MARKING SCHEME

SECTION - A
1 (a) The same 1
2 1

(d)

3 (d) Glycogen 1
4 (c) Eext > Ecell 1
5 (c ) treatment with pyridinium chlorochromate 1
6 (c ) Ethanol < Phenol < Acetic acid < Chloroacetic acid 1
7 (b) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent. 1
The reason is that if HCl is used, the oxygen produced from KMnO 4 + HCl is
partly utilized in oxidizing HCl to Cl, which itself acts as an oxidizing agent
and partly oxidises the reducing agent.
8 (d) infinite 1
9 (a) 3-methylphenol 1
10 (c ) Fehling’s solution 1
11 (d) Br2/NaOH 1
12 (c) 3.87 B.M 1
Exp: Electronic configuration of Cr 3+ is [Ar]3d3. The number of electrons
that contribute towards spin only magnetic moment is 3
13 (d) A is false but R is true. 1
Exp: The α-hydrogen atom in carbonyl compounds is more acidic.
14 (a) Both A and R are true and R is the correct explanation of A. 1
Exp: Conductivity depends on a number of ions per unit volume which
decreases on dilution of electrolytes.
15 (c) A is true but R is false. 1
Exp: According to the Arrhenius equation (k = Ae -Ea/RT); it is found almost
accurate for single as well as complex reactions.
However, orientation is essential for the reactant molecules participating in
 the reaction.
16 (a) Both A and R are true and R is the correct explanation of A. 1
Exp: p-Nitrophenol is more acidic than phenol because the nitro group
stabilizes phenoxide ion by dispersal of negative charge.
SECTION - B
17 (a) Deep-sea divers depend upon compressed air for breathing at high 1
pressure underwater. The compressed air contains N2 in addition to O2,
which are not very soluble in blood at normal pressure. However, at great
depths when the diver breathes compressed air from the supply tank, more
N2 dissolves in the blood and other body fluids because the pressure at that
depth is far greater than the surface atmospheric pressure, therefore,
increase pressure increases the solubility of atmospheric gases in the blood.
When the scuba divers comes towards the surface, the pressure decreases,
N2 comes out of the body quickly forming bubbles in the bloodstream.
These bubbles restrict blood flow, affect the transmission of nerve
impulses. The bubbles can even burst the capillaries or block them and
starve the tissues of O2 and create a medical condition called the bends
which are painful and life-threatening.
(b) At high altitudes, the partial pressure of O 2 is less than that at the
ground level. This results in a low concentration of oxygen in the blood and 1
tissues of the people living at high altitudes or climbers. The low blood
oxygen causes climbers to become weak and unable to think clearly,
symptoms of a condition known as anoxia.
18 (a) Equal proportion of dextro-rotatory and laevorotatory compound. 1
1/
(b) 2- Chlorobutane. As it is optically active molecule. 2

1/
2

19 (i) CH3CH2CH3 < CH3OCH3< CH3CHO< CH3CH2OH 2

(ii) Butanone < propanone < propanal < ethanol
OR
(i) Propanal Tollen’s test, Gives silver mirror on heating with ammoniacal AgNO3.
Propanone : Does not give silver mirror on heating with ammoniacal AgNO 3.

(ii) Phenol does not librate CO2 with sodium bicarbonate

Benzoic acid librates CO2 with sodium bicarbonate

20 (i) A reaction which is not truely of first order but under certain conditions 2
becomes a reaction of first order is called pseudo first order reaction, e.g.,
 acid hydrolysis of ethyl acetate.
CH3COOC2H5 + H2O ---------> CH3COOH + C2H5OH
Rate ∝ [CH3COOC2H5] as H2O is in excess.
ii) The half life (t1/2) of a reaction is the time in which the concentration of
reactant is reduced to one half of its initial concentration [R]0.
21 2

SECTION - C
22 (i) Wolff-kishner reduction- Reduction of the carbonyl group of aldehydes 3
and ketones to CH2 group on treatment with hydrazine followed by heating
with sodium or potassium hydroxide in high boiling solvent.
CH3CH2CHO+ NH2-NH2 CH3CH2CH=N-NH2-----KOH/HBS + heat - CH3CH2CH3
(ii) Aldol condensation---Aldehydes and ketones having at least one 𝛼 −H
atom undergo a reaction in the presence of dilute alkali as catalyst to form
β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol) respectively. 2
CH3-CHO----dil NaOH CH3-CH(OH)-CH2-CHO---Heat CH3-CH=CH-CHO
(iii) Cannizzaro reaction—Aldehydes having no α- H atom undergo self
oxidation and reduction in presence of conc. alkali.
C6H5-CHO + Conc KOH--------- C6H5-OH + C6H5-COOK
23 .(a) Due to lack of Vitamin A is Xerophthalmia and vitamin E fragility of RBC 3
and mascular weakness.
(b) Adenine , guanine and cytosine are present in both DNA and RNA.
DNA and RNA are considering as acid because they are from phosphate
group which readity remove proton so both are highly acidic.
(c) Both are reducing sugar because both contains free aldehydic & ketonic
group which undergo oxidation forming easily carboxylic acid and in the
process reactive reagents are reduce easily.
24 (a) O-nitrophenol shows intramolecular H- bonding 3
 (b)
OR
(b)

