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Comprehensive Study Notes

The MO approximation puts the electrons of a molecule in molecular orbitals, which extend over the whole molecule. As an approximation to the molecular orbitals, we usually use linear combinations of atomic orbitals. The VB method puts the electrons of a molecule in
atomic orbitals and constructs the molecular wave function by allowing for "exchange" of the valence electron pairs between the atomic orbitals of the bonding atoms. We compared the two methods for $\mathrm{H}_{2}$. We now consider other homonuclear diatomic molecules.

We begin with the ground state of $\mathrm{He}_{2}$. Each separated helium atom has the groundstate configuration $1 s^{2}$. This closed-subshell configuration does not have any unpaired electrons to form valence bonds, and the VB wave function is simply the antisymmetrized product of the atomic-orbital functions. In the notation of Eq. (10.47), the He VB ground state wave function is the Slater determinant

\(
\begin{equation}
\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}}\right| \tag{13.110}
\end{equation}
\)

The subscripts $a$ and $b$ refer to the two atoms, and the bar indicates spin function $\beta$. The $1 s$ function in this wave function is a helium-atom $1 s$ function, which ideally is an SCF atomic function but can be approximated by a hydrogenlike function with an effective nuclear charge. The VB wave function for $\mathrm{He}_{2}$ has each electron paired with another electron in an orbital on the same atom and so predicts no bonding.

In the MO approach, $\mathrm{He}{2}$ has the ground-state configuration $\left(\sigma{g} 1 s\right)^{2}\left(\sigma_{u}^{*} 1 s\right)^{2}$. With no net bonding electrons, no bonding is predicted, in agreement with the VB method. The MO approximation to the wave function is

\(
\begin{equation}
\left|\sigma{g} 1 s \overline{\sigma{g} 1 s} \sigma_{u}^{} 1 s \overline{\sigma_{u}^{} 1 s}\right| \tag{13.111}
\end{equation}
\)

The simplest way to approximate the (unnormalized) MOs is to take them as linear combinations of the helium-atom AOs: $\sigma{g} 1 s=1 s{a}+1 s{b}$ and $\sigma{u}^{*} 1 s=1 s{a}-1 s{b}$. With this approximation, (13.111) becomes

\(
\begin{equation}
\left|\left(1 s{a}+1 s{b}\right) \overline{\left(1 s{a}+1 s{b}\right)}\left(1 s{a}-1 s{b}\right) \overline{\left(1 s{a}-1 s{b}\right)}\right| \tag{13.112}
\end{equation}
\)

Using theorems about determinants, we can show (Prob. 13.34) that (13.112) is equal to

\(
\begin{equation}
4\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}}\right| \tag{13.113}
\end{equation}
\)

which is identical (after normalization) to the VB function (13.110). This result is easily generalized to the statement that the simple VB and simple LCAO-MO methods give the same approximate wave functions for diatomic molecules formed from separated atoms with completely filled atomic subshells. We could now substitute the trial function (13.110) into the variational integral and calculate the repulsive curve for the interaction of two ground-state He atoms.

Before going on to $\mathrm{Li}{2}$, let us express the Heitler-London valence-bond functions for $\mathrm{H}{2}$ as Slater determinants. The ground-state Heitler-London function (13.100) and Prob. 13.33a can be written as

\(
\begin{gather}
\frac{1}{2}\left(1+S{a b}^{2}\right)^{-1 / 2}\left{\left|\begin{array}{ll}
1 s{a}(1) \alpha(1) & 1 s{b}(1) \beta(1) \
1 s{a}(2) \alpha(2) & 1 s{b}(2) \beta(2)
\end{array}\right|-\left|\begin{array}{ll}
1 s{a}(1) \beta(1) & 1 s{b}(1) \alpha(1) \
1 s{a}(2) \beta(2) & 1 s{b}(2) \alpha(2)
\end{array}\right|\right} \
=\left(2+2 S{a b}^{2}\right)^{-1 / 2}\left{\left|1 s{a} \overline{1 s{b}}\right|-\left|\overline{1 s{a}} 1 s{b}\right|\right} \tag{13.114}
\end{gather}
\)

In each Slater determinant, the electron on atom $a$ is paired with an electron of opposite spin on atom $b$, corresponding to the Lewis structure $\mathrm{H}-\mathrm{H}$. The Heitler-London functions (13.101) for the lowest $\mathrm{H}{2}$ triplet state can also be written as Slater determinants. Omitting normalization constants, we write the Heitler-London $\mathrm{H}{2}$ functions as

\(
\begin{array}{ll}
\text { Singlet: } & \left|1 s{a} \overline{1 s{b}}\right|-\left|\overline{1 s{a}} 1 s{b}\right| \
\text { Triplet: } & \left{\begin{array}{l}
\left|1 s{a} 1 s{b}\right| \
\left|1 s{a} \overline{1 s{b}}\right|+\left|\overline{1 s{a}} 1 s{b}\right| \
\left|\overline{1 s{a}} \overline{1 s{b}}\right|
\end{array}\right. \tag{13.116}
\end{array}
\)

