Visual walkthrough — Octet rule — Lewis structures, exceptions (incomplete octet, expanded octet, odd-electron species)

Before line one, let me name every symbol I am about to use, in plain words:

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Step 1 — Count what you have (every dot must be real)

WHAT. We total up every valence electron the atoms bring to the table.

WHY. You cannot draw a dot that does not exist. Chemistry conserves electrons: the number you draw at the end must equal the number you started with. So first we count the "budget."

For $C{O}_2$: carbon is in group 14 → it brings 4 valence electrons. Each oxygen is in group 16 → 6 each. (Group number tells you valence count — that's a periodic fact.)

$N_{\text{total}}=\underset{\text{C brings}}{\underbrace{4}}+\underset{\text{two O’s bring}}{\underbrace{2\times 6}}=16$

Here $N_{\text{total}}$ is our whole electron budget — 16 dots, no more, no less.

PICTURE. Three separate atoms, each holding its own loose electrons like coins in a hand.

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Step 2 — Count what you'd need if nobody shared

WHAT. Pretend every atom is greedy and alone, demanding a full octet with its own electrons.

WHY. This gives us a "wish list." Comparing the wish list to the real budget will reveal how much sharing must happen — that is the whole trick of Step 3.

$N_{\text{need}}=\underset{\text{C, O, O each want 8}}{\underbrace{8\times 3}}=24$

The $8$ is "one full octet"; the $3$ is "how many non-hydrogen atoms." (Hydrogen would count as $2$, not $8$ — it only wants a duet. No H here.)

PICTURE. Each atom now demands a full ring of 8, side by side — a total demand of 24, larger than our 16-electron budget.

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Step 3 — The shortfall is the sharing (this is the key idea)

WHAT. Subtract the real budget from the wish list.

WHY — the heart of the whole method. The wish list (24) is bigger than the budget (16). We are "short" by 8 electrons. But here is the magic: a shared electron gets counted by both atoms at once. So the shortfall is not a problem — it's exactly the number of electrons that must be shared to make everyone's octet add up.

$N_{\text{shared}}=\underset{24}{\underbrace{N_{\text{need}}}}-\underset{16}{\underbrace{N_{\text{total}}}}=8$

Each bond is one shared pair = 2 electrons, so:

$\text{bonds}=\frac{N_{\text{shared}}}{2}=\frac{8}{2}=4$

$N_{\text{shared}}$ = electrons doing double-duty; dividing by 2 turns "shared electrons" into "shared pairs", i.e. lines.

PICTURE. The overlap zone: an electron in the shared region is claimed by the ring on the left and the ring on the right simultaneously.

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Step 4 — Place the bonds between the atoms

WHAT. We have 4 bonds to distribute across 2 carbon–oxygen connections (O–C–O).

WHY. Carbon is least electronegative, so it sits in the centre (Electronegativity decides this — the atom that "holds electrons weakest" is happiest sharing with many neighbours). There are only two links, O–C and C–O. Four bonds ÷ two links = two bonds per link = a double bond on each side.

$\underbrace{O}=\underset{\text{centre}}{\underbrace{C}}=\underbrace{O}$

PICTURE. Four bond-lines snapping into place: two on the left link, two on the right.

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Step 5 — Spend the leftover electrons as lone pairs, then verify

WHAT. We used $4\text{ bonds}\times 2=8$ electrons in bonds. Budget was 16, so $16-8=8$ electrons remain. Drop them onto the atoms that still need to reach 8.

WHY. Carbon already has 8 (four bonds × 2). Each oxygen only has 4 from its double bond, so each oxygen needs 4 more → two lone pairs each. $2\text{ O}\times 4=8$ — exactly the leftover. It balances perfectly, which is our sanity check.

Now verify every atom:

$\text{O: }\underset{\text{double bond}}{\underbrace{4}}+\underset{\text{2 lone pairs}}{\underbrace{4}}=8✓\text{C: }\underset{\text{four bonds}}{\underbrace{8}}=8✓$

PICTURE. The finished structure $\ddot{O}=C=\ddot{O}$, each atom circled with its 8.

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Step 6 — Edge case: too few electrons (incomplete octet, $B{F}_3$)

WHAT. Run the same machine on boron trifluoride.

WHY it breaks. Boron is group 13 → only 3 valence electrons. The arithmetic suggests a fourth bond (a B=F double bond) to give boron an octet — but check the formal charge:

$FC(\text{B in B=F})=3-0-4=-1,FC(\text{double-bonded F})=7-4-2=+1$

A positive charge on fluorine — the most electronegative atom in the whole periodic table — is absurd. So nature refuses the octet. Boron settles for only 6 electrons.

PICTURE. Boron with just three single bonds and a visibly empty slot — an electron-deficient atom, which is exactly why it behaves as a Lewis acid.

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Step 7 — Edge case: too many electrons (expanded octet, $S{F}_6$)

WHAT. Sulfur bonds to six fluorines → six S–F bonds → 12 electrons around sulfur.

WHY it's allowed. Sulfur is in period 3, so its valence level has extra empty orbitals it can use (Hypervalency & d-orbital participation). Period-2 atoms (N, O, F) sit too high up and too small — they physically cannot hold more than 8, ever. So the octet is a ceiling for period 2 but only a default for period 3 and below.

PICTURE. Sulfur surrounded by 12 electrons (6 bonds) — a ring stretched past 8.

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Step 8 — Edge case: an odd electron count (radical, $NO$)

WHAT. Nitric oxide: $N(5)+O(6)=11$ electrons — an odd number.

WHY it breaks. Octets and lone pairs come in pairs. With 11 electrons you can never pair every single one — one electron is left unpaired, no matter how you draw it. That lonely electron makes $NO$ a reactive free radical and paramagnetic (attracted by a magnet). See Free radicals.

PICTURE. The molecule with one visibly single, unpaired dot that has no partner.

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The one-picture summary

Everything above is one flow: count the budget → count the wish → the gap is the sharing → place bonds → fill lone pairs → verify → and know the three ways it can bend.

Recall Feynman retelling — the whole walkthrough in plain words

Every atom wants 8 marbles around it (hydrogen's happy with 2). First I count all the marbles the atoms actually brought — that's my budget. Then I ask, "If nobody shared, how many marbles would they all together demand?" That's always bigger, because shared marbles get double-counted. The difference between the demand and what I have is exactly how many marbles must be shared — divide by 2 and I know how many hand-holds (bonds) to draw. I snap the bonds between atoms (weakest-grabber in the middle), then sprinkle the leftover marbles as lone pairs until everyone shows 8, and I double-check each atom. Finally: three kids don't play by the rule — boron is chill with only 6, sulfur is greedy and grabs 12 because it's big enough (period 3), and NO got an odd number of marbles so one is stuck partnerless and twitchy. Same machine every time — I only watch for those three.

Connections

• VSEPR theory — once the Lewis picture is done, electron domains give the shape.
• Formal charge & resonance — the tie-breaker used in Steps 5 and 6.
• Electronegativity — decides the central atom and where charge belongs.
• Lewis acids and bases — the incomplete-octet $B{F}_3$ of Step 6.
• Hypervalency & d-orbital participation — the expanded octet of Step 7.
• Free radicals — the odd-electron $NO$ of Step 8.
• Periodic trends — why only period ≥3 atoms can expand.