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(Q) [Cr(NH 3 ) 5 (NO 2 )] 3 [Co(NO 2 ) 6 ] 2 3[Cr(NH 3 ) 5 (NO 2 )] 2+ + 2[Co(NO 2 ) 6 ] 3– No. of ions = 5 (R) K[Co(NH 3 ) 2 (NO 2 ) 4 ] K + + [Co(NH 3 ) 2 (NO 2 ) 4 ] – No. of ions = 2 (S) [Cr(NH 3 ) 3 (NO 2 ) 3 ] does not ionise, No. of ions = 0 Order of molar conductivity : (S) < (R) < (P) < (Q) It may be noted that (P) and (R) have same number of ions but charges on ions in (P) is double than that on (R). O – 10. (b) : RCH O NaOH DO 2 R C H R 11. (a) C OH DO 2 H RCH R C RCH + D transfer − →RCOO + RCHOD 2 12. (c) : Smelting is the process of reduction using carbon as a reducing agent. 13. (d): Energy required for the removal of second electron from He-atom 2 = + 13 6 = 136× 2 2 Z . . = 54. 4 eV 2 2 n 1 Hence, the total energy required for the removal of both the electrons = 24.6 + 54.4 = 79.0 eV 14. (b): For the first order reaction, A k = 2 . 303 0 log [ ] t [ A] t If [A] 0 = a, [ A] a a t = − × 15 = 085 . at , = 20min 100 a k = 2 . 303 log 2. 303 20 085 . a = × 0. 0706 20 = 8.13 × 10 –3 min –1 In second case, if [A] 0 = a, [ A] a a t = − × 60 = 040 . a and time, t = ? 100 A Now, t = 2 . 303 0 log [ ] 2. 303 a = k [ A] t 813× 10 − log 3 . 040 . a 2. 303 = × 0. 3979 = 112.7 min −3 813 . × 10 15. (b) HC 3 16. (d): HC 3 CH COOH HC 3 C COOH HC I COOH P/I 4 2 (HVZ reaction) 3 O PO/ 2 5 (–H O) 2 Na (Wurtz reaction) COOH O 17. (b): (A) Sn 2+ + 2e – Sn ; E° = –0.14 V ...(i) Sn 4+ + 2e – Sn 2+ ; E° = 0.13 V ...(ii) On adding equations (i) and (ii), Sn 4+ + 4e – Sn ; ° ° ° nE nE E 1 1+ 2 2 2 0 14 2 0 13 4+ = = ×− ( . ) + × . Sn / Sn n1+ n2 4 = –0.005 V (B) As, E° Sn 4+ /Sn = – 0.005 V E° Sn/Sn 4+ = –E° Sn 4+ /Sn = + 0.005 V (C) Disproportionation reaction : 2Sn 2+ Sn 4+ + Sn E° cell = E° Sn 2+ /Sn – E° Sn 4+ /Sn 2+ = –0.14 – 0.13 = – 0.27 V; the reaction is nonspontaneous. (D) Since E° Sn/Sn 4+ > 0, oxidation of Sn to Sn 4+ will be spontaneous. 18. (b): In the given ionic compounds, the oxidation states are + 2 − 2 + 1 − 2 + 3 −2 AB; AB 2 an d AB 2 3 Higher the oxidation state of A ion, higher will be its lattice energy. Hence, the correct order is as follows : A 2 B 3 > AB > A 2 B 19. (c) : Z for ccp, i.e., fcc = 4 Z× M d = N a A × 3 4× 100 d = − ( 400) × 10 × 6× 10 3 30 23 1000 = ≈10. 42 gcm 96 −3 O 30 4× 100× 10 = 64× 10 × 6× 10 6 23 CHEMISTRY TODAY | APRIL ‘17 15
20. (d) : CH 2 5 alk. KMnO 4 ( A) COOC H 2 5 ( D) Br COOH ( B) CHOH 2 5 H + Br 2/FeBr3 COOH 21. (d) 22. (b): No. of oxide ions per unit cell 1 = × + × = 8 8 1 2 6 4 Number of tetrahedral voids per unit cell = 2 × 4 = 8 Number of X 2+ 20 ions per unit cell = × = 100 8 8 5 Number of octahedral voids per unit cell = 1 × 4 = 4 Number of Y 3+ 50× 4 ions per unit cell = = 2 100 Hence, formula is X 8/5 Y 2 O 4 or X 4 Y 5 O 10 n 23. (a) : p = V RT = w RT M V 32 p CH4 = . 0. 0821× 300 × 16 9 ( C) = 055 . atm 44 p CO2 = . 0. 0821× 300 × = 027 . atm 44 9 P Total = 0.55 + 0.27 = 0.82 atm 24. (c) : COOH COCl CONH 2 Br m-Bromobenzoic acid SOCl 2 Br m-Bromobenzoyl chloride NH 3 Br Br m-Bromobenzamide Br , NaOH 2 27. (a) : Ph Ph CH 3 OH (enol) Hg 2+ /H+ HO 2 CH 3 Ph O CH 3 (etone) k 28. (a) : Path I is feasible since –OH group after I (a) and I (b) activate the benzene ring for nitration. –Cl and two —NO 2 groups deactivate the ring in path II hence, nitration is not possible. Cl OH NO 2 NO 2 (a) NaOH NO 2 2,4-Dinitrochlorobenzene (b) HO+ 3 NO 2 HNO / 3 HSO 2 4 OH ON 2 NO 2 NO 2 Picric acid In path I, due to —NO 2 group C—Cl bond weakens which makes nucleophilic substitution reaction possible. Also —NO 2 (deactivating group) prevents oxidation of 2,4-dinitrophenol when further nitrated. 