Precipitate X is

First recall the colours of the ferro/ferricyanide complexes of iron.
• $$K_4[Fe(CN)_6]$$ is potassium ferrocyanide and contains $$[Fe(CN)_6]^{4-}$$ (Fe in +2 state).
• When this ligand meets $$Fe^{2+}$$ (from $$FeSO_4$$) in the complete absence of air, no oxidation occurs, so every iron present stays in the +2 state.

Hence the only solid that can separate out is a compound in which both iron centres are $$+2$$, i.e. ferrous ferrocyanide.
The general composition of ferrous ferrocyanide is $$K_2Fe^{2+}[Fe^{2+}(CN)_6]$$.

The precipitation reaction is written as
$$K_4[Fe(CN)_6] + FeSO_4 \;\longrightarrow\; K_2Fe[Fe(CN)_6] \downarrow + K_2SO_4$$

Because every ion is in the +2 state, the lattice has no intense charge-transfer transition and appears almost white; this solid is named precipitate $$X$$.

When the white precipitate is exposed to air, some $$Fe^{2+}$$ is oxidised to $$Fe^{3+}$$. The mixed Fe(II)-Fe(III) complex $$Fe_4[Fe(CN)_6]_3$$ (Prussian/Turnbull’s blue) is then produced, giving the characteristic blue colour.

Therefore, the identity of the initial white precipitate $$X$$ is $$K_2Fe[Fe(CN)_6]$$.

Option C which is: $$K_2Fe[Fe(CN)_6]$$