What is the oxidation state for the ion Cr2O7^2-?

Oct 29, 2015

Oxygen $- 2$ and Chromium $+ 6$

Explanation:

The oxidation state of oxygen is always $- 2$ except in ${H}_{2} {O}_{2}$ it is $- 1$ and in $O {F}_{2}$ it is $+ 2$.

So the oxidation state of oxygen in $C {r}_{2} {O}_{7}^{2 -}$ is $\textcolor{red}{- 2}$. Assume that $x$ is the oxidation state of $C r$ in $C {r}_{2} {O}_{7}^{2 -}$, then to find $x$:

$\left(2 \times x\right) + 4 \times \left(- 2\right) = - 2 \implies x - 8 = - 2 \implies x = + 6$

Therefore, the oxidation state of $C r$ in $C {r}_{2} {O}_{7}^{2 -}$ is $\textcolor{b l u e}{+ 6}$.