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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$61$

$2$

$32$

$56$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.

JEE

NEET

Foundation JEE

Foundation NEET

CBSE