Octet rule exceptions

Lewis taught us to count to eight. Draw dots around an atom, pair them into bonds and lone pairs, and stop once every atom has eight electrons around it — a full octet, borrowed from the stability of the noble gases. It correctly predicts water, methane, ammonia and hundreds of everyday molecules. But BeCl $_2$ , BF $_3$ , PCl $_5$ , SF $_6$ and NO are all perfectly real, perfectly stable species, and not one obeys the octet rule. Some central atoms stop short of eight electrons; others comfortably exceed it; a few molecules cannot even pair every electron. Knowing these three exception categories — and, more usefully, predicting which one a new molecule falls into just by counting valence electrons — is exactly what the ISC Class 11 syllabus expects.

The core insight

Octet-rule exceptions split into three counted categories — incomplete octets (too few valence electrons, as in Be and B compounds), expanded octets (period 3+ central atoms using empty valence-shell $d$ orbitals to exceed eight electrons), and odd-electron molecules (an odd total valence-electron count that makes complete pairing structurally impossible).

The core idea

The octet rule is a bookkeeping shortcut, not a law of nature: atoms are stable when their valence shell resembles a noble gas's eight electrons, and most Lewis structures can be drawn to achieve exactly that. The three standard exceptions arise from three different reasons the bookkeeping breaks down.

Incomplete octet — too few electrons to reach eight. Beryllium (2 valence electrons) and boron (3 valence electrons) do not have enough electrons, or enough electronegative partners, to build four complete bond pairs. In $\text{BeCl}_2$ the central Be forms only 2 bond pairs (4 electrons); in $\text{BF}_3$ the central B forms 3 bond pairs (6 electrons). Both stop two electrons short of an octet and remain stable, because adding a fourth pair would give an electronegative halogen an unfavourable positive formal charge.

Expanded (extended) octet — more than eight electrons. From period 3 onward, the central atom's valence shell includes empty, energetically accessible $d$ orbitals ( $3d$ , $4d$ , ...) alongside its $s$ and $p$ orbitals. Electron promotion into these lets the atom support more than four bond pairs — variable covalency. $\text{PCl}_5$ (10 electrons around P, $sp^3d$ ) and $\text{SF}_6$ (12 electrons around S, $sp^3d^2$ ) are the standard examples. Period 2 elements (C, N, O, F) have no $d$ subshell at all — $n=2$ offers only $2s$ and $2p$ — so they can never expand past eight.

Odd-electron molecules — an odd total that cannot be fully paired. Most molecules have an even total valence-electron count, pairing neatly into bonds and lone pairs. A handful — nitric oxide $\text{NO}$ and nitrogen dioxide $\text{NO}_2$ are the textbook pair — have an odd total instead, so one electron is structurally forced to remain unpaired. The quantity that decides this is the total valence-electron count:

N_{\text{total}} = V_A + \sum_i n_i\, V_{X_i}

where: $N_{\text{total}}$ is the total number of valence electrons in the molecule (plus/minus one per unit of ionic charge, if any), $V_A$ is the valence-electron count of the central atom, $n_i$ is the number of atoms of a given peripheral element $X_i$ , and $V_{X_i}$ is that element's own valence-electron count (its group number). If $N_{\text{total}}$ is even it can be arranged as $N_{\text{total}}/2$ complete pairs; if odd, one electron is always left over, unpaired. Molecules with an unpaired electron are free radicals, and because an unpaired electron carries a magnetic moment, they are paramagnetic — unlike almost all even-electron molecules, which are diamagnetic.

Figure

Three panels showing an incomplete octet at boron with six electrons an expanded octet at sulphur with twelve electrons and an odd electron molecule with one unpaired dot on nitrogen oxide

Derivation APredicting an odd-electron molecule from the total valence-electron count

The goal is to show, from nothing more than adding up group numbers, that a molecule's total valence-electron count alone decides whether it is forced into the odd-electron exception category, so a student can classify NO and NO $_2$ as free radicals before drawing a single dot, purely by parity of a sum.

Step 1. List each atom's valence-electron contribution by group number: N (group 15) contributes 5, O (group 16) contributes 6, F (group 17) contributes 7, and so on.

Step 2. Add every atom's contribution using $N_{\text{total}} = V_A + \sum_i n_i V_{X_i}$ . For NO: one N and one O give $N_{\text{total}} = 5 + 6 = 11$ . For NO $_2$ : one N and two O give $N_{\text{total}} = 5 + 2(6) = 17$ .

Step 3. Test the parity of $N_{\text{total}}$ . A Lewis structure only places electrons into bond pairs or lone pairs — both are pairs of two — so the electron count used in any structure must be even, $2k$ . If $N_{\text{total}}$ is odd, as with 11 and 17 above, it cannot be written as $2k$ ; the closest a structure gets is $(N_{\text{total}}-1)$ electrons in pairs plus exactly one electron left over, unpaired.

