COMMON PRE-BOARD EXAMINATION CHEMISTRY (Marking Scheme)

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SET 3 Subject Code: 043 COMMON PRE-BOARD EXAMINATION 207-208 CHEMISTRY (Marking Scheme) 2 Colloid scatter light but true solutions do not( or any other valid difference) 3 CH 3 COCH 2 CH(Cl)CH 3 4 CH

1 SET 3 Subject Code: 043 COMMON PRE-BOARD EXAMINATION CHEMISTRY (Marking Scheme) 2 Colloid scatter light but true solutions do not( or any other valid difference) 3 CH 3 COCH 2 CH(Cl)CH 3 4 CH 3 COCH 3 + CH 3 MgBr (hydrolysis) (CH 3 ) 3 COH 5 Presence of F-centres 6 a b + 7 (a) Diethylamine, ethylamine, ammonia, aniline (b) p-toluidine,aniline, p-nitro aniline Primary amine sulphonamide soluble in alkali; secondary amine-sulphonamide insoluble in alkali; tertiary amine does not react with Hinsberg s reagent. RNH 2 + C 6 H 5 SO 2 Cl RNHSO 2 C 6 H 5 + HCl R 2 NH+C 6 H 5 SO 2 Cl R 2 NSO 2 C 6 H 5 + HCl 8 When the A-B interactions are stronger than A-A and B-B interactions, the escaping tendency into the vapour phase decreases.the total vapour pressure of the mixture is lesser than expected from Raoult slaw + Page of 6

2 9 (a) Hexacyanidoochromate (III) (b) Potassium hexacyanidoferrate(ii) 0 (a) order = zero molecularity=2 (b) unit of k = moll - s - (a)retention (b)inversion (c)racemization 2 Π=W 2 RT/M 2 V π=crt C=.8xx0/80= 0.M Π= 0. x x 298 = atm 3 (a) Because RNH - is less stabilized than RO - due to lesser electronegativity of N compared to O (b) Due to hydrogen bonding in primary amines. (c) Due to +I effect of alkyl group in aliphatic amines; availability of lone pairs for protonation. 4 (a) Coagulation or precipitation (b) Peptization resulting in a colloidal sol (c) Electrophoresis takes place (a) Due to weak vanderwaal s forces /due to strong chemical bonds. (b) Settling down of colloidal particles/ converting a precipitate to a colloid by adding electrolytes (c) Reversible /stable/ liquid loving; irreversible / unstable / liquid hating 5 (a) Reimer-Tiemann reaction Page 2 of 6

3 (b) Kolbe s reaction (c) Williamson synthesis 6 t /2 = /k k = 2.303/t x log[r 0 [/[R] k = 2.303/600 x log[00[/[75] ; 2.303/600 x log4/3 = sec - t /2 = 0.693/ = sec. 7 (a) Vapour phase refining: the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. (b) Zone refining: impurities are more soluble in the melt than in the solid state of the metal. (c) Chromatography: different components of amixture are differently adsorbed on an adsorbent. 8 d= zm /a 3 N A z=2; a=? a=3.306 x 0-8 cm r=43.5pm 9 (a) chlorine bleaches due to oxidation, SO 2 bleaches due to reduction. (b) Due to weak dispersion forces (c) Due to 2 P-H bonds in its structure 20 (a) thymine-dna; uracil-rna (b) Amino acids having -NH 2 groups more than COOH groups? (c) Vitamin C. Page 3 of 6

4 2 (a) [Co(NH 3 ) 5 Cl]SO 4 and [Co(NH 3 ) 5 SO 4 ]Cl or any other suitable example (b) Give reasons: (i) Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other. (ii) In case of strong ligands Co 3+ has more stable t 2 g 6 (splitting of d orbital) electronic distribution so Co 2+ is easily oxidized to Co (a) + (b) + (c) Hexamethylene diamine and adipic acid (a) Any two values. (b) They can act as poisons if taken without doctors consultation. (c) Neurologically active drugs / any two + 24 (a) (i) KMnO 4 /KOH(heat ); H+ (oxidation of ethyl benzene) (b) (ii) CrO 2 Cl 2 /CS 2 ;H 3 O + (Etard reaction) (i) (ii) (iii) Page 4 of 6

5 A= C 6 H 5 COCH 3 B=C 6 H 5 CH 2 CH 3 C=C 6 H 5 COOH D and E= C 6 H 5 COONa, CHI 3 25 (a) The decrease in atomic size with increase in atomic no. across the lanthanoid series due to the poor shielding effect of f orbitals. Consequences(any one) : Zr and Hf have identical radii ; separation of lanthanoids becomes difficult + (a) Cr 3+ due to 3d 3 half-filled stable t2 g (b) (i) due to varying degrees of stabilities of different configurations. (ii) high atomization enthalpy; low hydration enthalpy. (iii) 5f orbital has poorer shielding effect than 4f orbitals. (a) (b) (i) 5Fe 2+ +MnO H + Mn H 2 O + 5 Fe (ii) Cr 2 O H I - 2Cr I 2 + 7H 2 O (iii) 2CrO H + 2- Cr 2 O 7 + H 2 O 26 (a) because of higher value of reduction potential. (b) The limiting molar conductivity of an electrolyte is the sum of the individual contributions of the cation and anion of the electrolyte;. used to find the limiting molar conductivity of a weak electrolyte/ degree of dissociation / dissociation constant. (c) E 0 cell = 0.04 V Log Kc = ne 0 /0.059 = 2 x 0.04 /0.059 =.356 Kc = Page 5 of 6

6 (c) The conductivity of one mole of an electrolyte present in the solution. strong electrolytes - molar conductivity increases slowly with dilution weak electrolytes- molar conductivity increases steeply with dilution. (b) E 0 cell = 0.30 V E cell= V Page 6 of 6

COMMON PRE-BOARD EXAMINATION CHEMISTRY

COMMON PRE-BOARD EXAMINATION CHEMISTRY CLASS XII SET 1 Subject Code: 04 COMMON PRE-BOARD EXAMINATION 017-018 CHEMISTRY Time Allowed: hours Maximum Marks: 70 General Instructions: (i) There are 6 questions in all. All questions are compulsory.