Question

Oxidation number of V in $R{b}_{4}Na\left[H{V}_{10}{O}_{28}\right]$ is:

• A. +3
• B. +5
• C. +4
• D. Zero

Answer 1

The correct option is B +5

Oxidation number of Rb = +1
Oxidation number of Na = +1
Oxidation number of H = +1
Let oxidation number of V = x
Oxidation number of O = -2
$4\times 1+1+1+10x+28\times (-2)=0$
$6+10x-56=0$
$10x=50$
$x=\frac{50}{10}=+5$