G.C.E. (A. L.) Support seminar Chemistry - Paper I Answer Guide

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1 Question Number G..E. (A. L.) Support seminar - 04 hemistry - Paper I Answer Guide Answer Question Number Answer () 3 (6) () 5 (7) 5 (3) (8) 4 (4) (9) (5) 3 (30) 3 (6) 3 (3) 5 (7) 4 (3) 3 (8) 5 (33) (9) (34) 5 (0) 5 (35) () 3 (36) 3 () 3 (37) 5 (3) (38) (4) 3 (39) (5) 4 (40) (6) 4 (4) (7) 3 (4) 5 (8) (43) (9) 5 (44) 4 (0) (45) 5 () (46) 3 () 4 (47) 4 (3) (48) (4) 5 (49) 4 (5) (50) [see page

2 - - G..E. (A.L.) Support seminar - 04 hemistry - Paper II Answer Guide. (a) (i) e < < l < Mg (ii) Li (iii) chlorine (iv) 4 < < + (v) N < N < N 3 (4 5 0 marks) (b) (i) (+) (+) a b N (0 marks) () c N () () (ii) (+) (+) () (+) (+) N () As the charge distribution is equal, stabilities are similar to each other. But, stability decreases due to the presence of minus charges on more electronegative N and atoms. (5 3 5 marks) (iii) / arbon monoxide (0 marks) (c) Atom ybridization Electron geometry around the atom a SP 3 Tetrahedral N a SP Trigonal planar SP Linear (3 6 8 marks) (d) (i) Electron configuration Zn 4s 3d 0 o 4s 3d 7 onfiguration of Zn is relatively more stable. Less tendency for the de-localization of electrons. Strength of the metallic bond is low. ( 5 0 marks) (ii) has intermolecular bonds. But has only permanent dipole-permanent dipole attractions, which is relatively less stronger. (3 3 9 marks) (Total marks 9) [see page 3

3 (a) (i) A - Na B - Mg (ii) - Na E - Mg (6 marks) (iii) $ydrogen (04 marks) (iv) 4 Na + Na Na + Excess Na (v) Mg (s) + dil. S 4 (aq) MgS 4 (aq) + (g) Mg (s) + conc. S 4 (aq MgS 4(aq) + S + (l) (04 marks) (vi) Nal (aq), Nal (aq), (l) ( 3 6 marks) (vii) Na - strongly basic Mg() - basic Al () 3 - amphoteric Si 3 - very weakly acidic 3 P 4 - weakly acidic S 4 - strongly acidic l 4 - very strongly acidic ( 3 6 marks) (6 6) + (7 ) marks (b) (i) o (05 marks) (ii) Is s p 6 3s 3p 6 3d 7 4s (0 marks) (iii) + II o (0 marks) + III o 3 (0 marks) (iv) [o( ) 6 ] + (v) (a) o(), [o() ( ) 4 ] - blue precipitate [o(n 3 ) 6 ] + - yellow brown solution (06 marks) (05 marks) (05 marks) (b) [o( ) 6 ] + + o() + 6 / [o( ) 6 ] + + [o( ) ( ) 4 ] + (05 marks) o() + 6 N 3 [o(n 3 ) 6 ] + + (05 marks) (vi) olour intensity increases. / Yellow brown turns to orange brown. [o(n 3 ) 6 ] + complex oxidizes into [o(n 3 ) 6 ] 3+ ' (05 marks) (05 marks) 3. (a) (i) By PV nrt. V la. P la n T RT ; n T P la RT (ii) For the first instance, P la n T RT n T P la / RT For the second instance, P la n T RT n T P la / RT As the total number of gas moles does not change, P la P la RT RT P P [see page 4

4 - 4 - (iii) Fraction of molecules with a given energy T T Speed / ms - ^4.0 marks& (b) Vapour pressure P B P AB P A Vapour pressure P A Vapour pressure P A P P A P A Vapour pressure P X A 0 X B P B diagram () diagram () X A X c (v) Mixture AB is negatively deviated from the ideal condition whereas the mixture A is positively deviated from the ideal condition. ^6.0 marks& 4. (a) (i) N (ii) (iii) F 5 N B E Y N N (iv) N A D X (8 3 4) marks [see page 5

