Valence bond theory and orbital overlap
Heitler and London (1927) explained the H₂ bond; Pauling and Slater developed valence bond (VB) theory. The idea:
- As two atoms come close, each nucleus attracts the other atom's electron, and the two nuclei (and two electrons) repel.
- At one distance (74 pm for H₂) attraction wins the most and energy is lowest. 435 kJ mol⁻¹ is released. This is the bond.
- A bond forms when half-filled orbitals overlap and the electrons pair up with opposite spins.
- More overlap = stronger bond (the overlap concept).
Overlap can be positive (same phase, a bond forms), negative (opposite phase, no bond) or zero (orbitals at right angles).
Sigma (σ) and pi (π) bonds
| σ bond | π bond | |
|---|---|---|
| Overlap | head-on, along the line joining nuclei | side-on, above and below that line |
| Orbitals | s–s, s–p, p–p (end to end), hybrid orbitals | only p–p (or p–d) side by side |
| Strength | stronger (more overlap) | weaker |
| Rotation | free rotation about the bond | no free rotation |
| Exists alone? | yes | only with a σ bond |
Single bond = 1σ. Double bond = 1σ + 1π. Triple bond = 1σ + 2π.
Hybridisation
Carbon has 2s² 2p². Pure orbitals cannot explain four equal C–H bonds in CH₄ at 109.5°. Pauling's idea: the atom first promotes an electron (2s¹ 2p³) and then mixes orbitals of nearly equal energy into the same number of new, equal hybrid orbitals.
- Only orbitals of the same atom with similar energy mix.
- The number of hybrid orbitals = number of orbitals mixed.
- Hybrid orbitals are equal in energy and shape and point as far apart as possible.
- Hybrid orbitals make σ bonds (or hold lone pairs). Left-over pure p orbitals make π bonds.
| Type | Mixed | Shape | Angle | Examples |
|---|---|---|---|---|
| sp | s + p | linear | 180° | BeCl₂, C₂H₂, CO₂ (C) |
| sp² | s + 2p | trigonal planar | 120° | BF₃, C₂H₄ |
| sp³ | s + 3p | tetrahedral | 109.5° | CH₄, C₂H₆; NH₃ and H₂O (with lone pairs) |
| sp³d | s + 3p + d | trigonal bipyramidal | 120°, 90° | PCl₅ |
| sp³d² | s + 3p + 2d | octahedral | 90° | SF₆ |
s-character: sp 50 %, sp² 33 %, sp³ 25 %. More s-character → shorter, stronger bond, more electronegative carbon.
Examples: ethane, ethene, ethyne, PCl₅, SF₆
Ethane C₂H₆: each C is sp³. C–C σ (sp³–sp³) and six C–H σ (sp³–s). 7 σ bonds.
Ethene C₂H₄: each C is sp². C=C is one σ (sp²–sp²) + one π (p–p side on). Four C–H σ. 5σ + 1π. The π bond stops rotation, so ethene is flat.
Ethyne C₂H₂: each C is sp. C≡C is one σ + two π. 3σ + 2π. Linear molecule.
PCl₅: P (3s² 3p³) promotes one 3s electron to 3d → five half-filled orbitals → sp³d. The two axial bonds are longer than the three equatorial ones.
SF₆: S (3s² 3p⁴) promotes to 3s¹ 3p³ 3d² → six half-filled orbitals → sp³d², octahedral.
Shortcut: steric number (σ bonds + lone pairs on the atom) 2 → sp, 3 → sp², 4 → sp³, 5 → sp³d, 6 → sp³d².
Try it: count σ and π in food molecules
Draw the structures of ethanol (C₂H₅OH, from sanitiser) and acetic acid (CH₃COOH, in vinegar). Count: every bond has one σ; each double bond adds one π. Predict the hybridisation of each carbon (all single bonds → sp³; one double bond → sp²). Then use the free-play step to check the shape of each type.
Key formulas and definitions
- Number of hybrid orbitals = number of atomic orbitals mixed
- Steric number = σ bonds + lone pairs → 2 sp, 3 sp², 4 sp³, 5 sp³d, 6 sp³d²
- Single = 1σ; double = 1σ + 1π; triple = 1σ + 2π
- s-character: sp 50 %, sp² 33.3 %, sp³ 25 %
- Strength: σ > π; overlap ↑ ⇒ bond strength ↑
Worked examples
1. Find the hybridisation of C in CH₄.
Step 1: C forms 4 σ bonds, 0 lone pairs → steric number 4. Step 2: 4 → sp³. Answer: sp³, tetrahedral, 109.5°.
2. Find the hybridisation of B in BF₃ and of Be in BeCl₂.
Step 1: B: 3 σ bonds, 0 lone pairs → 3 → sp². Step 2: Be: 2 σ bonds, 0 lone pairs → 2 → sp. Answer: BF₃ sp² (trigonal planar); BeCl₂ sp (linear).
3. Find the hybridisation of N in NH₃ and O in H₂O.
Step 1: N: 3 σ bonds + 1 lone pair = 4 → sp³. Step 2: O: 2 σ bonds + 2 lone pairs = 4 → sp³. Answer: both sp³. Lone pairs sit in hybrid orbitals, which is why the angles (107°, 104.5°) are close to 109.5°.
4. Count σ and π bonds in ethene, C₂H₄.
Step 1: 4 C–H single bonds → 4σ. Step 2: One C=C → 1σ + 1π. Answer: 5σ and 1π.
5. Count σ and π bonds in ethyne, C₂H₂, and give the hybridisation of C.
Step 1: 2 C–H → 2σ. Step 2: C≡C → 1σ + 2π. Step 3: Each C has 2 σ bonds, 0 lone pairs → sp. Answer: 3σ, 2π; sp.
6. Explain the hybridisation of P in PCl₅.
Step 1: Ground state P: 3s² 3p³ (3 unpaired). Step 2: Excite one 3s electron into 3d: 3s¹ 3p³ 3d¹ (5 unpaired). Step 3: Mix 1 s + 3 p + 1 d → five sp³d orbitals. Step 4: Each overlaps a Cl 3p orbital → 5 σ bonds. Answer: sp³d, trigonal bipyramidal.
7. Count σ and π bonds in CH₂=CH–C≡N (acrylonitrile) and give the hybridisation of each C.
Step 1: Bonds: 3 C–H (3σ), C=C (1σ + 1π), C–C (1σ), C≡N (1σ + 2π). Step 2: Total σ = 3 + 1 + 1 + 1 = 6; π = 1 + 2 = 3. Step 3: CH₂ carbon and CH carbon: 3 σ each → sp²; C of C≡N: 2 σ → sp. Answer: 6σ, 3π; sp², sp², sp.
Common mistakes
- Thinking a π bond can exist alone. There is always a σ bond first; π bonds are the extra ones.
- Counting lone pairs out when finding hybridisation. Steric number = σ bonds + lone pairs (NH₃ is sp³, not sp²).
- Believing hybridisation happens in isolated atoms. It is a model for atoms that are forming bonds.
- Saying a π bond is stronger than a σ bond. Side-on overlap is smaller, so π is weaker.