# What is "VSEPR" in relation to molecular structure?

Aug 13, 2016

$\text{VSEPR}$ theory ($\text{vesper for convenience}$) is an acronym of $\text{VALENCE SHELL ELECTRON PAIR REPULSION THEORY.}$

#### Explanation:

$\text{VSEPR}$ allows us to make very good approximations of electronic AND molecular geometries simply by counting the number of electron pairs around a central atom. Now $\text{VSEPR}$ determines electronic geometries, but since we describe molecular geometries on the basis of the disposition of atoms not electron pairs, molecular and electronic geometries MAY differ. I will try to walk you thru with a couple of examples.

The key idea of VSEPR is that electron pairs around a central atom repel each other, and assume the shape that minimizes electronic interaction. It does not matter to a first approx. whether the electron pairs are bonding (i.e. the electron pair forms a direct element-element bond) or non-bonding.

The shapes, as we shall see, can correspond to the classic Platonic solids. If there are 2 electron pairs a central atom, the pairs should be disposed linearly; if 3 electron pairs, their shape should assume a trigonal plane; if 4 electron pairs, a tetrahedron is described; if 5, a trigonal bipyramid; and if 6, an octahedron.

Let's take 2 examples: methane, $C {H}_{4}$, and water, $O {H}_{2}$. Around the carbon centre in methane there are $4 \times C - H$ bonds, composed of the 4 electrons donated by the hydrogen atoms, and the 4 valence electrons of carbon. Their most symmetrical arrangement about the carbon centre is tetrahedral, and the energy minimum occurs when the $\angle H - C - H$ bond angles are all ${109.5}^{\circ}$. This is the prediction that is backed up experimentally. Now let's go the water.

In water there are $2 \times O - H$ bonds, and 2 oxygen-based lone pairs, distributed about the oxygen centre. Again, the most stable arrangement of these bonding and non-bonding pairs is a tetrahedron. But because we describe molecular geometry on the basis of ATOMS not lone pairs, we would describe the geometry of water as a bent molecule, where the $\angle H - O - H$ bond angle is compressed from the ideal tetrahedral angle of ${109.5}^{\circ}$ to approx. $104 - {5}^{\circ}$ due to the disproportionate influence of the oxygen lone pairs; the electron pairs are still distributed as a tetrahedron, but the symmetry has descended a little bit.

This same quasi tetrahedral arrangement pertains for both ammonia, $N {H}_{3}$, and phosphorus trichloride, $P C {l}_{3}$ but here the symmetry descends to trigonal pyramidal because there is one not two lone pairs.

Given what I have said, can you refer to a text, and predict the electronic geometries of say, $B C {l}_{3}$, and $P {F}_{3}$?