Exceptions to the Octet Rule
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You should be comfortable drawing Lewis structures, calculating formal charge, and identifying the total number of valence electrons in a molecule.
Why this matters
The octet rule is a powerful heuristic, but it fails for many stable, common molecules.
While the octet rule successfully predicts the bonding in compounds like water and methane, it is not a universal law of nature. Many main-group elements, particularly those in Period 3 and beyond, form stable compounds where the central atom is surrounded by fewer than eight electrons (electron-deficient) or more than eight electrons (electron-rich). Understanding these exceptions is crucial for predicting molecular geometry, reactivity, and the stability of inorganic compounds that are vital in industry and biology.
The Limits of the Octet Rule
The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their valence shell, giving them the same electron configuration as a noble gas. This works well for Period 2 elements (B to Ne) because their valence shell consists only of the 2s and 2p orbitals, which can hold a maximum of eight electrons. However, for elements in Period 3 and below, the valence shell includes d-orbitals (3d, 4d, etc.). Although the energetic availability of these d-orbitals in ground-state bonding is a subject of ongoing debate, the empirical fact remains that these atoms can accommodate more than eight electrons in their valence shell. Conversely, some atoms, like boron, are so electron-deficient that they form stable molecules with only six valence electrons. These are not "errors" in the Lewis model but rather cases where the simple octet rule is insufficient to describe the bonding.
Electron-Deficient Molecules
Electron-deficient molecules are those in which the central atom has fewer than eight electrons in its valence shell. The most common example is boron, which has three valence electrons. In boron trifluoride, BF3, boron forms three single bonds with fluorine atoms. Boron contributes three electrons, and each fluorine contributes one to the bond, resulting in six electrons around the boron atom. This molecule is stable because the empty p-orbital on boron can accept electron density from the filled p-orbitals of the fluorine atoms, creating a partial double-bond character that stabilizes the structure. Another class of electron-deficient species includes carbenes and radicals, though these are often transient. In the context of stable main-group compounds, boron and aluminium compounds are the primary examples. These molecules are often strong Lewis acids because the central atom has a vacant orbital that can accept an electron pair from a Lewis base.
Expanded Octets
Elements in Period 3 and beyond can form expanded octets, where the central atom is surrounded by more than eight electrons. This is possible because these elements have access to d-orbitals in their valence shell, allowing them to accommodate additional electron pairs. A classic example is sulphur hexafluoride, SF6. Sulphur has six valence electrons and forms six single bonds with fluorine atoms, resulting in twelve electrons around the sulphur atom. Similarly, phosphorus pentachloride, PCl5, has ten electrons around the phosphorus atom. It is important to note that expanded octets are generally limited to elements in Period 3 and below. Period 2 elements, such as carbon, nitrogen, oxygen, and fluorine, cannot form expanded octets because they do not have accessible d-orbitals in their valence shell. Any Lewis structure that shows a Period 2 atom with more than eight electrons is incorrect.
Odd-Electron Molecules
Some molecules contain an odd number of valence electrons, making it impossible for all atoms to satisfy the octet rule. These are known as free radicals. The most common example is nitric oxide, NO. Nitrogen has five valence electrons, and oxygen has six, giving a total of eleven valence electrons. When we draw the Lewis structure, we form a double bond between nitrogen and oxygen, and place the remaining electrons as lone pairs. Nitrogen ends up with seven electrons, and oxygen with seven. Neither atom has a full octet. These molecules are highly reactive because they seek to pair their unpaired electron. Another example is chlorine dioxide, ClO2, which has nineteen valence electrons. In such cases, the Lewis structure is drawn to minimise formal charge, even though the octet rule is violated. The presence of an unpaired electron is often indicated by a dot in the Lewis structure.
The result
Lewis structures and the octet rule
A molecule's total valence electrons can be distributed as bonding pairs (shared between two atoms) and lone pairs (localised on one atom) so that, wherever possible, every main-group atom is surrounded by eight electrons (two, for hydrogen). Where more than one such distribution is possible, the formal charge \(FC=V-L-\tfrac{B}{2}\) (valence electrons \(V\), lone-pair electrons \(L\), bonding electrons \(B\)) on each atom selects the best structure: minimise the magnitude of formal charges, and place any necessary negative formal charge on the more electronegative atom.
A common misconception is that expanded octets are always due to the use of d-orbitals. Modern computational chemistry suggests that the bonding in molecules like SF6 is better described by ionic character and hypervalent bonding models that do not require significant d-orbital participation. The "expanded octet" is a useful Lewis structure convention, but it does not necessarily imply that d-orbitals are the primary basis for bonding. Always check the period of the central atom: Period 2 elements cannot expand their octet, regardless of the formal charge calculation. C
Worked example
Problem. Draw the Lewis structure for the phosphate ion, PO43-, and determine the formal charge on the phosphorus atom and each oxygen atom. Assume that the structure minimizes formal charge.
- Calculate the total number of valence electrons. Phosphorus (Group 15) has 5, Oxygen (Group 16) has 6, and the 3- charge adds 3 electrons. Total = \(5 + 4(6) + 3 = 32\) electrons.
- Place phosphorus in the centre and four oxygen atoms around it. Connect them with single bonds. This uses \(4 \times 2 = 8\) electrons. Remaining electrons = \(32 - 8 = 24\).
- Distribute the remaining 24 electrons as lone pairs on the oxygen atoms to complete their octets. Each oxygen gets 3 lone pairs (6 electrons). \(4 \times 6 = 24\) electrons. All electrons are used.
- Check the octet on phosphorus. It has 4 single bonds, so it has 8 electrons. This satisfies the octet rule. However, let's check formal charges. For P: \(FC = 5 - 0 - \frac{8}{2} = 1\). For each O: \(FC = 6 - 6 - \frac{2}{2} = -1\). Total charge = \(1 + 4(-1) = -3\). This is a valid structure, but we can minimise formal charge by forming double bonds.
- Convert one lone pair from an oxygen into a double bond with phosphorus. Now P has 1 double bond and 3 single bonds. P has 10 electrons (expanded octet, allowed for Period 3). The double-bonded O has 2 lone pairs, the single-bonded O's have 3 lone pairs.
- Calculate formal charges for the new structure. P: \(FC = 5 - 0 - \frac{10}{2} = 0\). Double-bonded O: \(FC = 6 - 4 - \frac{4}{2} = 0\). Single-bonded O: \(FC = 6 - 6 - \frac{2}{2} = -1\). Total charge = \(0 + 0 + 3(-1) = -3\). This structure has lower formal charges (0 on P and one O, -1 on three O's) compared to the all-single-bond structure (1 on P, -1 on all O's). Therefore, the structure with one P=O double bond is preferred.
Units check. Formal charge is a dimensionless quantity, so no units are required.
A caution: the formal-charge-minimising P=O drawing is the traditional textbook answer, but modern calculations find little d-orbital involvement. The bonds are better described as four polar P–O single bonds with charge separation (P+, O−), and both drawings are bookkeeping for the same molecule.
Three quick tests
- Can a carbon atom form an expanded octet in a stable molecule?
Answer
No, carbon is a Period 2 element and does not have accessible d-orbitals to expand its octet beyond eight electrons. - How many valence electrons are in the nitrate ion, NO3-?
Answer
Nitrogen has 5, Oxygen has 6, and the negative charge adds 1. Total = \(5 + 3(6) + 1 = 24\) electrons. - Why is BF3 considered electron-deficient?
Answer
Boron has only six electrons in its valence shell (three bonding pairs), which is fewer than the eight required for a full octet.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.