Step 4. Conclude the classification. Since $N_{\text{total}} = 11$ for NO and $17$ for NO $_2$ , both are odd, so both must carry one unpaired electron in any correct structure — they are odd-electron molecules by arithmetic alone, independent of how the atoms are arranged. For a molecule whose atom contributions sum to an even total, $N_{\text{total}}$ is even and no such exception arises.

Worked example

Classify $\text{PCl}_5$ using the total valence-electron count, confirm it is an expanded-octet species (not an odd-electron one), and state its hybridisation.

Step 1. Total valence electrons: P (group 15) contributes 5, and each of 5 chlorine atoms (group 17) contributes 7, so $N_{\text{total}} = 5 + 5(7) = 40$ electrons, i.e. 20 pairs. Since 40 is even, PCl $_5$ is not a candidate for the odd-electron category.

Step 2. Build the skeleton: P is central, singly bonded to all 5 Cl atoms, using 5 bond pairs (10 electrons); the remaining $40-10=30$ electrons (15 pairs) sit as 3 lone pairs on each of the 5 Cl atoms, completing every chlorine's own octet.

Step 3. Count electrons around the central P: 5 bond pairs $= 10$ electrons — two pairs more than an octet's 4 pairs. This is an expanded octet, possible because phosphorus (period 3, $n=3$ ) has empty $3d$ orbitals available for electron promotion.

Step 4. Assign hybridisation from the electron-pair geometry: 5 bond pairs and 0 lone pairs on P means 5 hybrid orbitals, giving $sp^3d$ and a trigonal bipyramidal shape — the same pattern as $\text{SF}_6$ 's $sp^3d^2$ from Pass-1. Answer: $N_{\text{total}} = 40$ (even, so not odd-electron); P carries an expanded octet of 10 electrons; hybridisation is $sp^3d$ .

Competitive stretch

A quick check worth remembering for objective questions: run the same parity test on $\text{ClO}_2$ . Cl (group 17) contributes 7, and 2 O (group 16) contribute $2(6)=12$ , so $N_{\text{total}} = 7 + 12 = 19$ — odd. Chlorine dioxide is therefore also predicted, correctly, to be an odd-electron (paramagnetic) species, alongside NO and NO $_2$ , without needing to draw its structure first.

Common traps

Treating "fewer than eight electrons" as a drawing mistake rather than a genuine, stable exception — Be and B compounds are supposed to fall short of an octet.
Attempting to draw an expanded octet for a period 2 element (forcing structures like NCl $_5$ or OF $_4$ ) — no period 2 element has valence-shell $d$ orbitals to expand into.
Forgetting to check parity before assuming every molecule can be drawn with all electrons paired — an odd $N_{\text{total}}$ guarantees an unpaired electron no matter how the structure is arranged.
Mixing up the three categories in an answer — incomplete octet, expanded octet and odd-electron molecule are three separate reasons for failure, not shades of the same exception.

Ask the tutor: Want a quick round of classifying five more molecules — incomplete, expanded, or odd-electron — using just their formulas?

If BF3 has an incomplete octet, why doesn't it just react instantly with something to fix it?

It does react readily, but "instantly" depends on having a partner around. An incomplete octet means boron is short two electrons and has an empty orbital ready to accept a lone pair, which is exactly why BF3 is a strong Lewis acid — it reacts eagerly with electron-pair donors like ammonia to form a coordinate bond, F3B-NH3, completing its octet. In the pure gas with no such donor present, BF3 is perfectly stable on its own; the incomplete octet is a stable resting state, not a broken structure waiting to explode.

Why doesn't nitrogen ever form an expanded octet like phosphorus does in PCl5?

Nitrogen's valence shell is $n=2$ , which only contains $2s$ and $2p$ subshells — there is no $2d$ subshell at all, at any energy, for nitrogen to promote an electron into. Phosphorus sits one row down, in period 3, where the valence shell is $n=3$ and genuinely includes empty $3d$ orbitals close enough in energy to be used for bonding. So the difference is not about how many electrons nitrogen has, but about which orbitals physically exist in its shell — no amount of extra bonding demand can conjure a $2d$ orbital into existence.

Are odd-electron molecules like NO dangerous or unusual chemically, since they cannot be drawn with the octet rule?

They are unusual in structure but not exotic in real life — NO is produced continuously in car engines, and also by the human body as a signalling molecule that relaxes blood vessels. Because the unpaired electron makes such molecules paramagnetic and often reactive, free radicals do tend to combine readily (two NO $_2$ molecules can even dimerise into N $_2$ O $_4$ , pairing their two odd electrons together). But as isolated molecules NO and NO $_2$ are entirely stable and routinely studied; "cannot satisfy the octet rule" only means the electron-counting bookkeeping fails, not that the molecule is somehow broken.

Octet rule exceptions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads