Chemical Bonding — VSEPR, Hybridisation & MO Theory

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Shape questions look like memory work, but they are counting work. Count the electron pairs around the central atom, place them as far apart as possible, then look only at where the atoms are. That is VSEPR, and it answers most of what this chapter asks.

What you will be able to do

• Find the steric number and predict shape and bond angle for any molecule or ion
• Assign hybridisation from sp to sp³d³ and place lone pairs correctly
• Explain dipole moments (NH₃ vs NF₃, cis vs trans) and covalent character using Fajans’ rules
• Use formal charge and resonance to find bond orders in ions like CO₃²⁻ and O₃
• Fill MO diagrams for second-period diatomics and get bond order, bond length and magnetism

The big picture

Four questions, asked in order, solve almost every bonding problem.

In one line: Electron pairs around a central atom repel and spread as far apart as possible; lone pairs repel most. The arrangement fixes the hybridisation; MO theory then explains bond order and magnetism.

1. How many electron pairs surround the central atom? (steric number)
2. How are they arranged in space? (electron geometry → hybridisation)
3. Where are the lone pairs? (shape = where the atoms are)
4. Do the bond dipoles cancel? (dipole moment)

Think of it like this: Tie balloons together at their necks: two balloons point opposite ways, three make a flat triangle, four a tetrahedron. Electron pairs behave the same way. A lone pair is a fatter balloon — it squeezes the others closer together.

Counting electron pairs

Get this number right and the rest is a table look-up.

Steric number = bonded atoms + lone pairs on the central atom

Lone pairs = (valence electrons of central atom − one electron per bond to a monovalent atom ± charge) ÷ 2. Add one electron for each negative charge; remove one for each positive charge. Double-bonded oxygen uses two electrons of the central atom.

Key idea: Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. That is why CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°).

Solved example 01 — Counting for an ion

Find the steric number and shape of ICl₄⁻.

1. I has 7 valence electrons; add 1 for the negative charge = 8.
2. Four I–Cl bonds use 4 → 4 left = 2 lone pairs.
3. Steric number = 4 + 2 = 6 → octahedral electron geometry; the lone pairs sit opposite each other.

Answer: Square planar, sp³d².

VSEPR shapes

The whole chapter on one table.

SN	Lone pairs	Shape	Example	Angle
2	0	Linear	BeCl₂, CO₂	180°
3	0	Trigonal planar	BF₃, SO₃	120°
3	1	Bent	SO₂, O₃	about 117–119°
4	0	Tetrahedral	CH₄, NH₄⁺, SO₄²⁻	109.5°
4	1	Trigonal pyramidal	NH₃, XeO₃	107° (NH₃)
4	2	Bent	H₂O	104.5°
5	0	Trigonal bipyramidal	PCl₅	90°, 120°
5	1	See-saw	SF₄
5	2	T-shaped	ClF₃	about 87.5°
5	3	Linear	XeF₂, I₃⁻	180°
6	0	Octahedral	SF₆	90°
6	1	Square pyramidal	BrF₅, XeOF₄
6	2	Square planar	XeF₄, ICl₄⁻	90°
7	0	Pentagonal bipyramidal	IF₇	72°, 90°

Key idea: In a trigonal bipyramid, lone pairs always go equatorial (they have 120° of room there, against 90° in the axial positions). That single rule gives SF₄ (see-saw), ClF₃ (T-shape) and XeF₂ (linear).

Solved example 02 — Shape of ClF₃

Predict the shape of ClF₃.

1. Cl: 7 valence electrons; 3 used in Cl–F bonds; 4 left = 2 lone pairs.
2. Steric number 5 → trigonal bipyramidal arrangement, sp³d.
3. Both lone pairs equatorial; the three F atoms form a T.

Answer: T-shaped; F–Cl–F angle slightly under 90°.

Solved example 03 — Shape of XeO₃

Predict the shape of XeO₃.