5 - 5 - (b) (i) (4 marks) (ii) (4 4 6) marks (iii) Reaction 3 Reaction type E A E ( 3 6) marks 30 marks (c) (i) First step 3 +8 Br : Br : 3 (7 4) marks (ii) - Br is a primary alkyl halide' + It is formed by the carbocation ' As it is a primary carbocation, it is unstable. Breaking of the - Br bond, as well as the formation of the bond with the nucleophyle ( takes place at the same time, it is a single step reaction. (0) Br : 3 + Br (05) But, as ( ) 3 + is a tertiary carbocation, it is stable. (0) 30 marks 5. (a) (i) PV nrt P m 3 0.5mol 8.34JK mol 700K P Nm (ii) PV nrt Nm m 3 n 8.34JK mol 700K n mol [see page 6

6 - 6 - (iii) (g) + X (g) X (g) (x 0.08) + ( ) + (0.08) x 0.55 mol (iv) [ X (g) ] K [X (g) ] [ ] p (g) (v) K c ( P X (g) P (g) ) mol dm (.5 ) (.5 ) 4.7 mol dm 3 P X (g) [ X (g) ] [X (g) ] [ (g) ] mol dm 3 mol dm 3 mol dm 3 ( 0.08 ) K p Pa P ( ) ( 0. X (g) ( ) 8.5 ) P ( ) 05 (g) Pa P ( X(g) ) Pa Pa Pa (b) (i) K w [ + (aq) ] [ (aq) ] 5a - ^60 marks& (ii) log K w log {[ + ] (aq) [ ]} (aq) log [ + ] log (aq) [ ] (aq) pkw p + p (iii) + - B (aq) B (aq) + (aq) (iv) K b [B + (aq) ][ (aq) ] [B (aq) ] [ K b ] (aq) [ ] / (K (aq) b ) log[ ] log K log (aq) b p pkb log 4 p pkb log p 4 pkb + log P 4 + log ( 0 5 mol dm 3 ) + log 0. mol dm ^30) [see page 7

7 - 7 - (v) Bl + Na B + Nal Na + l Nal + B + l Bl + (3 4 ) marks (vi) (I) p } ph n } Me V l ( 6 ) marks (II) Number of l moles reacted with Na Number of l moles reacted with B Na moles in X } mol } (35 5) 0 3 mol ( ) mol moldm moldm 3 5 (8 6) marks 5b - ^90 marks& 6. (a) (i) i. Substances which are completely dissociated into ions in the aqueous solution. or Substances in which ions exist in its solid, liquid and in the aqueous solution. ii. Ion concentration is very high in the saturated solutions of such compounds. There are attractive forces between the ions. (Ions are not independent on each other). [see page 8

8 - 8 - (ii) Ú First, Ag r 4 (s) dissociates. Ú At a particular instance, the rate of the forward and backward reactions becomes equal to each other. Ag r 4 (s) Ag + (aq) + r 4 (aq) ^0) Ú From that instant, concentration of ions remain constant, the ionic product, [Ag + (aq)] [r 4- (aq)] is equal to K SP of the compound at the given temperature. When N 3 is added, Ú As the equilibrium,ag + (aq) +N 3(aq) [Ag(N 3 ) ]+ (aq) () reaches, Ag + ion concentration decreases. There, according to Le hatterlier's principle, as the reaction () above, gets shifted towards the products, [r 4 (aq)] increases. (iii) Ag r 4 (s) equlibrium concentration mol dm 3 Ag + + r (aq) (aq) ( 0 0) marks K SP [Ag + (aq) ] [r 4(aq) ] ( 0 4 mol dm 3 ) ( 0 4 mol dm 3 ) 4 0 mol 3 dm 9 ^0 marks& (iv) New [r4 ] ( ) mol dm 3 mol dm 3 / [Ag + ] K sp (aq) [r 4 (aq) ( / 4 0 mol 3 dm 9 ) moldm moldm 3 Ag r 4 moles precipitated ( ) mol 0 4 mol ^0 marks& (v) K [Ag(N 3 ) + (aq)] [Ag + (aq) ] [N 3 (aq) ] 0 8 dm 6 mol ( 0 4 ) mol dm 3 ( 0 6 mold m 6 ) [N 3 (aq) ] [N 3 (aq) ] Total N 3 moles 0 3 mol dm 3 ( 0 3 ) + ( 0 4 ) mol mol ^0 marks& [see page 9

