JEE Advanced 2021 · Paper 2, Chemistry question 15

Paragraph

The reaction of $\ce{K3[Fe(CN)6]}$ with freshly prepared $\ce{FeSO4}$ solution produces a dark blue precipitate called Turnbull's blue. Reaction of $\ce{K4[Fe(CN)6]}$ with the $\ce{FeSO4}$ solution in complete absence of air produces a white precipitate $\mathbf X$, which turns blue in air. Mixing the $\ce{FeSO4}$ solution with $\ce{NaNO3}$, followed by a slow addition of concentrated $\ce{H2SO4}$ through the side of the test tube produces a brown ring.