Kinetics — Hard JEE chemistry MCQ
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
- A. $\left[\mathrm{Fe}(\mathrm{NO})_{2}\left(\mathrm{SO}_{4}\right)_{2}\right]^{2-}$
- B. $\left[\mathrm{Fe}(\mathrm{NO})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{3+}$
- C. $\left[\mathrm{Fe}(\mathrm{NO})_{4}\left(\mathrm{SO}_{4}\right)_{2}\right]$
- D. $\left[\mathrm{Fe}(\mathrm{NO})\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2+}$
Solution
The correct option is **D**.
