Carbon has two different kinds of valence orbital, yet all four C–H bonds in methane are identical. Hybridization explains how. Before it, valence bond theory explains why atoms bond at all: their orbitals overlap.
Chapter 3 of the PECTAA Chemistry 11 book is Chemical Bonding. After bond energy, bond length and the comparison of bonds (sections 3.3 to 3.5), section 3.6 covers valence bond theory (VBT) and section 3.7 atomic orbital hybridization. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.
What to know
Why VBT (section 3.6). VSEPR predicts molecular shapes but does not say why bonds form. VBT deals with both bond formation and shape, treating a molecule as a combination of atoms.
Postulates of VBT.
- A covalent bond forms when half-filled valence orbitals of similar energy on two atoms overlap.
- Greater overlap gives a stronger bond.
- Covalent bonds are directional; the direction depends on the shape and mode of overlap.
Sigma (σ) bond (section 3.6.1). Formed by linear, head-on overlap of two half-filled orbitals along the nuclear axis. Both s and p orbitals can do this.
- s–s overlap: H₂. Each H has a half-filled s orbital; electron density is symmetrical around both nuclei.
- s–p overlap: HCl. The H 1s orbital overlaps a half-filled Cl p orbital. Electron density is higher near Cl, which is more electronegative, so HCl is polar.
- p–p overlap: Cl₂. Two Cl p orbitals overlap on the nuclear axis; density is symmetrical because both atoms have the same electronegativity.
- H₂S: two half-filled 3p orbitals of sulfur overlap with the 1s orbitals of two H atoms, giving a V-shaped molecule with bond angle 92°. The book contrasts this with 104.5° in H₂O.
Pi (π) bond. In O₂ each O atom has two unpaired electrons in perpendicular p orbitals, px and py. The px orbitals overlap end-to-end to give a σ bond. The py orbitals lie parallel and overlap sideways, one pair of lobes above the plane of the nuclei and one below, giving a π bond. So O=O is one σ plus one π bond.
Hybridization (section 3.7). Atomic orbitals of slightly different energies and shapes mix to form a new set of equivalent orbitals of equal energy and shape. Keep in mind:
- Mixing happens within a single atom or ion.
- Only orbitals of comparable energy can mix.
- The number of hybrid orbitals equals the number of orbitals mixed.
Types (section 3.7.1).
| Type | Orbitals mixed | Hybrids | Geometry | Book example |
|---|---|---|---|---|
| sp | one s + one p | 2 | linear, 180° | BeCl₂ |
| sp² | one s + two p | 3 | trigonal planar, 120° | BF₃ |
| sp³ | one s + three p | 4 | tetrahedral | CH₄ |
- BeCl₂: Be promotes a 2s electron to 2p, then 2s and one 2p mix into two sp orbitals, each overlapping a half-filled Cl p orbital.
- BF₃: B mixes 2s and two 2p into three sp² orbitals, each overlapping a half-filled F 2p orbital.
- CH₄: C mixes 2s and three 2p into four sp³ orbitals; each overlaps an H 1s orbital, giving four σ bonds pointing to the corners of a regular tetrahedron.
Each sp³ orbital has one large and one small lobe; the small lobe is usually left out of drawings.
Key points
- VBT: bonds come from overlap of half-filled orbitals; more overlap, stronger bond.
- σ is head-on; π is sideways. A double bond is σ + π.
- Number of hybrids = number of orbitals mixed.
- sp → 180°, sp² → 120°, sp³ → tetrahedral.
Worked examples
Example 1 (counting bonds). How many σ and π bonds are in O₂ and in CH₄?
- O₂: one σ (px–px) and one π (py–py).
- CH₄: four σ bonds (sp³–s), no π bonds.
Example 2 (predicting hybridization). Which hybridization does the central atom use in BeCl₂, BF₃ and CH₄?
- Be forms two bonds in a line → sp.
- B forms three bonds in a plane → sp².
- C forms four bonds in a tetrahedron → sp³.
Example 3 (polarity from overlap). Why is HCl polar but Cl₂ not?
- In HCl, Cl is more electronegative, so shared density shifts towards Cl.
- In Cl₂, both atoms are identical, so the density stays symmetrical.
Common mistakes in the exam
- Saying hybridization mixes orbitals of different atoms. It happens in one atom.
- Calling a π bond a head-on overlap. π is sideways overlap.
- Giving BF₃ a tetrahedral shape. It is trigonal planar, 120°.
- Forgetting promotion of an electron before hybridization in Be, B and C.
- Counting a double bond as two σ bonds. It is one σ and one π.
Exam technique
- Draw box diagrams for ground, excited and hybridized states, as the book does.
- In overlap diagrams, label the orbitals used (for example sp³–s).
- Use the table above for quick-recall questions.
- Check your board's notice for the sections in your paper.
Practice questions
- State two postulates of VBT. Answer: Bonds form by overlap of half-filled orbitals of similar energy; greater overlap gives a stronger bond.
- How many hybrid orbitals form from one s and two p orbitals? Answer: Three sp² orbitals.
- What is the bond angle in an sp-hybridized molecule? Answer: 180°.
- Which overlap forms the H–Cl bond? Answer: s–p overlap.
- What kind of bond forms from sideways overlap of parallel p orbitals? Answer: A π bond.
Ready to practise this? Continue in 1st Year Chemistry.