What is the oxidation number of chromium in the chromate ion, CrO42-
The correct answer and explanation is:
The oxidation number of chromium in the chromate ion (CrO₄²⁻) is +6.
To determine the oxidation number of chromium in CrO₄²⁻, we need to apply the rules for oxidation numbers.
- Oxidation number of oxygen: In most compounds, oxygen has an oxidation number of -2. Since there are four oxygen atoms in the chromate ion, the total oxidation number contributed by oxygen is 4 × (-2) = -8.
- Charge on the ion: The chromate ion (CrO₄²⁻) has a total charge of -2. This means that the sum of the oxidation numbers of all elements in the ion must equal -2.
- Oxidation number of chromium: Let the oxidation number of chromium be x. The sum of the oxidation numbers must equal the charge of the ion. Therefore, we can write the equation:
x+(−8)=−2x + (-8) = -2
Solving for x: x−8=−2x – 8 = -2 x=+6x = +6
Thus, the oxidation number of chromium in CrO₄²⁻ is +6.
This oxidation state indicates that chromium is in a highly oxidized form. Chromates are often used in various chemical processes, including as oxidizing agents. The +6 oxidation state of chromium is also a key feature in the chemistry of chromium compounds, particularly in its toxicological impact, as hexavalent chromium compounds are more dangerous compared to other chromium oxidation states.