Cr2O7 2− + 14 H+ + 6 e− 7 H2O + 2 Cr3+
X = 1.34 V Y = −0.408 V
Correct E calculation and inference also awarded marks
Problem 1 17 Marks
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Name of Student Roll No.
potential.
in change the
is 0.27V
−
oxygen.
of m 3.50 ) v(O
hydrogen.
of m 6.69 2)
v(H
. - O e 4 O H 4 O H 6 is anode at the reaction The
H O H - 2 e 2 O H 2 is cathode at the
reaction The
3 2
3
2 3
2
2 2
3
=
=
+ +
→
+
→ +
+ +
V 1.140 Esy =
ionates.
disproport (IV)
Cr
0 J 222915
∆G
⇒
<
−
=
% 6.87 Efficiency
%
2O H 4 - Cr
6e H - 8 2 CrO4
=
+
→ + +
+
% 3 2 ) O H (C Cu of
% 18 33 2 2 =
Problem 2 11 Marks
2.1 B2O3
Al2O3 Tl2O3
2.2
b) Three centered 2e− bond 2.3
2.4
2.5
2.6
acidic amphoteric basic
X
B4H10
B H
B H
H H
H H
B H
B H
B H
B H
H
B
H
H H H
Y
or
H H
H H H B
H H
B H
H B
B H H
H H
H H
H H
B
H B
H B H
B
BnHn+6 or BnHn+42− or BnHn+24− or BnHn6−
B2H6 + 6H2O →2 B(OH)3 + 6 H2 or 2H3BO3 + 6H2
2.7
2.8
C2B10H12
Z
.
. . . .
. .
. .
.
. .
. .
.
.
. . .
. . .
. . . .
. .
. .
. . . . . . .
C C
C C
C C
Problem 3 18 Marks Thermodynamics of a sustainable bio process
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Efficiency = 32.6%
Decrease in the level of CO2 (in ppm) = 2.56 x 105 ppm
∆Hf = − 1271.4kJ
E = 1.24V
dE/dT = 7.82 x 10−5 VK−1
Time = 5.597 or 6 .0 min
Change in current produced= 0.672 x 10−3 mA
C6 H12O6 + 6H2O 24e− + 24H+ + 6CO2
6O2 +24H+ + 24e− 12H2O C6H12O6 + O2 H2O2 + C6H10O6
3.10
3.11
3.12
3.13
3.14 Show that ∆S step3 = 0
3.15
dU S = TSdS dG S = 0
(∆S)step1 = 60NAhc/λ ( 1/ TPO − 1/TS ) or (∆S)step1 = 60hc/λ ( 1/ TPO − 1/TS )
(∆S)step2 = −∆GPO/T
In step 3 sun does not participate.
∆U Po in step 3 =(∆U Po in step 1 − ∆U Po in step 2) = the energy transmitted to the earth
(∆UPO (step1) − ∆GPo)/ TPo is the change in entropy of PO Change in entropy of earth = - (∆UPO (step1) −∆GPo)/ TE
Adding ∆SE and ∆SPO and since TE= TPO ∆S step3 = 0
∆S (overall) = (60NAhc/λ - ∆GPO)/TPO
or ∆S (overall) = (60hc/λ - ∆GPO)/TPO
(∆U) S = −60NAhc/ λ. (∆U) PO = 60NAhc/ λ Or (∆U) S = −60hc/ λ. (∆U) PO = 60hc/ λ
Problem 4 23 marks Organosulphur Compounds
4.1
4.2
4.3
O
O
S O
:
: :
..
..
..
_
_ S
(H) (C6H10O)
O
C H3 C H3
(D) Ph
Ph S+ Cl
(E) Intermediate _
S+ Cl
Ph Ph
(F) (C15H15S) S+
Ph H Ph
(G) C6H10O
O C H3
C H3 _
_ _
_
_ _
_ _
_
_ _
_
_
_
_ _
_
_ _
S S
_
O
O O S+ S
S O
O S+ O
O
O S O
S2+
S O O
O O
O
S O
(J) 4.4
4.5
4.6
SEt
SEt
H
O P(C9H9N)
(an intermediate) L N H
O S
M (C10H13NSO)
O N H2 S
K
S+ O
NH
N
SMe
N
C N H
SMe
NH
4.7
T (C16H16O2)
U (C16H12) Q (C14H18S4)
R (C22H24S4)
S (C16H12O2)
H
S S
S S
H S
C H2
S
C H2 S S
O O
O
H O H
Problem 5 17 marks A. Chemistry of Main Group Elements
5.1 a)
b)
c).
5.2 a)
b)
c) No
T-Shaped
Lone pairs in the axial position Lone pairs in the Equatorial position
Bent /V shaped
X . ::
. .
Cl F
F F
X X
X
. . . .
Cl F
F F
or
. .
O ..
O S
O
O S
. . . .
Cl
F F F
3s 3p 3d
i) Ground state
ii) Excited state
iii) Hybridized state
sp2 Two π bonds (having gained four electrons
from two oxygen atoms)
..
. .
Cl F
F F
::
X
Cl
X
X X
X X X X
X X
X X X XX
X X X
F F F
[IrCl6]2- Spherical field
Ir4+
5.3 i) b) Sn4+ is more stable than Sn2+
c) Pb2+ is more stable than Pb4+
ii) oxidizing agent
iii)
B. Chemistry of d and f- block elements
5.4
5.5
5.6 i) a) the central metal ion is in higher oxidation state.
b) Ir belongs to third transition series.
ii)
5.7 2Ln + 6H2O → 2 Ln(OH)3 + 3H2
2Ln + 6H+ → 2 Ln3+ + 3H2↑ Ce + O2 → CeO2
Ce + 2F2 → 2 CeF4
0.0 B.M
X X
X X
X
Complex No of unpaired electrons Spin state
[Fe(CN)6]4− 0 Diamagnetic/low spin
[Fe(CN)6]3− 1 Low spin
[FeCl4]− 5 High spin
[Fe(H2O)6]2+ 4 High spin
This sub part was found to be Ambiquous – Hence Omitted
5 or
3
X
Problem 6 16 Marks
Chemistry of unusual organic compounds
6.1
6.2
6.3
6.4 (i) carbonyl group (b) has conjugated double bond
(syn) (anti)
+
2
H
H
H H
H H
O
O O
H
H
H H
H H
X 6.5
6.6
. 6.7
6.8 (ii) pentacyclic compound 6.9
6.10
CHBr3 + NaOH → :CBr2 + NaBr + H2O (16)
17 18 19
CH2Cl ClH2C
CH2 CH2
HOH2C CH2
CH2OH
OHC CHO
12 13 14
15
O Br O
Br
Br Br
O
Br O O
Br
X
O
Cl Cl Cl
Cl
Cl
Cl
or 6.11
6.12
Br Br
Cl
Cl
20
Propellane
C C C H2
C H2 C C H2