1. Xe: 8 valence electrons. Each Xe=O double bond uses 2 → 6 used, 2 left = 1 lone pair.
2. Steric number = 3 + 1 = 4 → sp³.

Answer: Trigonal pyramidal (like NH₃).

Check yourself

• C1. Give the shapes of SF₄ and I₃⁻.
• C2. Why is the bond angle in H₂O smaller than in NH₃?

Answers at the end.

Hybridisation

Mixing orbitals to point where the electron pairs are.

Steric number	Hybridisation	Orbitals mixed	Electron geometry
2	sp	s + p	Linear
3	sp²	s + 2p	Trigonal planar
4	sp³	s + 3p	Tetrahedral
5	sp³d	s + 3p + d(z²)	Trigonal bipyramidal
6	sp³d²	s + 3p + d(z²) + d(x²−y²)	Octahedral
7	sp³d³	s + 3p + 3d	Pentagonal bipyramidal

• Solid PCl₅ is ionic: [PCl₄]⁺ (sp³, tetrahedral) and [PCl₆]⁻ (sp³d², octahedral).
• NO₂⁺ is linear (sp); NO₂⁻ is bent (sp²); NO₃⁻ is trigonal planar (sp²).
• In ethyne each carbon is sp (linear, 180°), in ethene sp² (120°), in ethane sp³ (109.5°).
• s character: sp (50%) > sp² (33%) > sp³ (25%). More s character → shorter bond and more electronegative carbon: ethyne’s C–H is the most acidic of the three.

Formal charge and resonance

Choosing the best Lewis structure, and averaging the ones that tie.

Formal charge = valence electrons − non-bonding electrons − ½ (bonding electrons)

The best structure has formal charges closest to zero, with negative charge on the more electronegative atom

Resonance: when two or more valid Lewis structures differ only in where electrons sit, the real molecule is an average (resonance hybrid). Bond order = total bonds between the atoms ÷ number of positions.

Species	Resonance structures	Bond order of each bond
O₃	2	1.5
CO₃²⁻	3	1.33
NO₃⁻	3	1.33
Benzene (C–C)	2 (Kekulé)	1.5

Solved example 04 — Formal charges in O₃

Find the formal charge on each oxygen in one Lewis structure of O₃ (O=O–O).

1. Central O: 6 − 2 − ½(6) = +1.
2. Double-bonded terminal O: 6 − 4 − ½(4) = 0.
3. Single-bonded terminal O: 6 − 6 − ½(2) = −1.

Answer: +1 on the centre, 0 and −1 on the ends; the two resonance forms average to a bond order of 1.5 for both O–O bonds.

Dipole moment and Fajans’ rules

When charges separate, and when ionic bonds turn partly covalent.

Dipole moment (μ = charge × distance, in debye, D) is a vector: bond dipoles add or cancel according to the shape.

Zero dipole moment (symmetric)	Non-zero dipole moment
CO₂, BF₃, CH₄, CCl₄, PCl₅, SF₆, XeF₄, trans-1,2-dichloroethene	H₂O, NH₃, SO₂, CHCl₃, NF₃, cis-1,2-dichloroethene

Key idea: NH₃ (1.47 D) vs NF₃ (0.23 D): in NH₃, N is more electronegative than H, so the bond moments point toward N — the same way as the lone pair’s moment. They add. In NF₃ the bond moments point toward F, away from the lone pair, and partly cancel it.

Fajans’ rules: covalent character in ionic compounds

• Small, highly charged cation polarises more: Al³⁺ > Mg²⁺ > Na⁺. So AlCl₃ is far more covalent than NaCl.
• Large anion is more easily polarised: I⁻ > Br⁻ > Cl⁻ > F⁻. So LiI is more covalent than LiF.
• 18-electron cations (Cu⁺, Ag⁺) polarise more than noble-gas cations of similar size: CuCl is more covalent than NaCl.
• Covalent character lowers melting point and solubility in water.

