$K F e^{I I} [F e^{I I I} (C N)_{6}]$ is ionize in aqueous solution. What are the ions ?

No ionization

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K^{+}, F e^{2 +}, [F e (C N)_{6}]^{3 -}

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K^{+}, F e^{3 +}, [F e (C N)_{6}]^{3 -}

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None of the above

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Solution

The correct option is B $K^{+}, F e^{2 +}, [F e (C N)_{6}]^{3 -}$
$K F e^{I I} [F e^{I I I} (C N)_{6}]$ on ionization in aqueous solution gives $K^{+}, F e^{2 +}, [F e (C N)_{6}]^{3 -}$ .
Hence option B is correct.

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Similar questions

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

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

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}

E^{\circ}

[F e (C N)_{6}]^{4 -} \to [F e (C N)_{6}]^{3 -} + e^{-} E^{\circ} = - 0.35 V

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

[F e {(C N)}_{6}]^{4 -} \to [F e {(C N)}_{6}]^{3 -} + e^{- 1}, E^{0} = - 0.35 V

F e^{2 +} \to F e^{3 +} + e^{- 1}, E^{0} = - 0.77 V

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