Valence-Shell Electron-Pair Repulsion (VSEPR) Model

Considerations in Apply the Model

The bond angles for highly symmetric molecules are usually nice round numbers such as 90° or 120°. If one or more of the electron groups in the VSEPR Model is a lone pair, the molecule may have lower symmetry. In this case, the bond angles might be slightly different from those for highly symmetric molecules.

For example, water has N_EG = 4, which corresponds with a tetrahedral arrangement of electron groups. Two of the electron groups are lone pairs and two are σ bonds to hydrogen atoms. For perfect tetrahedral geometry, the bond angle is cos^-1(-⅓) = 109.5°. However, the experimental H-O-H bond angle in water is 104.5°.

Fortunately, such deviations are easily understood and predictable based upon the three considerations explained below. Note that these concepts lead to only qualitative (not quantitative) predictions.

1. Lone Pairs Occupy More Space than Bonding Pairs

When a pair of electrons is confined to a σ bonding orbital, the electron density is concentrated in the region between the bonded atoms.

A lone pair of electrons, however, is associated with a single atom and occupies more space around the atom.

Because a lone pair of electrons requires more space than a bonding pair, the molecular geometry distorts to afford the lone pair more room, with the result that bond angles are smaller than those of the ideal geometry.

In the case of water, the two lone pairs push the bonding pairs closer together, yielding a smaller bond angle. The white lines shown at right display the ideal 109.5° angle. The experimental H-O-H bond angle is only slightly smaller, but the effect is noticeable.

This effect is clearly observed by comparison of the F-P-F bond angles in PF₃ and OPF₃. The white lines in the boxes below show the ideal bond angle of 109.5°.

In PF₃ the lone pair on the phosphorus pushes the P-F bonding electrons away from itself, resulting in a F-P-F bond angle of 97.8°, which is appreciably smaller than the ideal bond angle of 109.5°.

In OPF₃, the lone pair is replaced with a P-O bond, which occupies less space than the lone pair in PF₃. Consequently there is less distortion of the P-F bonds, and the F-P-F bond angle is 107°. (The P=O bond occupies only slightly more space than a P-F bond.)

2. Avoid 90° Interactions between Lone Pairs

Because lone pairs occupy more space than σ bonding pairs, locate lone pairs in positions with the most space. The greatest source of repulsion is from 90° interactions. 120° interactions involve very little repulsion, even for lone pair - lone pair interactions.

First Priority:     Minimize 90° lone pair - lone pair repulsions.
Second Priority: Minimize 90° lone pair - σ bonding pair repulsions.

The trigonal bipyramidal geometry (N_EG = 5) has two different sites: axial and equitorial. There is more room in the equitorial positions, because there are only two 90° interactions (with the axial positions). Each axial position has three 90° interactions. 90° interactions are very strong; 120° interactions are much weaker. For this reason, lone pairs occupy equitorial positions.

PCl₅ has N_EG = 5 and no lone pairs, consequently the molecule has a perfect trigonal bipyramidal geometry, as seen at left below.

SF₄ also has N_EG = 5, but in this case there is a lone pair. The lone pair is placed in an equitorial position. Because the lone pair occupies more space than bonding pairs in the molecule, the bond angles are all smaller than for PCl₅.

3. Think Creatively about Hybridization

The most basic hybridization scheme for ammonia (NH₃) involves mixing the nitrogen atomic 2s, 2p_x, 2p_y, and 2p_z orbitals to yield a new set of sp³ hybrid atomic orbitals. Each orbital is equivalent and oriented at the corners of a tetrahedron. One can think of a sp³ hybrid orbital as having 25.0% s character and 75.0% p character.

However, it is not necessary for each of the hybrid orbitals to be equivalent or that the s and p orbitals need be combined in integer proportions. (Recall that hybrid orbitals are just mathematical functions constructed from the the mathematical functions for the native s, p and d orbitals.)

For a molecule like CH₄, it makes sense that each hybrid orbital has identical s and p character, because each hybrid orbital is involved in an identical C-H bond. But for NH₃, the hybrid orbital accommodating the lone pair might be expected to have a different composition than the hybrid orbitals associated with the N-H bonds. As the formulation of the hybrid orbitals differ from sp³, the geometries of the orbitals differ from tetrahedral.