Worked example 1
Explain why BF₃ is stable with only six electrons on boron, while NF₃ is not drawn that way.
Try it first: Count valence electrons for each molecule.
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What you'll be able to do: Recognise incomplete octets, odd-electron species and expanded octets, and explain when each is allowed.
Best after: Valence Electrons and the Octet Rule
The octet rule works for most molecules you meet, but BF₃, NO and SF₆ all break it for very different reasons.
These are recommended, not required. You can start this lesson at any time.
Beryllium and boron form stable compounds with fewer than eight electrons around the central atom. BeCl₂ leaves beryllium with 4 electrons and BF₃ leaves boron with 6. These molecules are strong Lewis acids precisely because the centre is short of electrons, which is why BF₃ reacts eagerly with ammonia.
For BF₃, the budget is 3 + 3(7) = 24 electrons. Three B–F bonds use six; completing the fluorine octets uses eighteen, leaving none for B. Boron therefore has six surrounding electrons in the usual neutral Lewis drawing. It can accept a lone pair from NH₃ to form an adduct, illustrating Lewis acidity. Calling BF₃ electron deficient does not mean you may add an uncounted pair to its original neutral structure.
An odd total cannot be partitioned entirely into pairs. NO has 5 + 6 = 11 valence electrons and NO₂ has 5 + 2(6) = 17, so each requires an odd-electron Lewis description. Electron parity identifies the radical before deciding exactly where the unpaired electron is represented.
For a useful NO drawing, a double bond contains four electrons, two lone pairs on O contain four, and one lone pair plus one unpaired electron on N contain three: eleven total. O has an octet and N has seven surrounding electrons. Adding an electron to finish nitrogen would change neutral NO into a different charged species. Radical placement and delocalization must be assessed from valid structures; the unpaired electron is not universally assigned by electronegativity alone.
Some Lewis drawings for period-3 and heavier central atoms, including PCl₅ and SF₆, assign more than eight electrons around the center. These are electron-bookkeeping models. The historical explanation invoking readily accessible d orbitals is not a required modern account of hypervalent bonding; use delocalized bonding models for that purpose. Second-period C, N, O and F cannot exceed eight electrons in ordinary Lewis structures.
For SF₆ the budget is 6 + 6(7) = 48 electrons. Six S–F bonds use twelve and eighteen fluorine lone pairs use thirty-six, exhausting the budget. The drawing assigns twelve bonding electrons around sulfur. It does not require extra electron pairs invented after counting. Period-2 N, O and F do not have corresponding ordinary ten- or twelve-electron Lewis centers. Period 3 is permission to consider expanded-valence drawings, not an instruction to expand every atom from that period.
| Diagnostic | Likely treatment | Example |
|---|---|---|
| Odd total valence count | Retain an unpaired electron | NO |
| Typical electron-deficient center after terminal octets | Allow an incomplete octet | B in BF₃ |
| More than four bonded neighbors on an allowed heavier center | Consider expanded-valence bookkeeping | S in SF₆ |
| Ordinary period-2 center short of eight with suitable lone pairs nearby | Test multiple bonds instead of expanding the budget | C in CO₂ |
An electron shortage at carbon is not the same exception as boron's usual electron deficiency. Complete the count and check formal charges before declaring a proposed structure an exception. Stable species do not all obey the octet rule, but “exception” cannot excuse an incorrect total electron count.
An odd total number of valence electrons means at least one electron remains unpaired in any Lewis structure. Such a radical cannot give every atom a conventional pair-only octet.
This is a different exception from an electron-deficient central atom such as B or an expanded-valence central atom from period 3 or below. Classify the reason for the exception before repairing a drawing.
C, N, O and F can never hold more than eight valence electrons.
A species with one or more unpaired electrons; an odd total guarantees at least one, while an even total does not rule out radicals.
BF₃ leaves boron with only six electrons and readily accepts a lone pair.
SF₆ places twelve electrons around sulfur in six bonding pairs.
Explain why BF₃ is stable with only six electrons on boron, while NF₃ is not drawn that way.
Try it first: Count valence electrons for each molecule.
0 of 3 steps revealed.
Can the central atom in ClF₃ hold more than eight electrons? Support your answer.
Try it first: Check the period of the central atom.
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Why must neutral NO be an odd-electron exception?
Try it first: Count N and O valence electrons.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Expanding the octet of nitrogen or oxygen to remove a formal charge. | Period 2 atoms have only four valence orbitals and cannot exceed eight electrons. | Keep period 2 centres at eight and accept the formal charges. |
| Treating any odd count as a mistake in arithmetic. | Radicals such as NO genuinely have an odd number of valence electrons. | Recount once, and if the total is still odd, draw a radical. |
| Assuming every period 3 central atom must expand. | Expansion happens only when there are surplus electrons or a clear formal-charge benefit. | Fill all octets first, then place leftovers on the central atom. |
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After the lesson · Question bank
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Recognize electron-deficient centers, odd-electron radicals and permitted expanded-valence Lewis drawings on suitable heavier atoms. Preserve the total electron budget in every case. Second-period centers do not exceed eight surrounding electrons in ordinary Lewis structures; modern hypervalent descriptions do not require literal d-orbital hybridization.
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