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We solve the secular equation for the case when a1 = a2. This is a situation we get when two atoms, molecules, or molecular fragments interact that are the same; i.e., two H atoms or two He atoms.
Rewrite the secular equation as
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This equation can only have non-trivial solutions for c1 and c2 (i.e., c1 and c2 not both equal to zero) if the matrix is singular. This means
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= (a-e)2-(b-eS)2 = 0. |
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This is a quadratic equation in e. It can be solved by using the general expression for solutions of quadratic equation. However, it often is wise, less error-prone, to have a look at the equation to see if a simpler approach might be better. The equation above equates two squares. This means that
| a-e = +(b-eS), or a-e = -(b-eS) |
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must hold, or
| e = |
a-b
1-S
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, or e = |
a+b
1+S
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We have in general a < 0, bS < 0, |S| < 1 and |aS| < |b|. For identical orbitals we also have S > 0, b < 0 and aS > b. This means that (a-b)/(1-S) is a higher energy than (a+b)/(1+S).
We can determine the coefficients by substituting a solution for the energy in the secular equation. With (a+b)/(1+S) we find
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aS-b
1+S
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which yields c1 = c2. With (a-b)/(1-S) we find
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aS-b
1-S
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which yields c1 = -c2. We can only determine the relative magnitude of the coefficients from the secular equation, not their absolute value. This is normal for an eigenvalue equation. We can determine absolute values by normalizing the orbitals. We have
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dr |c1c1+c2c2|2 = |c1|2+|c2|2 +(c1c2+c1c2)S. |
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Equating this to one and using c1 = c2 yields
The sign of c1 is undetermined, but, as we know from quantum mechanics, it has no physical relevance anyway. So
is a normalized orbital with energy (a+b)/(1+S). Similarly we find that
is a normalized orbital with energy (a-b)/(1-S).
Assuming that the energy of an atom, molecule, or molecular fragment can be approximated by the sum of the MO-energies, we can use the energy expressions above to calculate bonding energies. Suppose that each of the interacting orbitals is occupied by just one electron. After a bond has been formed both electrons end up in the orbital with the lower energy. The energy change is then
| 2 |
a+b
1+S
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-2a = -2 |
aS-b
1+S
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< 0. |
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The energy decreases, which means that the bond is a stable one. If each of the interacting orbitals is doubly occupied, then the energy changes by
| 2 |
a+b
1+S
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a-b
1-S
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-4a = 4S |
aS-b
1-S2
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> 0. |
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Now the energy increases, which means that the atoms, molecules, or molecular fragments repel each other. It is this difference between single or double occupation of the interacting orbitals that makes H2 a stable molecule, and He2 not.
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