Question

In which compound does vanadium have an oxidation number of $+4$?

• A. $NH_{4}VO_{2}$
• B. $K_{4}V C N_{6} $
• C. $VSO_{4}$
• D. $VOSO_{4}$

Answer 1

Answer: (D)

Oxidation state, also referred to as oxidation number, describes the degree of oxidation of an atom in a chemical compound.
Oxidation state of Vanadium in $NH_{4}VO_{2}$ is $+3$ .
Oxidation state of Vanadium in $K_{4}\left[\right.V\left(\right. CN \left.\right)_{6}\left]\right.$ is $+2$ .
Oxidation state of Vanadium in $VSO_{4}$ is $+2$ .
Oxidation state of Vanadium in $VOSO_{4}$ is:
$\left(\text{VOSO}\right)_{4}= \, \left(VO\right)^{2 +}+ \, SO_{4}^{2 -}x+\left(- 2\right)=+2$
Hence, $x=+4$