# What are the effects of the repulsive forces between electron on the potential energy graph for the dihydrogen molecule?

This diagram is from my textbook. It states that as long atoms get closer, the energy decreases due to the attractive forces between the electrons and the nuclei. When the nuclei get too close, the nuclei’s repulsive forces drives the energy upward. However, what of the repulsive forces between the electrons? As the atoms get closer and closer from infinity, where are the effects of the increasing repulsive forces between the electrons?

## 1 Answer

First of all, you have to understand that the potential energy depicted on the graph in your question already accounts for repulsive forces between the electrons. This potential energy is the sum of two terms, namely, the sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$, where the first term contains contributions from kinetic energy of electrons motion $T_{\mathrm{e}}$, potential energy of their interactions with the nuclei $V_{\mathrm{en}}$, as well as potential energy of their interactions with each other $V_{\mathrm{ee}}$. This sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$ plays a role of potential energy for the motion of the nuclei, hence its name.

All these components of the potential energy, $E_{\mathrm{e}} = T_{\mathrm{e}} + V_{\mathrm{en}} + V_{\mathrm{ee}}$ and $V_{\mathrm{nn}}$, behave differently with respect to internuclear separation. For instance, as you have mentioned, the last term $V_{\mathrm{nn}}$ grows when nuclei get closer to each other. The same has to be true for $V_{\mathrm{ee}}$ term since the closer the nuclei are the closer are the electrons. However this effects of increasing the energy could be overcompensated by a decrease in $V_{\mathrm{en}}$ term thanks to attractive nature of the corresponding forces between nuclei and electrons.

When a decrease in $V_{\mathrm{en}}$ overcompensates an increase in $V_{\mathrm{nn}} + V_{\mathrm{ee}}$ the bond is formed. Not that an overall decrease in potential energy is accompanied by an increase in kinetic energy in accordance with the virial theorem. But the increase in kinetic energy is only half of the lowering in the potential energy, so that the total energy goes down when a bond is formed.

Here is the graph of all these terms for $\ce{H2}$ molecule obtained from quantum chemistry calculations1:

1. Wilson, C. W. & Goddard, W. A. The role of kinetic energy in chemical binding. Theor. Chim. Acta 26, 195–210 (1972). PDF from caltech.edu.
• Very nice answer. I have just one point I'd like to discuss: In your last sentence you state: However this effect of increasing the energy could be compensated [...] by a decrease in $T_{\mathrm{e}}$.. But wouldn't the kinetic energy of the electrons rise as the atoms come closer together? My argument for that would be as follows: When the $\ce{H}$ atoms come closer the orbitals get, sort of, compressed. This compression leads to a higher gradient of the wave function near and inbetween the nuclei and since the kinetic energy of the electrons is proportional to said gradient it rises. – Philipp Sep 20 '15 at 9:56
• The argument about the proportionality between the gradient of the wave function and the kinetic energy of the electrons I discussed in the last paragraph of this answer of mine in more detail. – Philipp Sep 20 '15 at 9:58
• @Philipp, yeah, you're perfectly right. My mistake. – Wildcat Sep 20 '15 at 11:26
• Very interesting links. – Philipp Sep 20 '15 at 12:40