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A complex of iron and cyanide ions is 100% ionized at 1m (molal). If its elevation in b.p. 2.08° $(K_{b} = 0 . 52^{o} {mol}^{- 1} Kg),$ then the complex is

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

$Fe {(CN)}_{2}$

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

$Fe {(CN)}_{4}$

answer is A.

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Detailed Solution

$∆ T_{b} = 0.52 \times 1$

=0.52

$i = \frac{2.08}{0.52} = 4$

$i = 1 - α + nα$

4=1-1+n(1)

$∴ n = 4$

$K_{3} [Fe {(CN)}_{6}] \to 3 k^{+} + {[Fe {(CN)}_{6}]}^{- 3} = 4 ions$

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