CBSE Board XII Chemistry Paper Sol
CBSE Board XII Chemistry Paper Sol
CBSE Board XII Chemistry Paper Sol
1. Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason. [1]
Sol. Since in FeO, Fe present in both +2 & +3 O.S.. Hence FeO has non-Stoichometric composition
2. CO (g) and H2 (g) react to give different products in the presence of different catalysts. Which ability of the
catalyst is shown by these reactions ? [1]
Sol. It specify selectivity of a catalyst in this reaction which means a catalyst for one reaction can be inhibitors for
another
3. Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2]. [1]
Sol. Complex given –
[Pt (en)2Cl2]
Coordination no. = denticity × number of ligand
Coordination number = 2 × 2 + 2 × 1 = 6
Charge on ligand + O.S. of metal ion = charge on complex
–2+x=0
⇒ x = +2
4. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why ?
[1]
Sol.
Cl CH2Cl
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
CH3 C CH CH3
C2H5 OH
CH3
3 2 1
Sol. CH3 C CH CH3 3, 3-dimethyl pentan-2-ol
C2H5 OH
4 5
6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of
water. (Kf of water = 1.86 K kg mol–1) [2]
Sol. W1 = 250 g, w2 = 60 g, mw2 = 180 g/mol, Kf = 1.86 k kg mol–1
Δtf = kfm
w 2 × 1000
or Δtf = kf ×
m.w 2 × w1 (g )
60 × 1000
= 1.86 ×
180 × 250
1.80 × 600
=
18 × 25
1116
=
450
= 2.48 K
Δtf = t(solvent) – t(Solution)
Or t(solution) = t(solvent) – Δtf
= 273.15 – 2.48
= 270.67 K
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
ΔN 2O5 1 ΔNO2
or =
Δt 2 Δt
ΔN 2O5 1
= × 2.8 × 10 −3 = 1.4 × 10–3 M/sec
Δt 2
Benzoic acid
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
OR
(a) In case of aromatic carboxylic acid –COOH attach to the Benzene ring having electron with drawing
effect and deactivated the benzene ring, hence do not exhibit friedel craft reaction.
(b) pKa value of 4-nitrobenzoic acid is lower then benzoic acid is due to e– with drawing nature of –NO2
attach at para position of Benzene due to which tendency to loose H+ ion increases and acidic character
increases.
12. An element 'X' (At. mass = 40 g mol–1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate
the density of 'X' and the number of unit cells in 4 g of 'X'. (NA = 6.022 × 1023 mol–1) [3]
Sol. Atomic mass = 40 g/mol
A = 400 pm = 400 × 10–10 cm or 4 × 10–8 cm
z×M 4 × 40
⇒ So d = =
a3 × N A (4 × 10 −8 ) 3 × 6.023 × 10 23
160 160
d= –1
= ≅ 4.18gm / cc
64 × 6.023 × 10 6.4 × 6.023
6.023 × 10 23 6.023 × 10 22
90 = atoms or = 1.50×1022 unit cell
10 4
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
13. A first order reaction is 50 % completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the
activation energy of the reaction. (Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK–1 mol–1) [3]
Sol. T1 = 300 K t1/2 = 40 min
T2 = 320 K t1/2 = 20 min
.693
We know t1/2 =
K
log K 2 Ea ⎛ 1 1 ⎞
= ⎜⎜ − ⎟
K1 2.303 R ⎝ T1 T2 ⎟⎠
1
K∝
t1 / 2
log (t1 / 2 )1 Ea ⎛ 1 1 ⎞
or = ⎜ − ⎟
( t1 / 2 ) 2 2.303 R ⎜⎝ T1 T2 ⎟⎠
⎛ 40 ⎞ Ea ⎛ 20 ⎞
⇒ log⎜ ⎟ = ⎜ ⎟
20
⎝ ⎠ 2 . 303 × 8 . 314 ⎝ 320 × 300 ⎠
20 Ea
0.3010 =
2.303 × 8.314 × 320 × 300
15. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN
and Zn in this process. [3]
Sol. Chemical reaction during extraction of gold –
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
&
OH OH
(i) (ii)
(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of
sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with
alcoholic KOH [3]
Sol. (a) The chiral molecule
OH
Cl CH3
Na
(b) + CH3–Cl Dry ether
Toluene
Br CH3 CH3 CH2
Alc KOH/Δ
(c) +
(Major) (Minor)
18. (A), (B) and (C) are three non-cyclic functional isomers of carbonyl compound with molecular formula C4H8O.
Isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ Test but gives positive
Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/cons.HCl give the same product (D).
(a) Write the structures of (A), (B), (C) and (D)
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards addition of HCN ? [3]
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CH3 O
(C) CH3–CH–C–H
(D) CH3–CH2–CH2–CH3
(b) Since (B) is ketons so it will be less reactive towards nucleophilic addition reaction with HCN due to +I
effect & sterric hinderence.
19. Write the structures of the main products in the following reactions :
O
CH2–C–OCH3 NaBH4
(i) O
O
CH=CH3
H+
(ii) + H2O
OC2H5
NaBH4
(iii) + HI
[3]
O OH
CH2–C–OCH3 CH2–C–OCH3
NaBH4
Sol. (i) O O
CH=CH2 CH–CH3
H+
(ii) + H2O
OH
OC2H5 OH
(iii) + HI + C2H5-I
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
22. (a) Write the formula of the following coordination compounds: Iron(III)Hexacyanoferrate(II)
(b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4 ?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3–
(Atomic No. of Co = 27) [3]
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23. Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the item in polythene
bags and gave them to Shyam. But Shyam refused to accept the polythese bags and asked the shopkeeper to
pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the
government for using polythene bags. The shopkeeper promised that he would use paper bags in future in
place of polythene bags.
