Question

What will be the n-factor for $F{e}_2(HP{O}_4{)}_3$ ?

• A. 3
• B. 4
• C. 2
• D. 6

Answer 1

The correct option is D 6

For salts the n-factor is
Total positive charge or $\parallel \text{Total negative charge}\parallel$
$F{e}_2(HP{O}_4{)}_3\rightleftharpoons 2F{e}^{3+}+3HP{O}_4^{2-}$
Charge of $F{e}^{3+}=+3$
Total $2Fe$ atoms is present.
n-factor $=2\times 3=6$

Alternative way :
Charge of $HP{O}_4^{2-}=-2$
Total $3HP{O}_4^{2-}$ atoms are present.
n-factor =$\parallel -2\parallel \times 3=6$
So , n-factor of $F{e}^{3+},HP{O}_4^{2-}$ is 6.