Question

Oxidation number of V in $R{b}_{4}Na\left[H{V}_{20}{O}_{18}\right]$ is-

• A. 3
• B. $+\frac{3}{2}$
• C. $+\frac{2}{3}$
• D. zero

Answer 1

The correct option is B $+\frac{3}{2}$

O.N. of $Rb=+1$
O.N. of $Na=+1$
O.N. of $H=+1$
Let O.N. of $V=x$
O.N. of $O=-2\therefore 4+1+1+20x-36=0$
$\Rightarrow 20x=30$
or $x=+\frac{3}{2}$