JEE Advanced 2021 — Paper 2 — Question 42

The reaction of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ with freshly prepared $\mathrm{FeSO}_{4}$ solution produces a dark blue precipitate called Turnbull's blue. Reaction of $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ with the $\mathrm{FeSO}_{4}$ solution in complete absence of air produces a white precipitate $\mathbf{X}$, which turns blue in air. Mixing the $\mathrm{FeSO}_{4}$ solution with $\mathrm{NaNO}_{3}$, followed by a slow addition of concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$ through the side of the test tube produces a brown ring.

Among the following, the brown ring is due to the formation of