29. (b) : The value of equilibrium constant of a reaction depends only on temperature and does not depend upon concentration, pressure or presence of catalyst. 30. (a) : The equation representing the decomposition of NH 4 Cl by slaked lime, i.e., Ca(OH) 2 is Ca(OH) 2 + 2NH 4 Cl CaCl 2 + 2NH 3 + 2H 2 O 74 g 107 g 111 g 34 g 36 g From the above equation, 107 g of NH 4 Cl is decomposed by 74 g of Ca(OH) 2 . NH 2 Br m-Bromoaniline 25. (a) 26. (d): A compound is soluble if hydration enthalpy (released) is greater than lattice enthalpy. MPP CLASS XII ANSWER KEY 1. (b) 2. (c) 3. (d) 4. (d) 5. (d) 6. (c) 7. (d) 8. (c) 9 . (c) 10. (a) 11. (d) 12. (c) 13. (a) 14. (b) 15. (c) 16. (c) 17. (d) 18. (b) 19. (c) 20. (c,d) 21. (b,c) 22. (a,c,d) 23. (a,b,c,d) 24. (4) 25. (2) 26. (7) 27. (c) 28. (c) 29. (c) 30. (b) 16 CHEMISTRY TODAY | APRIL ‘17
Page 2 and 3: Volume 26 No. 4 April 2017 Managing
Page 5 and 6: (R) K[Co(NH ) (NO ) 4 ] 1. The volu
Page 7 and 8: 24. m-Bromoaniline can be prepared
Page 9: (a) (b) (c) (d) CH CH C CH CH CH C
Page 13 and 14: 41. (a) : According to Gibbs’-Hel
Page 15 and 16: 5. Compound 'X' C 7 H 8 O, is insol
Page 17 and 18: (c) (d) O HC 2 O O HC 2 O O CH CH C
Page 19 and 20: 4. As 2 S 3 sol carries a negative
Page 21 and 22: 15. Cl CHONa + EtOH 2 5 Heat Possib
Page 23 and 24: (c) If θ = 10°, the charge underg
Page 25 and 26: (c) (d) NH 2 O heat NH HO 2 C CO 2
Page 27 and 28: (c) Dil. HCl followed by reaction w
Page 29 and 30: 33. Which of the following reaction
Page 31 and 32: Balz-Schiemann reaction : tetrauoro
Page 33 and 34: 26. (c) : HC 3 HC 3 -H O CH 2 3 H +
Page 35 and 36: 12. At equimolar concentration of F
Page 37 and 38: (a) 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d
Page 39 and 40: Fraction of M present as M 2+ = 88
Page 41: 26. (b) : According to Gay-Lussac
Page 47 and 48: CONCEPT MAP PERIODICITY IN PROPERTI
Page 49 and 50: Class XII This specially designed c
Page 51 and 52: JEE MAIN / JEE ADVANCED / PETs Only
Page 53 and 54: CHEMISTRY MUSING SOLUTION SET 44 24
Page 55 and 56: Isothermal reversible expansion :
Page 57 and 58: (a) H 2 O/H + , BH 3·THF/H 2 O 2·
Page 59 and 60: 22. H 2 can be obtained from 28. Th
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7. Predict the product for the foll
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7. (d) : Cyclopropane is under seve
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46. If the first and the (2n - 1) t
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1 (a) 6 2 ms − (b) 4 2 (c) 7 m s
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tension, viscous force, weight but
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(a) HN 2 SO 3 H (b) (NH 2 ) 2 CO (c
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SOLUTIONS PAPER-I 1. (a) : The phot
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11. (a, d) : As, F 1 = k 1 x, F 2 =
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Given that, at equilibrium, ⎛ n
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2 ∫ 53. (3) : (| x− 2| + [ x])
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percentage change in v = 1 2 perce
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Again, c 2 a 2 b 2 = + + 2a⋅b
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