The n-factor for Ferrocyanide in the reaction given below: $[F e {(C N)_{6}}^{4 -}] ⟶ {F e}^{3 +} + {C O}_{2} + {N O}_{3}^{- 1}$ is

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$61$

$2$

$32$

$56$

answer is D.

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Detailed Solution

The oxidation state of iron in reactant is $+ 2$ and in product is $+ 3$ , net loss of electrons is $1$ .

The oxidation state of carbon in reactant is $+ 2$ and in product is $+ 4$ , net loss of electrons is $2$ .

The oxidation state of nitrogen in reactant is $- 3$ and in product is $+ 5$ , net loss of electrons is $8$ .

Overall reaction is oxidation.

There are $6$ carbons and nitrogen in the reactant, therefore the $n$ -factor is

${[F e (C N)_{6}]}^{4 -} = [(1 \times 1) + (6 \times 2) + (6 \times 8) = 61$

Hence, the correct option is D.

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