Lewis Structure of Ozone (O₃) and Resonance, Worked Step by Step

Ozone is only three oxygen atoms, yet it is one of the best molecules in chemistry for learning what a Lewis structure can and cannot tell you. Draw it once and you get a structure with a double bond on one side and a single bond on the other. Measure the real molecule and the two bonds turn out to be identical. Resolving that contradiction is the whole point of resonance.

The six problems below build from a plain electron count to predictions of bond length, shape and polarity, and finish by comparing ozone with two molecules that share its electron count. Each step is shown so you can check your own working against it.

Problem 1: count the electrons and build the skeleton

Question. How many valence electrons does O₃ have, and how are the atoms connected?

Electron count. Oxygen is in group 16, so each atom brings 6 valence electrons (the oxygen element page shows its 2s² 2p⁴ outer shell). Total = 3 × 6 = 18 electrons, which is 9 pairs.

Skeleton. With three identical atoms, there are two sensible options: a chain (O–O–O) or a triangle. We try the chain first, since that is what experiment shows. (Problem 4 looks at why the triangle is rejected.)

O–O–O uses two single bonds = 4 electrons. 14 electrons remain.

Problem 2: complete the octets and find the formal charges

Outer atoms first. Each terminal oxygen has one bond (2 electrons) and needs 6 more: three lone pairs each. That uses 12 electrons, leaving 2.

Central atom. The last 2 electrons go on the middle oxygen as one lone pair. Now count around it: two single bonds (4) + one lone pair (2) = 6 electrons. It is short of an octet. ✗

Fix. Turn one lone pair on a terminal oxygen into a second bond to the centre. Call the atoms Oa, Ob (central) and Oc:

• Oa=Ob: Oa now has two lone pairs and a double bond.
• Ob–Oc: Oc keeps three lone pairs and a single bond.
• Ob has a double bond, a single bond and one lone pair.

Recount. Bonds: 3 pairs = 6 electrons. Lone pairs: Oa 2, Ob 1, Oc 3 = 6 pairs = 12 electrons. Total = 18. ✓

Octets. Oa: 4 + 4 = 8. Ob: 4 + 2 + 2 = 8. Oc: 2 + 6 = 8. ✓

Formal charges (valence − lone-pair electrons − ½ bonding electrons):

Atom	Lone-pair e⁻	Bonding e⁻	Formal charge
Oa (double-bonded, terminal)	4	4	6 − 4 − 2 = 0
Ob (central)	2	6	6 − 2 − 3 = +1
Oc (single-bonded, terminal)	6	2	6 − 6 − 1 = −1

Sum = 0, as it must be for a neutral molecule. ✓

So even though ozone is neutral overall, its best single Lewis structure has a +1 on the central atom and a −1 on one end. This is a real feature of the molecule, not a mistake in the working. For more practice with the formula, see formal charge.

Problem 3: resonance, bond order and bond length

Question. The double bond in Problem 2 could just as well have been drawn on the other side. What does that mean for the real molecule?

Two resonance structures. Swap the roles of Oa and Oc and you get a second structure, equally valid, with the same formal charges on mirror-image atoms:

• Form I: Oa=Ob–Oc (−1 on Oc)
• Form II: Oa–Ob=Oc (−1 on Oa)

They are drawn with a double-headed arrow between them. Neither form exists on its own. The molecule does not flip back and forth between them; it is a single structure, a resonance hybrid, that the two drawings describe together. The general idea is covered in resonance structures.

Bond order. Across the two forms, each O–O link is double in one and single in the other. Average: (2 + 1) ÷ 2 = 1.5. Equivalently, there are 3 bonding pairs spread over 2 links, which is 1.5 per link.

Charge. The −1 sits on Oc in one form and Oa in the other, so each terminal atom carries on average −½, while the centre stays at +1.

Bond length prediction. A bond order of 1.5 should give a length between a typical O–O single bond and the O=O double bond. The numbers bear this out:

Bond	Example	Length (approx.)
O–O single	hydrogen peroxide, H₂O₂	148 pm
O–O in ozone	O₃	128 pm
O=O double	oxygen gas, O₂	121 pm

Both bonds in ozone measure about 128 pm: equal to each other, and between the other two values. That is the experimental fingerprint of resonance. For how bond order links to length and strength more generally, see bond order calculations.

Problem 4: why not a different structure?

Question. Test two alternatives: (a) the chain with a triple bond, O≡O–O, and (b) a triangle of three O–O single bonds. Why do chemists reject each?

