The big idea: A covalent bond is a shared pair of electrons between two non-metal atoms. By sharing, each atom gets a stable, full outer shell (usually an octet — eight electrons).
A Lewis (electron-dot) structure is the picture of this: every bond is a shared pair (drawn as a line), and every lone pair (a non-bonding pair) is drawn as two dots.
Each line is a shared (bonding) pair; each pair of dots is a lone pair.
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Two kinds of pair: - Bonding pair — shared between two atoms → drawn as a line. - Lone pair — stays on one atom → drawn as two dots.
Every valence electron must appear somewhere: in a bond or in a lone pair.
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Atoms can share more than one pair. Sharing one pair is a single bond, two pairs a double bond, three pairs a triple bond. More shared pairs means a shorter, stronger bond.
Two double bonds (O=C=O). The carbon has no lone pairs.
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A triple bond (N≡N) plus one lone pair on each nitrogen.
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Bond order: The number of shared pairs is the bond order:
- C–C single → bond order 1 - C=C double → bond order 2 - C≡C / N≡N triple → bond order 3
Higher bond order → shorter, stronger bond. (That is why N≡N in nitrogen gas is so hard to break.)
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The method
- Count the total valence electrons (add up each atom's group number; add/subtract for ions).
- Put the least electronegative atom in the centre (never hydrogen).
- Join the outer atoms to the centre with single bonds first.
- Add lone pairs to the outer atoms to complete their octets.
- Any electrons left go on the central atom; if it is short of an octet, make double/triple bonds.
Worked example — methane, CH₄
Draw the Lewis structure of methane, CH4.
Solution
- Count electrons: C contributes 4, each H contributes 1 → 4 + 4 = 8 valence electrons (4 pairs).
- Centre: carbon is central; the four H atoms surround it.
- Bonds: four C–H single bonds use all 4 pairs — carbon now has its octet and each H has 2 electrons (full for H). No lone pairs left.
Final answer
Four C–H single bonds, no lone pairs — shown below.
Four single bonds, no lone pairs on carbon.
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How this is tested: 'Deduce / draw the Lewis structure of …' is a near-guaranteed Paper 2 mark, and Paper 1A often asks which molecule obeys the octet rule or compares bond order.
The markers want: the right number of bonds, the correct bond order (single/double/triple), and every lone pair shown.
Octet exceptions: A few central atoms do not reach a full octet:
- BF3 — boron has only 6 electrons (three bonds, no lone pair). - BeCl2 — beryllium has only 4.
These 'electron-deficient' molecules are common exam traps.
IB-style question — hydrogen cyanide (a)
(a) Deduce the Lewis structure of hydrogen cyanide, HCN. [2]
How to score the marks
- Count: H = 1, C = 4, N = 5 → 10 valence electrons (5 pairs).
- Centre: carbon (least electronegative, not H). Arrange H–C–N.
- Bonds: H–C is a single bond. To give both C and N an octet, C≡N must be a triple bond, leaving one lone pair on N.
Final answer
H–C≡N, with one lone pair on the nitrogen (shown below).
H–C≡N: a single bond to H and a triple bond to N (lone pair on N).
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IB-style question — sulfur dioxide (b)
(b) Deduce a Lewis structure of sulfur dioxide, SO2. [2]
How to score the marks
- Count: S = 6, each O = 6 → 6 + 12 = 18 valence electrons (9 pairs).
- Centre: sulfur, with the two oxygens either side (bent).
- Bonds & pairs: one S=O double and one S–O single bond, lone pairs completing each oxygen's octet, and one lone pair on sulfur.
Final answer
A bent molecule with one S=O and one S–O bond and a lone pair on S (shown below).
One double and one single S–O bond, with a lone pair on sulfur.
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