In the previous unit, we applied our quantum mechanical approach to solve a problem which was notorious in 19th century; a structure for benzene in which all C-C-bonds of molecule were equal. But describing a system of electrons in the proximity of a framework of nuclei by combining wave functions will also prove to be valuable in the most fundamental cases of molecular bonding: Molecular hydrogen which consists of a pair of electrons moving in the surroundings of two protons. For a start, we will consider the simplest molecule we could imagine, which is theoretically obtained by adding one proton to a single Hydrogen atom. Electrostatics states that there is only repulsion between two protons and it would be interesting to find out whether one electron could serve as a kind of glue, thereby forming a stable system of three particles. Finally, our approach will support the idea of a stable cationic molecule H2+.
If we imagine a proton pushed into the sphere of one hydrogen atom, there is a certain point where the electron does not belong to one proton anymore. Such an exchange goes along with the exchange energy or interaction
Basic states | Linear combination | |
|I> | ![]() |
In complete analogy to the benzene, the
electron's wave function in the H2 cation is described by |I>
and |II> which are combination of the basic states. ![]() |
|II> | ![]() |
The solution with the lowest energy for the whole system is
characterized by a wavefunction with equal amplitude in the proximity of both protons. The total energy is equal to EI =
Eo- ΔE in state |I>. The dependency between the two
protons' distance and this energy is shown in the diagramm to the left.
The interaction ΔE increases if the protons get closer to each
other.
It further suggests that the previously mentioned quantum mechanical exchange
energy overcompensates the repulsive forces between the two protons and is
fundamental for chemical bonding. Note that the same exchange energy would
destabilize the system of the particles in state |II>.
Basic states | ![]() |
![]() |
If we use |I> = 1/√2 (|1> + |2>) as linear combination, we will obtain |I'> = 1/√2 (|2> + |1>) = |I>. No change of sign occurs. In consequence, if both electrons' spin are equal, there is only one possible state |II> = 1/√2 (|1> − |2>) since electron exchange yields: |II'> = 1/√2 (|2> − |1>) = −|I>.
Because of the higher energy of this state |II> both H atoms will repulse each other. There is no H2 molecule with parallel spins. If both electrons have spins pointed to opposite directions, then there is obviously stable state of molecule. The total spin of this molecule is essentially equal to zero for this state.
Such two electron binding leads to the quite usual valence bond. Nevertheless, the above-mentioned considerations hold even true in situations where the electron attraction of one of the nuclei is greater (for instance, NaCl). All possible gradations of bindings between covalent and ionic can be explained using the considered model of two electron system.