WHAT IS THE OXIDATION NUMBER OF (NH4)2CrO4?

The oxidation number of ammonium ion (NH4+) is +1, and the overall charge of (NH4)2CrO4 compound is 0. The oxidation number of chromium (Cr) can be calculated as follows: 2(+1) + 2(x) + 4(-2) = 0 2 + 2x - 8 = 0 2x - 6 = 0 Therefore, the oxidation number of chromium (Cr) in (NH4)2CrO4 is +3.