Wikipedia 10K Redux by Reagle from Starling archive. Bugs abound!!!

<-- Previous | Newer --> | Current: 984115152 Josh Grosse at Fri, 09 Mar 2001 05:19:12 +0000.

The stationary states electrons can have within [[molecules]]. It's next to impossible to find out what the orbitals of a molecule are directly. Instead, one approximates the molecular orbitals as linear combinations of some basis for the electron's state space, usually what each atom's orbitals would be if it was on its own. Some qualitative rules: * There must be as many molecular orbitals as there were basis orbitals. * Basis orbitals mix more (ie contribute more to the same molecular orbitals) when they are closer in energy. * Molecular symmetries map stationary states to stationary states, so any collection of degenerate molecular orbitals must transform according to some [[representation]] of the [[symmetry group]]. As a result, basis orbitals that transform according to different representations don't mix. As a simple example consider H_{2}, with the atoms labelled H' and H". The lowest energy atomic orbitals, 1s' and 1s", don't transform according to the symmetries of the molecule. However, the following linear combinations do: 1s' - 1s" Antisymmetric combination: negated by reflection, unchanged by other ops 1s' + 1s" Symmetric combination: unchanged by all symmetry ops Since these are of very different energy than all the other atomic orbitals, we would expect these two combinations to be close approximations to the lowest two molecular orbitals. In general, the symmetric combination (called a bonding orbital) is lower in energy than the basis orbitals, and the antisymmetric combination (called an antibonding orbital) is higher. Since the H_{2}molecule has two electrons, they can both go in the bonding orbital, making the system lower in energy (and thence more stable) than two free hydrogen atoms. This is called a [[covalent bond]]. On the other hand consider a hypothetical molecule of H_{3}, with the atoms labelled H, H', H". Then we would expect three low energy combinations: 1s + 1s" - 1s' Symmetric 1s - 1s" Antisymmetric 1s + 1s' + 1s" Symmetric The bonding and antibonding orbitals end up with roughly the same energy, but now there is a third orbital between them, of roughly the same energy as one of the basis orbitals. Since the bonding electron can only hold two electrons, the third will have to go into the middle level, so there is no energy advantage for the molecule to stay together except at high pressures. Under those conditions, we can expect to see an effectively infinite number of hydrogens packed together, so the molecular orbitals form continuous bands. Now let's move to larger atoms. Considering a hypothetical molecule of He_{2}, we find that both the bonding and antibonding orbital are filled, so there is no energy advantage to the pair. HeH would have a slight energy advantage, but not so much as H_{2}+ 2 He, so the molecule exists only a short while. In general, we find that atoms like He that have completely full energy shells rarely bond with other atoms. The same sort of thing applies for the lower energy shells in larger molecules: although they still mix with other orbitals, there is no real energy advantage gained as a result, so they can be neglected from consideration. Molecular structure relies on the outermost (valence) electrons of the different atoms, which are usually of comparable energy. When there is a fair difference between them, though, one gets the case where one atom's orbitals contribute almost entirely to the bonding orbitals, and the other's almost entirely to the antibonding orbitals. Thus the situation is effectively that some electrons have been transferred between the atoms. This is called a (mostly) [[ionic bond]].