IB Chemistry HLTopic 2 — Models of Bonding & StructurePaper 1 & 2Higher level~10 min read
Expansion of the Octet
Nitrogen can never make five bonds. Phosphorus, sitting directly below it, makes five without complaint. That single difference opens up seven new shapes, and the good news is that you already know how to work all of them out — the counting method has not changed at all.
📘 What you need to know
Elements in period 3 and below can hold more than eight electrons in their outer shell.
This is possible because they have vacant d orbitals close enough in energy to be used for bonding.
Period 2 elements (C, N, O, F) have no d orbitals in their valence shell, so they can never expand the octet.
Five domains give a trigonal bipyramidal electron domain geometry.
Six domains give an octahedral electron domain geometry.
In a trigonal bipyramid, lone pairs always take the equatorial positions, where there is more room.
In an octahedron, two lone pairs sit opposite each other, which is why XeF4 ends up flat.
Why only some elements can do it
An atom in period 2 has only 2s and 2p orbitals available. Together those hold a maximum of eight electrons, and there is nowhere else to put any more — the 3s orbital is far too high in energy to bother with. The octet rule is not a preference for these elements; it is a hard limit.
From period 3 onwards there are also 3d orbitals, and they sit close enough in energy to be used. That gives sulfur, phosphorus, chlorine, iodine and xenon somewhere to put extra pairs, so they can accommodate ten or twelve electrons around them.
The rule that decides it
period 2 → no d orbitals → maximum 8 | period 3 and below → vacant d orbitals → 10 or 12 possible
If your Lewis formula ever puts ten electrons on nitrogen or oxygen, stop and recount. It is always a mistake. The same structure on phosphorus or sulfur is perfectly fine.
Counting works exactly as before
🧩 The method, unchanged
Count the valence electrons, adjusting for any charge.
Draw the skeleton and put one bonding pair between each joined pair of atoms.
Give every outer atom its octet first. Fluorine and oxygen can never expand, so they always take exactly three lone pairs each (fluorine) or two (a doubly bonded oxygen).
Everything left over goes on the central atom, even if that takes it past eight.
Count the domains on the central atom and read off the geometry.
Five electron domains
Five domains spread out into a trigonal bipyramid: three in a flat triangle round the middle (equatorial, 120° apart) and two pointing straight up and down (axial, at 90° to the triangle). Swap bonds for lone pairs and you get four different molecular shapes.
The trigonal bipyramidal family
Notice the pattern going right: every lone pair you add removes one equatorial atom, and the name changes even though the underlying arrangement is identical.
🤔 Why do lone pairs choose equatorial positions?
An axial position has three neighbours at 90°. An equatorial position has only two at 90° (the axial ones), with the other two sitting comfortably at 120°. Since 90° contacts are the crowded ones, and lone pairs repel more strongly than bonding pairs, the lone pairs take the roomier equatorial spots and leave the tighter axial positions to the bonds.
Six electron domains
Six domains arrange themselves into an octahedron: four in a square round the middle, plus one above and one below, all at 90°. Removing bonds and adding lone pairs gives three shapes.
The octahedral family
XeF4 is a nice test of understanding: it has two lone pairs but is still nonpolar, because they sit directly opposite each other and the four bond dipoles cancel in the square plane.
The full list
Molecule
Valence electrons
Bonding pairs
Lone pairs on centre
Molecular geometry
PCl5
40
5
0
trigonal bipyramidal
SF4
34
4
1
seesaw
ClF3
28
3
2
T-shaped
I3−
22
2
3
linear
SF6
48
6
0
octahedral
BrF5
42
5
1
square pyramidal
XeF4
36
4
2
square planar
🧠 A shortcut for the lone pairs on the centre
For a molecule AXn made of a central atom and n halogens, work out (valence electrons − 8n) ÷ 2. Every outer halogen takes 8 electrons in total (one bonding pair plus three lone pairs), so whatever is left over belongs to the central atom. For SF4: (34 − 32) ÷ 2 = 1 lone pair. Quick, and it always works.
Worked examples
WE 1
Draw the Lewis formula for ClF3 and deduce its shape
Step 1: countCl: 7 + (3 × 7) = 28 electronsStep 2: bonds and outer octets
Three Cl–F bonds use 6. Each F needs 3 lone pairs: 3 × 6 = 18.
28 − 6 − 18 = 4 electrons = 2 lone pairs on ClStep 3: count domains3 bonding + 2 lone = 5 domains → trigonal bipyramidalStep 4: place the lone pairs equatorially
That leaves the two axial fluorines and one equatorial fluorine.
T-shaped, with bond angles slightly under 90°chlorine ends up with 10 electrons — correct, because it is in period 3
WE 2
State the electron domain geometry, molecular geometry and F–Xe–F bond angle in XeF2
Step 1: count the valence electronsXe: 8 + (2 × 7) = 22 electronsStep 2: use them up
Two Xe–F bonds use 4. Each F takes 3 lone pairs: 2 × 6 = 12.
22 − 4 − 12 = 6 electrons = 3 lone pairs on XeStep 3: domains and placement2 bonding + 3 lone = 5 domains → trigonal bipyramidal
All three lone pairs go equatorial, leaving the two fluorines axial — directly opposite.
Trigonal bipyramidal domains, linear molecule, 180°three answers were asked for, so give all three explicitly
WE 3
Explain why SF6 exists but OF6 does not
Step 1: what SF6 requires
Six bonding pairs means 12 electrons around the central atom.
Step 2: why sulfur can manage it
Sulfur is in period 3, so it has vacant 3d orbitals available to hold the extra pairs.
Step 3: why oxygen cannot
Oxygen is in period 2, with only 2s and 2p orbitals. Those hold a maximum of 8 electrons and there is no accessible d subshell.
Only period 3 and below can expand the octetthe same argument explains why NCl₅ does not exist while PCl₅ does
💡 Exam tips
Give the outer atoms their octets first. Whatever is left over is the lone pairs on the central atom.
Always justify expansion with vacant d orbitals and period 3 or below.
Lone pairs go equatorial in a trigonal bipyramid and opposite each other in an octahedron.
Learn the seven names as pairs of geometries: domain geometry then molecular geometry.
Bond angles here are usually slightly less than 90°, 120° or 180°, because lone pairs squeeze them.
For polarity, check symmetry: PCl5, SF6 and XeF4 are nonpolar; SF4, ClF3 and BrF5 are polar.
⚠ Common mix-ups
Expanding the octet on a period 2 atom. Nitrogen with five bonds or oxygen with three bonds and no charge is always wrong.
Putting lone pairs in axial positions. They always take equatorial places in a trigonal bipyramid.
Assuming a lone pair makes a molecule polar. XeF4 has two and is still nonpolar.
Confusing seesaw with T-shaped. Seesaw is four bonds and one lone pair; T-shaped is three and two.
Forgetting the three lone pairs on every outer fluorine when counting electrons.
Quoting exact bond angles. Say “slightly less than 90°” where lone pairs are present.
Up next: Formal Charge — the tie-breaker you use when two Lewis formulas both look valid and you have to decide which one the examiner wants.
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