JEE Chemistry Practice Question

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. Precipitate $\mathbf{X}$ is

1. A.$\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$
2. B.$\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
3. C.$\mathrm{K}_{2} \mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
4. D.$\mathrm{KFe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$