Which of the following reactions does not involve the change in oxidation state of metal?

Let us assume $\mathrm{X}$ be the oxidation number of vanadium. Oxygen oxidation number is $-2$. There is a charge $-2$ on the complex. 

            $\mathrm{X} + 1\times (-2) = -2$

             $\mathrm{X}-2 = -2$

           $\mathrm{X} = -2+2$

         $\mathrm{X} = 0.$ 

Whereas in the case of ${\mathrm{V}}_2{\mathrm{O}}_3$, vanadium exists in $+3$.

Let us assume $\mathrm{X}$ be the oxidation number of vanadium. Oxygen oxidation number is $-2$.

So, $2\mathrm{X}+3\times (-2)=0$

  $2 \mathrm{X}+(-6)=0$ 

 $2 \mathrm{X}=+6$

$\mathrm{X}=6 / 2$

$\mathrm{X}=3$

In case of option (B) elemental sodium oxidation state is $0$, whereas $N{a}^{+}$ ion oxidation state is $+1$.

In case of option (C) $Cr{O_4}^{2-}$ oxidation state of chromium ($+6$) is $+6$ and in the product $C{r}_2{O_7}^{2-}$ oxidation state of chromium ($+6$) is $+6$.

In case of option (D) $\mathrm{Z}{{\mathrm{n}}^{2+}}^{ }$ ion oxidation state is $+2$, whereas elemental zinc oxidation state is $0$.