# Why are bonding orbitals more stable?

Jul 25, 2015

Bonding orbitals minimize the nuclear repulsion energy.

Let us consider the following equation which describes the energy of a quantum mechanical system via the Particle-in-a-Box model for the helium atom:

$E = {\overbrace{- \frac{1}{2} {\nabla}_{1}^{2} - \frac{1}{2} {\nabla}_{2}^{2}}}^{\text{Kinetic Energy" overbrace(- e^2/(4piepsilon_0vecr_1) - e^2/(4piepsilon_0vecr_2))^"1-electron terms" overbrace(+ (2e^2)/(4piepsilon_0vecr_(12)))^"2-electron term" + overbrace(h_(n uc))^"Nuclear repulsion energy}}$

The first two terms indicate kinetic energy. Let's ignore that since that is not our focus.

The 1-electron terms describe the coulombic attractions of each individual electron to the nucleus of the atom, whereas the 2-electron term describes the coulombic repulsions between the pairwise electron interactions in the atom. (Note: this term is why solving for the exact ground-state energy of helium is impossible)

You can tell from the equation that to maintain the equality, if the third and/or fourth term increases, the sixth term decreases (if it changes), and if the third and/or fourth term decreases, the sixth term increases (if it changes). The fifth term changes randomly.

Using the Born-Oppenheimer Approximation, the nuclei stay still, and so if the electrons move, interactions between the electrons change (2-electron term) and interactions between the nucleus and electron changes (1-electron terms).

The point is, the more nuclear repulsion, the higher in energy the molecular orbital is.

Bonding orbitals minimize the nuclear repulsion energy.