Answers to INChO-2008 problems
Question No. 1 Subdivisions
1.1) F2 -1.6, F2
- 1.1,1.3
1.2) F atom -3.4, F2 molecule - 3.0,3.2 1.3) F2: -1.4 F2
: -1.9
1.4) 4.15
1.5)
1.6) F2 – 1.0 F2
- 0.5 1.7) F2
1.8) I1= 18.9 eV I2= 15.6 eV 1.9) I1= 2p I2 = 2p
F2
2s 2s
2p 2p
F2
2p
2p
2p
2p
T.S.2 T.S.1
Int
Reaction Coordinate
Free energy
product
O O
OO OO O O
HH
H H
H
CH2 C C CH3
OH O CH2COOH
COOH
O O
O OH O
Br CH2 CH CH CH2 Br
CH2 CH CH2
Br CH
Br
Question No. 2 Subdivisions
2.1) a) Z-5-methyl hex-2-en-1-al
b) Z-2-methyl-1-phenyl hept-1-en-6-yne
2.2) B C
2.3)
2.4) i) E
ii) D iii) E
2.5) a) iv b) iii c) ii d) v e) i
2.6)
2.7) D.
E. F.
2.8) G.
H.
H
H Br
Br CH2Br
CH2Br
R
S
H
Br H
Br CH2Br
CH2Br
R R
O
O
O O
O
O
O
O
K
O O
O
L
COOH
NO2 O
H
COOH
NO2 H7C3O
H7C3O
COCl
NO2
N(C2H5)2 OH
OC3H7 C
O
N(C2H5)2 NO2
O OC
3H7 C
O
N(C2H5)2 NH2 O
I. J.
2.9)
2.10)
Question No. 3 Subdivisions 3.1)
i) aromatic ii) aromatic iii)
iv) yes v) acidic vi) a)
3.2) I < III < II
3.3) K L
M N
O P
N Br +
+ Br
N
Br NO2
OMe Br
NO2
OMe Br Br
O O
3.4)
3.5) A. B. C. D.
3.6) iv.
Question No. 4 Subdivisions
4.1) 8.17 10-8 bar 4.2) 304 K
4.3) 641.02 K
4.4) Yes the reaction will proceed towards NOCl 4.5) Rate = k [NO]2 [Cl2]
4.6) Ea = 98.82 kJ / mol
4.7) Mechanism I and II both are possible.
4.8) Extent of the reaction = 0.1
4.9) Extent of the reaction at completion = 0. 195
Question No. 5 Subdivisions 5.1) a.
5.2) b.
5.3) X1 Solid phase
X2 Solid –Liquid equilibrium phase X3 Liquid – Gas equilibrium phase X4 Gas Phase
2 1 1
1 2 1
1 2 1 1
1 2 1
1 1
/
1
p p
p p
or p p
p
p p
p
D n
C K C
2
1 n KC A] 2
[
5.4)
5.5) b
5.6) Volume will increase on melting
5.7) Single Phase system called Supercritical Fluid 5.8) c
5.9)
5.10) 56
5.11) a. nA =An
b.
Liquid
Temp T0
Tm
Liquid –gas gas equilibrium Solid-liquid
equilibrium solid
Time
P
P P
P
H P
O H
O Na
H P
ONa H
or O +
O P O H
O Na
ONa P
O H
ONa or
2 + 2
P O O
O
O P
O
P O
O O
O P O
POCl3 +3EtOH O P(OEt)3 + 3HCl Question No. 6
Subdivisions 6.1)
6.2) P4 (s) + 3 NaOH + 3H2O PH3(g) + 3 H2PO2
Na+
6.3) a.
b. hypophosphite-reducing agent phosphate –reducing agent
c. They are reducing agents due to presence of P-H bond and lower oxidation state of P 6.4) Ca5(PO4)3F + 5H2SO4 3H3PO4 + 5CaSO4 + HF
6.5)
or 6.6) 504 gs of CaO.
6.7) PCl3 + O2 2Cl3PO
6.8)
O
P P P
O O O
O P
O
O O O O
3s 3p 3d
Ground state
Excited State
Or trigonal bipyramidal P
Cl Cl Cl Cl
Cl
6.9) 3PCl5 + 3NH4Cl (Cl2PN)3 + 12 HCl
P Cl Cl
Cl Cl
N N
P P
N Cl
Cl
[Co(NH3)6]2+ = Co2+ [Co(NH3)6]3+ = Co3+
Loss of electron easy Question No. 7
Subdivisions
7.1) Co3+ : 3d6 4S0
7.2) A Pink: [Co(NH3)5.H2O]Cl3: Pentaamine aqua cobalt(III)chloride B Purple: [CoCl(NH3)5]Cl2 : Pentaamine chlorocobalt(III)chloride 7.3)
7.4) a.
b. diamagnetic.
7.5)
Co3+ = 3d6
3d 4s 4p
d2sp3 hybridisation Octahedral
XX XX XX XX XX
XX
Co3+=d6
t2g
eg
Cl Co Cl
NH3 NH3
NH3
Cl
Cl Co Cl
Cl NH3
NH3
NH3
Co O
O O
O
Cl Cl
Co O
Cl O
O
Cl O
01 . 0 ] [
97 . 2
3
Fe
K 7.6)
Facial Meridional
7.7) Facial isomer - one peak due to ammonia (with all similar environments) Meridional isomer – two peaks
7.8)
Inactive plane of
symmetry Chiral
Question No. 8 Subdivisions
8.1) Fe(s) + 2OH(aq) FeO(s) + H2O(l) + 2e Ni2O3 (s) + H2O(l) + 2e 2NiO(s) + 2OH(aq) Fe(s) + Ni2O3 (s) FeO(s) + 2NiO(s)
8.2) iii.
8.3) E0cell = 0.83633 V Ecell = 0.80283 V
8.4) Ag(s)+ Fe3+(aq) = Fe2+(aq) + Ag+(aq) Fe(s) + 2Ag+(aq) = Fe2+(aq) + 2Ag(s) 8.5)
H2NCH2COOH
H2N CH2 NH2 (CH2)6 COOH COOH
+
CH2 S S
CH2C6H5 CH2 CH3 H3N-CH-COO H3N-CH-COO
H3N-CH-COO
H3+N-CH-COO +
+ +
+ +
2 2
CH2 S S CH2 H3N-CH-COO
H3N-CH-COO
SH CH2 H3N-CH-COO (H)
+
+
+ 2
N H3 C
CHCOOCH H
C O
NH C COO
H CHCH +
Question No. 9 Subdivisions 9.1)
9.2) i) Nylon ii)
9.3)
9.4) a)
b)
9.5) i) 4
ii)
Peptide
Nature Charge
Acidic Basic Neutral Positive Negative Zero
Gly-Leu-Val X X
Leu-Trp-Lys-Gly-Lys X X
Arg-Ser-Val X X
9.6) A – aspartic acid B – alanine C- arginine
