Question

Determine the oxidation state of the underlined element in ${Rb}_{4}Na\left[H\underset{–}{V_{10}}{O}_{28}\right]$.

• A. $+5$
• B. $+3$
• C. $+1$
• D. $+4$

Answer 1

The correct option is A $+5$

Let $x$ be the oxidation number of $V$ in ${Rb}_{4}Na\left[H\underset{–}{V_{10}}{O}_{28}\right]$.
The oxidation numbers of Rb, Na, H and O are $+1,+1,+1$ and $-2$ respectively.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore, $4(+1)+1+1+10x+28(-2)=0$
$4+2+10x-56=0$
$10x-50=0$
$x=+5$
Hence, the oxidation number of V in ${Rb}_{4}Na\left[H\underset{–}{V_{10}}{O}_{28}\right]$ is $+5$.