Question

Find the n-factor for Zn in the following chemical changes.
$(Zn+4HN{O}_3(dil)\to Zn(N{O}_3{)}_2+2NO+2H_{2}O)$

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

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 $|O.S{.}_{Product}-O.S{.}_{Reactant}|$ for the valency factor
$(Zn\to Zn(N{O}_3{)}_2)$
O.S. of Zn in product side
$=x+2\times (-1)=0$
$x=+2$
O.S. of Zn in reactant side
$=2$

$\text{valency factor}=|O.S{.}_{Product}-O.S{.}_{Reactant}|$
$\text{valency factor}=|+2-0|$
$\text{valency factor}=2$

Balanced chemical equation for the above change,
$Zn+4HN{O}_3\left(dil\right)\to Zn(N{O}_3{)}_2+2H_{2}O+2NO$
So, 1 mole of $Zn$ loses 2 electrons.


for $Zn=\text{number of Zn atom}\times |O.S{.}_{Product}-O.S{.}_{Reactant}|=1\times 2$
So, n factor =$2$