In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
(b) $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$
(c) $\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
(d) $\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{4}$
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as
$6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
$3\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+4 \mathrm{Fe}^{3+} \stackrel{\mathrm{xH}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3} \cdot x \mathrm{H}_{2} \mathrm{O}$
Prussian blue
Hence, the Prussian blue colour is due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$.