Example Question 1

Question: Work out the oxidation number of (a) manganese in MnO₄⁻, (b) chromium in Cr₂O₇²⁻ and (c) sulfur in S₂O₃²⁻.Method: give each element you know a fixed value first. Oxygen is −2 (none of these is a peroxide). Then use the rule that the oxidation numbers in an ion add up to the charge on the ion. Call the unknown x.(a) MnO₄⁻: four oxygens give 4 × (−2) = −8.x + (−8) = −1, so x = +7
Manganese is +7, which is why KMnO₄ is called potassium manganate(VII).(b) Cr₂O₇²⁻: there are two chromium atoms, so the unknown is 2x. Seven oxygens give −14.2x + (−14) = −2, so 2x = +12 and x = +6
Each chromium is +6: this is the dichromate(VI) ion.(c) S₂O₃²⁻: three oxygens give −6.2x + (−6) = −2, so 2x = +4 and x = +2
Each sulfur is +2 (an average over the two sulfur atoms).Common mistakes: forgetting to multiply by the number of atoms of the unknown element, and setting the total to zero for an ion instead of to the charge on the ion. Always write the sign in front of an oxidation number, for example +6, not 6.