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Valency and Oxidation State

Find the n-fa...

Question

Find the n-factor for Zn in the following chemical changes.
$(Z n + 4 H N O_{3} (c o n c .) \to Z n (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O)$

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Solution

The correct option is B 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 $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$ for the valency factor
$(Z n \to Z n (N O_{3})_{2})$
O.S. of Zn in product side
$= x + 2 \times (- 1) = 0$
$x = + 2$
O.S. of Zn in reactant side
$= 2$

$valency factor = | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = | + 2 - 0 |$
$valency factor = 2$

Balanced chemical equation for the above change,
$Z n + 4 H N O_{3} (c o n c .) \to Z n (N O_{3})_{2} + 2 H_{2} O + 2 N O_{2}$
So, 1 mole of $Z n$ loses 2 electrons.

for $Z n = number of Zn atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | = 1 \times 2$
So, n factor = $2$

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