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

1 Answer
Aug 13, 2016

#"VSEPR"# theory (#"vesper for convenience"#) is an acronym of #"VALENCE SHELL ELECTRON PAIR REPULSION THEORY."#


#"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 #"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, #CH_4#, and water, #OH_2#. Around the carbon centre in methane there are #4xxC-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 #/_H-C-H# bond angles are all #109.5^@#. This is the prediction that is backed up experimentally. Now let's go the water.

In water there are #2xxO-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 #/_H-O-H# bond angle is compressed from the ideal tetrahedral angle of #109.5^@# to approx. #104-5^@# 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, #NH_3#, and phosphorus trichloride, #PCl_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, #BCl_3#, and #PF_3#?