# Valence Bond Theory and Hybrid Atomic Orbitals

However, the application of VSEPR theory can be expanded to complicated molecules such as

  H H H O | | | // H-C-C=C=C-C=C-C-C | | \ H N O-H / \ H H 

By applying the VSEPR theory, one deduces the following results:

• $$\ce{H-C-C}$$ bond angle = 109o
• $$\ce{H-C=C}$$ bond angle = 120o, geometry around $$\ce{C}$$ trigonal planar
• $$\ce{C=C=C}$$ bond angle = 180o, in other words linear
• $$\ce{H-N-C}$$ bond angle = 109o, tetrahedral around $$\ce{N}$$
• $$\ce{C-O-H}$$ bond angle = 105 or 109o, 2 lone electron pairs around $$\ce{O}$$

## Confidence Building Questions

1. In terms of valence bond theory, how is a chemical bond formed?

Hint: A chemical bond is due to the overlap of atomic orbitals.

Discussion -
Molecular orbital theory considers the energy states of the molecule.

2. When one s and two p atomic orbitals are used to generate hybrid orbitals, how many hybrid orbitals will be generated?

Hint: Using three atomic orbitals generates three hybrid orbitals.

Discussion -
Number of orbitals does not change in hybridization of atomic orbitals.

3. In the structures of $$\ce{SO2}$$ and $$\ce{NO2}$$, what are the values of the bond angles?

Hint: The bond angles are expected to be less than 120 degrees.

Discussion -
Since the lone electron pair in $$\ce{:SO2}$$ and lone electron in $$\ce{.NO2}$$ take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds.

4. What is the geometrical shape of the molecule $$\ce{CH4}$$, methane?

Hint: Methane molecules are tetrahedral.

Discussion -
The 4 $$\ce{H}$$ atoms form a tetrahedron, and methane has a tetrahedral shape.

5. What do you expect the bond angles to be in the $$\ce{NH4+}$$ ion?

Hint: All bond angles are 109.5 degrees, the ideal value for a symmetric tetrahedral structure.

Discussion -
The structure of this ion is very similar to that of $$\ce{CH4}$$.

6. What hybrid orbitals does the $$\ce{C}$$ atom use in the compound $$\ce{H-C\equiv C-H}$$, in which the molecule is linear?

Hint: The sp hybrid orbitals are used by the $$\ce{C}$$ atom.

Discussion -
Sigma (s) bonds are due to sp hybrid orbitals, and 2 p orbitals are used for pi (p) bonds. The two sigma bonds for each $$\ce{C}$$ are due to overlap of sp hybrid orbitals of each $$\ce{C}$$ atom.

7. What hybrid orbitals does $$\ce{C}$$ use in the molecule:
        H
O=C<
H

This is a trigonal planar molecule. It is called formaldehyde, a solvent for preserving biological samples. The compound has an unpleasant smell.
Hint: The $$\ce{C}$$ atom uses sp2 hybrid orbitals.

Skill -
The $$\ce{C}$$ atom has 3 sigma (s) bonds by using three sp2 hybrid orbitals and a pi (p) bond, due to one 2p orbital.

8. What is the shape of the molecule $$\ce{SF6}$$?

Hint: Its shape is octahedral.

Discussion -
Since the $$\ce{S}$$ atom uses d2sp3 hybrid orbitals, you expect the shape to be octahedral. The $$\ce{F}$$ atoms form an octahedron around the sulfur.

9. Phosphorus often forms a five coordinated compound $$\ce{PX5}$$. What hybrid orbitals does $$\ce{P}$$ use in these compounds?

Hint: The $$\ce{P}$$ atom uses dsp3 hybrid orbitals.

Discussion -
A total of 5 atomic orbitals are used in the hybridization: one 3d, one 3s and three 3p orbitals. The dsp3 hybrid orbitals of $$\ce{P}$$ give rise to a trigonal bipyramidal coordination around the $$\ce{P}$$ atom.

The energy of d orbitals in $$\ce{N}$$ is not compatible with 2s and 2p orbitals for hybridization. Thus, you seldom encounter a compound with formula $$\ce{NX5}$$ with $$\ce{N}$$ as the central atom.