# What is the oxidation number of CR_2 in (NH_4)_2Cr_2O_7?

Jun 12, 2017

Each chromium ($C r$) atom in ${\left(N {H}_{4}\right)}_{2} C {r}_{2} {O}_{7}$ has an oxidation number of +6.

#### Explanation:

It doesn't really make sense to talk about the oxidation number of '$C {R}_{2}$' (should be written as $C {r}_{2}$), just the oxidation number of $C r$ (chromium).

The oxidation number of the $N {H}_{4}^{+}$ group is +1, and there are 2 such groups for a total of +2.

The oxidation number of ${O}^{2 -}$ is -2, and there are seven oxide ions, for a total of -14.

Since the total oxidation number for the compound must be 0, the two $C r$ atoms must account for +12 (since +2 +12-14 = 0). That means each must have an oxidation number of +6.