Question

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

Answer 1

Each chromium (`Cr`) atom in `(NH_4)_2Cr_2O_7` has an oxidation number of +6.

Explanation:

It doesn't really make sense to talk about the oxidation number of '`CR_2`' (should be written as `Cr_2`), just the oxidation number of `Cr` (chromium).

The oxidation number of the `NH_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 `Cr` atoms must account for +12 (since `+2 +12-14 = 0)`. That means each must have an oxidation number of +6.