The oxidation number of Fe in $K_{4} F e (C N)_{6}$ is?

+ 6

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+ 4

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+ 3

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+ 2

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Solution

The correct option is A $+ 2$
Let oxidation state of Fe is x,
$4 \times (1) + x + 6 \times (- 1) = 0 x = + 2$

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The oxidation number of Fe in K₄Fe(CN)₆ is

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

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

{F e}_{3} O_{4}

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

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