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CHEM 121 Studioby Learn4Less · UBC CHEM 121

1.3 · Bonding & Lewis

Exceptions to the octet rule

Electron-deficient atoms, expanded octets and radicals.

By the end you should be able to:

  • Recognise octet exceptions: incomplete octets, expanded octets and odd-electron species

Key idea

Incomplete octets: boron and beryllium

Compounds of B and Be (and gas-phase Al halides) often leave the central atom with fewer than 8 electrons.

  • BFX3\ce{BF3} (24 e⁻): three B–F single bonds; B has 6 electrons and every formal charge is 0.
  • BeClX2\ce{BeCl2} (16 e⁻): two Be–Cl single bonds; Be has 4 electrons.

A B=F double bond would give boron an octet but put +1 on F, the most electronegative element, so the all-single-bond structure is preferred. Electron-deficient atoms act as Lewis acids that accept a lone pair to complete the octet: BFX3+NHX3→FX3B−NHX3\ce{BF3 + NH3 -> F3B-NH3}.

Key idea

Expanded octets: period 3 and beyond

A central atom from period 3 or below (P, S, Cl, Br, I, Xe, …) can hold more than 8 electrons. Period-2 atoms (C, N, O, F) never can.

SpeciesValence e⁻Electrons on the central atom
PClX5\ce{PCl5}4010 (5 bonds)
SFX4\ce{SF4}3410 (4 bonds + 1 lone pair)
SFX6\ce{SF6}4812 (6 bonds)
XeFX2\ce{XeF2}2210 (2 bonds + 3 lone pairs)
XeFX4\ce{XeF4}3612 (4 bonds + 2 lone pairs)
IX3X−\ce{I3-}2210 (2 bonds + 3 lone pairs)

The tell-tale sign: after every terminal atom has its octet, electrons are left over. They go on the central atom.

Key idea

Sulfate, phosphate and SO₂: two accepted pictures

For some period-3 species, textbooks draw different "best" structures:

SpeciesOctet pictureExpanded picture (formal charges minimised)
SOX4X2−\ce{SO4^2-}four S–O single bonds; S +2, every O −1two S=O and two S–O; S 0, two O −1
POX4X3−\ce{PO4^3-}four P–O single bonds; P +1, every O −1one P=O and three P–O; P 0, three O −1
SOX2\ce{SO2}O=S–O; S +1, one O −1O=S=O; all 0, S has 10 electrons

Formal-charge reasoning favours the expanded picture; many modern bonding analyses favour the octet picture with very polar bonds. Both describe the same real species (all four S–O bonds in sulfate are identical either way). This app accepts both and never asks a question whose answer depends on the choice.

Key idea

Odd-electron species (radicals)

If the valence-electron total is odd, one electron must stay unpaired and some atom falls short of an octet. Such species are radicals and are usually reactive.

  • NO\ce{NO} (11 e⁻): N=O with the unpaired electron on N; N has 7 electrons and both formal charges are 0.
  • NOX2\ce{NO2} (17 e⁻): two resonance forms, O=N–O and O–N=O, usually drawn with the odd electron on N. NOX2\ce{NO2} pairs up through an N–N bond to form NX2OX4\ce{N2O4}.
  • ClOX2\ce{ClO2} (19 e⁻): a stable radical used to bleach paper pulp and disinfect water.

Method

Spotting an exception while drawing

  1. Odd electron total? It is a radical: leave exactly one electron unpaired, conventionally on the less electronegative atom.
  2. Central B or Be with single bonds and zero formal charges? Stop there; do not form multiple bonds to halogens.
  3. Electrons left over after every terminal octet is complete, and the central atom is in period 3 or below? Put them on the central atom as lone pairs (expanded octet).
  4. Seem to need more than 8 electrons on C, N, O or F? Recheck the electron count or the skeleton; period-2 atoms cannot expand.

Common mistake

Exception traps

Wrong: giving N five bonds in NOX3X−\ce{NO3-} to remove its formal charge. Right: period-2 atoms never exceed an octet; N in nitrate keeps its +1.

Wrong: adding a B=F double bond so that boron has an octet. Right: BFX3\ce{BF3} has three single bonds and a 6-electron boron.

Wrong: placing the leftover electrons of XeFX4\ce{XeF4} or IX3X−\ce{I3-} on terminal atoms. Right: terminal atoms stop at 8; extra electrons go on the central atom.

Worked example

Worked example: XeF₄ and I₃⁻

Xenon tetrafluoride. 8+4(7)=368 + 4(7) = 36 e⁻. Four Xe–F bonds (8 e⁻) plus three lone pairs on each F (24 e⁻) use 32; the remaining 4 e⁻ become 2 lone pairs on Xe. Xe carries 12 electrons and every formal charge is 0.

Triiodide. 3(7)+1=223(7) + 1 = 22 e⁻. Two I–I bonds (4 e⁻) plus three lone pairs on each terminal I (12 e⁻) use 16; the remaining 6 e⁻ become 3 lone pairs on the central I, which carries 10 electrons and a formal charge of 7−6−2=−17 - 6 - 2 = -1.

Answer: XeFX4\ce{XeF4} has 2 lone pairs and IX3X−\ce{I3-} has 3 lone pairs on the central atom; both are expanded octets.

Check yourself

Fresh questions every time you visit. Answers count toward your progress.