Barium ion,, ${CN}^{-}$ and ${Co}^{2 +}$ form an ionic complex. If that complex is supposed to be 75% ionised in water with van't Hoff factor 'i' equal to four, then the coordination number of ${Co}^{2 +}$ in the complex can be :

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four

six and Four both

five

six

answer is B.

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Detailed Solution

Given compound is,

${Ba}_{x - 2} {[Co (CN)_{x}]}_{2}$

Here,

$\begin{array}{l} i = 1 + (n - 1) \times \\ = 1 + (n - 1) \times 0.75 \\ 3 = 0.75 n - 0.75 \\ 3.75 = 0.75 n \\ n = \frac{3.75}{0.75} \\ n = 5 \end{array}$

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