Chemistry-79.Phase-5-Subjective-Solutions

FACULTY TRAINING TEST PHASE - V CHEMISTRY SOLUTIONS 1. a) Reaction involved with the ‘Brown Ring’ test is – +4H2SO4 + 6Fe+2  6Fe3+ + 2NO + + 4H2O Fe+2 + NO + 5H2O  [Fe(H2O)5NO]+2 The brown ring is due to the formation of [Fe(H¬2O)5NO]+2 b) Hybridization of Co+3 is d2sp3, shape is octahedral Hybridization of Ni+2 is dsp2 so shape is square planar. Hybridisation of Ni is sp3 so shape is tetrahedral c) The transition metal ion in Cu+2 the blue colour solution is CuSO4. CuSO4 on reaction with H2S gives black precipitate of CuS Cu+2 + H2S  + 2H+ CuS is insoluble in warm aqueous solution of KOH CuSO4 on treatment with KI in weakly acidic medium undergoes following reaction 2Cu+2 + 4I–   + I2 2. a) i) Argentite is Ag2S Ag2S + 4NaCN 2Na[Ag(CN)2] + Na2S 2Na[Ag(CN)2] + Zn 2Ag + Na2Zn(CN)4 ii) + AgBr The developed image is made stable to light by fixing. AgBr + 2Na2S2O3 = Na3[Ag(S2O3)2] + NaBr b) Borax bead test for Cobalt (II) oxide The chemical reaction involved with the borax bead test of Cobalt (II) oxide is Na2B4O7  2NaBO2 + B2O3 Cobalt salts when heated strongly in a bunsen flame give CoO, which reacts with B2O3 to give the blue coloured anhydrous Co(BO2)2. Co - salt CoO + gas CoO + B2O3 Co(BO¬2)2 In the reducing flame the cobaltous meta borate, Co(BO2)2 does not change hence the colour of the bead is blue in the oxidising as well as in the reducing flame. c) XeF2 XeOF2 3. a) Cu2 + e  Cu+, E0 = 0.15, G0 = – nFE0 = – 1  0.15F = – 0.15F Cu2 + e  Cu : E0 = 0.50; G0 = 1  0.50 F = – 0.50 F On adding, Cu2 + 2e  Cu; G0 = – 0.65F = 0.325 volt x = 0.325 volt Further Cu+ + e  Cu E0RP = 0.050 Cu+ – e  Cu++ E0OP = – 0.15 –––––––––––––––––––––––––––– 2Cu+  Cu + Cu++ E0 = 0.35 V Since E0 of above reaction is positive it means G0 = is negative i.e. the process is spontaneous. So Cu+ disproportionates to Cu and Cu++. b) Anode Zn – 2e Zn2+ Cathode Cu2+ 2e Cu ––––––––––––––––––––––––– Zn + Cu2+ Zn2+ + Cu Ecell = E0Cell – = E0¬cell + Comparing with Y = mx + C Intercept = C = E0Cell = 1.1 Ecell = 1.1 + = 1.0705 V 4. In case of nitrogen intramolecular multiplicity is possible leading to completion of octet via  bond formation. Hence octet of N2, gets fulfilled by N  N. In case of P intramolecular multiplicity is ruled out because of longer P – P single bond. Hence P has to complete its octet by undergoing sigma bond formation with other phosphorus. Hence formula is P4. 5. Compound X is BCl3 and Y is B2H6 (% hydrogen is 21.72) 4BCl3(X) + 3Li[AlH4]  3LiCl + 2B2H6 (Y) + 3AlCl3 B2H6 (Y) + 3O2  B2O3 + 3H2O + 80 kcal, 6. Formula of A is [Cr(NH3)4ClBr]Cl + AgNO3  AgCl (white) Formula of B is [Cr(NH3)4Cl2]Br + AgNO3  AgBr (yellow) Cr is in +3 oxidation state i.e. 3d3 system. Therefore the hybridisation of Cr is d2sp3 and spin only magnetic moment. 7. i) ii) iii) SiO2 + 2NaOH  Na2SiO3 + H2O 8. A = ZnS, B = H2S, C = ZnSO4, D = S, E = SO2 9. (1) assumed compound (A) (NH¬4)2Cr2O7 Reactions: N2 + 3Mg (OH)2 + 10. i) The compound (B) reacts with NaCl solution (brine) to give white precipitate (D) soluble in NH4OH so (D) is AgCl ii) Thus, (B) must contain Ag+ ion iii) (B) is obtained from (A) and dilute HNO3, so (B) is Ag. Reactions 11. d = d = d = 2.1805g/cc Observed density is less than calculated density due to some missing ions. Actual number of atoms per unit cell = = 3.969 Missing unit = 4 – 3.969 = 0.031 % missing = = 0.775% 12. a) No. of Meqs. of metal that should be deposited = 30  0.1  3 = 9 9  10–3 =  I = = 0.4825 amp b) ACl(s)  A+(aq) + Cl–(aq) A+ (aq) + H2O AOH (aq) + H+ (aq) At equilibrium C (1 –) C For salts of weak base and strong acid  = antilong (– 4.827) =  M -= 53.5 For CsCl type structure, P = 2.2 =  a = 3.43Å a = 0.866a = 0.866  3.43 =2.97Å = 0.732 on solving = 1.255Å and = 1.715Å 13. A solution containing one mole of HgI2 and two mole of NaI is orange in colour due to the partial solubility of HgI2. On addition of excess of NaI, the colourless complex Na2HgI4 is fomred. 3Na2HgI4 + 2NaOCl + 2H2O  3HgI2 + 2NaCl + 4NaOH + 2NaI3