# What is the metal oxidation state in "dichromate ion", Cr_2O_7^(2-)?

##### 1 Answer
May 28, 2017

In $\text{dichromate}$, $C {r}_{2} {O}_{7}^{2 -}$? We got $C r \left(V I +\right)$.

#### Explanation:

The SUM of the individual oxidation states of chromium and the oxygen atoms in dichromate anion is equal to the charge on the ion.

And thus $2 \times C {r}_{\text{oxidation number"+7xxO_"oxidation number}} = - 2$.

Now oxygen generally has an oxidation of $- 2$ in its compounds, and it does so here, so.............

$2 \times C {r}_{\text{oxidation number}} + 7 \times \left(- 2\right) = - 2$.

Add +14 to both sides..............

$2 \times C {r}_{\text{oxidation number}} = + 12$.

$C {r}_{\text{oxidation number}} = + \frac{12}{2} = + V I$, as required................