JEE Advanced · Chemistry · 21. p Block (G15-18)
Paragraph:
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.
Question:
Among the following, the brown ring is due to the formation of
- A
- B
- C
- D
Answer & Solution
Correct Answer
(D)
Step-by-step Solution
Detailed explanation
\(3 Fe ^{2+}+ NO _3^{-}+4 H ^{+} \rightarrow NO +3 Fe ^{3+}~+\) \(2 H _2 O\)
\(\left[ Fe \left( H _2 O \right)_6\right]^{2+}+ NO \rightarrow \underset{\text { Brown ring }}{\left[ Fe \left( H _2 O \right)_5 NO \right]^{2+}}\) \(+~ H _2 O\)
\(\left[ Fe \left( H _2 O \right)_6\right]^{2+}+ NO \rightarrow \underset{\text { Brown ring }}{\left[ Fe \left( H _2 O \right)_5 NO \right]^{2+}}\) \(+~ H _2 O\)
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