(a) O≡O–O. 18 electrons: a triple bond (6) and a single bond (2) use 8, leaving 10. The single-bonded end needs three lone pairs (6), the triple-bonded end needs one lone pair (2), and the centre gets the last pair (2). Check the centre: 6 + 2 + 2 = 10 electrons. Oxygen is a period 2 element and cannot hold more than 8. Even ignoring that, the formal charges are poor: triple-bonded O 6 − 2 − 3 = +1, centre 6 − 2 − 4 = 0, single-bonded O 6 − 6 − 1 = −1. Rejected, because it breaks the octet limit.

(b) The triangle. Three single bonds use 6 electrons, leaving 12, which is two lone pairs on each atom. Each oxygen: 2 bonds (4) + 2 lone pairs (4) = 8. ✓ Formal charges: 6 − 4 − 2 = 0 on every atom.

On formal charges alone the triangle looks better than the bent chain. But forcing three atoms into a triangle means 60° bond angles, far from what the electron pairs on each oxygen would prefer, and that strain makes the ring much less stable. The measured structure of ozone is the open, bent chain.

The lesson: formal charge is a useful guide for choosing between structures that have the same connections, but it is not the only factor. Always check the answer against geometry and experiment where you can.

Problem 5: predict the shape and polarity

Question. Use the Lewis structure to predict ozone’s shape. Is the molecule polar?

Electron groups on the central atom. In either resonance form, Ob has one double bond, one single bond and one lone pair. For shape, a multiple bond counts as one group. So Ob has three electron groups: two bonding, one lone pair.

Shape. Three groups arrange themselves in a trigonal planar pattern, but with one position taken by a lone pair the atoms form a bent molecule. The lone pair pushes the bonds a little closer than 120°; the measured O–O–O angle is about 117°. The VSEPR guide covers this bent-from-trigonal case.

Polarity. Every bond joins two oxygen atoms, so you might expect no polarity at all. But the atoms are not in the same situation: the centre carries a formal +1 and each end, on average, −½. Combined with the bent shape, those charges do not cancel, and ozone has a small overall dipole moment. It is a polar molecule, although only weakly so, which is unusual for a substance made of a single element.

Problem 6: the same 18 electrons in other molecules

Question. The nitrite ion, NO₂⁻, and sulfur dioxide, SO₂, both have 18 valence electrons. Compare their Lewis structures with ozone’s.

Nitrite, NO₂⁻. N 5 + O 2 × 6 + 1 (charge) = 18. ✓ The working mirrors ozone’s exactly: O–N–O, octets on the ends, a lone pair on N, then one N=O double bond. Formal charges: N 5 − 2 − 3 = 0, double-bonded O 0, single-bonded O 6 − 6 − 1 = −1. Two resonance forms, equal N–O bonds, and a bent shape. The only difference from ozone is that the central atom is now neutral, and the whole ion carries the −1. It is worked in full in Lewis structures of polyatomic ions.

Sulfur dioxide, SO₂. S 6 + O 2 × 6 = 18. ✓ Following the octet rule gives the same pattern as ozone: O=S–O, with S carrying a formal charge of +1 and the single-bonded O −1, plus a second resonance form. Because sulfur is in period 3, some textbooks also draw O=S=O, which gives zero formal charge on every atom but puts 10 electrons around sulfur. Syllabuses differ on which to accept; the octet-rule version with resonance is closer to modern descriptions of the bonding. Either way the molecule is bent, with an angle a little under 120°, for the same reason ozone is: three electron groups on the central atom, one of them a lone pair.

What the three have in common: same electron count, same arrangement of bonds and lone pairs, two equivalent resonance forms, equal bond lengths and a bent shape. Species like these, with the same number of valence electrons in the same arrangement, are called isoelectronic, and recognising them is a quick shortcut in exams.

Key takeaways

• O₃ has 18 valence electrons. The octet-rule structure is O=O–O with formal charges 0, +1, −1.
• Two equivalent resonance forms exist. The real molecule is a hybrid with two equal O–O bonds of bond order 1.5, about 128 pm long, between the O–O single (≈148 pm) and O=O double (≈121 pm) bonds.
• The central oxygen has three electron groups, one a lone pair, so ozone is bent, with an angle of about 117°, and weakly polar.
• Formal charges alone would favour a triangular ring, but geometry rules it out: always sanity-check a structure against shape and experiment.
• NO₂⁻ and SO₂ have the same 18-electron pattern and the same bent, resonance-stabilised structure.