CHEMISTRY MARKING SCHEME
Bhubaneswar – 2015
Set 2 - Code No. 56/2/B
Ques. Value points Marks
1. HOCl , HOClO, HOClO2 , HOClO3 ½ +½
(Any two of these)
2. CH3 – CH – CH2 – CH2 – Br 1
|
CH3
3. Negative charge 1
4. XY3 1
5. 1-Phenylpropan-2-ol 1
6. Formula: w z i t
w Valance 96500 ½
timetakenin sec
Mol Mass Current in Amp
Substituting the values in the formula we get:
1.17 g 2 96500 C mol 1 1
timetakenin sec
58.5 g mol 1 5 amp
225810
timetakenin sec
292.5
t=772 s ½
( Or by any other correct method)
7. (i) Potassium hexacyanidoferrate (III) 1
(ii) [Co(NH3)5 NO2]2+ 1
8. (i) Due to comparable energies of 5f, 6d and 7s orbitals . 1
(ii) Because 5f electrons have poorer shielding effect than 4f electrons. 1
9. (i) Positive deviation, lowering of temperature or absorption of heat. ½ ,½
(ii) By applying an external pressure greater than the osmotic pressure on the solution
or P > π ½,
Reverse osmosis is used in desalination of hard water / sea water. ½
10. (i) H2 / Pd-BaSO4 1
(ii) NaOH/CaO, ∆ 1
OR
10. i) C6H5 CO C6H5 < CH3COCH3< CH3CHO 1
ii) Cl – CH2 – COOH < Cl2CH – COOH < CCl3 – COOH 1
11. (i) Distillation 1
(ii) Collector/ enhancing the non-wettability of mineral particles. 1
(iii) As ∆S is positive /∆G is more negative 1
12. (i) Stoichiometric Defect 1
(ii) Frenkel Defect 1
(iii) Due to small size of Ag+ ion 1
13. (i) CH3 – CH(OH) – CN 1
(ii) C6H5 – COOH 1
(iii) CH3 – CH2NH2 1
14. (ii) t2g3 eg1 1½
(iii) Hybridization dsp2, Shape Square planar or diagram 1½
(Marks of (i) part is merged into (ii) and (iii) part )
15. (i) Due to the stability of benzyl carbocation/resonance/Diagram 1
(ii) Because 2-Bromobutane has a chiral centre. 1
(iii) Due to – I effect of halogen. 1
16. NaNO2 HCl H 2O H 1
(i) C6H5NH2
o C6H5N2Cl
Or Hydrolysis
C6H5OH
0 5 C
o
CH3 – CH = CH2 CH3 – CH2 – CH2Br CH3CH2CH2OH
HBr KOH
(ii) Organic peroxide
Aq
1
(iii)
(Or any correct method)
16. OR
(i) CH3 – CH2 – CH2OH Cu /573 K
Dehydrogenation / Oxidation
CH3CHO + H2 1
(ii)
(iii) C2H5Cl + NaOCH3 C2H5-O-CH3 + NaCl 1
17. (i) Maltose 1
(ii) Sugar Present in DNA is Deoxyribose whereas in RNA it is Ribose 1
Thymine is present in DNA whereas in RNA Uracil is present (Any one)
(iii) Beri-Beri 1
18. 1
2.6805 = E 0cell - 0.059 V log [ 0.0001]
1
2 [0.001]
2.6805 = E 0cell - 0.059 V log 10-1 = E 0cell - 0.059 V (-1)
2 2
2.6805 = E 0cell + 0.0295 V
E 0cell = 2.6805 – 0.0295
1
E 0cell =2.6510 V
19. (i) Glyptal: 1
and HO-CH2- CH2-OH (ethylene glycol)
(ii) Teflon: 1
Monomer: 1,1,2,2-Tetrafluoroethene
1,1,2,2-Tetrafluoroethene
(iii) Nylon-6
Monomer: Caprolactum 1
(Note : half mark for structure/s and half mark for name/s)
20. (i) Because of higher oxidation state of Mn in Mn2O7. 1
(ii) Due to almost similar atomic size / comparable size. 1
(iii) 2MnO2 + 4KOH + O2 2K2MnO4 + 2H2O 1
21. (i) Solution is homogeneous colloid is heterogeneous 1
In solution the size of particles (solute) is less than 1 nm whereas in colloids the range
of size of particles is 1 – 1000 nm (10 – 9 to 10 – 6 m)(Any one point)
(ii) In homogeneous catalysis the reactant and catalyst are in the same phase whereas in 1
heterogeneous catalysis they are in different phase.
