A complex of iron and cyanide ions is 100 % ionised at 1 m (molal). If its elevation in b.p. is 2.08. Then the complex is (Kb = 0.52° mol-1 kg) :

Question

A complex of iron and cyanide ions is 100 % ionised at 1 m (molal). If its elevation in b.p. is 2.08. Then the complex is (Kb = 0.52° mol-1 kg) : (A) K3[Fe(CN)6] (B) Fe(CN)2 (C) K4[Fe(CN)6] (D) Fe(CN)4

Tardigrade Admin · Accepted Answer

Δ Tb =iKb m So, i=(2.08/0.52× 1)=4 So the complex is K3[Fe(CN6)] K3[Fe(CN)6] leftharpoons 3K++[Fe(CN)6]3-

Q. A complex of iron and cyanide ions is $100%$ ionised at $1 m$ (molal). If its elevation in b.p. is $2.08$ . Then the complex is $(K_{b} = 0.5 2^{\circ} m o l^{- 1} k g)$ :

$K_{3} [F e (CN)_{6}]$

$F e (CN)_{2}$

$K_{4} [F e (CN)_{6}]$

$F e (CN)_{4}$

$Δ T_{b} = i K_{b} m$
So, $i = \frac{2.08}{0.52 \times 1} = 4$
So the complex is $K_{3} [F e (C N_{6})]$
$K_{3} [F e (CN)_{6}] ⇌ 3 K^{+} + [F e (CN)_{6}]^{3 -}$

Q. A complex of iron and cyanide ions is 100% ionised at 1m (molal). If its elevation in b.p. is 2.08. Then the complex is (Kb​=0.52∘mol−1kg) :

K3​[Fe(CN)6​]

Fe(CN)2​

K4​[Fe(CN)6​]

Fe(CN)4​