Answer the following :
(a) Write the values (at least two) shown by Shyam
(b) Write one structural difference between low-density polythene and high-density polythene
(c) Why did Shyam refuse to accept the items in polythene bags ?
(d) What is a biodegradable polymer ? Give an example. [4]
Sol. (a) The value display by Shyam is awareness and concern about the consequences of using polythene bags.
(b) Low density polyethese form by free radical polymerization and posses highly branch structure while
high density polythene posses a linear structure due to which it has high density due to closed packing
(c) As he knew about the consequences that they get into soil and slowly release toxic chemical as they are
non-biodegradable polymers.
(d) Polymers broken down rapidly by enzyme catalysed reaction are called biodegradable polymers
Eg. Poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV) etc.
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(b) Arrange the following in the decreasing order of their reducing character
HF, HCl, HBr, HI
(c) Complete the following reaction :
Δ –3
4H3PO3 3H3PO4 + PH3↑
+3 +5
(ii) Due to high electronegativity & small size and absence of higher vaccant orbital flourine exibit only –1
O.S. so due to which FCl3 is not possible.
(iii) Electron in oxygen atom, owing to small size of atom and tightly held hence less induced dipole-induced
dipole attraction while electron of sulphur atom, owing to large size of the atom reach farther and cause.
Strong induced dipole-induced dipole attraction.
Structure of compound –
..
F F OH
(A) Xe (B) O=Cl
F .. F O
(HClO3)
OR
(a) (i) A → NO2 (nitrogen dioxide)
B → N2O4 (dimitrogen tetra oxide)
(ii) Structures of A and B –
Θ
O O
•
⊕ ⊕
N N N
O O O O
Angular Θ
N2O4
(Planar)
(iii) “NO2 contain odd number of valence electron hence behave as odd molecule and dimerized to
convert into stable N2O4 molecule
(b) Decreasing order of reducing character -
HI > HBr > HCl > HF
(c) Complete following reaction
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25. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298K:
Sn (s) | Sn2+ (0.004 M) || H+ (0.020 M) | H2 (g) (1 bar) | Pt (s)
0
(Given : ESn 2+
/ Sn
= 0.14)
2 AgCl (s) + H2 (g) (1 atm) ⎯⎯→ 2Ag (s) + 2H+ (0.1 M) + 2Cl– (0.1 M)
ΔGº = – 43600 J at 25ºC
Calculate the e.m.f. of the cell
[log 10–n = –n]
(b) Define fuel cell and write its two advantages. [5]
Overall reaction :
E ocell = ESn
o
/ sn + 2
– EoH + / H
2
= – (–.14)
= 0.14 V
0.0591 [Sn +2 ]
Ecell = Eocell – log10
2 [ H + ]2
0.0591 [0.004 ]
= 0.14 – log10
2 [0.020 ]2
= 0.14 – 0.0295
= + 0.1105 V
(b) (i) EoValue of O2 is higher then Cl2 but O2 will lie liberate from H2O only when the higher voltage is
applied so because of this Cl2 liberate instead of O2.
Eg. NaCl → Na+ + Cl–
H2O H+ + OH–
(ii) Conductivity changes as the concentration of the electrolyte changes. The number of ions per unit
volume carrying the current decreases on dilution so conductivity will always decreases with
decrease in concentration hence CH3COOH conductivity decreases on dilution.
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OR
A reaction is given –
(a) 2 AgCl(s) + H2(g) (a atm) ⎯⎯→ 2Ag(s) + 2H+ (0.1M) + 2Cl– (0.1M)
ΔGº = – n F Eocell
− ΔG º − (– 43600 )
So Eocell = = = 0.23 V
nF 2 × 96500
0.0591
Ecell = Eocell – log10[H+]2[Cl–]2
n
0.0591
⇒ 0.23 – log10[10–1]2[10–1]2
2
⇒ 0.23 + 0.12 = 0.35 volt
(b) They are galvanic cell that are designed to convert the energy of combustion of fuel like hydrogen,
methane, methanol directly into electrical energy.
Eg. Hydrogen-oxygen fuel cell
Two advantages
(1) Do not cause any pollution like thermal plant
(2) Due to continuous supply of fuels, these cells never become dead.
(CH3CO)2O
(i)
(CH3)2NH
(ii) SO2Cl
N2+Cl–
CH3CH2OH
(iii)
(b) Give a simple chemical test to distinguish between Aniline and N,N-dimethylaniline.
(c) Arrange the following in the increasing order of their pKb values. :
C6H5NH2, C2H5NH2, C6H5NHCH3 [5]
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
H2O OH–/Δ
COOH
C2H5NH2 +
Ethanamine COOH
CH3
.. ⊕
(CH3)3–N + H CH3–N–H
⊕
3 amine (acidic)
CH3
(Salt)
Less stable
Since 2 amine salt form are more stable then 3 amine salt due to inductive effect (+I effect) and higher
degree of hydration, so higher the state of salt more will be the reactivity of corresponding compound.
(ii) The aromatic diazonium salt more stable then aliphatic diazonium salt is due to resonance.
⊕ ⊕ Θ ⊕ Θ ⊕ Θ
N≡N N=N N=N N=N
⊕ ⊕
⊕
R.S1 R.S2 R.S4
R.S3
OR
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CHEMISTRY CBSE-XII-2018 EXAMINATION CAREER POINT
SO2Cl O
(CH3)2NH S–N–CH3
(ii)
O CH3
N, N-dimethyl Benzene sulphonamide
N2+Cl–
CH3CH2OH
(iii) + N2↑ + HCl + CH3CHO
O N(CH3)2
S–Cl
+ No reaction
O
Benzene N,N-di methyl
sulphonamide aniline
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