Topic 8 · page 3 of 29

Roman numerals and writing formulae

A Roman numeral in a name is the oxidation number of that element. Read it to find the charge on an ion, and use it to write a formula whose numbers add up to zero.

12 min read8.38.4

Read first: Calculating oxidation numbers

Roman numerals in names 8.3

Some elements form compounds with more than one oxidation number: iron is +2 or +3, copper +1 or +2, chlorine −1, +1, +5 and more. The name has to say which. A Roman numeral in brackets gives the oxidation number of that element.

Numeral I II III IV V VI VII
Number 1 2 3 4 5 6 7
Formula Name Oxidation number
FeCl₂ iron(II) chloride Fe +2
FeCl₃ iron(III) chloride Fe +3
Cu₂O copper(I) oxide Cu +1
MnO₄⁻ manganate(VII) Mn +7
Cr₂O₇²⁻ dichromate(VI) Cr +6
SO₄²⁻ sulfate(VI) S +6
NO₃⁻ nitrate(V) N +5
ClO⁻ chlorate(I) Cl +1
ClO₃⁻ chlorate(V) Cl +5

The numeral is not the charge. Manganate(VII) is MnO₄⁻: the numeral says Mn is +7, and the ion's charge is 1−.

You may leave the numeral out when the name is unambiguous, which is why sodium chloride and magnesium oxide have none: Group 1 and 2 metals have only one oxidation number. A name for an ion such as sulfate is often shortened for the same reason, but with the numeral it is sulfate(VI), and the sulfite ion SO₃²⁻ is sulfate(IV).

Chlorine shows the pattern most clearly. Each step up the scale is a different species.

A vertical scale of chlorine oxidation numbers from minus 1 at the bottom to plus 7 at the top. Each number has its species beside it: Cl minus (chloride) at minus 1, Cl2 (chlorine) at 0, ClO minus (chlorate(I)) at plus 1, ClO2 minus (chlorate(III)) at plus 3, ClO3 minus (chlorate(V)) at plus 5 and ClO4 minus (chlorate(VII)) at plus 7. Chlorate(I) and chlorate(V) are marked as the two used in this topic.
Chlorine in six oxidation numbers. Chlorate(I) and chlorate(V) come back in the reactions of chlorine with water and alkali.

Worked exampleFrom a name to an oxidation number

Give the oxidation number of manganese in potassium manganate(VII), KMnO₄, and show it adds up.

  1. The numeral gives it: manganese is +7.
  2. Check: K +1, Mn +7, 4 × O (−2) = −8. The total is +1 + 7 − 8 = 0. ✓

Writing formulae from oxidation numbers 8.4

A formula must have oxidation numbers that add up to zero (or to the charge, for an ion). Work in three steps.

MethodFrom a name to a formula

  1. Find the number or charge of each part: the oxidation number of a metal, the charge on any polyatomic ion.
  2. Choose how many of each part make the total zero. Swap the numbers if it helps, then cancel.
  3. Write the formula. Put brackets round a polyatomic ion if you need more than one of it: Cu(NO₃)₂, not CuNO₃₂.

Worked exampleIron(III) oxide

  1. Fe is +3. O is −2.
  2. The smallest numbers that cancel: 2 × (+3) = +6 and 3 × (−2) = −6.
  3. The formula is Fe₂O₃.

Worked exampleCopper(II) nitrate

  1. Copper is +2. The nitrate ion is NO₃⁻ (N +5, three O at −2: a charge of −1).
  2. For a total of zero: (+2) + 2 × (−1) = 0, so two nitrates.
  3. The formula is Cu(NO₃)₂.

Worked exampleSodium chlorate(V)

  1. Sodium is Na⁺. Chlorate(V) means Cl is +5, so ClO₃⁻ has +5 + 3 × (−2) = −1.
  2. (+1) + (−1) = 0, so one of each.
  3. The formula is NaClO₃.
Name Working Formula
vanadium(V) oxide V +5, O −2: 2 × (+5) + 5 × (−2) = 0 V₂O₅
manganese(IV) oxide Mn +4, O −2: (+4) + 2 × (−2) = 0 MnO₂
iron(III) sulfate Fe +3, SO₄²⁻: 2 × (+3) + 3 × (−2) = 0 Fe₂(SO₄)₃
potassium dichromate(VI) K⁺ and Cr₂O₇²⁻: 2 × (+1) + (−2) = 0 K₂Cr₂O₇
chromium(III) chloride Cr +3, Cl −1 CrCl₃
copper(I) oxide Cu +1, O −2: 2 × (+1) + (−2) = 0 Cu₂O

Check yourself

  1. Give the oxidation number of the underlined element: (a) Fe in iron(II) sulfate, (b) N in nitrate(III), (c) Cr in chromate(VI), CrO₄²⁻.
  2. Name (a) FeCl₃, (b) NaClO, (c) K₂Cr₂O₇ using Roman numerals.
  3. Write the formula of (a) vanadium(V) oxide, (b) iron(II) sulfate, (c) copper(II) hydroxide.
  4. Write the formula of (a) manganese(IV) oxide, (b) sodium chlorate(I), (c) ammonium dichromate(VI).
  5. The ion with formula ClO₃⁻ is chlorate(V). Show that its charge is 1− when chlorine is +5, then deduce the formula of the chlorate(III) ion using the same oxygen number.
  6. A student writes "the formula of iron(III) sulfate is Fe₃(SO₄)₂". Explain the error and write the correct formula.

Answers cover these until you have tried the questions

  1. (a) Iron(II): +2. (b) Nitrate(III): +3. (c) The numeral is VI: +6; check Cr + 4 × (−2) = −2 gives Cr = +6. ✓
  2. (a) Cl is −1 so Fe must be +3: iron(III) chloride. (b) Na⁺ and ClO⁻ (Cl +1): sodium chlorate(I). (c) K₂Cr₂O₇ has Cr₂O₇²⁻ with Cr +6: potassium dichromate(VI).
  3. (a) V +5, O −2: V₂O₅. (b) Fe²⁺ and SO₄²⁻: FeSO₄. (c) Cu²⁺ and OH⁻: Cu(OH)₂.
  4. (a) MnO₂. (b) Na⁺ and ClO⁻: NaClO. (c) NH₄⁺ and Cr₂O₇²⁻: 2 × (+1) + (−2) = 0, so (NH₄)₂Cr₂O₇.
  5. +5 + 3 × (−2) = −1, so the charge is 1−. For chlorate(III): Cl +3 with two oxygens gives +3 + 2 × (−2) = −1, so the ion is ClO₂⁻.
  6. The numbers of ions are swapped: the charges are Fe³⁺ (3+) and SO₄²⁻ (2−). Total: 2 × (+3) = +6 and 3 × (−2) = −6. So two iron ions and three sulfate ions: Fe₂(SO₄)₃.