$Fe (SCN)_{n}^{3 - n}$ absorb color because of the complex $Fe (SCN)_{n}^{3 - n}$ only, whereas ${Fe}^{3 +}$ and ${SCN}^{-}$ both are colorless. In the given table, the equilibrium concentration of ${Fe}^{3 +}$ and ${SCN}^{-}$ and corresponding absorbance are given.

Considering absorbance proportional to concentration of the complex, the value of n is

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Detailed Solution

$\begin{array}{l} {Fe}^{3 +} + n {SCN}^{-} ⇄ Fe (SCN)_{n}^{3 - n} \\ K_{form} = \frac{[Complex]}{[{Fe}^{3 +}] {[{SCN}^{-}]}^{n}} \end{array}$

$∵ [complex] = K_{form} [{Fe}^{3 +}] {[{SCN}^{-}]}^{n}$

$∴ \log [complex] = \log K_{form} + \log [{Fe}^{3 +}] + n \log [SCN]$

$[{Fe}^{3 +}] is constant, hence,$

$\log \frac{[Complex]_{1}}{[Complex]_{2}} = n \log \frac{{[{SCN}^{-}]}_{1}}{{[{SCN}^{-}]}_{2}}$

$∴ \log 3 / 0.3 = n \log \frac{10}{(10)^{2 / 3}} \Rightarrow 1 = n \times \frac{1}{3} or n = 3$

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