Chemistry_Today_April_2017_vk_com_stopthepress

O
O
(c) HN C (CH ) C NH
2 2 4 2
(d) NC (CH )
24 CN H/Ni/heat
2
(Nitriles are
reduced to
1° amines)
Br /NaOH 2
(Hofmann
bromamide
reaction)
HN (CH ) NH
2 2 4 2
HNCH
2 2 (CH 24 ) CH2NH2
29. (a) : Let number of a-particles emitted be m and
number of b-particles emitted be n.
Hence, 232 90 Th 82
208 Pb + m 4 2 He + n 0 –1 e ...(i)
On equalising mass numbers on both sides of eq.
(i), we get
232 = 208 + (m × 4) + n × 0 4m = 232 – 208
m = 24 = 6 (number of a-particles emitted)
4
Similarly, on equalising atomic numbers on both
sides of eq. (i), we get
90 = 82 + (m × 2) + [n × (–1)] = 82 + 2m – n
or, 2m – n = 90 – 82 = 8
or, n = 2m – 8 = 2 × 6 – 8 = 4
(number of b-particles emitted)
30. (a, b, c)
31. (a, b) : H 2 S + O 3 H 2 O + S + O 2
PbS + 4O 3 PbSO 4 + 4O 2
32. (a, b, d) 33. (c) 34. (a)
35. (a) : [Cr(NH 3 ) 3 Cl 3 ] gives two geometrical isomers
facial (fac) and meridional (mer).
36. (b): When an octahedral complex contains all the
three bidentate ligands, it shows optical isomerism
because it lacks plane of symmetry.
( 1−
x )
⎛ 1+
x ⎞
+ )( 1−
x)
⎛ 1+
x ⎞ 1+
x
37. (c) : lim ( 1−
x ) lim
x→
⎝
⎜
+ x⎠
⎟ =
1
⎜
x→
⎝ + x⎠
⎟ =
⎛ 1
+ 2x
⎞ ( 1+ x)( 1−
x)
⎛ 1
+ 2x
⎞ + x
lim⎜
lim
x→
⎝ + x
⎟ =
⎠
⎜
x→
⎝ + x⎠ ⎟ 1
=
1 2
1 2
38. (a) : If 0 < x < 1, then
1
x > π/
2 2
x > ⎡ π
x
⎣
⎢
2
= 157 ⎤
. ( app.)
⎦
⎥
π/
⇒ 1+ x> 1+ x 2 > 1 + x
2
1 1 1
⇒
1 + < <
x π/ 2 2
1 + x 1 + x
1 1
1
dx dx dx
⇒ ∫
1 + x
< ∫
1 x
2 <
π/ ∫
1 x
2
0 0
+
0
+
1 −1
1
π
⇒ ⎡⎣ log( 1 + x) ⎤ ⎦ < I