Topic 8 · page 2 of 29

Calculating oxidation numbers

Write the known numbers under the atoms, multiply by the number of atoms, set the total equal to the overall charge, and solve for the unknown. Peroxides and metal hydrides are the two exceptions the rules must be bent for.

14 min read8.2

Read first: Oxidation number and the rules

A method that always works 8.2

MethodFinding an unknown oxidation number

  1. Write the formula and put the known numbers under each atom: H +1, O −2, Group 1 +1, Group 2 +2.
  2. Multiply each known number by how many atoms of that element there are.
  3. Set the total of all the numbers equal to the overall charge (0 for a compound).
  4. Solve for the one unknown. If there are n atoms of the unknown element, divide by n. The answer may be negative.

Worked exampleSulfur in the sulfate ion, SO₄²⁻

  1. Oxygen is −2. There are four, so 4 × (−2) = −8.
  2. The total must be the ion charge, −2: S + (−8) = −2.
  3. S = −2 + 8 = +6.

Worked exampleChromium in the dichromate ion, Cr₂O₇²⁻

  1. Seven oxygens: 7 × (−2) = −14.
  2. Total is −2: 2Cr + (−14) = −2, so 2Cr = +12.
  3. There are two chromium atoms, so each is +12 ÷ 2 = +6.

Worked exampleNitrogen in the ammonium ion, NH₄⁺

  1. Four hydrogens: 4 × (+1) = +4.
  2. Total is +1: N + 4 = +1.
  3. N = −3.

Nitrogen is negative here because hydrogen is less electronegative than nitrogen. A positive ion does not mean every atom in it is positive.

Species Known numbers Working Unknown
Fe₂O₃ O −2 2Fe + 3 × (−2) = 0 Fe +3
NO₃⁻ O −2 N + 3 × (−2) = −1 N +5
MnO₄⁻ O −2 Mn + 4 × (−2) = −1 Mn +7
SO₂ O −2 S + 2 × (−2) = 0 S +4
H₂S H +1 2 × (+1) + S = 0 S −2

Notice the same element can have many numbers. Sulfur is −2 in H₂S, 0 in S₈, +4 in SO₂ and +6 in SO₄²⁻. The more electronegative the partner, and the more bonds to it, the higher the oxidation number.

Two exceptions: peroxides and metal hydrides 8.2

The usual numbers for H (+1) and O (−2) fail in two families. In both, applying the usual rule gives a number that does not fit.

Peroxides contain an O–O bond. In hydrogen peroxide, H–O–O–H, the two oxygen atoms share their bonding pair equally, because they are identical. So each oxygen gains only one electron from its bond to hydrogen. The oxidation number of oxygen is −1, not −2.

Metal hydrides are compounds of hydrogen with a metal, such as NaH and CaH₂. Hydrogen is more electronegative than the metal, so it takes the electrons: it is hydride, H⁻. The oxidation number of hydrogen is −1, not +1.

Worked exampleOxygen in Na₂O₂

  1. Group 1 is always +1, so 2 × (+1) = +2.
  2. The total is 0: +2 + 2 × O = 0, so 2 × O = −2.
  3. Each O is −1. This is a peroxide, with the O₂²⁻ ion.

Worked exampleHydrogen in calcium hydride, CaH₂

  1. Group 2 is always +2.
  2. The total is 0: +2 + 2 × H = 0, so 2 × H = −2.
  3. Each H is −1.
Species Hydrogen Oxygen
H₂O +1 −2
H₂O₂ +1 −1
NaH −1 –
CaH₂ −1 –
Na₂O₂ – −1

Check yourself

  1. Calculate the oxidation number of sulfur in (a) SO₂, (b) H₂SO₄, (c) SO₃²⁻, (d) H₂S.
  2. Calculate the oxidation number of manganese in (a) MnO₂, (b) K₂MnO₄, (c) MnO₄⁻, (d) Mn²⁺.
  3. Give the oxidation number of nitrogen in NH₃, N₂O, NO, NO₂ and NO₃⁻.
  4. State the oxidation number of oxygen in BaO₂ and in H₂O₂. Explain your answer for BaO₂.
  5. State the oxidation number of hydrogen in LiH and in HCl, and explain the difference.
  6. Give the oxidation number of carbon in CH₄, CH₃OH, HCHO\ce{HCHO}, HCOOH\ce{HCOOH} and CO₂. What happens to the number as the carbon is oxidised?

Answers cover these until you have tried the questions

  1. (a) S + 2 × (−2) = 0: S = +4. (b) 2 × (+1) + S + 4 × (−2) = 0: S = +6. (c) S + 3 × (−2) = −2: S = +4. (d) 2 × (+1) + S = 0: S = −2.
  2. (a) Mn + 2 × (−2) = 0: +4. (b) 2 × (+1) + Mn + 4 × (−2) = 0: Mn = +6. (c) Mn + 4 × (−2) = −1: +7. (d) simple ion, +2.
  3. NH₃: N + 3 × (+1) = 0, −3. N₂O: 2N + (−2) = 0, +1. NO: +2. NO₂: +4. NO₃⁻: N + 3 × (−2) = −1, +5.
  4. Both −1. In BaO₂, barium is Group 2 so is +2. The total is 0: +2 + 2 × O = 0, so each O is −1. It is a peroxide, with an O–O bond.
  5. In LiH hydrogen is −1: lithium is a metal and less electronegative than hydrogen, so the compound is a hydride, Li⁺H⁻. In HCl hydrogen is +1 because chlorine is more electronegative than hydrogen.
  6. Using H +1 and O −2: CH₄ −4; CH₃OH −2 (C + 4 − 2 = 0); HCHO\ce{HCHO} 0 (C + 2 − 2 = 0); HCOOH\ce{HCOOH} +2 (2 + C − 4 = 0); CO₂ +4. The oxidation number of carbon increases by 2 at each step, so each step is an oxidation.