an applied electric potential. / 1
1
1 / 3
23 / a) d = Z x M
a3 x NA
Z = d x a3 X NA
M
= 7.2 x (289 x 10─10)3 x 6.023 x 1023
52
= 2.01
Structure is bcc
b)Schottky defect: Cations and anions missing from the normal lattice site.
Frenkel Defect: Smaller ion usually cation dislocated from its normal site to an interstitial site. / ½
½
1
½
½ / 3
24 / a) Any 1 equation – NCERT Text page 186
b)i)

ii)

OR
a) Decreases down the group from NH3 to BiH3.Lone pair availability on the central atom decreases with increase in size.
b) SO2 reduces KMnO4 to colourless Mn2+ ions.
c)
/ 1
1
1
1
1
1 / 3
25 / a) H- bond
b) Amide
c) condensation reaction / 1
1
1 / 3
26 / Mg CH3CHO
a) CH3Br CH3MgBr
dry ether H2O
PCC
CH3CH(OH)CH3 CH3COCH3
b) CH3─CH=CH2 HBr CH3CH2CH2Br
org.peroxide
AgNO2
CH3CH2CH2NO2
c) CH3CH2Cl KCN CH3CH2CN H2O/H+
CH3CH2COOH / 1
1
1 / 3
27 / a)i)antihistamine,antacid
ii) tranquiliser-antidepressant and for hypertension.
b) Quarternary ammonium salts with acetates, chlorides or bromides as anions. They have long hydrocarbon chain with positive charge on N atom. e.g page 452 NCERT Text / 1
1
1 / 3
28 / a)i) Cr3+ is more stable than Cr2+ due to half filled t2g level so Cr2+ gets oxidized.
Mn2+ is the more stable state of Mn due to half filled d-subshell so Mn3+ is easily reduced.
ii) Due to the presence of maximum no. of unpaired electrons(5) in the 3d subshell along with 2 electrons in the 4s orbital.
iii) The high energy needed to convert Cu to Cu2+ is not compensated by hydration energy.
b)i) Cr2O72─ + 2 OH─ → 2 CrO42─ + H2O
ii) 3MnO42─ + 4H+ → 2MnO4─ + MnO2 + H2O
OR
a)i)Due to their small size, high ionic charge and availability of vacant d-orbitals for bond formation.
ii) In Mn2O7 Mn is in +7 state with covalent bonds.In MnO bonding is ionic since Mn is in +2 state.
iii)The 5f electrons of actinoids penetrate less into the atom and are more effectively shielded from the nucleus than the 4f electrons. Hence the outer electrons of actinoids are more readily available for bonding.
b)i)2MnO4─ +6H+ + 5SO32─→ 2Mn2+ + 5SO42─ +
3H2O ii) Cr2O72─ +3C2O42─ + 14H+ → 2Cr3+ + 6CO2 +
H2O / 1
1
1
1
1
1
1
1
1
1 / 5
29 / a)anode: Zn(s) → Zn2+(aq) + 2e
cathode: Fe2+(aq)+ 2e → Fe(s)
Ecell = Eocell + 0.059 log[Fe2+]
n [Zn2+]
n = 2
= 0.32 + 0.059 log [1]
2 [0.05]
= 0.32 + 0.0295 log 20
= 0.32 + 0.0295 x 1.3010
= 0.32 + 0.0384
= 0.358 V
b) Molar conductivity is the conducting ability of ions produced by 1 mole of the electrolyte. Λm = k/C
For strong electrolytes it increases slowly with dilution due to decrease in interionic forces.
For weak electrolytes it increases steeply on dilution due to increase in degree of dissociation. So no. of ions in solution increases.
OR
a) Mg(s)+ Sn2+(aq) → Mg2+(aq) + Sn(s)
Ecell = Eocell + 0.059 log[Sn2+]
n [Mg2+]
n = 2
= 2.22 + 0.059 log [0.1]
2 [0.01]
= 2.22 + 0.0295 log 10
= 2.22 + 0.0295 x 1.0
= 2.22 + 0.0295
= 2.2495 V
b) Mercury cell: NCERT Text page 86 / ½
½
½
½
1
1
½
½
1
½
½
1
1+1 / 5
30 / a) 4-Methylpent-3-en-2-one
b)i) Cross aldol condensation-NCERT Text page 364
ii) HVZ reaction- NCERT Text page 375
c)i) Due to resonance the positive charge on carbonyl C is less.
ii) Lower the pKa value stronger the acid. Electron donating methyl group reduces stability of the acetate ion.
OR
a) 3-Bromo-3-methylbutanal
b) Decarboxylation:
NaOH+CaO
RCOONa R─H + Na2CO3
Heat
c)i) The lone pair of electrons on the N is in conjugation with the C=O, hence less available for donation.
ii)-COOH is an electron withdrawing group, hence it deactivates the benzene ring towards electrophilic substitution.
iii)Due to absence of intermolecular H-bonding in ethanal it has lower boiling point than ethanol. However it is more polar than ether, hence has a higher boiling point than ether. / 1
1
1
1
1
1
1
1
1
1 / 5

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