Check yourself

• C3. Which has a larger dipole moment: cis- or trans-1,2-dichloroethene?
• C4. Arrange in increasing covalent character: NaCl, MgCl₂, AlCl₃.

Answers at the end.

Molecular orbital theory

The only model that explains why O₂ is magnetic.

Bond order = (Nb − Na) / 2

Nb = electrons in bonding MOs; Na = electrons in antibonding MOs

• Bond order 0 → no stable molecule (He₂, Ne₂).
• Higher bond order → shorter, stronger bond.
• Any unpaired electron → paramagnetic.

Filling order

Up to N₂ (B₂, C₂, N₂), s–p mixing: σ1s, σ∗1s, σ2s, σ∗2s, π2p (×2), σ2p, π∗2p (×2), σ∗2p

O₂ and F₂: σ1s, σ∗1s, σ2s, σ∗2s, σ2p, π2p (×2), π∗2p (×2), σ∗2p

Species	Electrons	Bond order	Unpaired e⁻	Magnetism
H₂	2	1	0	Diamagnetic
He₂	4	0	0	Does not exist
B₂	10	1	2	Paramagnetic
C₂	12	2	0	Diamagnetic
N₂	14	3	0	Diamagnetic
N₂⁺	13	2.5	1	Paramagnetic
O₂	16	2	2	Paramagnetic
O₂⁺	15	2.5	1	Paramagnetic
O₂⁻	17	1.5	1	Paramagnetic
O₂²⁻	18	1	0	Diamagnetic
NO	15	2.5	1	Paramagnetic
CO	14	3	0	Diamagnetic

Key idea: Bond length follows bond order in reverse: O₂²⁻ (1) > O₂⁻ (1.5) > O₂ (2) > O₂⁺ (2.5). Liquid O₂ sticks between the poles of a magnet — the two unpaired π∗ electrons that Lewis structures can’t show.

Solved example 05 — Superoxide ion

Find the bond order and magnetism of O₂⁻.

1. 17 electrons. Bonding: σ1s², σ2s², σ2p², π2p⁴ = 10.
2. Antibonding: σ∗1s², σ∗2s², π∗2p³ = 7.
3. Bond order = (10 − 7)/2 = 1.5; one unpaired electron.

Answer: Bond order 1.5; paramagnetic.

Solved example 06 — Why B₂ is paramagnetic

Use the filling order to explain why B₂ is paramagnetic but C₂ is not.

1. B₂ has 10 electrons; after σ1s², σ∗1s², σ2s², σ∗2s² (8), two remain.
2. With s–p mixing, π2p (two orbitals) comes before σ2p: one electron in each π orbital (Hund’s rule).
3. C₂ has two more electrons, which pair up in the π orbitals.

Answer: B₂: 2 unpaired (paramagnetic). C₂: all paired (diamagnetic).

Hydrogen bonding

The weak bond with big consequences.

• Forms when H is bonded to F, O or N and is attracted to a lone pair on another F, O or N.
• Intermolecular H bonding raises boiling points: H₂O (373 K) vs H₂S (213 K); HF boils far above HCl.
• Intramolecular H bonding (within one molecule) lowers boiling point and water solubility: o-nitrophenol is steam-volatile, p-nitrophenol is not.
• Ice is less dense than water because H bonds lock molecules into an open cage.
• Explains DNA base pairing and protein folding.

Exam trap gallery

Where marks are lost in bonding questions.

