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A polyatomic ion is a small group of atoms held together by covalent bonds that, as a whole, carries an electric charge. Sulfate, nitrate, ammonium and hydroxide are the ones you meet in almost every practical, and each of them has a Lewis structure you can work out with the same method you use for neutral molecules. There is just one extra step at the start and one extra piece of notation at the end.
The method, adapted for ions
- Count valence electrons. Add up the valence electrons of every atom. Then add one electron for each negative charge and subtract one for each positive charge. This is the step people forget.
- Build a skeleton. The least electronegative atom (never hydrogen) usually goes in the centre. Join every outer atom to it with a single bond (2 electrons each).
- Complete the outer atoms’ octets with lone pairs (hydrogen only needs 2 electrons).
- Place any leftover electrons on the central atom.
- Fix any central atom short of an octet by turning a lone pair on a neighbour into a double or triple bond.
- Work out formal charges. Formal charge = valence electrons − lone-pair electrons − ½ × bonding electrons. The formal charges must add up to the charge on the ion.
- Draw the ion in square brackets with the overall charge written outside, top right: [ … ]⁻.
The brackets matter. They tell the reader that the charge belongs to the ion as a whole, even though the bookkeeping places it on particular atoms. If you want a refresher on the neutral-molecule version first, see Lewis dot structures, and for the arithmetic behind step 6, see formal charge.
Example 1: hydroxide, OH⁻
Electron count. O has 6 valence electrons, H has 1, and the 1− charge adds 1. Total = 6 + 1 + 1 = 8 electrons.
Skeleton. There are only two atoms, so O–H. That single bond uses 2 electrons, leaving 6.
Lone pairs. Hydrogen is already satisfied with its bond. The remaining 6 electrons go on oxygen as three lone pairs. Oxygen now has 2 bonding + 6 lone-pair electrons = 8. ✓
Formal charges.
- O: 6 − 6 − ½(2) = −1
- H: 1 − 0 − ½(2) = 0
- Sum = −1, matching the ion. ✓
Final structure: [H–O]⁻ with three lone pairs on oxygen. The negative charge lives on the oxygen, which fits: oxygen is the more electronegative atom and the one that picks up a proton when hydroxide acts as a base.
Example 2: ammonium, NH₄⁺
Electron count. N has 5, each H has 1 (4 in total), and the 1+ charge removes 1. Total = 5 + 4 − 1 = 8 electrons.
Skeleton. N in the centre, four N–H single bonds. That uses all 8 electrons. Nothing is left for lone pairs.
Octets. Nitrogen has four bonds = 8 electrons. ✓ Every H has 2. ✓
Formal charges.
- N: 5 − 0 − ½(8) = +1
- each H: 1 − 0 − ½(2) = 0
- Sum = +1. ✓
Final structure: [NH₄]⁺ drawn as N at the centre with four single bonds to H and no lone pairs, all inside brackets with a + outside.
Notice what happened compared with ammonia, NH₃. Ammonia’s nitrogen had a lone pair. When it accepts H⁺, that lone pair becomes the fourth N–H bond. Once formed, all four bonds are identical, and the ion is a regular tetrahedron with angles of about 109.5°. The article on coordinate covalent bonds looks at this type of bond formation in more detail.
Example 3: cyanide, CN⁻
Electron count. C has 4, N has 5, plus 1 for the charge. Total = 4 + 5 + 1 = 10 electrons.
Skeleton. C–N uses 2, leaving 8.
Lone pairs first. Give N three lone pairs (6 electrons) to complete its octet. The last 2 electrons go on C as one lone pair.
Check octets. N: 2 + 6 = 8. ✓ C: 2 + 2 = 4. ✗ Carbon is 4 short.
Fix. Move two of nitrogen’s lone pairs into bonds, making a triple bond. Now each atom has a triple bond (6 bonding electrons) plus one lone pair (2 electrons) = 8. ✓
Formal charges.
- C: 4 − 2 − ½(6) = −1
- N: 5 − 2 − ½(6) = 0
- Sum = −1. ✓
Final structure: [:C≡N:]⁻, one lone pair on each atom.
This one catches people out, because the negative formal charge sits on carbon, the less electronegative atom. There is no better arrangement with 10 electrons that gives both atoms octets, so it is the correct structure. It also makes chemical sense: cyanide typically binds to metal ions through its carbon, using that carbon lone pair.
Example 4: nitrite, NO₂⁻
Electron count. N has 5, each O has 6 (12 in total), plus 1. Total = 5 + 12 + 1 = 18 electrons.
Skeleton. O–N–O. Two single bonds use 4, leaving 14.
Outer octets. Each O needs 6 more electrons as three lone pairs: 12 used, 2 left.
