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.
If the total valence electron count is odd, no arrangement can pair every electron. Nitrogen monoxide NO has 11 valence electrons and nitrogen dioxide NO₂ has 17. One unpaired electron is left on the least electronegative atom, and the species is a radical, which is why these molecules are so reactive.
Central atoms in period 3 and below, such as P, S, Cl and Xe, can hold 10 or 12 electrons in structures like PCl₅ and SF₆. The traditional explanation is that empty 3d orbitals are energetically accessible. Period 2 atoms have only 2s and 2p orbitals, giving four orbitals and a hard ceiling of eight electrons.
Count the total valence electrons first. An odd total means a radical. An even total with too few electrons to complete the centre points to an electron-deficient atom. An even total with electrons left over after every octet is filled means the extras go on a period 3 or heavier central atom.
C, N, O and F can never hold more than eight valence electrons.
A species with an odd number of valence electrons and therefore one unpaired electron.
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.
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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 it's wrong: Period 2 atoms have only four valence orbitals and cannot exceed eight electrons.
Check instead: Keep period 2 centres at eight and accept the formal charges.
Why it's wrong: Radicals such as NO genuinely have an odd number of valence electrons.
Check instead: Recount once, and if the total is still odd, draw a radical.
Why it's wrong: Expansion happens only when there are surplus electrons or a clear formal-charge benefit.
Check instead: Fill all octets first, then place leftovers on the central atom.
No practice questions are available for this topic yet. You can still practice the whole unit.
Three families of species break the octet rule: electron-deficient centres such as Be and B, odd-electron radicals such as NO and NO₂, and expanded octets on period 3 and heavier central atoms that have accessible d orbitals. Period 2 elements can never expand beyond eight because their valence shell has only four orbitals.
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