(iii) In O/W emulsion oil is the dispersed phase while in W/O water is dispersed in oil
The O/W type emulsion can be diluted with water whereas the W/O emulsion can’t 1
be diluted with water. (Any one point)
22. p p w M1
0 1
Formula 1 0 1 2
p1 M 2 w1
23.75 mm 23.375 mm 5.0 g 18 g / mol
23.75 mm M 2 95.0 g
5.0 g 18.0 g / mol 23.75 mm 1
M2
95 g 0.375 mm
M2 = 60.0 g/mol 1
23. (i) Concern for students health, Application of knowledge of chemistry to daily life, ½, ½
empathy , caring or any other (Any two)
(ii) Through posters, nukkad natak in community, social media, play in assembly or any
other (Any two) 1
(iii) Tranquilizers are drugs used for treatment of stress or mild and severe mental
½,½
disorders . Eg: equanil (or any other suitable example)
(iv) Aspartame is unstable at cooking temperature. 1
24. a)
(i) The +3 Oxidation state of Bi is more stable than Sb(III) . 1
(ii) Because the electronegativity of Cl is greater than that of I . 1
(iii) Due to decrease in electronegativity and increase in the atomic size. 1
1+1
(b)
OR
i) Due to formation of fumes of HCl or equation
PCl5 + H2O POCl3 + 2HCl
24.
ii) Rhombic sulphur or -Sulphur
1
iii) Due of loss of Chlorine. The yellow colour is due to dissolved Cl2. On
1
standing the Cl2 is consumed in reacting with water to form colourless products:
Cl2 + H2O HOCl + HCl
1
2HOCl 2HCl + O2
iv) 4H3PO3 3H3PO4 + PH3
Oxidation state of P is +3 Oxidation state of P is +5 Oxidation state of P is – 3
1
v) 2F2 + 2H2O 4HF + O2
1
25. 1x5
OR
25.
b) (CH3)3N < CH3NH2< (CH3)2NH
c) Dye Test:
1
On treating with benzene diazonium Chloride at low temperature C6H5-NH2
will form coloured dye while CH3-NH2 will not form.
1
(or any other correct distinguishing test)
26. (a)
2.303 CH 3COOCH 3 1 ½
Formula: k log
t CH 3COOCH 3 2
2.303 0.4M
k1 log 1
20s 0.2M
k1 = 0.03 s – 1
1
2.303 0.4M
k2 log
40s 0.1M
–1
k2 = 0.03 s
Since constant values of rate constants are obtained by applying 1st Order integrated rate ½
law, the reaction is pseudo first order reaction.
total changein concentration ½
(b) Av rate
total changeintime
OR
[CH 3COOCH 3 ] final [CH 3COOCH 3 ]initial
Av rate
Time( f ) Time(i)
0.10 M 0.20 M
Av rate 1
40 Sec 20 Sec
Av rate = 0.0005 M sec – 1 or 5.0 x 10 – 3 mol L – 1 sec – 1 ½
OR
26. a) i) Collision frequency: No of collisions taking place per second per unit volume. 1
ii) Rate Constant: It is the rate of reaction when the concentration of reactants
is unity i.e. 1 M. It is temperature dependent 1
k Ea T2 T1
b) log 2 1
k1 2.303R T1T2
k2 Ea T2 T1
log
k1 2.303R T1T2
Ea 50
log 6
19.147 105000
1
Ea 50
0.7782
19.147 105000
Ea
0.7782 0.00047619
19.147
0.7782 19.147
Ea =31290.44 J
0.00047619 1
Ea = 31.29 kJ/mol
