The reaction of K 3 [Fe(CN) 6 ] with freshly prepared FeSO 4 solution produces a dark…
Chemistry · JEE Advanced · NTA Exams — d- and f-Block Elements
The reaction of K3[Fe(CN)6] with freshly prepared FeSO4 solution produces a dark blue precipitate called Turnbull’s blue. Reaction of K4 [Fe(CN)6] with the FeSO4 solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO4 solution with NaNO3, followed by a slow addition of concentrated H2SO4 through the side of the test tube produces a brown ring.
Precipitate X is (2021)
- \(\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}\)
- \(\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\)
- \(\mathrm{K}_{2} \mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\)
- \(\mathrm{KFe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\)
Answer
(C) K _ 2 Fe [ Fe ( CN )_ 6 ]
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