Question

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

• A. +5
• B. 6
• C. $+\frac{7}{5}$
• D. 4

Answer 1

The correct option is A +5

Let oxidation state of ${}^{\prime }{V}^{\prime }$ in $R{b}_{4}Na\left[H{V}_{10}{O}_{28}\right]$ is $x$:
Charge on Rb=+1
Charge on Na=+1
Charge on O=-2
Charge on H=+1
$\Rightarrow 4(+1)+(+1)+(+1)+10x+28(-2)=0\Rightarrow 10x=50\Rightarrow x=+5.$