9 - 9 - (b) (i) (ii) I. ( ) (0 + 89) kj mol 99 kj mol () II. S ( 7 + 7) (4 37)JK mol 4 JK mol () III. G S T 99 kj mol 4 JK mol 98 K 99 kj mol kjmol ( )kkJmol 3.48 kjmol () (+) ve value obtained for G. The reaction given, is not spontaneous. According to that, X should act as the oxidizing agent and Y should act as the reducing agent. Therefore, Y is above X in the electro-chemical series. (0) 7' (a) (i) (iii) I. Anode - Y athode X II. Anode Y(s) Y + (aq) + e III. athode X + + e (aq) X(s) IV. Y(s) Y + (aq, moldm 3 ) X + (aq, moldm 3 ) / X(s) ^4 marks& 6b - (60 marks) NNa + + Na + (0) (ii) NNa + + NNa + (iii) g + dil. S 4 Na NNa + NNa + PBr 3 Br Br Br Br Br Br / l 4 Br (5 55) marks 7a - (75 marks) [see page 0

10 - 0 - (b) (i) a, h and f (0) (ii) a - acid base h - acid base f - nucleophylic substitution (5) ^5 marks& 7b - (5 marks) (c) (i) N A (i) NaN (ii) dil. l 0-5 NN l UN KN Room temperature N dil. S 4 conc. S 4 (9 4 36) marks B (ii) The compound given below is heated with aqueous Na. NN Na (aq) N + NNa + he resultant solution is treated with dil.l solution. N + dil. l N 3 + l (aq) NNa + + dil.l + Nal The resultant white precipitate is filtered and separated. It is benzoic acid. (4) Alternative answer : N dil. l N 3 + l (aq) + The resultant white precipitate is filtered and separated. It is benzoic acid. 7c - ^50 marks& [see page

11 - - 8' (a) (i) Presence of PbI is confirmed as the dark yellow precipitate is dissolved in water when heated and precipitated back when cooled. As the pink compound dissolves in a acid, it is a basic compound. The species which forms blue with conc.l, is o +. ontain PbI and o(). ( 8) + (4 + 4) 4 marks (ii) [ol 4 ] Tetrachlorocobaltate(II) ion (3 6) marks 8a - 30 marks (b) () The white precipitate of lead, which turns black when heated, is PbS 3. (3 3 9) marks () Filtrate Y contains a reducing agent. That reducing agent is removed with a + ions by forming a precipitate. This precipitate should be a 4. (3 3 9) marks (3) There are a + ions remain in the filtrate Z. As a precipitate was formed on heating, 3 ions have dissociated to 3 ions. a 3 precipitates. (3 3 9) marks - (4) Possible to be N 3 or N. As an oxidation is not possible in the filtrate Z, it can not - be N. ( 4 8) marks Four anions are, 4,S 3, 3 and N 3. (4 4 6) marks 8b - ^50 maximum marks& (c) (A) Determination of the concentration of l Agl mol Ag + mol 0.87g 43.5gmol 0.00mol oncentration of l 0.08moldm 3 ^7 marks& (B) Fe S FeS + S (4) Fes + 7 Fe 3 + 4S (4) (0) S + S + 5S + Mn 4 5S 4 + Mn + (4) Mn 4-4 mol moles Mn 4 remain Mn (4) (4) mol Mn 4 moles reacted with S mol. 0 3 ^3 5 5 marks&. 0 3 mol Fe 3+ mol Fes mol S mol. 0 3 mol 5 3 Fe 3+ concentration mol dm 3 (3 8 4 marks) [see page

12 - - () Determination of the concentration of + : Na mol overall + mol in the filtrate (B) mol + mol from S 3 Fe 3+ mol mol mol + mol in the initial solution ( ) mol concentration of + in the initial solution moldm 3 (3 8 4) marks 9' (a) (i) (i) a(p 4 )/ a(p 4 ) ax (05) (ii) As its water solubility is low. (05) (iii) 3a 3 (P 4 ) ax + 4N 3 3a( P 4 ) + 7a(N 3 ) + X (0) (iv) Due to high water solubility, can be applied for short term crops, too. Those ions are easily absorbed by the roots of plants. (05) (v) Petroliem (g) X P 50 atm T 450 catalyst Fe N (g) fractional distillation air N 3(g) Y P - 9 atm T pt/ Rd catalyst N (g) electrolysis N (g) water Apatite Super phosphate partial N 3 (aq) [see page 3

