CHEMISTRY MARKING SCHEME 2015 BLIND SET -56/B. Value points

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Claude Houston
5 years ago
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1 Q ue s. CHEMISTRY MARKING SCHEME 05 BLIND SET -56/B Value points With increase in temperature extent of adsorption decreases. Marks (Either name or structure) 3 Phosphorus is already in its highest oxidation state (+5) in H 3 PO 4 whereas in H 3 PO 3, phosphorus is in its intermediate oxidation state, can be oxidized as well as reduced. 4 Because of non- involvement of d electrons / Due to small splitting energy gap, electrons are not forced to pairup. 5 Diazotization 6 Cationic vacancies are formed/ non-stoichiometric defect/ Impurity defect. 7 Density = MZ a 3 N A a 3 = MZ N A 99 g mol - x 4 = 3.4 g cm 3 x 6.03 x0 3 mol - = 3.4 x 0-3 cm 3 a = 5.78 x 0-8 cm 7 Insulators Semiconductors. Large energy gap occurs between Small energy gap occurs between valence valence band and conduction band band and conduction band.temperature has no effect. Conduction increases with temperature. 8 K 4 [Fe(CN) 6 ] 4 K + + [Fe(CN) 6 ] 4- Initial Conc. 0 0 Conc. After dissociation α 4α α Total no.of moles after dissociation = - α + 4 α + α = + 4 α Van t Hoff s factor = No. of moles after dissociation No. of moles before dissociation i = ( +4 α) / = +4 α α = 50 / 00 = 0.5 i = + 4 x 0.5 = (a)saline water contains many electrolytes which favour formation of more no. of electrochemical cells. (b) Mg acts as sacrificial anode / Mg is more reactive than iron/ cathodic protection/ Mg prevents,

2 the oxidation of steel.. 0 Sulphur; Due to small size and greater inter electronic repulsions in oxygen., Molality of the solution,m = Moles of solute / Mass of solvent in Kg = 5 / / 000 T f = K f m; K f = T f / m K f = (73.5 7)K x 95 x 34 5 x 000 K f = 3.97 K kg mol - Molality of the glucose = (5/80) / (95 / 000) T f = K f m = (3.97 x 5000) / (95 x 80) = 4.08 K f.p of glucose = = K E cell = E o cell V log [Mg + ] n [Cu + ] =.7 (0.059/) V log(0./ 0.0) = V log (0) = =.684 V If the conc. of Mg + ions increases, E cell deceases If the conc. of Cu + ions increases, E cell increases (or by any other method) 3 (a)depending upon size of the particles. (b)it causes coagulation of the colloidal particles of cloud. The colloidal particles of clay get coagulated by the ions of the electrolytes. 4 (a)the impurities are more soluble in the melt than in solid state of the metal, eg., Germanium/Silicon. (b)low melting metals with high melting impurities are heated, and made to flow on sloping surface. eg-tin, lead 4 Calcination: Heating of concentrated ore in limited supply or absence of air or oxygen Heat Eg., ZnCO 3 (s) ZnO(s) + CO (g) Roasting : Heating of concentrated ore in presence of air or oxygen Eg., ZnS(s) + 3O (g) heat ZnO(s) + SO (g) (or any other correct example) 5 (a)because O-H bond is stronger than S-H bond / Due to strong H-bonding in water. (b)bi +3 more stable due to inert pair effect, so Bi +5 gets reduced to +3 state. Due to its affinity for water. 6 (a) 5 (b) [Co(NH 3 ) 6 ] (SO 4 ) 3, octahedral. (c) It forms a Copper complex, not having free Cu + ions 7 (a) Due to resonance the C-Cl bond develops double bond character / The C in Chlorobenzene is in sp hybridised state but it is in sp 3 state in chloromethane (b) (CH 3 ) CHCl < CH 3 CH Cl < CH 3 Cl < CH 3 Br (c) -Chloro-6-methylcyclohexene,,,

