Valence Bond Theory and Hybrid Atomic Orbitals

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

Confidence Building Questions

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.
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.
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.
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.
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 \(\ce{CH4}\).
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.
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.
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.
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.

Contributors and Attributions

Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)