Question

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

• A. $N{H}_{4}V{O}_2$
• B. $K_{4}VC{N}_6$
• C. $VS{O}_4$
• D. $VOS{O}_4$

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 $N{H}_{4}V{O}_2$ is $+3$ .
Oxidation state of Vanadium in $K_4[V(CN{)}_6]$ is $+2$ .
Oxidation state of Vanadium in $VS{O}_4$ is $+2$ .
Oxidation state of Vanadium in $VOS{O}_4$ is:
${\left(\text{VOSO}\right)}_4={\left(VO\right)}^{2+}+S{O}_4^{2-}x+\left(-2\right)=+2$
Hence, $x=+4$