What is the oxidation number of K2CrO4?

Answer 1

K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. The oxidation number of oxygen is always -2 (except when chemically bound to fluorine, which is the only element higher in electronegativity/ionization energy than oxygen) K is an alkali metal, (in column 1A of the Periodic Table) and henceforth will almost always have an oxidation number of +1. With this information, we can deduct that...

2 K atoms will each have +1 as an oxidation number (totaling +2)

4 O atoms will each have -2 as an oxidation number (totaling -8)

We set the equation to zero (because the compound has no charge)

so.

2K+Cr+4O=0

or

2(+1)+Cr+4(-2)=0

or

+2+cr-8=0

simple algebra from here.

cr=+6

each, individually, would be, k=+1, cr=+6, o=-2.

Generally, you'll only be looking to figure out the oxidation number of cr in this problem.