Foundations · Section 7 of 64

Lewis structures

Practice this — interactive lesson

A Lewis structure is the working drawing of organic chemistry. It shows which atoms are joined to which, where every valence electron sits, and which atoms carry charge — and from those three things you can predict geometry, polarity, acidity and reactivity. Nearly every mistake in a mechanism can be traced to a Lewis structure drawn carelessly at the start.

The recipe

The octet rule says most atoms are most stable with 8 electrons in their outer shell (2 for hydrogen). A Lewis structure arranges a molecule's valence electrons — some as shared bonding pairs, some as lone pairs on a single atom — to get every atom as close to that as possible.

  1. Count total valence electrons. Add up each atom's group-number valence count. Then adjust for charge: add one electron per negative charge, subtract one per positive charge.
  2. Connect the atoms with single bonds. The least electronegative atom (never hydrogen) is usually central; carbon is essentially always central when present. Each bond uses 2 of your electrons.
  3. Give the outer atoms full octets with lone pairs, working from the most electronegative outward.
  4. Put any electrons still left over on the central atom.
  5. If the central atom is still short of an octet, convert a lone pair on a neighbor into a second shared pair — a double bond. This gives the central atom more electrons without changing the total count, which is why step 1 is the one you must not get wrong.

Then check formal charges, and check that they sum to the overall charge. That check is the whole reason the previous section exists.

Worked example — CO₂

Count: C (4) + O (6) + O (6) = 16 valence electrons.

Connect: O–C–O with two single bonds uses 4, leaving 12.

Fill the outer atoms: three lone pairs on each oxygen uses all 12, leaving 0.

Check the center: carbon has only 2 bonds — 4 electrons, four short of an octet.

Fix it: move one lone pair from each oxygen into a second bond, giving O=C=O. Carbon now has 4 bonds (8 electrons) and each oxygen has 2 bonds plus 2 lone pairs (8 electrons). Formal charges: carbon 4 − 0 − 4 = 0; each oxygen 6 − 4 − 2 = 0. All zero, summing to 0 for a neutral molecule. ✓

1. COUNT16 valence electronsC 4 + O 6 + O 616 e⁻2. CONNECTO–C–O uses 412 leftCOO3. FILL THE OUTER ATOMS3 lone pairs each uses 120 leftCOOcarbon: 2 bonds = 4 electrons — four short4. FIX THE CENTREshare two more pairsevery atom at 8 ✓COO000all formal charges zero, summing to 0 ✓
The recipe run once, on CO₂. Nothing is ever added or removed after step 1 — step 4 only moves a pair that is already on the page, turning a lone pair on each oxygen into a second shared pair. That is why miscounting step 1 wrecks everything downstream, and why the formal-charge sum at the end is a real check rather than a formality.
Worked example — a charged species, NO₃⁻

Count: N (5) + 3 × O (6) = 23, plus 1 for the negative charge = 24.

Connect: three N–O single bonds uses 6, leaving 18 — exactly three lone pairs for each oxygen. All 24 placed.

Check the center: nitrogen has 3 bonds, 6 electrons, two short.

Fix it: move one lone pair from one oxygen into a double bond. Now nitrogen has 4 bonds: FC = 5 − 0 − 4 = +1. The doubly-bonded oxygen is 6 − 4 − 2 = 0; the two singly-bonded oxygens are 6 − 6 − 1 = −1 each. Sum: +1 + 0 − 1 − 1 = −1. ✓

Note that the choice of which oxygen gets the double bond was arbitrary — three equally good structures exist. That is not a defect in the method; it is the molecule telling you it has resonance, which is the subject of Module 2.

OO−O−N+sum = −1 ✓O−OO−N+sum = −1 ✓O−O−ON+sum = −1 ✓↔↔NO₃⁻ — three equally good answers
Nitrate, drawn three times. The recipe says to promote one oxygen lone pair into a double bond, but it cannot say which oxygen — and no experiment can either, because all three N–O bonds in real nitrate are measurably identical. The method has not failed here; it has detected something. When one Lewis structure is not enough, the molecule has resonance, and Module 2 is about what that means.+1 on nitrogen, −1 on each singly-bonded oxygen, whichever one you pick

Using formal charge as the tiebreaker

When more than one arrangement satisfies the octet rule, the realistic structure is the one that keeps formal charges closest to zero, and puts any negative charge on the most electronegative atom. For CO₂, the symmetric O=C=O with every formal charge at zero beats the alternative with one single and one triple bond, which forces a −1 on one oxygen and a +1 on the other. Same atom count, same octets, worse charge distribution.

Three real exceptions to the octet rule

Incomplete octets. Boron and beryllium routinely stop short. BF₃ has only 6 electrons on boron, and that is the correct structure as drawn, not an error — which is exactly why BF₃ is such a powerful electrophile, or in the vocabulary of Module 3, a strong Lewis acid: an electron-pair acceptor with a genuinely empty orbital waiting. Carbocations belong in this family too.

Expanded octets. Sulfur, phosphorus and other period 3+ elements really can hold more than 8 electrons — SF₆, H₂SO₄, PCl₅. The reason is that these atoms are physically large enough to accommodate extra neighbors, and the extra bonds carry substantial ionic character, so the central atom is not truly sharing a full covalent octet with each one. It is not because empty d orbitals mix into the bonding, as older textbooks claimed; modern computation has retired that explanation. Carbon, nitrogen, oxygen and fluorine can never do this — period 2 atoms are simply too small — and "I gave carbon five bonds" is the single most common structural error in a first organic course.

Odd-electron species. When the total valence electron count is odd, a full octet on every atom is arithmetically impossible. Nitric oxide (NO) has 11 valence electrons and one unpaired electron; so does every free radical. These are real, they are reactive, and they get their own treatment when radical mechanisms come up.

FFFBIncomplete octetBF₃ — boron has only 6and a genuinely empty orbitalso it is a strong Lewis acidSFFFFFFExpanded octetSF₆ — sulfur holds 12a period-3 atom has the roomC, N, O and F never canNOone unpairedOdd electron countNO — 11 valence electronsso a full octet everywhereis arithmetically impossiblethis is a radical
The three places the octet rule genuinely stops applying — and none of them is a licence to give carbon five bonds. Boron stops short of eight and leaves an empty orbital, which is precisely what makes BF₃ hungry for an electron pair. Sulfur goes past eight because a period-3 atom is physically big enough to seat six neighbours. And an odd electron count simply cannot be paired off, which is what a radical is.
Carbon never gets five bonds. If a structure you have drawn has a carbon with five bonds, it is wrong, full stop — there is no exception, no resonance form, and no clever argument that rescues it. The same applies to nitrogen: four bonds maximum, and a nitrogen with four bonds always carries a +1 charge. When arrow-pushing goes wrong later in the course, this is usually the symptom, and the cause is almost always forgetting to break a bond while forming a new one.
Practise the count-and-check loop until it is automatic: total electrons, place them, verify octets, verify formal charges, verify the sum. It takes under a minute for any molecule in this course, and it catches errors that would otherwise propagate through an entire mechanism. Most structural mistakes in organic exams are not conceptual — they are a miscount in step 1.

What carries forward

Everything. Molecular geometry is read off a completed Lewis structure; resonance is the observation that some molecules need more than one; arrow pushing is the practice of converting one Lewis structure into another legally; and every mechanism in the course is a sequence of Lewis structures. If one section of Module 1 is worth over-practising, it is this one.