SRJC 2008 JC2 Prelim H2 Paper1 Solutions

Solution H2 MCQ Prelim 2008 1) Solution: B Amount of H3PO3 used in the experiment = 30 1000 x 0.05 = 1.50 x 10-3 mol Amount of NaOH neutralised = 15 1000 x 0.2 = 3 x 10-3 mol H3PO3 ≡ 2 NaOH 2 mol of NaOH neutralizes 2 mol of H + Ö 1 mol of H + is left (since H3PO3 ≡ 3H+) Ö Salt formed will contain 1 mol of H + 2) Solution: C Large jump in 2nd Ionisation energy, 2nd electron removed is found in inner shell. G has 1 valence electron; G is a Group I element. Thus, the oxide of G has the formula: G 2O 3) Solution: A Using pV = nRT, At constant T, pV = constant, so graph should be a straight line at all density 4) Solution: B O H OC H C H N H C OH O I I I I H H Tetrahedral about C Bent about O Trigonal planar about C Trigonal pyramidal about N 5) Solution: C To show an energy profile diagram with two activation energies and the first activational energy is greater than the 2nd one as the 1st step is the slow step.

6) Solution: B Kc = [] [] [ ]2 2 2 I H HI no units 7) Solution: B Initial pH of 13 indicated that a strong base was used as it dissociates fully to give OH -. The final pH of 2.5 indicates a weak acid buffer region due to presence of CH 3COOH with CH 3COO-Na+. The end point is basic due to hydrolysis of the salt CH3COO-Na+. CH3COO-Na+ (s) Æ CH3COO- (aq) + Na+ (aq) CH 3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq) 8) Solution: B Let [Ba 2+] be w mol dm-3 , [L-] will be 2w mol dm-3 q = (w) (2w) 2 q= 4w3 w = 31 4 q ⎟ ⎠ ⎞⎜ ⎝ ⎛ [L-] = 31 4 q 2 ⎟ ⎠ ⎞⎜ ⎝ ⎛ 9) Solution: D There are 5 C* and number of optical isomers = 25 = 32 10) Solution: C KMnO 4 will cleave the 2 double bonds present as well as oxidised the secondary alcohol to ketone CH 3 CH 2CH=C(CH 3)2 HO Cleave the double bonds 2oROH will be oxidised to ketone

11) Solution: C 2 “OD” will be added across the 2 C=C in Linalo-ol C CH3 H3C OD C H Br C H C H H H C CH 3 OH C H C OD H H Br 12) Solution: A CH3OCH2CH2Br Æ CH3OCH2CH2NH2 (nucleophilic substitution by ammonia) CH 3OCH2CN Æ CH3OCH2CH2NH2 (reduction of nitriles) 13) Solution: B Only ethanol undergoes oxidation with K2Cr2O7 in dilute H2SO4 whereas phenol does not. Orange K2Cr2O7 will turn green. Both ethanol and phenol will react with alkaline aqueous iodine and decolourised the brown iodine. Ethanol und ergoes iodoform test with alkaline aqueous iodine. Phenol undergoes easy electrophilic substitution with alkaline aqueous iodine. 14) Solution: D CH(CH COCH3 3)OH 2oROH which can be oxidised by acidified KMnO4 to ketone Chiral C resulting in optical isomers No aldehyde group present to react with Tollen’s reagent. 15) Solution: C C6H5CH2CN undergoes acidic hydrolysis to give C6H5CH2COOH C6H5CH2CH2OH undergoes oxidation to give C6H5CH2COOH

16) Solution: B C3H7CO2H + C2H5OH Æ C3H7CO OC2H5 OH-(aq) K2Cr2O7/H+ Tollen’s reagent Acid Alcohol 17) Solution: C C 3H7Br X Y Z + Silver mirror C3H7Br undergoes nucleophilic substitution with OH- to form X, CH3CH2CH2OH is a primary alcohol. CH 3CH2CH2OH undergoes mild oxidation with K2Cr2O7 to form Y, CH3CH2CHO which is an aldehyde. CH 3CH2CHO, an aldehyde undergoes oxidation with Tollen’s reagent to form Z and silver mirror. Z is CH3CH2COOH 18) Solution: B Propanone gives positive iodoform test as it contain the group C O CH3 19) Solution: B H2NCC O OH H CH2 C OC l H2NCC O OH H CH2 CHO H CH2 CH2 NH2 2 1 3

