For the reaction $K_{4} [F e {(C N)}_{6}] ⟶ {F e}^{3 +} + C O_{2} + N O_{3}^{⊖}$ , the n-factor is:

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\frac{5}{3}

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Solution

The correct option is B 61
$_{4} {[{F e}^{2 +} {(- 1 \times 6 C N)}_{6}]}^{- 4}$ Oxidation state of $F e$ is +2.
${F e}^{2 +} ⟶ {F e}^{3 +} + e^{-}$
$6 {(C N)}^{⊖} ⟶ 6 C O_{2} + 12 e^{-}$
$6 (x - 3 = - 1) 6 (x - 4 = 0)$
$6 (x = 2) 6 (x = 4)$
$6 {(C N)}^{⊖} ⟶ 6 N O_{3}^{⊖} + 48 e^{-}$

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Similar questions

K_{4} [F e {(C N)}_{6}] ⟶ {F e}^{3 +} + C O_{2} + N O_{3}^{-}

n

(60 + x)

x

K_{4} [F e {(C N)}_{6}] O x i d a t i o n - ------------- \to {F e}^{3 +} + C O_{2} + N O_{3}^{⊖}

F e S_{2} ⟶ {F e}^{3 +} + S O_{2}

n

[F e (C N)_{6}]^{4 -} \to F e^{3 +} + C O_{2} + N O_{3}^{-}

3.74

= \frac{M o l . w t .}{61}

\begin{matrix} List - I & List - II (I) For the reaction & (P) \frac{3}{5} K_{4} [F e (C N)_{6}] \to F e^{3 +} + C O_{2} + N O_{3}^{-} the n-factor is: (II) For the reaction & (Q) 28 F e S_{2} \to F e^{3} + + S O_{2} the n-factor is: (III) For the reaction & (R) 61 B r_{2} + 2 N a O H \to N a B r O_{3} + N a B r + H_{2} O the n-factor is: (IV) For the reaction & (S) 1 A s_{2} S_{3} \to A s^{5 +} + S O_{4}^{2 -} the n-factor is: (T) \frac{5}{3} (U) 11 \end{matrix}

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