25 (a) i) Second order reaction (ii) first order reaction. 3

(b) log k2 - log k1 = Ea/2.303 R [1/T1 – 1/T2]
logk2/k1 = Ea/2.303 R [1/T1 – 1/T2]
log2 = (Ea/2.303 x 8.314) [1/298 – 1/308]
Ea = log2 x 2.303 x 8.314 x 298 x 308
308 - 298
Ea = 52.898 kJ
26 (i) (CH3 )3C–I is more reactive towards SN 1 as I is better leaving group than 3
Cl.
(ii) p-Nitrophenol
(iii) Primary alkyl halides prefer to undergo substitution reaction by SN2
mechanism whereas tertiary halides undergo elimination reaction due to
the formation of stable carbocation.
27 (i) As isomer A reacts with AgNO3 to give a white precipitate, Cl must be 1
present in the ionisation sphere. As it does not react with BaCl2, SO42− must
be present in the coordination sphere.Therefore the formula of A =
[Co(NH3)5SO4]Cl (coordination no. of Co = 6)
As reactions are reversed for isomer B, formula B = [Co(NH 3)5Cl]SO4
 (ii) The type of isomerism involved is Ionisation isomerism 1
(iii) The IUPAC name of A = Pentaamminesulphatocobalt (III) chloride , 1
B = Pentaamminechloridocobalt (III) sulphate.
28 (a) Limiting Molar conductivity -limiting value of molar conductivity when 1
concentration approaches to zero.
(b) cell constant = conductivity x resistance = 1.29 cm –1, Conductivity, k =
2
Cell constant/Resistance = 0.00248 ohm –1 cm–1, Λm = k X 1000/M = 124
ohm–1 cm2 mol–1
SECTION - D
29 (a) Europium (Eu) is well known to exhibit +2 Oxidation state due to its half 1
filled f orbital in +2 oxidation state.
1
+4
(b) Ce has the tendency to attain the +3 oxidation state by accepting one
1
electron and hence it is used as an oxidising agent in volumetric analysis.
(c) (i) due to attainment of stable empty (4f0), half filled (4f7)and fully 1
filled(4f14) sub shell.
(ii) Lanthanoids showing +4 oxidation state are Ce, Pr, Tb,Dy.
OR
(i) As the size decreases covalent chracter increases therefore La 2O3 is more
ionic and Lu2O3 is more covalent.
(ii) Radii of 4d and 5d elements will be almost same. It is associated with
intervention of the 4f orbitals which must be filled before 5d series of
elements begins.the filling of 4f before 5d results in lanthanoids contraction
hence 4d and 5d series have almost same size.
30 (a) Mercury cell 1
(b) A fuel, such as hydrogen, is fed to the anode, and air is fed to the 1
cathode
2
(c) Anode: Pb(s) + SO4−4(aq) → PbSO4(s) + 2e−
Cathode: PbO2(s) + 4H+(aq) + SO2−4(aq) + 2e− → PbSO4(s) + 2H2O(l)
Overall: Pb(s) + PbO2(s) + 4H+(aq) + 2SO4−4(aq) → 2PbSO4(s) + 2H2O(l)

SECTION - E
31 A = C6H5COOH Benzoic acid. 1
 B = C6H5CONH2 Benzamide. 1
1
C = C6H5NH2 Aniline. 1
(i) aq NH3 (ii) Heat 1

Reaction - C6H5COOH C6H5CONH2

(A) (B)
Br2 | KOH
C6H5CONH2 C6H5NH2
(B) (C)
OR

1,
1,
1,

(ii)
1
(a) Methyl amine being more basic than water, accepts a proton from water
and liberate OH ion. Which combine with Fe3+ to form brown precipitate of
hydrated ferric oxide.
(b) The diazonium salt of aromatic amines are more stable than those of
1
aliphatic amines due to dispersal of the positive charge on the benzene
ring.
32 (a) due to unpaired electrons. 1,
1,
(b) Because of large number of unpaired electrons in their atoms they have
1,
stronger interatomic interaction and hence stronger bonding between 1,
atoms resulting in higher enthalpies of atomization.
1
(c) d-d transition
(d) Due to the presence of vacant orbitals or their tendency to form
variable oxidation state.
(e) On the basis of incompletely filled 3d orbitals in case of scandium atom
in its ground state (3d1), it is regarded as a transition element. On the other
hand, zinc atom has completely filled d orbitals (3d10) in its ground state as
well as in its oxidised state; hence it is not regarded as a transition element.
 (f) Large third ionization energy of Mn (3d5) is mainly reason for this.
(g) This is due to poorer shielding by 5f- electrons in actinoids than 4f-
electrons in the lanthanoids.
33 a) Mole fraction of gas in the solution is directly proportional to partial 1
pressure of gas in the vapour phase +
1
b) Chilled as solubility of CO2 is more at low temp.
+
(C) MB = 241.98 g mol-1, Molecular mass of C6H5COOH = 122 g mol-1 3

2C6H5COOH⇌(C6H5COOH)2,
Degree of association of benzoic acid in benzene=99.2%
OR
(a) Azeotropes are the binary mixtures of solutions that have the same 1
composition in liquid and vapour phases and that have constant boiling +
points. It is not possible to separate the components of azeotropes by 1
fractional distillation. showing a large positive deviation from Raoult’s law +
at a speci_c composition. For example an ethanolwater mixture containing 1
approximately 95% ethanol by volume +
2
(b) If K2SO4is completely dissociated, Here i= 3
ᴨ= i c RT =i (nB/V)RT =i(wB/ MB V)RT=5.27 x10-3 atm