Question

What will be the overall n-factor for the disproportionation reaction ?
$P_4\left(s\right)+3NaOH\left(aq\right)+3H_{2}O\left(l\right)\to 3Na{H}_{2}P{O}_2\left(aq\right)+P{H}_3\left(g\right)$

• A. 4
• B. 3
• C. 2
• D. 6

Answer 1

The correct option is B 3

For the given reaction :
${\stackrel{0}{P}}_4\left(s\right)+3NaOH\left(aq\right)+3H_{2}O\left(l\right)⟶3Na{H}_2\stackrel{+1}{P}{O}_2\left(aq\right)+\stackrel{-3}{P}{H}_3\left(g\right)$
Oxidation :
${\stackrel{0}{P}}_4⟶Na{H}_2\stackrel{+1}{P}{O}_2$
$n_f=n_1=|0-(+1)|\times 4=4$
Reduction :
${\stackrel{0}{P}}_4⟶\stackrel{-3}{P}{H}_3$
$n_f=n_2=|0-\left(3\right)|\times 4=12$
$(n_f{)}_T=\frac{n_1\times n_2}{n_1+n_2}$
$(n_f{)}_T=\frac{4\times 12}{4+12}=\frac{4\times 12}{16}=3$