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26. (c) :
HC 3
HC 3
–H O
CH 2
3
H
+
CH CH
+ 3 1,2-Methyl
3
shi
CH 3
HC 3 HC 3
HO:
: OH 2
CH 3 H +
CH 3
CH 3
CH 3
3°carbocation
2°carbocation
–H + CH 3
(more stable)
(less stable)
CH 3
:
HC 3
27. (d): From 1 st reaction,
− dB [ ] = k1 [A][B] – k
dt
–1 [C]
From 2 nd reaction, − dB [ ] = k2 [C][B]
dt
∴ Total − dB [ ] = k1 [A][B] – k –1 [C] + k 2 [C][B]
dt
28. (c) : 2Al + Na 2 CO 3 + 3H 2 O 2NaAlO 2 + CO 2 + 3H 2
29. (d) : I 2 being the weakest oxidising agent cannot
displace stronger oxidising agents such as F 2 , Cl 2 or
Br 2 from their salts.
30. (c)
31. (b): Gd(Z = 64) : [Xe]4f 7 5d 1 6s 2
32. (a) : Energy released (∆E) = ∆m × 931.5 MeV
−13
16 . × 10 1
= 0.02 × 931.5 × × × 6.023 × 10 23
4.
184 1000
= 428 × 10 6 kcal/mole = 428 million kcal/mole
33. (a) : Reactions I and II give 2-propanol, i.e.,
I. CH 3 CH
H
CH 2 + H 2 O
Markovnikov’s addition
CH 3 CHOHCH 3
2-Propanol
(i) CH3MgI
II. CH 3 CHO CH
+
3 CH(OH)
(ii) H /HO 2
2-Propanol
CH 3
In contrast, reaction III gives 1-propanol and IV
gives 1,2-propanediol.
III. CH 2 O
(i) C2HMgI
5
(ii) H + /HO 2
C 2 H 5 CH 2 OH
1-Propanol
Neutral
KMnO 4
IV.
O O
1, 2-Propanediol
34. (b): The salt can be Pb(NO 3 ) 2 . The reactions are,
Pb(NO 3 ) 2 + 2NH 4 OH Pb(OH) 2 + 2NH 4 NO 3
White ppt.
Pb(NO 3 ) 2 + 2NaCl PbCl 2 + 2NaNO 3
(dil.) White ppt.
Pb(NO 3 ) 2 + H 2 S PbS + 2HNO 3
Black ppt.
35. (b): All given compounds have cis and trans
isomers but only cis-isomer of [Co(en) 2 Cl 2 ] + will
also have optical isomers (d and l) due to presence
of symmetrical didentate ligand, (en).
36. (a) : IE 1 of Mg is higher than that of Na because of
increased nuclear charge and also than that of Al
because in Mg an s-electron has to be removed while
in Al it is the 3p-electron that has to be removed.
Mg also has stable fully filled configuration. The
IE 1 of Si is, however, higher than those of Mg and
Al because of increased nuclear charge. Thus, the
overall order is Na < Mg > Al < Si.
37. (a)
38. (a) : Let the amount of Na 2 CO 3 present in the
mixture be x g. Na 2 SO 4 will not react with H 2 SO 4 .
Then
N 1 V 1 = N 2 V 2
25× N1
= 01 . × 20
( Solution)
HSO 2 4
N 1 = 2/25
2 x /53×
1000
Normality
25
=
250
2× 250×
53
⇒ x = = 1.06 g
25×
1000
106 . × 100
∴ Percentage of Na 2 CO 3 = = 84.8%
125 .
O
OH
OH
OH
39. (c) :
NaBH 4
O O CH O O
3OH
HIO
HC 3 CH 3 HC CH 4
3 3
CHO
O O
HC 3 CH 3
40. (c)
38 CHEMISTRY TODAY | APRIL ‘17