Now consider $\mathrm{Li}{2}$. The ground-state configuration of Li is $1 s^{2} 2 s$, and the Lewis structure of $\mathrm{Li}{2}$ is $\mathrm{Li}-\mathrm{Li}$, with the two $2 s \mathrm{Li}$ electrons paired and the $1 s$ electrons remaining in the inner shell of each atom. The part of the valence-bond wave function involving the $1 s$ electrons will be like the $\mathrm{He}{2}$ function (13.110), while the part of the VB wave function involving the $2 s$ electrons (which form the bond) will be like the Heitler-London $\mathrm{H}{2}$ function (13.115). Of course, because of the indistinguishability of the electrons, there is complete electronic democracy, and we must allow every electron to be in every orbital. Hence we write the ground-state VB function for $\mathrm{Li}_{2}$ using $6 \times 6$ Slater determinants:

\(
\begin{equation}
\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}} 2 s{a} \overline{2 s{b}}\right|-\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}} \overline{2 s{a}} 2 s{b}\right| \tag{13.117}
\end{equation}
\)

We have written down (13.117) simply by analogy to (13.110) and (13.115). For a fuller justification of it, we should show that it is an eigenfunction of the spin operators $\hat{S}^{2}$ and $\hat{S}_{z}$ with eigenvalue zero for each operator, which corresponds to a singlet state. This can be shown, but we omit doing so. To save space, (13.117) is sometimes written as

\(
\begin{equation}
\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}} \widehat{2 s{a} 2 s{b}}\right| \tag{13.118}
\end{equation}
\)

where the curved line indicates the pairing (bonding) of the $2 s{a}$ and $2 s{b} \mathrm{AOs}$.
The MO wave function for the $\mathrm{Li}_{2}$ ground state is

\(
\begin{equation}
\left|\sigma{g} 1 s \overline{\sigma{g} 1 s} \sigma_{u}^{} 1 s \overline{\sigma{u}^{*} 1 s} \sigma{g} 2 s \overline{\sigma_{g} 2 s}\right| \tag{13.119}
\end{}
\)

If we approximate the two lowest MOs by $1 s{a} \pm 1 s{b}$ then the same procedure used in Prob. 13.34 to show that $(13.111)$ is the same wave function as $(13.110)$ shows that $(13.119)$ is the same as

\(
\left|1 s{a} \overline{1 s{a}} 1 s{b} \overline{1 s{b}} \sigma{g} 2 s \overline{\sigma{g} 2 s}\right|
\)

Recall the notation $K K\left(\sigma{g} 2 s\right)^{2}$ for the $\mathrm{Li}{2}$ ground-state configuration.
Now consider the VB treatment of the $\mathrm{N}{2}$ ground state. The lowest configuration of N is $1 s^{2} 2 s^{2} 2 p^{3}$. Hund's rule gives the ground level as ${ }^{4} S{3 / 2}$, with one electron in each of the three $2 p \mathrm{AOs}$. We can thus pair the two $2 p{x}$ electrons, the two $2 p{y}$ electrons, and the two $2 p_{z}$ electrons to form a triple bond. The Lewis structure is $: \mathrm{N} \equiv \mathrm{N}:$. How is this Lewis structure translated into the VB wave function? In the VB method, opposite spins are given to orbitals bonded together. We have three such pairs of orbitals and two ways to give opposite spins to the electrons of each bonding pair of AOs. Hence there are $2^{3}=8$ possible Slater determinants that we can write. We begin with

\(
D{1}=\left|1 s{a} \overline{1 s{a}} 2 s{a} \overline{2 s{a}} 1 s{b} \overline{1 s{b}} 2 s{b} \overline{2 s{b}} 2 p{x a} \overline{2 p{x b}} 2 p{y a} \overline{2 p{y b}} 2 p{z a} \overline{2 p_{z b}}\right|
\)

In all eight determinants, the first eight columns will remain unchanged, and to save space we write $D_{1}$ as

\(
\begin{equation}
D{1}=\left|\cdots 2 p{x a} \overline{2 p{x b}} 2 p{y a} \overline{2 p{y b}} 2 p{z a} \overline{2 p_{z b}}\right| \tag{13.120}
\end{equation}
\)

Reversing the spins of the electrons in $2 p{x a}$ and $2 p{x b}$, we get

\(
\begin{equation}
D{2}=\left|\cdots \overline{2 p{x a}} 2 p{x b} 2 p{y a} \overline{2 p{y b}} 2 p{z a} \overline{2 p_{z b}}\right| \tag{13.121}
\end{equation}
\)

There are six other determinants formed by interchanges of spins within the three pairs of bonding orbitals, and the VB wave function is a linear combination of eight determinants (Prob. 13.35). The following rule (see Kauzmann, pages 421-422) gives a VB wave function that is an eigenfunction of $\hat{S}^{2}$ with eigenvalue 0 (as is desired for the ground state): The coefficient of each determinant is +1 or -1 according to whether the number of spin interchanges required to generate the determinant from $D{1}$ is even or odd, respectively. Thus $D{2}$ has coefficient -1 . [Compare also (13.115).] Clearly, the single-determinant ground-state $\mathrm{N}_{2} \mathrm{MO}$ function is easier to handle than the eight-determinant VB function.