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

1 Answer
Jun 12, 2017

Answer:

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.