2 compounds can be formed; amide bonds are formed between the following functional groups: 1 and 2 as well as 1 and 3. 20) Solution: B 2Al3+(g) + 3O2-(g) 3(EA(O)) 2Al3+(g) + 3O(g) + 6e 2(3rd IE(Al)) 2Al 2+(g) + 3O(g) + 4e 2(2nd IE(Al)) 2Al+(g) + 3O(g) + 2e 2(1st IE(Al)) 2Al(g) + 3O(g) ∆Hlatt ө(Al2O3) 3∆Hat ө(O) 2Al(g) + 3/2O2(g) 2∆Hat ө(Al) 2Al(s) + 3/2O2(g) ∆Hf ө(Al2O3) Al2O3(s) ∆Hat ө(Al) = ∆Hrxn ө(Al3+) – 1st IE(Al) – 2nd IE(Al) – 3rd IE(Al) = 5467 – 577 – 1820 – 2740 = +330 kJ mol-1 Total amount of energy required to form Al3+ and O2- ions = 2 (5467) + 3 (897) = 13625 kJ mol -1 Lattice energy of aluminium oxide is definitely more exothermic than –13625 kJ mol-1 as the question has already stated that aluminium oxide is a stable compound. 21) Solution: B ∆Gө f (O3) = ∆Hө f (O3) – T∆Sө (O3) = 142.67 – 298(-68.7/1000) = +163.14 kJ mol -1

22) Solution: C Q = I x t = 0.15 x 3 x 60 x 60 2e + Pb2+ Æ Pb 2 Faradays are required nPb = 2 96500 60 60 3 15 . 0 × ×× × Î Mass of Pb = 2 96500 60 60 3 15 . 0 207 × × ××× 23) Solution: D Basic Amphoteric Acidic Na2O, MgO Al2O3 SiO2 , P4O10, SO3 24) Solution: C From Figure 1, since the greatest increase in IE occurs between the 1 st and 2nd IE, U is in group I. From Figure 2, A-B have low boiling point s, hence they are molecular structure with weak intermolecular van der Waals’ forces of attraction. C is the start of next period and hence is group I with has giant metallic structure and strong metallic bonds, hence high boiling point. Therefore, C corresponds to element U. 25) Solution: C Ca(OH)2 is less soluble than Ba(OH)2 due to its higher lattice energy. 26) Solution: D 2nd IE decreases down the group due to dec reasing effective nuclear charge. The solubility of sulphates decreases down the group as decrease in hydration energy is more significant than the decr ease in the lattice energy. The pH of oxides in solution increases down the group due to decreasing polarizing power of the cations due to larger ionic radius.

27) Solution: B The graph shows the boiling point of the HX down the group. HF has the highest boiling point due to intermolecular hydrogen bonds present. For the rest of the HX, they are held by weak inte rmolecular Van der Waals’ forces of attraction. Strength of Van der Waals’ is proportional to the size of the molecule, hence boiling point increases down the group. 28) Solution: D The top layer is the aqueous layer of sodium chloride and the iodine is found in the bottom layer. (Iodine is purple in organic solvent and brown in aqueous solution due to I3 -) 29) Solution: A Beryllium cannot form 6 coordinated co mplexes as there are no d orbitals present for dative bond to occur. 30) Solution: A The mass of Copper is larger than calcium and copper has a smaller radius than Ca. Hence Cu has higher density than Ca. 31) Solution: D The largest jump in IE is the 5th ionisation, hence its group number is (5 -1 ), Gp IV. It does not have similar chemical properties as Al as it is not from GP III. W is not from GP VI. 32) Solution: B FH F Hydrogen bond Covalent bond

33) Solution: B At low pH, equilibrium position shifts left to form [Ga(H2O)6]3+ Increasing temperature will favour the forward endothermic reaction. Kc remains unchanged when co ncentration of [Ga(H 2O)6]3+ is increased as K c only changes with temperature. 34) Solution: B When [Y] is doubled and the [H +] is doubled, the rate of the experiment increases by 8 times not 6 times. 35) Solution:C O2N NH O Cl Cl OH OH Oxidation of alkyl side chain Basis hydrolysis of amide linkage 36) Solution: C Reaction with (1) will not have any observ able reactions as there is no reaction between NaOH and hexane-1,6-diamine, which are both bases. CH3CH2 C O O-NH4 + + NaOH CH3CH2 C O O-Na+ + NH3 H2O + room temperature H3C C NH2 O H3C C O - Na + O + +NaOH(aq) reflux NH3

37) Solution: A Magnesium ion has a higher charge density t han