Question

What will be the value of n-factor $\left(n_f\right)$ or valence factor of $P$ in $C{a}_3{P}_2$ and $H_{3}P{O}_4$ respectively for the following change:

$C{a}_3{P}_2⟶H_{3}P{O}_4$

• A. 10,5
• B. 4,8
• C. 16,8
• D. 3,3

Answer 1

The correct option is C 16,8

Oxidation state of $P$ in $C{a}_3{P}_2:$
$(2\times 3)+2x=0x=-3$
Oxidation state of $P$ in $H_{3}P{O}_4$
$(3\times 1)+x+(4\times -2)=0x=+5$

$C{a}_3{\stackrel{-3}{P}}_2⟶H_3\stackrel{+5}{P}{O}_4$
Difference of Oxidation number $=5-(-3)=8$
$n_f=\text{Difference in oxidation number}\times \text{Number of atoms}$
$n_f$ of $C{a}_3{P}_2=8\times 2=16$
$n_f$ of $H_{3}P{O}_4=8\times 1=8$