Bonding and anti-bonding orbitals

We look, first, at the form of the orbitals that correspond to the energies $\Delta E_{\pm}$, respectively. These can be obtained by solving for the variational coefficients, $C_1$ and $C_2$. These will be given by the matrix equation:

$$\left(\matrix{H_{11} & H_{12} \cr H_{21} & H_{22}}\right) \l... ...{12} \cr S_{21} & 1}\right) \left(\matrix{C_1 \cr C_2}\right)$$

For example, using $E_+=(H_{11}-H_{12})/(1-S)$, the following equations for the coefficients are obtained:

$$H_{11} C_1 + H_{12}C_2 = {H_{11}-H_{12} \over 1-S}(C_1+SC_2)$$

$$H_{12} C_1 + H_{11}C_2 = {H_{11}-H_{12} \over 1-S}(SC_1+C_2)$$

which are not independent but are satisfied if $C_1=-C_2\equiv C_+$. Similarly, for $E=E_-$, we obtain the two equations:

$$(H_{11}C_1 + H_{12}C_2) = {H_{11}+H_{12} \over 1+S}(C_1+SC_2)$$

$$(H_{12}C_1 + H_{11}C_2) = {H_{11}+H_{12} \over 1+S}(SC_1+C_2)$$

which are satisfied if $C_1=C_2\equiv C_-$. Thus, the two states corresponding to $E_{\pm}$ are

$$\vert\psi_-\rangle = C_-\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)$$

$$\vert\psi_+\rangle = C_+\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)$$

The overall constants $C_{\pm}$ are determined by requiring that $\vert\psi_{\pm}\rangle$ both be normalized. For $C_-$, for example, we find

$$\langle \psi_-\vert\psi_-\rangle = \vert C_-\vert^2 \left(\langle \psi_1\vert + \langle \psi_2\vert\right) \left(\vert\psi_1\rangle + \psi_2\rangle \right)$$

$$= \vert C_-\vert^2\left(1+1+\langle \psi_1\vert\psi_2\rangle + \langle \psi_2\vert\psi_1\rangle \right)$$

$$= 2\vert C_-\vert^2(1+S)$$

$$= 1$$

which requires that

$$C_- = {1 \over 2\sqrt{1+S}}$$

Similarly, it can be shown that

$$C_+ = {1 \over 2\sqrt{1-S}}$$

Thus, the two states become

$$\vert\psi_-\rangle = {1 \over 2\sqrt{1+S}}\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)$$

$$\vert\psi_+\rangle = {1 \over 2\sqrt{1-S}}\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)$$

Notice that these are orthogonal:

$$\langle \psi_+\vert\psi_-\rangle = 0$$

Projecting onto a coordinate basis, we have

$$\psi_-({\bf r}) = {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) + \psi_2({\bf r})\right)$$

$$\psi_+({\bf r}) = {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) - \psi_2({\bf r})\right)$$

The state $\psi_-$, which corresponds to the energy $E_-$ admits a chemical bond and is, therefore, called a bonding state. The state $\psi_+$, which corresponds to the energy $E_+$ does not admit a chemical bond and is, therefore, called an anti-bonding state. $\psi_+({\bf r})$ and $\psi_-({\bf r})$ are examples of what are called molecular orbitals. In this case, they are constructed from linear combinations of atomic orbitals.

The functional form of the two molecular orbitals for H$_2^+$ within the current approximation scheme is

$$\langle {\bf r}\vert\psi_-\rangle = \psi_-({\bf r}) = {1 \over 2\sqrt{1+S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}+e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$$

$$\langle {\bf r}\vert\psi_+\rangle = \psi_+({\bf r}) = {1 \over 2\sqrt{1-S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}-e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$$

The contours of these functions are sketched below (the top plot shows the two individual two atomic orbitals, while the middle and bottom show the linear combinations $\psi_-({\bf r})$ and $\psi_+({\bf r})$, respectively):

図 3:

Since $\psi_1({\bf r})$ and $\psi_2({\bf r})$ have the same value at the origin and very similarly values near the origin, it is clear that the electron probability density in this region will intensify for $\psi_-({\bf r})$ which is the sum of $\psi_1({\bf r})$ and $\psi_2({\bf r})$. This clearly corresponds to a chemical bonding situation. In contrast, for $\psi_+({\bf r})$, which is the difference between $\psi_1({\bf r})$ and $\psi_2({\bf r})$, there will be a deficit of electron density in the region between the two protons, which is indicative of a non-bonding situation.