Question

The reaction of $K_3\left[Fe\right(CN{)}_6]$ with freshly prepared $FeS{O}_4$ solution produces a dark blue precipitate called Turnbull’s blue. Reaction of $K_4\left[Fe\right(CN{)}_6]$ with the $FeS{O}_4$ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the $FeS{O}_4$ solution with $NaN{O}_3,$ followed by a slow addition of concentrated $H_{2}S{O}_4$ through the side of the test tube produces a brown ring.

Precipitate X is:

• A. $F{e}_4\left[Fe\right(CN{)}_6{]}_3$
• B. $Fe\left[Fe\right(CN{)}_6]$
• C. $K_{2}Fe\left[Fe\right(CN{)}_6]$
• D. $KFe\left[Fe\right(CN{)}_6]$

Answer 1

The correct option is C $K_{2}Fe\left[Fe\right(CN{)}_6]$

Potassium hexacyanoferrate(II) $K_4\left[Fe\right(CN{)}_6{]}^{-4}$ is a yellow coloured solid and can be made by the action of $C{N}^{-}$ on an iron(II) saltin solution. $F{e}^{2+}$ ions give a white precipitate of $K_{2}Fe\left[Fe\right(CN{)}_6]$.
$FeS{O}_4+K_4\left[Fe\right(CN{)}_6]\underset{\text{While ppt.}}{\to K_{2}Fe\left[Fe\right(CN{)}_6]}$