Question

Write the wavefunctions describing all 4 bonds in methane molecule using the VB theory.

Answer 1

In valence bond theory, promotion of a paired electron is its excitation to an empty orbital at higher energy during bond formation. This energy input is worthwhile so long as it is less than the energy released through the formation of the extra bonds. In the case of carbon we have the electron configuration 2s²2p_x¹2p_y¹. Promotion of one of the paired electrons in the 2sorbital into the vacant 2p_z orbital results in us having four unpaired electrons. We are now able to form four bonds by pairing the spins of these electrons with unpaired electrons of four other atoms. The simplest example is the formation of four bonds to four hydrogen atoms in the methane molecule, CH₄. The energy put in to promote the 2s electron is more than made up for by the formation of two extra C-H bonds, that would not have otherwise been possible.

The picture is still incomplete. The bonding picture in CH₄ implies that we have three ?-bonds of one type, by pairing three of the H1s with the C2p electrons, and one ?-bond of another type from pairing the remaining H1s and C2s electrons. However, we know experimentally that all four C-H bonds in the methane molecule are identical, having the same bond strengths and length. This discrepancy is overcome because quantum mechanics allows us to describe the same electron density in a variety of equivalent ways. With the excited carbon atom, we can consider the electron distribution as being derived from having four unpaired electron in a 2sorbital and three 2p orbitals. We can also describe it, however, as arising from having the four unpaired electrons in four equivalent orbitals that result from mixing the original atomic orbitals. Mixed orbitals derived from the atomic orbitals on the same atom are termed hybrid orbitalsand are formed by the interference between the wave functions of the individual atomic orbitals. The combinations that give the four hybrid orbitals for the carbon atom in methane are described mathematically as:

h₁ = s + p_x + p_y + p_z
h₂ = s - p_x - p_y + p_z
h₃ = s - p_x + p_y - p_z
h₄ = s + p_x - p_y - p_z

What does this actually look like? The interference between the positive and negative parts of the wave functions for the s and p-orbitals results in four hybrid orbitals with large lobes pointing towards the corners of a regular tetrahedron centred on the nucleus of the carbon atom. As these hybrids are constructed from one s and three p orbitals, they are called sp³hybrid orbitals. Using the concepts of promotion and hybridisation we can now rationalise the tetravalent nature of carbon and the equivalency of the four C-H bonds in the methane molecule.

We also have molecules containing carbon atoms which are bonded to only three atoms, for example ethene H₂CCH₂. This is a planar molecule with HCH and HCC bond angles of 120