INDIAN SCHOOL MUSCAT THIRD PRELIMINARY EXAMINATION CHEMISTRY ANSWER KEY

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1 Roll Number Code Number / 1 INDIAN SCHOOL MUSCAT THIRD PRELIMINARY EXAMINATION CHEMISTRY ANSWER KEY CLASS: XII Sub. Code: Ferromagnetism (C H 5 ) N > (C H 5 ) NH > C H 5 NH > NH i) The solubility of a gas in a liquid is directly proportional to the pressure of the gas at constant temperature. It is the elevation in boiling point when molality of the solution is unity. 7. i) Due to unpaired electrons in d-orbital and undergoes d-d transition. Oxygen and fluorine are strong oxidizing agents, highly electronegative and small size. i) It oxidizes iodides to iodates - MnO + H O + I - MnO + OH IO

2 It oxidizes iodide ion to iodine. MnO H I - Mn + + 8H O + 5I 8. i) Movement of colloidal particles towards oppositely charged electrodes under the influence of an electric field. It is a process of removal of soluble impurities from the colloid by means of diffusionthrough a semi- permeable membrane. 9. i) One(½ x ) S -1 i t 1/ = 0.69/k iv) slope = -k 10. i) Coagulation takes place due to neutralisation of charges. The path of the light gets illuminated due to the scattering of light by the colloidal particles. 11. i) 1. Z = Z = d x a x N 0 M 7. x (89 x ) x 6.0 x 10 5 Z = (bcc) (½ + ½ + ½) r = a = 1.7 = 15.1pm x 89 (½ + ½ + ½) MnO + KOH + O K MnO + H O MnO - + H + MnO - + MnO + H O MnO + 5Fe + + 8H + Mn + + 5Fe + + H O (or any other reaction) 1. i) It is due to the symmetry of para-isomers that fits in the crystal better as compared to ortho and meta-isomers. Resonance effect / Difference in hybridization of carbon atom in C-X bond / Instability of phenyl cation / because of the repulsion, it is less likely for the electron rich nucleophile to approach electron rich arenes. i Alkoxide ion present in alcoholic KOH, is not only a strong nucleophile but also a strong base. 1. i = (½)

3 ΔT f = ik f x W B x 1000 (½) W A x M B ΔT f = x 1.86 x 7 x 1000 (½) 100 x Δ T f = 1.1 K (1) Freezing point of solution = = K (½) 15. NH Br + i) CH COOH CH CONH CH NH + K CO + KBr + H O KOH Sn/HCl CH Cl CH NO CH NH CH NHCH Conc i Cl ClCl HSO + conc HNO Sn/HCl NO NH 16. i) van- Arkel method Reducing agent i Process of extraction of metal from the ore by reducing it at high temperature k =.0 log [Ao] (½ + ½ + ½) t [A] k =.0 log k =.0 x = min -1 0 t 1/= 0.69 (½ + ½ + ½) k

4 t 1/ = t 1/ = 77.7min i) 1.5 x 10 - = k(0.1) x (0.1) y (1).0 x 10 - = k(0.) x (0.) y () 6.0 x 10 - = k(0.) x (0.) y () Divide () by () 1 = 1 y = y y = 1 (½ ) Divide (1) by () 1 = 1 x 1 y x y x = x = 0 x = 0 (½ ) rate law = k[a] 0 [B] 1 (1) 1.5 x 10 - = k(0.1) 1 (1) k = 1.5 x 10 - min i) Phospho diester linkage β- D-Glucose and β- D-Galactose i a) n- hexane b) Saccharic acid 0. i) [PtNH BrClNO ] 1 Because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other. i e g t g 0 1. i) The electron arrangement is trigonal bipyramidal. The shape is linear because the lone pairs prefer the equatorial positions. The moleculexef has lone-pairs and

5 bond-pairs. Low bond dissociation enthalpy and high hydration enthalpy of fluorine. i Because of strong bond strength of P P than N N bond as there is repulsion between the valence electrons of two nitrogen atom due to its small size.. i) Addition polymers: Polyvinyl chloride, Polythene. (½ + ½) Condensation polymers: Terylene, Bakelite. (½ + ½) Buna- N: 1,-Butadiene + Acrylonitrile. (½ + ½) Buna -S: 1,-Butadiene + Styrene. (½ + ½). i) Stable only at low temperature/ unstable at cooking temperature. ( ) Its sweetness is difficult to be controlled. i Sucralose, stable at cooking temperature and does not provide calories. iv) Protect themselves from diabetics and heart ailments. i) a) 5 b) a) Because sulphur is sterically protected by six F atoms b) Bond dissociation enthalpy of F is lower than that of Cl involved in the process. c) Bond dissociation enthalpy of HCl is lower than that of HF i) N + H NH (½) Catalyst iron (½) Promoter - K O + Al O (½) Conditions: low temperature / 700 K and high pressure (½) a) NH Cl (aq.)+ NaNO (aq. ) N (g) +H O(l) +NaCl(aq.) b) P + NaOH + H O NaH PO + PH i

6 H SO is a very strong acid in water because of its first ionisation to H O + and HSO -. The ionization of HSO - to H O + and SO - is very small (it is difficult to remove a proton from a negatively charged ion). 5. i) a) The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolytic solution. b) Molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of contribution from individual ions. (1+1) Cell constant = G* = 1cm -1 5 K = 1/R x l/a ( ½ ) = 1/00 = 5 x 10 - Scm -1 ( ½ + ½) λ = 1000 x K ( ½ ) C = 5 x 10 - x 1000 ( ½ ) 0.01 = 500 Scm /mol (1) i) a) It is the conducting power of all the ions produced by dissolving 1gm mole of the electrolyte in solution. (1) b) Cells that can be recharged by passing direct current through it. (1) Ecell = E 0 cell log [Sn + ] [Zn + ] (½) n [Sn + ] = log 0.5 x (½) 1.5 = V (½) If [Sn + ] is increased, E cell will increase. (½) 6. i) A = CH CHOHCH CH CH B = CH COCH CH CH C = CH CH CH CH CH [O Zn- CH CHOHCH CH CH CH COCH CH CH HCl 5 CH CH CH CH CH a) As Cl acts as electron withdrawing group/ CH shows +I effect. b) The carbonyl carbon atom in carboxylic acid is resonance stabalised.

7 i) Di-tert-butyl ketone < Acetone < Acetaldehyde a) Add neutral FeCl to both - Phenol gives a violet colour b) Add NaOH + I to both. - Acetaldehyde yellow ppt of iodoform i a) b)

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 >