Find the n-factor in the following chemical changes.
$K_{2} C r_{2} O_{7} H^{+} - -- \to C r^{3 +}$

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Solution

The correct option is D 6
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
$K_{2} C r_{2} O_{7} H^{+} - -- \to C r^{3 +}$
O.S. of Cr in reactant side
$= + 2 + 2 x + 7 \times (- 2) = 0$
$x = + 6$
O.S. of Cr in product side=+3
$valency factor = number of Cr atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = 2 \times | + 3 - 6 |$
$valency factor = + 6$

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