In which of the following compounds is the oxidation number of chromium the greatest? (A) \(\mathrm{CrO}_{4} 2^{-}\) (B) \(\mathrm{CrO}\) (C) \(\mathrm{Cr}^{3+}\) (D) \(\operatorname{Cr}(s)\)

An oxidation number is calculated with the understanding that the sum of the oxidation states in a neutral compound is zero, while in an ion it equals the ion's charge. (A) For the ion \(\mathrm{CrO}_{4}^{2-}\) - Here, oxygen has an oxidation number of -2. We know that the total oxidation number for the species (compound or ion) is the charge of the species, which in this case is -2. Hence, the oxidation number of Cr is +8 (since there are four oxygen atoms). (B) In \(\mathrm{CrO}\) - The sum of oxidation states is 0 as it is a molecule (and hence neutral). With oxygen being -2, the oxidation state of chromium is +2. (C) For the ion \(\mathrm{Cr}^{3+}\) - The oxidation number is the same as the charge, so it is +3. (D) In \(\operatorname{Cr}(s)\) - This is pure chromium, so the oxidation number is 0.

With the oxidation numbers for chromium in each compound, the following ensemble is found: (A) +8, (B) +2, (C) +3, (D) 0. Comparing these, the highest oxidation number of chromium is found in \(\mathrm{CrO}_{4}^{2-}\).