Central atom. The last 2 electrons go on N as a lone pair. N now has 2 + 2 + 2 = 6 electrons. ✗
Fix. Turn one lone pair on one oxygen into an N=O double bond. N now has 4 (double bond) + 2 (single bond) + 2 (lone pair) = 8. ✓
Formal charges.
| Atom | Lone-pair e⁻ | Bonds | Formal charge |
|---|---|---|---|
| N | 2 | 3 | 5 − 2 − 3 = 0 |
| O (double-bonded) | 4 | 2 | 6 − 4 − 2 = 0 |
| O (single-bonded) | 6 | 1 | 6 − 6 − 1 = −1 |
Sum = −1. ✓
Resonance. Either oxygen could have been the one with the double bond. So nitrite has two equivalent resonance structures, and the real ion is an average of them: both N–O bonds are the same length, somewhere between single and double, and the negative charge is shared equally between the two oxygens. Draw both forms with a double-headed arrow between them. The lone pair on nitrogen makes the ion bent rather than linear. The same 18-electron pattern turns up in ozone, covered in the Lewis structure of ozone.
Example 5: carbonate, CO₃²⁻
Electron count. C has 4, three O have 18, and the 2− charge adds 2. Total = 4 + 18 + 2 = 24 electrons.
Skeleton. C in the centre, bonded to three O. Three single bonds use 6, leaving 18.
Outer octets. Each O takes three lone pairs: 18 used, 0 left.
Central atom. C has only three bonds = 6 electrons. ✗ Move one lone pair from any O into a C=O double bond. Now C has 4 + 2 + 2 = 8. ✓
Formal charges.
- C: 4 − 0 − ½(8) = 0
- O (double): 6 − 4 − ½(4) = 0
- each O (single): 6 − 6 − ½(2) = −1 (two of these)
- Sum = −2. ✓
Resonance. The double bond could sit on any of the three oxygens, giving three equivalent resonance structures. In reality all three C–O bonds are identical, each with a bond order of 1⅓ (four bonding pairs shared over three bonds), and each oxygen carries on average −⅔ of a charge. The ion is flat and triangular, with 120° angles. Nitrate, NO₃⁻, is isoelectronic with carbonate (also 24 valence electrons) and has the same pattern, except that the central N carries a formal charge of +1 (5 − 0 − 4), so each form reads +1 − 1 − 1 = −1 overall.
Example 6: sulfate, SO₄²⁻
This is the hardest of the set, because there are two structures you will see in books, and you should understand both.
Electron count. S has 6, four O have 24, plus 2. Total = 6 + 24 + 2 = 32 electrons.
Skeleton. S in the centre with four S–O single bonds: 8 electrons used, 24 left.
Outer octets. Each O takes three lone pairs: 24 used, 0 left. S has four bonds = 8 electrons. ✓ Every atom already has an octet.
Formal charges (structure A, all single bonds).
- S: 6 − 0 − ½(8) = +2
- each O: 6 − 6 − ½(2) = −1 (four of these)
- Sum = +2 − 4 = −2. ✓
Structure B (two S=O double bonds). Sulfur is in period 3, so older formal-charge rules allow it to hold more than eight electrons. Convert a lone pair on each of two oxygens into S=O bonds:
- S: 6 − 0 − ½(12) = 0
- each O (double): 6 − 4 − ½(4) = 0
- each O (single): 6 − 6 − ½(2) = −1 (two of these)
- Sum = −2. ✓
Structure B has smaller formal charges, and many school and first-year textbooks prefer it for that reason. Structure A keeps sulfur to an octet, and modern calculations of the electron distribution suggest the S–O bonds are best described as highly polar single bonds with a strong ionic attraction added on, with no real involvement of sulfur d orbitals. That is closer to structure A. Both give the same shape: a tetrahedron with four equal S–O bonds.
Which should you draw? Follow your syllabus. If a mark scheme asks you to “minimise formal charges”, draw B and mention its resonance forms (the two double bonds can sit on any pair of the four oxygens, six arrangements in all). If it insists on the octet rule, draw A. Either way, state the formal charges and show that they sum to −2. For more on atoms that go past eight electrons, see exceptions to the octet rule.
Key takeaways
- For an ion, add one electron per negative charge and subtract one per positive charge before you draw anything.
- Put the finished structure in square brackets with the charge outside.
- Formal charges on individual atoms must add up to the charge of the ion. This is your built-in check.
- Ions such as NO₂⁻, NO₃⁻ and CO₃²⁻ need resonance structures. The real ion is an average, with equal bonds and the charge spread out.
- The negative formal charge usually sits on the most electronegative atom, but not always (CN⁻ puts it on carbon).
- Sulfate can be drawn with all single bonds (S +2) or with two double bonds (S 0). Know both and why each is used.
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