Question

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 sides of the test tube produces a brown ring.

The precipitate ‘X’ is:

पर्याय

• Fe₄[Fe(CN)₆]₃

• Fe[Fe(CN)₆]

• K₂Fe[Fe(CN)₆]

• KFe[Fe(CN)₆]

Answer 1

K₂Fe[Fe(CN)₆]

Explanation:

• Reaction of K₃[Fe(CN)₆] with FeSO₄ (freshly prepared):
  When K₃[Fe(CN)₆] (potassium ferrocyanide) reacts with freshly prepared FeSO₄ (iron(II) sulfate), a dark blue precipitate called Turnbull’s blue is formed.
  \[\ce{Fe^{2+} + K3Fe(CN)6 -> \underset{Trunbull’s blue ppt}{Fe3[Fe(CN)6]2}}\]
  Fe(III) from \[\ce{Fe(CN)^{3-}_6}\] gets reduced by Fe(II) from FeSO₄, resulting in the formation of Fe₃[Fe(CN)₆]², which is Turnbull’s blue.
• Reaction of K₄[Fe(CN)₆] with FeSO₄ (absence of air):
  When K₄[Fe(CN)₆] (potassium ferricyanide) reacts with FeSO₄ in the complete absence of air, a white precipitate X is formed.
  \[\ce{Fe^{2+} + K4[Fe(CN)6] ->[In absence][of air] \underset{white ppt (X)}{K2Fe[Fe(CN)6]}}\]
  The Fe(II) from FeSO₄ reacts with the \[\ce{Fe(CN)^{4-}_6}\] to form K₂Fe[Fe(CN)₆], which is a white precipitate.

The white precipitate X formed when K₄[Fe(CN)₆] reacts with FeSO₄ in the absence of air is K₂Fe[Fe(CN)₆].