Introduction
By now you know that many transition metals (and a few other metals like lead and tin) don't stick to just one charge — they can form more than one stable cation. In this lesson we go a step further: instead of just naming compounds when you're told the charge, you'll learn how to work out the charge yourself from the formula, and how to name binary compounds confidently using Roman numerals (Stock notation).
Quick recap: why Roman numerals?
For metals with a fixed charge (like group 1, group 2, Al³⁺, Zn²⁺, Ag⁺), we never need a Roman numeral — the charge is already obvious. But for metals like iron, copper, cobalt, lead, and tin, the charge can vary, so we must state it in brackets after the metal's name (Talbot, p.130).
- Example: CoCl₂ → cobalt(II) chloride
- Example: BaCl₂ → just barium chloride (no Roman numeral needed, since Ba is always 2+) (Talbot, p.130)
Working out the charge from the formula
Since binary ionic compounds are electrically neutral overall, you can use the anion's charge (which you do know) to calculate the cation's charge.
Method:
- Identify the anion and its charge (from its position on the periodic table or its "-ide" ending).
- Use the formula's subscripts to balance total positive and negative charge to zero.
- Divide to find the charge on a single cation.
- Write the charge as a Roman numeral in brackets straight after the metal name.
Worked Example 1
Name Fe₂O₃
- Oxide ion = O²⁻, and there are 3 of them → total negative charge = 6−
- There are 2 iron atoms, so each must contribute 3+ (since 2 × 3+ = 6+)
- Name: iron(III) oxide
Worked Example 2
Name PbO₂
- Oxide ion = O²⁻, there are 2 → total negative charge = 4−
- Only 1 Pb atom, so it must be 4+
- Name: lead(IV) oxide
Worked Example 3
Name Cu₂S
- Sulfide ion = S²⁻ (charge 2−), only 1 S atom → total negative charge = 2−
- There are 2 copper atoms sharing that charge, so each Cu is 1+
- Name: copper(I) sulfide
Going the other way: formula from name
You can also reverse the process. If you're told the name, use the Roman numeral to fix the cation charge, then balance it against the anion charge using the criss-cross method.
Worked Example 4
Write the formula for tin(IV) oxide
- Tin ion: Sn⁴⁺
- Oxide ion: O²⁻
- Criss-cross charges: Sn⁴⁺ needs 2 O²⁻ to balance (4+ = 2 × 2−)
- Formula: SnO₂
A note on consistency
IUPAC's whole purpose is to make sure chemists worldwide use names and symbols unambiguously (Talbot, p.172). That's exactly why Roman numerals matter here — without them, "iron oxide" could mean FeO or Fe₂O₃, and confusion in chemistry (unlike in everyday language) can lead to real lab mistakes.
Key takeaways
- Variable-charge metals need a Roman numeral in brackets to show their oxidation state in a compound name.
- To find the charge: use the total negative charge from the anion, then divide by the number of cations present.
- Fixed-charge metals (groups 1 & 2, Al, Zn, Ag) never need a Roman numeral.
- The criss-cross method works both ways — from formula to name, and from name to formula.
- Clear, standardized nomenclature (via IUPAC) prevents ambiguity in chemical communication.