#	What students write	What is actually correct	Why
1	H₂O is tetrahedral.	Its electron geometry is tetrahedral; its shape is bent.	Shape describes atoms only, not lone pairs.
2	In SF₄ the lone pair is axial.	The lone pair is equatorial → see-saw.	Equatorial positions have more room (120° neighbours).
3	XeF₄ is tetrahedral.	XeF₄ is square planar (SN 6, 2 lone pairs opposite).	Count Xe’s 8 valence electrons: 4 bonds + 2 lone pairs.
4	O₂ is diamagnetic because all electrons pair in the Lewis structure.	O₂ is paramagnetic: two unpaired π∗ electrons.	Lewis structures can’t show this; MO theory can.
5	B₂ is diamagnetic.	B₂ is paramagnetic.	s–p mixing puts π2p below σ2p up to N₂.
6	A molecule with polar bonds has a dipole moment.	Not if the shape cancels them: CO₂, BF₃, CCl₄ have μ = 0.	Dipole moment is a vector sum.
7	NF₃ has a larger dipole than NH₃ because F is more electronegative.	NH₃ (1.47 D) > NF₃ (0.23 D).	In NF₃ the bond moments oppose the lone pair’s moment.
8	Bond order of C–O in CO₃²⁻ is 2.	1.33 (three resonance structures, four bonds).	The double bond is shared over three positions.

Memory hook: “Lone pairs like the equator” — in trigonal bipyramids, LPs go equatorial. · “Up to N, pi first” — B₂, C₂, N₂ fill π2p before σ2p. · “FON makes H bonds” — only F, O, N. · “Small and charged, big and soft” — Fajans: small, high-charge cation + large anion = covalent.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The shape of NH₃ is: (Easy)

• (a) tetrahedral
• (b) trigonal planar
• (c) trigonal pyramidal
• (d) T-shaped

Q02. Hybridisation of carbon in ethyne is: (Easy)

• (a) sp
• (b) sp²
• (c) sp³
• (d) sp³d

Q03. Which molecule has zero dipole moment? (Easy)

• (a) NH₃
• (b) H₂O
• (c) BF₃
• (d) SO₂

Q04. Bond order of N₂ is: (Easy)

• (a) 1
• (b) 2
• (c) 2.5
• (d) 3

Q05. Which can form hydrogen bonds? (Easy)

• (a) CH₄
• (b) HCl
• (c) NH₃
• (d) PH₃

Q06. The shape of XeF₂ is: (Moderate)

• (a) bent
• (b) linear
• (c) T-shaped
• (d) see-saw

Q07. Which is paramagnetic? (Moderate)

• (a) N₂
• (b) C₂
• (c) O₂²⁻
• (d) B₂

Q08. The correct order of bond angles is: (Moderate)

• (a) H₂O > NH₃ > CH₄
• (b) CH₄ > NH₃ > H₂O
• (c) NH₃ > CH₄ > H₂O
• (d) CH₄ > H₂O > NH₃

Q09. Hybridisation of P in the [PCl₆]⁻ ion is: (Moderate)

• (a) sp³
• (b) sp³d
• (c) sp³d²
• (d) sp³d³

Q10. Most covalent among these is: (Moderate)

• (a) NaCl
• (b) MgCl₂
• (c) AlCl₃
• (d) KCl

Q11. Bond order of each C–O bond in CO₃²⁻ is: (Moderate)

• (a) 1
• (b) 1.33
• (c) 1.5
• (d) 2

Q12. Which species has the shortest bond? (Moderate)

• (a) O₂²⁻
• (b) O₂⁻
• (c) O₂
• (d) O₂⁺

Q13. ClF₃ has the shape: (Tough)

• (a) trigonal planar
• (b) trigonal pyramidal
• (c) T-shaped
• (d) see-saw

Q14. Which pair has the same shape? (Tough)

• (a) XeF₄ and SF₄
• (b) XeF₂ and I₃⁻
• (c) NH₃ and BF₃
• (d) SO₂ and CO₂

Q15. o-Nitrophenol is more volatile than p-nitrophenol because of: (Tough)

• (a) intermolecular H bonding in the ortho isomer
• (b) intramolecular H bonding in the ortho isomer
• (c) larger dipole moment of the ortho isomer
• (d) higher molar mass of the para isomer