3 8 (CH 3 CO) O C H 5 OH CH 3 COOH + CH 3 COOC H 5 A B C CH 3 COOC H 5 HO / H+ CH 3 COOH + C H 5 OH D CH 3 COOH Ca(OH) / Heat CH 3 COCH 3 E (Note: award full marks if the identification of the compounds is correct) 9 (a) (i) CH 3 CHO dil NaOH CH 3 CH(OH)CH CHO (ii) R C=O + 4(H) Zn-Hg/ Conc,HCl R CH + H (b) CH 3 CH CH 3 < CH 3 CHO < CH 3 CH OH 0 (a) (i) Due to resonance in the benzene ring the lone pair of nitrogen gets delocalized in aniline (ii) Methylamine forms water soluble complex with Ag + ions (b) NH 3 < R 3 N < RNH < R NH (a) H 3 N + -CH -COO - (b) COOH-(CHOH) 4 -COOH (c) Denaturation of albumin occurs, water soluble globular protein gets converted to water insoluble fibrous protein which absorbs the water. (a) Buna-S < Polythene < Nylon-66 (b) CH =CH-CH=CH and C 6 H 5 CH=CH (either name or structure) (c) Thermoplastic polymer 3 (a) Synthetic detergents work even in hard water (b) C 7 H 35 COONa + Ca + (aq) (C 7 H 35 COO) Ca (s) + Na + (aq) (or any other correct reaction of soap) (c) General awareness, use of knowledge of chemistry, helping,caring, social concern. 4 (a)(i) First order (ii) Due to low atmospheric pressure, water boils at low temperature. (iii) k = Ae -Ea /RT, if E a = 0 then k =A, so the rate constant does not depend on temperature. (b)rate(r) = k [A][B] (i)rate(r ) = k[a][b] (R ) = 9R, so the rate increases 9 times. (ii) R = k[a][b] R = 8R, rate increases 8 times 4 (a) (i)rate = k(x) n ; 3Rate = k(7x) n Solving the two n = /3, so order of reaction = /3 (ii)rate = k[a] 0 [B] 0 = k (iii)the activation energy for combustion of fuels is generally very high, and not achieved at room temperature. (b) t =.303 log [R 0 ] k [R] t 3/4 =.303. log [R 0 ].54 x 0-3 ¼[R 0 ], =.303 log 4.54 x0-3 = 5.46 x 0 s 3

4 5 5 6 a) MnO + 4KOH + O Fusion K MnO 4 + H O (A) (B) K MnO 4 + H O + (O) KMnO 4 + KOH ( C) 3MnO 4 + 4H + - MnO 4 + MnO + H O (award full marks for identification only) (b)(i)electronic configuration of M = (Ar) 3d 7 4s Magnetic moment of M + = n(n+) = 3(3+) = 3.87 BM (ii) Metal-metal interactions are strong in Cr due to large no of unpaired d-orbital electrons, but in Hg no unpaired d-orbital electrons hence metal-metal interactions are weak. (a) Cr + changes to Cr 3+ with stable t 3 g configuration, but Mn 3+ changes to Mn + with stable Half filled d 5 configuration. (b) Mn + with d 5 stable configuration has high third I.E, whereas Fe + with d 6 configuration loses electron easily / Mn + is more stable than Mn 3+ whereas Fe 3+ is more stable than Fe +. (c) Dichromate ion and chromate ion are interconvertible with change in ph / Cr O 7 + H O CrO 4 + H + Orange Yellow ph< 7 ph > 7 (d) This is due to relatively poor shielding effect of 5f electrons as compared to 4f. (e) Ti (III) is less stable than Ti (IV) is more stable. a) Step : b) i) C 6 H 5 OH Zn dust / Heat C 6 H 6 CH 3 Cl / Anhy AlCl 3 C 6 H 5 CH 3 ii) C 6 H 5 NH NaNO +HCl C 6 H 5 N Cl H O / H + C 6 H 5 OH C (or by any other correct method) c) bromo-3-methylbut-en--ol 4

5 6. a) (i) Due to electron withdrawing effect of NO group (ii) Due to resonance, C-O bond in phenol acquires a partial double bond character. In ethanol, resonance is not possible / carbon in phenol is sp hybridised whereas in ethanol it is sp 3 hybridised. b) i) Add neutral FeCl 3 to both the compounds. Phenol gives violet complex whereas ethanol does not. ii) Heat both the compounds with I and NaOH. Propan-ol gives yellow ppt of iodoform whereas methanol does not. (or any other correct distinguishing test) c) 3-Phenoxyheptane 5

INDIAN SCHOOL MUSCAT THIRD PRELIMINARY EXAMINATION CHEMISTRY ANSWER KEY

Roll Number Code Number 4 / INDIAN SCHOOL MUSCAT THIRD PRELIMINARY EXAMINATION CHEMISTRY ANSWER KEY CLASS: XII Sub. Code: 04 08.0.018 1. P type semiconductor 1. 1. (C H 5 ) N > (C H 5 ) NH > C H 5 NH >