Consider the reversible reaction, $\underset{Yellow}{{Fe}^{+ 3} (aq)} + \underset{colourless}{{SCN}^{-} (aq)} ⇌ \underset{deep red}{{[Fe (SCN)]}^{+ 2} (aq)}$
A red colour appears on adding two drops of 0.002 M potassium thiocyanate solution to 1 ml of 0.2 M of Iron(III) nitrate solution due to the formation of ${[Fe (SCN)]}^{+ 2}$ . At equilibrium intensity of red colour remains constant. Intensity of red colour decreases by the addition of
(A) $H_{2} C_{2} O_{4} (aq)$
(B) ${HgCl}_{2} (aq)$
(C) $KSCN (aq)$

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A, B, C

A, B only

B, C only

A, C only

answer is B.

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Detailed Solution

A) Addition of $H_{2} C_{2} O_{4} (aq)$

${Fe}^{+ 3} + 3 C_{2} O_{4}^{- 2} \to {[Fe {(C_{2} O_{4})}_{3}]}^{- 3}$

This decreases $[{Fe}^{+ 3}]$ thus favours backward reaction. Intensity of red colour decreases.

B) Addition of ${HgCl}_{2} (aq)$

${Hg}^{+ 2} + 4 {SCN}^{-} \overset{}{\to} {[Hg {(SCN)}_{4}]}^{- 2}$

This decreases $[{SCN}^{-}]$ thus favours backward reaction. Intensity of red colour decreases.

C) Addition of $KSCN (aq)$

${Fe}^{+ 3} + {SCN}^{-} \to {[Fe (SCN)]}^{+ 2}$

This increases ${[Fe (SCN)]}^{+ 2}$ thus favours forward reaction. Intensity of red colour increases.

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