Chapter II. The Two Orbital Problem

In this section of material we are going to take a general situation for two AOs on adjacent atoms c₁ and c₂ on atoms A and B, respectively.

The two LCAOs are;

y₁ = c₁₁c₁ + c₂₁c₂

y₂ = c₁₂c₁+ c₂₂c₂

Recall that for the mixing coefficients, c_mi, we use the convention that the first subscript, m, refers to the atomic orbital and the second, i, to the molecular orbital.

And we have:

<c₁|c₁> = <c₂|c₂> = 1

<c₁|c₂> = S₁₂ = (+)

<c₁|H^eff|c₁> = H₁₁ = = (-)

<c₂|H^eff|c₂> = H₂₂ = = (-)

<c₁|H^eff|c₂> = H₁₂ = (-)

H₁₂ µ -S₁₂ <0

The secular determinant for this situation is:

Expansion of the secular determinant leads to -

( - e_i)( - e_i) - (H₁₂ - e_iS₁₂)² = 0

A. The Degenerate Case: =

In terms of an orbital interaction diagram:

One orbital is lowered in enrgy and the other is raised. What is important is that because of the denominator for the e_i solutions:

The lower orbital is stabilized less than the upper orbital is destabilized.

So putting one or two electrons into the system is stabilizing but putting four electrons is destabilizing - the system is unstable, i.e.

In quantitative terms the stabilization and destabilization is

E⁽²⁾ = 2e₁ - 2 2(H₁₂ - S₁₂)(1 - S₁₂) µ H₁₂ µ -S₁₂

E⁽⁴⁾ = 2(e₁ + e₂) - 4 -4S₁₂(H₁₂ - S₁₂) µ S₁₂²

We have the two secular equations and the normalization condition to use for solving for the orbital mixing coefficients -

( - e_i)c_1i + (H₁₂ - e_iS₁₂)c_2i = 0 (eq. 1)

(H₁₂ - e_iS₁₂)c_{1i +} (-e_i)c_2i = 0 (eq. 2) <y_i|y_i> = + 2c_1ic_2iS₁₂ + = 1

let us arbitrarily put e ₁ into eq. 1

The only way for this to be true is if

c₁₁ - c₂₁ = 0 or c₁₁ = c₂₁

thus,

y₁ = c₁₁(c₁ + c₂)

If you put e₂ into eq. 2 you will get:

c₁₁ + c₂₁ = 0 or c₁₁ = -c₂₁

thus,

y₂ = c₂₂(c₁ - c₂)

We can take this one step further by using the normalization condition;

Likewise for y₂ we have

The orbital interaction diagram is:

A plot of these two MOs for H₂ is -

B. The Nondegenerate Interaction:

Going back to the expansion of the secular determinant:

(1+)e_i² + (2H₁₂S₁₂ - - )e_i + ( - ) = 0

This is nothing more than a quadratic equation which can easily be solved by a pocket calculator! One only has to put in values for the two atomic orbital energies, the reasonance integral and the overlap integral. One can get a physically more meaningful algebraic solution by solving the quadractic equation:

where

a = 1 -

b = 2H₁₂S₁₂ - -

D = b² -4ac c = -

One can also use series expansions to simplify these expressions to -

The orbital interaction diagram is now more complicated (and we assumed that < )

Notice again that e₁ is stabilized less than e₂ is destabilized. Therefore stabilization occurs with one or two electrons in the interaction diagram and destabilization results in the occupation with four electrons. In a more quantitative basis:

E⁽²⁾ = 2e₁ - 2

E⁽⁴⁾ = 2(e₁ + e₂) - 2(+)

The orbital mixing coefficients are determined to be:

y₁ (1-tS₁₂ t²)c₁ + tc₂

where t is

And

y₂ t'c₁ + (1 - t'S₁₂ t'²)c₂

where t' is

Since invariably

In pictorial terms this is shown in terms of an orbital interaction diagram -

Each MO most strongly resembles that starting orbital closest to it in energy

C. So let us summarize the two cases:

E⁽⁴⁾ µ

E⁽²⁾ µ -S₁₂

E⁽⁴⁾ µ

So how do we determine the relevant parameters - they are the starting AO energies, the , the H₁₂ and S₁₂ terms? We have discussed the relationship between H₁₂ and S₁₂ and how to determine S₁₂ in a qualitative sense. The quanities are listed for the main group elements -

To a first approximation they are determined in a qualitative sense by electronegativity.

D. Dividing up the electron density - Mullikan population analysis.

Let us take the general two orbital wavefunction -

y₁ = c₁₁c₁ + c₂₁c₂

and the normalization condition -

The electron density then if y₁ has one electron is:

In the Mullikan approximation the shared electron density - called overlap population - is divided equally between the two atoms.

We can generalize this to ;

where n = the orbital occupation number (0, 1 or 2)

Here P_mn is called the overlap population. The electron density assign to a particular AO, m, is then

Summing over all of the AOs on an atom and subtracting the nuclear charge gives the charge on the atom.