Find the n-factor of $C_{2} O_{4}^{2 -}$ in the following chemical changes.
$C_{2} O_{4}^{2 -} \to C O_{2}$

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Solution

The correct option is D 2
Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$ for the valency factor
$C_{2} O_{4}^{2 -} \to C O_{2}$
O.S. of C in reactant side
$= 2 x + 4 \times (- 2) = - 2$
$x = + 3$
O.S. of C in product side
$= x - 4 = 0$
$x = + 4$
$valency factor = number of C atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = 2 \times | + 3 - 4 |$
$valency factor = 2$

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