Chemical Bonding and Molecular Structure • Topic 2 of 3

VSEPR & Valence Bond Theory

Lewis structures tell us which atoms are bonded, but not the shape of a molecule. Two theories fill this gap: VSEPR predicts geometry from electron repulsion, and Valence Bond Theory explains bonding through orbital overlap.

VSEPR theory (Valence Shell Electron Pair Repulsion, Sidgwick and Powell; Gillespie) rests on one idea: electron pairs in the valence shell of the central atom repel one another and arrange themselves as far apart as possible to minimise repulsion. The key rules are:

  • Count the number of electron pairs (bond pairs + lone pairs) around the central atom; this fixes the basic geometry: 2 pairs - linear, 3 - trigonal planar, 4 - tetrahedral, 5 - trigonal bipyramidal, 6 - octahedral.
  • Repulsion order: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Lone pairs occupy more space and push bond pairs closer together, reducing bond angles.

This explains why CH4 (4 bond pairs) is a perfect tetrahedron at 109.5 degrees, NH3 (3 bond pairs + 1 lone pair) is pyramidal at 107 degrees, and H2O (2 bond pairs + 2 lone pairs) is bent at 104.5 degrees - the angle shrinks as lone pairs increase.

Valence Bond Theory (VBT) (Heitler-London, Pauling, Slater) describes a covalent bond as the overlap of two half-filled atomic orbitals with electrons of opposite spin. Greater overlap gives a stronger bond. End-on (axial) overlap forms a sigma bond (strong); sideways overlap of p-orbitals forms a pi bond (weaker). A single bond is one sigma; a double bond is one sigma + one pi; a triple bond is one sigma + two pi.

Hybridisation is the mixing of atomic orbitals of nearly equal energy to form new, equivalent hybrid orbitals that point in definite directions, explaining observed shapes:

  • sp - one s + one p; linear, 180 degrees; e.g. BeCl2, C2H2.
  • sp2 - one s + two p; trigonal planar, 120 degrees; e.g. BF3, C2H4.
  • sp3 - one s + three p; tetrahedral, 109.5 degrees; e.g. CH4, NH3, H2O.
  • sp3d - trigonal bipyramidal; e.g. PCl5.
  • sp3d2 - octahedral; e.g. SF6.

Resonance. When a single Lewis structure cannot describe a molecule, we draw two or more canonical (resonance) structures and the real molecule is a resonance hybrid of them. In O3 and CO32- the bonds are identical in length, intermediate between single and double. Resonance lowers the energy and stabilises the molecule; resonance energy is the difference between the hybrid and the most stable contributing structure.

VSEPR shapes: CH4 tetrahedral, NH3 pyramidal, H2O bentCCH4 sp3109.5 degNlone pairNH3 sp3107 degOH2O sp3104.5 deg
Solved Examples
1
Worked Example
Using VSEPR theory, predict the shape and bond angle of methane, CH4.
Solution
  1. Step 1: Carbon is the central atom with 4 valence electrons, forming 4 C-H bonds.
  2. Step 2: There are 4 bond pairs and 0 lone pairs around carbon.
  3. Step 3: Four electron pairs arrange tetrahedrally to minimise repulsion.
  4. Step 4: With no lone pairs the angle is the ideal 109.5 degrees.

Answer: CH4 is tetrahedral with bond angle 109.5 degrees (sp3 hybridised).

2
Worked Example
Explain why the bond angle decreases in the order CH4 > NH3 > H2O.
Solution
  1. Step 1: All three central atoms are sp3 hybridised (4 electron pairs each).
  2. Step 2: CH4 has 0 lone pairs, NH3 has 1, H2O has 2.
  3. Step 3: Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.
  4. Step 4: More lone pairs squeeze the bond pairs together, so the angle falls: 109.5 to 107 to 104.5 degrees.

Answer: Increasing lone pairs increase repulsion on bond pairs, reducing the angle: 109.5 > 107 > 104.5 degrees.

