The chemical formula of iron(III) hexacyanoferrate(II) is:

F e [F e {(C N)}_{6}]

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F e_{3} [F e {(C N)}_{6}]

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F e_{3} {[F e {(C N)}_{6}]}_{4}

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F e_{4} {[F e {(C N)}_{6}]}_{3}

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Solution

The correct option is C $F e_{4} {[F e {(C N)}_{6}]}_{3}$
We proceed from left to right, iron (III) means $F e^{3 +}$ , hexacyano means $6 C N^{-}$ ions are present and ferrate (II) means $F e^{2 +}$ is the central metal atom.

So, the complex ion is $[F e (C N)_{6}]^{4 -}$ and the counter ion is $F e^{3 +}$ . So, after balancing the charge, the formula becomes $F e_{4} [F e (C N)_{6}]_{3}$ .

Option D is correct.

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Similar questions

In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]₄

F e^{3} + {[F e {(C N)}_{6}]}_{6}^{3 -} \to \times H_{2} O_{2} S o l u t i o n - ------- \to Z

E^{\circ}

[F e (C N)_{4}]^{4^{-}} \to [F e (C N)_{6}]^{3 -} + e^{-}

\begin{matrix} E^{\circ} = - 0.35 V F e^{2 +} \to F e^{3 +} + e^{-} & E^{\circ} = - 0.77 V \end{matrix}

F e_{3} {[F e {(C N)}_{6}]}_{2}

E^{o}

{[F e {(C N)}_{6}]}^{4 -} ⟶ {[F e {(C N)}_{6}]}^{3 +} + e^{-}, E^{o} = - 0.35 V

{F e}^{2 +} \to {F e}^{3 +} + e^{- 1}

{F e}^{2 +} \to {F e}^{3 +} + e^{- 1}; E^{o} = - 0.77 V

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