VALENCE BOND (VB) THEORY & Hybridization

Why do we need hybridization? What was wrong with atomic orbitals?


Let's use the example of BH₃.
The electron configuration for each atom is:
  • B: 1s²2s²2p¹
  • H: 1s¹

I've drawn the atomic orbital (AO) diagram for boron (B) on the left.
If we think about covalent bonding as the sharing of electron, then it would seem that the AO model would only allow one B-H bond to form. One hydrogen atom would share its 1s¹ electron with the unpaired electron in the 2p orbital on the boron atom.

But we need 3 B-H bonds, so boron needs to share 2 more electrons somehow. How and where do these come from?
Hybridization is the process by which an atom's atomic orbitals mix to form new hybrid orbitals. This is the basis of valence bond theory.
The one 2s and two of the 2p orbitals on boron will hybridize (mix) to form three new sp² hybrid orbitals (shown in red above). Since boron had three electrons in its valence shell, we can populate these three electrons into the new hybrid orbitals. Now there are three unpaired electrons on boron for which three hydrogen atoms can share their 1s¹ electron.

atomic orbital hybridization

Example: H₂O and SF₄
Example: NO₂⁻ (nitrite ion)
Determine the hybridization for the central carbon in each of the following compounds.
  1. HCN
  2. CCl₄
  3. HCHO (formaldehyde)
hint
Start by drawing the Lewis structures for both molecules.

Then determine the electron geometry and the molecular shape of the molecules. Remember that electron geometry takes into account all the atoms and lone pairs around the central atom, whereas molecular shape only takes into account the atoms around the central atom.
Lewis Structures
Solution 1
The central carbon in HCN has a steric number SN = 2, one is from the single bond to H and the other is the triple bond to the N.

When SN = 2, we have sp hybridization.
solution 2
The central carbon in CCl₄ has a steric number SN = 4, one for each of the single bonds to Cl.

When SN = 4, we have sp³ hybridization.
solution 3
The central carbon in HCHO has a steric number SN = 3, one for each of the single bonds to H and one for the double bond to O.

When SN = 3, we have sp² hybridization.
Using the skeleton for methyl isocyanate (right) ...
  1. Complete the Lewis structure,
  2. Determine the hybridization for the C, N, and O atoms,
  3. Determine the total number of σ-bonds and π-bonds in the structure,
  4. Determine the electron geometry and molecular shape of the two C atoms and the N atom.
solutions 1 & 2
For any atom with a SN = 2  → sp hybridization.

For any atom with a SN = 3  → sp² hybridization.

For any atom with a SN = 4  → sp³ hybridization.
solution 3
Remember:
  • A single bond is 1 σ-bond
  • A double bond is 1 σ-bond + 1 π-bond
  • A triple bond is 1 σ-bond + 2 π-bonds

In this structure, there are 4 single bonds and 2 double bonds, so there are:
  • 6 σ-bonds and
  • 2 π-bonds.
Solution 4
The carbon atom on the far-left (the sp³-hybridized carbon with a SN = 4) has a tetrahedral electron geometry and tetrahedral molecular shape.

The nitrogen in the middle (which is sp²-hybridized with a SN = 3) has a trigonal planar electron geometry and a bent (or V-shaped) molecular shape.

The carbon atom on the far-right (the sp-hybridized carbon with a SN = 2) has a linear electron geometry and a linear molecular shape.