3
Worked Example
Identify the hybridisation of the central atom in BF3 and predict its shape.
Solution
  1. Step 1: Boron has 3 valence electrons and forms 3 B-F bonds.
  2. Step 2: There are 3 bond pairs and 0 lone pairs.
  3. Step 3: Mixing one s and two p orbitals gives three sp2 hybrid orbitals.
  4. Step 4: These point to the corners of a triangle at 120 degrees.

Answer: Boron is sp2 hybridised; BF3 is trigonal planar with 120 degree angles.

4
Worked Example
Describe the bonding in ethyne (C2H2) in terms of hybridisation and sigma/pi bonds.
Solution
  1. Step 1: Each carbon in C2H2 is sp hybridised (linear, 180 degrees).
  2. Step 2: One sp orbital of each C overlaps to form a C-C sigma bond; the other overlaps with H (1s) for a C-H sigma bond.
  3. Step 3: The two unhybridised p orbitals on each carbon overlap sideways.
  4. Step 4: This gives two pi bonds, so the C-C bond is a triple bond (1 sigma + 2 pi).

Answer: Each C is sp hybridised; the molecule is linear with a C≡C triple bond made of 1 sigma and 2 pi bonds.

5
Worked Example
Predict the geometry and hybridisation of PCl5 and SF6.
Solution
  1. Step 1: P in PCl5 has 5 bond pairs and no lone pairs - 5 electron pairs.
  2. Step 2: Five pairs need sp3d hybridisation, giving a trigonal bipyramidal shape.
  3. Step 3: S in SF6 has 6 bond pairs and no lone pairs - 6 electron pairs.
  4. Step 4: Six pairs need sp3d2 hybridisation, giving an octahedral shape.

Answer: PCl5 is sp3d, trigonal bipyramidal; SF6 is sp3d2, octahedral.

6
Worked Example
Why are all three C-O bonds in the carbonate ion CO32- equal in length?
Solution
  1. Step 1: A single Lewis structure shows one C=O double bond and two C-O single bonds.
  2. Step 2: Three equivalent resonance structures can be drawn by shifting the double bond.
  3. Step 3: The real ion is a resonance hybrid of these three structures.
  4. Step 4: Therefore each C-O bond has the same bond order (about 1.33) and identical length.

Answer: Resonance delocalises the bonding equally, so all three C-O bonds are identical in length.

Key Points

  • VSEPR: electron pairs around the central atom arrange to minimise repulsion (2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
  • Repulsion order lone pair-lone pair > lone pair-bond pair > bond pair-bond pair explains CH4 (109.5) > NH3 (107) > H2O (104.5 degrees).
  • Valence Bond Theory: a covalent bond is the overlap of half-filled orbitals; axial overlap gives sigma, sideways p-overlap gives pi.
  • Hybridisation: sp (linear), sp2 (trigonal planar), sp3 (tetrahedral), sp3d (trigonal bipyramidal), sp3d2 (octahedral).
  • Resonance: the real molecule is a hybrid of canonical structures; it equalises bond lengths and lowers energy (resonance stabilisation).
Quick Check — 4 MCQs
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Q1.The shape and bond angle of NH3 are:
Explanation: NH3 has 3 bond pairs and 1 lone pair; the lone pair pushes the bonds to a pyramidal shape at 107 degrees.
Q2.The hybridisation of carbon in ethene (C2H4) is:
Explanation: Each carbon forms 3 sigma bonds and one pi bond, so it is sp2 hybridised (trigonal planar, 120 degrees).
Q3.A pi bond is formed by:
Explanation: Pi bonds arise from lateral overlap of unhybridised p-orbitals and are weaker than sigma bonds.
Q4.Which molecule has sp3d2 hybridisation and octahedral geometry?
Explanation: SF6 has 6 bond pairs requiring sp3d2 hybridisation, giving a regular octahedral shape.
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