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 $(K_{b} = 0.52^{\circ} mol^{-1}\,kg)$ :

• A. $K_{3}[Fe(CN)_{6}]$
• B. $Fe(CN)_{2}$
• C. $K_{4}[Fe(CN)_{6}]$
• D. $Fe(CN)_{4}$

Answer 1

Answer: (A)

$\Delta\,T_{b} =iK_{b}\,m$
So, $i=\frac{2.08}{0.52\times 1}=4$
So the complex is $K_{3}[Fe(CN_{6})]$
$K_{3}[Fe(CN)_{6}] \rightleftharpoons 3K^++[Fe(CN)_{6}]^{3-}$