Chemistry - Complete d-block Question with Solution | TestHub

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{OH}}^{-}⟶2{\mathrm{CrO}}_4^{2-}$

Let the oxidation number of chromium be $\mathrm{x}$

The oxidation state of chromium in ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$$\begin{gathered} =2x+(-2)7=-2 \\ x=+6 \end{gathered}$$

Oxidation state of $\mathrm{Cr}$ in ${\mathrm{CrO}}_4^{2-}$  

$$\begin{gathered} =x+(-2)4=-2 \\ x=+6. \end{gathered}$$

So, the oxidation number of $\mathrm{Cr} \mathrm{in} {\mathrm{CrO}}_4^{2-} \mathrm{is} \mathrm{also} +6.$

Hence, the difference in oxidation number between ions is $0$