Answer key with explanations

1. (c) trigonal pyramidal — SN 4 with one lone pair.
2. (a) sp — Triple bond: two π bonds, linear.
3. (c) BF₃ — Trigonal planar BF₃: bond dipoles cancel.
4. (d) 3 — (10 − 4)/2 = 3.
5. (c) NH₃ — H bonded to N.
6. (b) linear — SN 5 with 3 equatorial lone pairs.
7. (d) B₂ — s–p mixing leaves two unpaired π2p electrons.
8. (b) CH₄ > NH₃ > H₂O — More lone pairs → stronger compression: 109.5° > 107° > 104.5°.
9. (c) sp³d² — SN 6, no lone pairs.
10. (c) AlCl₃ — Al³⁺ is the smallest, most highly charged cation.
11. (b) 1.33 — Four bonds shared over three positions.
12. (d) O₂⁺ — Highest bond order (2.5) → shortest.
13. (c) T-shaped — SN 5, two equatorial lone pairs.
14. (b) XeF₂ and I₃⁻ — Both linear (SN 5, three lone pairs).
15. (b) intramolecular H bonding in the ortho isomer — Intramolecular H bonding means the ortho molecules don’t hold each other.

Advanced problems

Each combines two ideas from the chapter.

Solved example 07 — Shape and polarity together

Which of SF₄, XeF₄, CF₄ and SF₆ have a non-zero dipole moment?

1. CF₄ (tetrahedral), SF₆ (octahedral), XeF₄ (square planar): symmetric → bond dipoles cancel.
2. SF₄: SN 5, one equatorial lone pair → see-saw → bond dipoles don’t cancel.

Answer: Only SF₄.

Solved example 08 — Ranking by MO theory

Arrange N₂, N₂⁺ and N₂²⁻ by bond length and identify which are paramagnetic.

1. N₂: 14 e⁻, bond order 3, diamagnetic.
2. N₂⁺: 13 e⁻ (one fewer bonding), bond order 2.5, one unpaired.
3. N₂²⁻: 16 e⁻ (two in π∗), bond order 2, two unpaired.

Answer: Bond length N₂²⁻ > N₂⁺ > N₂; N₂⁺ and N₂²⁻ are paramagnetic.

Check-yourself answers

• C1. SF₄: see-saw. I₃⁻: linear.
• C2. H₂O has two lone pairs; the extra lone pair–lone pair repulsion squeezes the H–O–H angle more.
• C3. cis (the C–Cl dipoles add); trans cancels to zero.
• C4. NaCl < MgCl₂ < AlCl₃.

One-page revision

Read this page the night before the exam.

Counting

• SN = bonded atoms + lone pairs
• Add e⁻ for − charge, remove for + charge
• LP–LP > LP–BP > BP–BP
• CH₄ 109.5° > NH₃ 107° > H₂O 104.5°

Shapes to know

• SN 5: TBP, see-saw (SF₄), T (ClF₃), linear (XeF₂)
• SN 6: octahedral, sq. pyramidal (BrF₅), sq. planar (XeF₄)
• Lone pairs go equatorial in TBP
• XeO₃ pyramidal; ICl₄⁻ square planar

Hybridisation

• sp linear, sp² 120°, sp³ 109.5°
• sp³d TBP; sp³d² octahedral; sp³d³ IF₇
• More s character → shorter bond
• Solid PCl₅ = [PCl₄]⁺[PCl₆]⁻

Polarity

• μ is a vector; symmetric shapes → 0
• NH₃ 1.47 D > NF₃ 0.23 D
• Fajans: small charged cation + big anion = covalent
• AlCl₃ > MgCl₂ > NaCl covalent

MO theory

• BO = (Nb − Na)/2
• Up to N₂: π2p below σ2p
• O₂ paramagnetic (2 unpaired); B₂ paramagnetic
• O₂⁺ 2.5 > O₂ 2 > O₂⁻ 1.5 > O₂²⁻ 1

Resonance and H bonds

• CO₃²⁻, NO₃⁻: BO 1.33; O₃, benzene: 1.5
• H bonds only via F, O, N
• Intramolecular H bond → more volatile (o-nitrophenol)
• Ice floats: open H-bonded cage