13 - 3 - (I) A - ((g) B - N (g) - N 3(g) D - N (g) E - N (g) F - Apatite (II) P - Fractional distillation Q - Electrolysis R - Partial acidulation (6 4 4) marks (3 3 09) marks (III) X - 50 atm, 450, Fe - catalyst Y , - 9 atm, atalyst Platinum / Rhodium (6 ) marks 9a - ^70 marks& (b) (i) Anode - Titanium athode - Nickel (3 06) marks (ii) l l (aq) (g) + e ( At the anode ) (04) (l) + e (aq) + (g) (At the cathode) (04) (iii) (iv) As charge removes at the anode and charge is supplied at the cathode / By moving Na + ions from anode to cathodic compartment. (04) As anions are not exchanged through the membrane, it is prevented the moving of ions, by that prevents the collision with l. (04) (v) For the absorption of and other acidic gases ; production of soap, paper, dyes ; refining of water ; removing heavy metals ; removing the acidity of water. (4 08) 9b - ^30 marks& (c) (i) In the photosynthesis, breaking of water molecules by sunlight (photo-lysis) (3 06) (ii) + N 4, N, N 3 (3 06) (iii) Acceptable answers such as, Ú Bringing down the human health Ú Growth retarding in plants Ú Destroying the structures made of marble and metals. Ú Weathering of rocks. Ú Weather changing etc. (3 4 ) (iv). Acids such as sulphuric, nitric etc. dissolve the the substance of alumino-silicates in the soil, and release Al 3+ ions into water. This disturbs the function of the gills in fish.. Acidic rain water which flows through soil, removes nutrients and releases Al 3+ ions. Therefore instead of the essential nutrients such as a + and Mg +, plants absorb Al 3+ ions. (4 08) [see page 4

14 - 4 - (v) zone, aldehyde, peroxy acetyl natrates (PAN), peroxy benzyl natrates (PBN) (3 06) (vi) Reforestation and conservation of forests. Minimizing the combustion of fuels. (3 06) (vii) Tuning the engine Using catalytic converters. (3 06) 9c- ^50 marks& + 0. (a) n+ X X + (5 n) 0 (4) + + I 3 + 0e I + 6 (5 n) (4) (5 n) + + (5 n) I 3 I 3 : X n+ (0 n) : 0 + 0X n+ 0 X + (5 n) I + 6(5 n) (4) (5 n) : 5 () Moles of X n+ consumed mol () I 3 (5 n). 0 3 mol X n mol (4) (5 n) 0 3 () n n n (3) 0a - ^5 marks& (b). P - S Q - S R - r (S 4 ) 3 X - Y - Na S 3 Z - S (6 4 4) marks. (i) 3 S (g) + 4 S 4(aq) + K r 7(aq) K S 4 (aq) + r (S 4 ) 3 (aq) + 3S (s) + 7 (l) (ii) S (g) + excess Na (aq) Na S (s) + (l) (iii) 4S (s) + 6Na (aq) (iv) Na S 3(aq) + l Na S (s) + Na S 3(aq) + 3 Nal (aq) + (l) + S (s) + S (g) (v) S (s) + hot conc.' S 4 (aq) 3S (g) + (l) (5 4 0) marks [see page 5

15 S (g) + (l) S 4 (aq) e X + n + + ne X n flower petals/ coloured dyes colourless compounds (04) 0b - ^50 marks& (c) (i) I R K [A] x [B] Y [] Z (0) II 0 3 mol dm 3 s K (0.05mol dm 3 ) x (0.05 mol dm 3 ) Y (0.05 mol dm 3 ) Z... () mol dm 3 s K (0.5mol dm 3 ) x (0.05 mol dm 3 ) Y (0.05 mol dm 3 ) Z... () mol dm 3 s K (0.5mol dm 3 ) x (0.5 mol dm 3 ) Y (0.05 mol dm 3 ) Z... (3) mol dm 3 s K (0.5mol dm 3 ) x (0.5 mol dm 3 ) Y (0.05 mol dm 3 ) Z... (4) ^5 4 0) marks 0 3 (0.05) X X (0.5) x (0) (0) y ( ) Y (0) (0) z 0 ( ) Z (0) III R K [A] [B] (0) (ii) I Amount of A consumed 0. mol (0) Amount of B formed 0. mol (0) n (0) As, 0. mol of A is consumed, when 0. mol of B is formed. (0) II K c [P] n (0.4 ) mol dm3 [A] (0.4) mol dm 3 (0) (0) (0) III R K [P] [B] IV K c [P] [A] [P] K c [A] R K K c [A] [B] (0) (0) (0) (0) 0c - ^75 marks& * * *