Identify the relevant salt-analysis reactions

We are told:

• $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ gives Turnbull's blue.
• $K_4[Fe(CN)_6]$ with $FeSO_4$ in complete absence of air gives a white precipitate $X$, which turns blue in air.
• $FeSO_4$ gives a brown ring test with $NaNO_3$ and conc. $H_2SO_4$, confirming the presence of $Fe^{2+}$.

So the reacting iron ion in $FeSO_4$ is ferrous ion, $Fe^{2+}$.

Reaction of ferrous ion with ferrocyanide

$K_4[Fe(CN)_6]$ is potassium ferrocyanide, containing the complex ion $[Fe(CN)_6]^{4-}$.

In absence of air, $Fe^{2+}$ reacts with $[Fe(CN)_6]^{4-}$ to form a white precipitate known as potassium ferrous ferrocyanide:

$Fe^{2+} + K_2Fe(CN)_6^{2-} \longrightarrow K_2Fe[Fe(CN)_6] \downarrow$

More conventionally, from the reagent $K_4[Fe(CN)_6]$:

$FeSO_4 + K_4[Fe(CN)_6] \rightarrow K_2Fe[Fe(CN)_6] \downarrow + K_2SO_4$

Thus,

$X = K_2Fe[Fe(CN)_6]$

Why does it turn blue in air?

On exposure to air, some $Fe^{2+}$ gets oxidized to $Fe^{3+}$, and the white ferrous ferrocyanide is converted into a blue ferric-ferro/ferricyanide complex (Prussian/Turnbull type blue).

Check the options

• A: $Fe_4[Fe(CN)_6]_3$ → blue ferric ferrocyanide, not the initial white precipitate.
• B: $Fe[Fe(CN)_6]$ → not the correct white potassium-containing precipitate formed here.
• C: $K_2Fe[Fe(CN)_6]$ → correct; this is the white precipitate.
• D: $KFe[Fe(CN)_6]$ → incorrect stoichiometry.

Final answer

Therefore, the precipitate $X$ is:

$\boxed{K_2Fe[Fe(CN)_6]}$

So, the correct option is C.