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Antonia Berry
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1 CHEMISTRY MARKING SCHEME //RU. NO. Value points MARKS Q. H 2 SO 3 H 2 SO 4 H 2 S 2 O 8,H 2 SO 5 ( any two formulae) + Q.2 -ethoxy-2-methylpropane Q.3 Due to coagulation of colloidal clay particles Q.4 CH 3 -CH(Br)-CH 3 Q.5 X 4 Y 3 Q.6 Similarity : Both show contraction in size /Both show irregularity in their electronic configuration/both are stable in +3oxidation state (any one) Difference :Actinoids are mainly radioactive but lanthanoids are not/ Actinoids show wide range of oxidation states but lanthanoids do not /Actinoid contraction is greater than lanthanoid contraction. ( any other one similarity and one difference) Q.7 (i) Pentaamminechloridocobalt(III) ion (ii) K 2 [NiCl 4 ] Q.8 (i) PCC / Cu at 573 K 8. (ii) (i) NH 3, Δ (heat) C 6 H 5 COCH 3 < CH 3 COCH 3 < CH 3 CHO (ii) CH 3 COOH<Cl-CH 2 -COOH < F-CH 2 -COOH Q.9 (i) Negative deviation,temperature will increase. + Q.0 (ii) Blood cell will swell due to osmosis, water enters into the cell. Cu e Cu 63.5 g Cu is deposited = 2x96500 C.27 g Cu is deposited = 2x96500x.27/63.5 C = ixt (Q = ixt) t = 2x96500x.27/63.5 x 2 = 930s Or +

2 by Faraday First law m = zx i xt z = atomic mass/valencyxf.27 = 63.5x2xt/2x96500 t = 930 s Q. p 0 p = w s x Msolvent, s = solute p 0 M s x Wsolvent ( )/32 = 0 x 8/ Ms x 200 M s = 80 g/mol Q.2 (i) Zone refining (ii) SiO 2 act as flux to remove the impurity of Iron oxide (iii) Depressants prevent one type of sulphide ore forming the froth with air bubbles. Q.3 (i) Starch. (ii) α- Helix polypeptide chains are stabilized by intramolecular H-bonding whereas β- pleated sheet is stabilized by intermolecular H-bonding. (or any other difference) (iii) Pernicious anaemia Q.4 (i) Hydration isomerism (ii) Electronic configuration ist 2g 4 / or by diagram (iii) Hybridization is sp 3 d 2 Q.5 (i) and shape is octahedral. + (where X=Br) (ii)

3 iii)ch 3 CH 2 Cl CH 3 CH 2 CH 2 CH 3 5 (I) (ii) CH 3 CH 2 Cl + AgNO 2 CH 3 CH 2 NO 2 + AgCl (iii) CH 3 CH 2 CH 2 CH(Br)CH 3 + KOH (alc.) CH 3 CH 2 CH=CHCH 3 Q.6 (i) Stoichiometric defect (ii) Schottky defect e.g.nacl(or any other example) (iii) Density of crystal decreases Q.7 Λ m = Scm 2 mol - + Λ m = Scm 2 mol - = 20Scm 2 mol - Λ m 0 HCOOH = λ 0 HCOO - + λ 0 H + ( ) S cm 2 mol - = 400 S cm 2 mol - α = Λ m / Λ m 0

4 α = 20 /400 = Q.8 Physisorption : adsorbate is held by weak van der Waals force non-specific It forms multimolecular layer Chemisorption : adsorbate molecules are held by strong forces like a chemical bond It is specific It forms unimolecular layer (or any correct three points) Q.9 (i) Phenoxide ion is stabilized by resonance as compared to CH 3 O - / In phenol, oxygen acquires + ve charge due to resonance and releases H + ion easily whereas there is no resonance in methanol. (ii) Due to lone pair- lone pair repulsion on oxygen.,, Q.20 (iii) (CH 3 ) 3 C + is 3 0 carbo-cation which is more stable than CH 3 + for S N reaction. ++ i) (CH 3 ) 2 C= N-NH 2 ii) / benzoic acid iii) / m-bromobenzoic acid Q.2 (a) (i) Because Cu + undergoes disproportionation as 2Cu + Cu + Cu 2+ (ii) Because of small size of metal, high ionic charge and availability of vacant d orbital. 2- (b) Cr 2 O 7 + 8H + + 3NO - 2 2Cr NO 3 Q.22 (i) ethylene glycol HO-CH 2 -CH 2 -OH + 4H 2 O (Balanced equation only) + Terephthalic acid (ii),3- butadiene CH 2 =CH-CH=CH 2 + Styrene (iii) Chloroprene CH 2 =C(Cl)-CH=CH 2 (Note: Half mark for name/s and half mark for structure/s in each case), Q.23 (i) Social awareness,health conscious, Caring, empathy, concern.(or any other two values),

(ii)oxygen exists as diatomic O 2 molecule while sulphur as polyatomic S 8 (iii)due to non- availability of d orbitals (b) + + i) ii) 24 Q.

5 Q.24 (ii) Cartoon display / street display/poster making (or any other correct answer) (iii) Wrong choice and over dose may be harmful. (iv) Saccharin, Aspartame (or any other example) (a) (i) Due to decrease in bond dissociation enthalpy from HF to HI, there is an increase in acidic character observed. (ii)oxygen exists as diatomic O 2 molecule while sulphur as polyatomic S 8 (iii)due to non- availability of d orbitals (b) + + i) ii) 24 Q.25 (i) White Phosphorus, because it is less stable due to angular strain (ii)nitrogen oxides emitted by supersonic jet planes are responsible for depletion of ozone layer. Or NO+O 3 NO 2 + O 2 (iii)due to small size of F, large inter electronic repulsion / electron- electron repulsion among the lone pairs of fluorine (iv)helium (v) XeF 2 + PF 5 [XeF] + [PF 6 ] -, each A = B= C= D= E=

6 25,, i) ii) iii) (b) C 2 H 5 NH 2 <(C 2 H 5 ) 3 N < (C 2 H 5 ) 2 NH (c) Add CHCl 3 and alc KOH, C 6 H 5 -NH 2 gives foul smell of isocyanide whereas C 6 H 5 -NH-CH 3 does not ( or any other correct test) Q.26 (a) [A] 0 = 0.0 mol/l [A] = 0.05 mol/l at time t = 0s k = log[ A 0 ] t [A] k = log0.0 0 s 0.05 k =0.0693s - t = 20s k = log[ A 0 ] t [A] k = log s k =0.0693s - As the rate constant is same so it follows pseudo first order reaction. (b) Average rate of reaction = - Δ*R+/ Δt = - [ / 20-0] = mol L - s -

7 26 (a) (i) Rate of reaction becomes 4 times (ii) Over all order of reaction = 2 (b) t /2 = k 30min =0.693 k k = 0.023min - k = log [ A 0 ] t [A] t = 2.303log t = 2.303min t = 99.7min

CHEMISTRY MARKING SCHEME /2/RU. NO. Value points MARKS Q.1 Due to coagulation of colloidal clay particles 1 Q.2 X 4 Y 3 1

CHEMISTRY MARKING SCHEME /2/RU. NO. Value points MARKS Q.1 Due to coagulation of colloidal clay particles 1 Q.2 X 4 Y 3 1 CHEMISTRY MARKING SCHEME 205 56/2/RU. NO. Value points MARKS Q. Due to coagulation of colloidal clay particles Q.2 X 4 Y 3 Q.3 H 2 SO 3 H 2 SO 4 H 2 S 2 O 8,H 2 SO 5 ( any two formulae) + Q.4 -ethoxy-2-methylpropane