What is the oxidation number of CR_2CR2 in (NH_4)_2Cr_2O_7(NH4)2Cr2O7?

1 Answer
David G.
Jun 12, 2017

Each chromium (CrCr) atom in (NH_4)_2Cr_2O_7(NH4)2Cr2O7 has an oxidation number of +6.

Explanation:

It doesn't really make sense to talk about the oxidation number of 'CR_2CR2' (should be written as Cr_2Cr2), just the oxidation number of CrCr (chromium).

The oxidation number of the NH_4^+NH+4 group is +1, and there are 2 such groups for a total of +2.

The oxidation number of O^(2-)O2− 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 CrCr atoms must account for +12 (since +2 +12-14 = 0)+2+12−14=0). That means each must have an oxidation number of +6.

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