Is the hydrogen molecule bonding really covalent?

[eoghan]

Hi there!
I'm studying the hydrogen molecule. According to the MO technique I start from one-electron molecular orbitals to construct the wave function of the entire molecule.
Afa hydrogen is concerned, I can construct two ^1\Sigma_g orbitals:
\Phi_A=\Phi_g(1)\Phi_g(2)\chi_{0,0}
\Phi_B=\Phi_u(1)\Phi_u(2)\chi_{0,0}
Where \Phi_g(i) means that the i-th electron is in the gerale orbital obtained from the one-electron molecule.
Computing the energies for \Phi_A and \Phi_B it is found that only \Phi_A is a bonding orbital, so that the wave function of the hydrogen molecule is \Phi_A where the two electrons are both in the gerale orbital with antiparallel spins.

To improve the accuracy of the model, a linear combination of \Phi_A and \Phi_B can be considered:
\Phi_T=\Phi_A+ \lambda \Phi_B
using this wave function as the trial function for a variational method, \lambda can be found.
Now, this wavefunction can be rewritten as a sum of a covalent bonding wavefuntion and a ionic bonding wavefunction:
\Phi_T=(1- \lambda )\Phi_A^{COV}+(1+ \lambda )\Phi_A^{ION}
and it is found that the coefficient for the ionic bonding is 20% of the coefficient of the covalent bonding.

So.. can I say that the hydrogen molecule is covalent, but with a little part of ionic bonding?

 

[cgk]

I think homonuclear dimers, like hydrogen, are some of the few species where one cannot say that. Ionic bonding implies the existence of persistent partial charges, and these you cannot have in a such molecules. Also I believe that the weight of the leading determinant in a CI expansion for H2 is much higher than 80% if one does a full variational calculation with more degrees of freedom in the basis set (I seem to recall something like 98%).

In general, however, "covalent bonding" and "ionic bonding" are more qualitative of a concept, so whenever you have some bonding/antibonding orbital pair where the bonding orbital has a higher occupation, you can say there is some covalent bonding, and whenever there are partial charges, you can say there is some ionic bonding (at the same time).

 

[DrDu]

You should take in mind that the value of lambda is basis set dependent. And changes drastically e.g. by going from a basis of Hydrogen 1s orbitals to orthogonalized s orbitals.
On the other hand, ionic structures can be avoided at all in a valence bond calculation of H2 by using a basis of orbitals in which some of the orbital on the other atom is hybridized, i.e. phi_1= a*s_1 +b*s_2 and phi2=-b*s_1 +a*s_2.
See, e.g.,
@article{mcweeny1988hybridization,
title={{Hybridization in valence bond theory: The water molecule}},
author={McWeeny, FE and others},
journal={Journal of Molecular Structure: THEOCHEM},
volume={169},
pages={459--468},
issn={0166-1280},
year={1988},
publisher={Elsevier}
}

Probably the best measure would be some Bader analysis
http://en.wikipedia.org/wiki/Bader_analysis
to assess in a somewhat objective way the amount of ionicity.

A more qualitative proceedure would be the following: The electronic wavefunction is purely covalent in the limit of infinite separation of the nuclei and the purely ionic structure is always (infinitely) above the covalent in that limit. When the atoms are brought close, the ground state is mainly covalent if there is no avoided intersection between the potential curves.

 

[eoghan]

Thank you!