![\[ \hat {H}_{elec} \Psi (r)=E_{elec}\Psi (r) \]](formulas/lcao_mo_1.gif)
The problem in theoretical and computational studies of Molecular Electronic Structure is to solve the electronic Schrödinger equation for a system of N nuclei and n electrons:
![\[ \hat{H}_{elec}= \sum_{i=1}^{n} {\frac{-\nabla_i^{2}}{2}}
+ \sum_{i}^{n}\sum_{A}^{N} \frac{-Z_{A}}{r_{iA}}
+ \sum_{i}^{n}\sum_{j>i}^{n} \frac{1}{r_{ij}} + \sum_{A}^{N}\sum_{B>A}^{N}
\frac{Z_{A}Z_{B}}{R_{AB}} \]](formulas/lcao_mo_2.gif)
The Schrödinger Equation has analytical solutions for some simple systems:
In most cases, one needs instead to use Numerical approaches. This involves assuming that the real wavefunction, which is too complicated to be found directly, can be approximated by a simpler function. For some types of function, it is then possible to solve the electronic Schrödinger Equation numerically. Provided the assumption one has made about the form of the function is not too drastic, one obtains a good approximation to the correct solution. Molecular electronic structure computations consist in choosing sensible approximations to the wavefunction. A variety of approximate forms can be suggested; the simpler ones will lead to an easier solution of the Schrödinger Equation, with more complex approximations being harder to solve but leading to wavefunctions which are closer to the correct solution.
The most common starting point for solving the Electronic Schrödinger Equation for many electron systems is the (Molecular) Orbital Approximation. The wavefunction is taken to be a product of one electron wavefunctions:
![\[ \Psi(r_1,r_2,\ldots r_n)=\prod_{i=1}^{n}\psi_i(r_i) \;\;\;
[\text{that is,} \;\; \psi_1(r_1)\psi_2(r_2)\ldots \psi_n(r_n) ] \]](formulas/lcao_mo_3.gif)
These one-electron wavefunctions are also called Orbitals, and, in the case of molecules, are generally approximately expanded in a basis set of atomic functions:
![\[ \psi_i(r)=\sum_{j=1}^{n_{basis}}c_{ij}\phi_j(r) \]](formulas/lcao_mo_4.gif)
Molecular orbitals are of course familiar to you already - you have
discussed them in all sort of courses you have already followed. In the level
2 lecture courses of Prof. Ashfold and Dr. Mulholland, you have learned more
about molecular orbitals in diatomic molecules and in conjugated
(
-electron) systems.
The molecular orbital approximation is equivalent to assuming that the electrons behave independently from each other. The probability of finding an electron in orbital i at a given place ri does not depend on where the other electrons are:
![\[ P(e_i,r_i)=\psi_i^2(r_i)\int \left \{ \prod_{j \neq i} \psi^2_j(r_j)
\right \} \; dr = \psi_i^2(r_i) \]](formulas/lcao_mo_5.gif)
As a result, the probability of finding ANY electron at a given point in space is just the sum of the probabilities of finding one there in each of the orbitals.
The Orbital Approximation is also equivalent to assuming that the electronic Hamiltonian can be expressed as a sum of one-electron hamiltonians:
![\[ \hat{H}_{elec} \approx \sum_{i=1}^{n}h_{elec}^{i} \]](formulas/lcao_mo_6.gif)
For comparison, the full, correct electronic Hamiltonian is shown below. As can be seen, the first and second terms, which correspond to the kinetic energy of individual electrons, and to the interaction of each electron with the nuclei, respectively, are indeed sums over all the electrons. The final term is a constant, which is added separately to the sum of one-electron terms. The third term, which corresponds to the repulsion between electrons, cannot be separated exactly into one-electron terms. By imposing the separability, the Molecular Orbital Approximation thereby inevitably involves an incorrect treatment of the way in which the electrons interact with each other. For more details, see the section on the Hartree-Fock Method.
![\[ \hat{H}_{elec}= \sum_{i=1}^{n} {\frac{-\nabla_i^{2}}{2}} +
\sum_{i}^{n}\sum_{A}^{N} \frac{-Z_{A}}{r_{iA}} + \sum_{i}^{n}\sum_{j>i}^{n}
\frac{1}{r_{ij}} + \sum_{A}^{N}\sum_{B>A}^{N}
\frac{Z_{A}Z_{B}}{R_{AB}} \]](formulas/lcao_mo_7.gif)
How can one determine the shape of the molecular orbitals? In general, one needs to solve the Schrödinger equation. One can however often predict the qualitative shape of the orbitals without computation. In this section, we will examine how molecular orbitals arise in this way from linear combinations of atomic orbitals. This will be seen to be intimately related to the question of molecular symmetry.
As already mentioned above, molecular orbitals are obtained as linear combinations of atomic orbitals:
![\[ \psi_i(r)=\sum_{j=1}^{n_{basis}}c_{ij}\phi_j(r) \]](formulas/lcao_mo_8.gif)
This may at first sight appear to be a significant restriction on the form
molecular orbitals can take. In fact, if enough atomic orbital functions
are taken, any function can be written as a linear
combination of the form given. In practice, it is often found that the
qualitative shape of molecular orbitals can be reproduced from a very
small number of atomic orbitals. In fact, just taking the occupied
valence and core orbitals of the constituent atoms is usually enough to
understand the shape of molecular orbitals.
There are three qualitative rules for predicting how atomic orbitals mix to give molecular orbitals:
The first two rules above should be familiar to you from Prof. Ashfold's course. The third rule is in essence a consequence of the second, but may be less familiar to you so we shall now discuss it in some more detail, in the context of the molecular orbitals of the water molecule.
The water molecule belongs to the C2V point group. This means that it has four symmetry elements:
(To help you visualise these symmetry elements, you can use the following link to see a 3D representation of a water molecule. Click Here to view the water molecule (this link will open in a new window).
Let us now examine the core and valence atomic orbitals of the O and H atoms. These are shown on the following picture (Note that following the usual convention, the symmetry axis of the molecule is chosen as the z axis, and the plane of the molecule is chosen as the yz plane):
The individual H 1s orbitals do not belong to any of the irreps of the molecular point group. (Reminder: to belong to one of the irreducible representations, the object or function needs to be either symmetric or antisymmetric with respect to all the symmetry operations of the point group.)
Because the MOs have to belong to one of the irreps, this implies that there are constraints on the values of the cij for the H 1s orbitals in the linear combination. These coefficients have to match one of the two following conditions:
![\[
\begin{array}{rr}
\mbox{either:} \\
\mbox{or:} \\ \end{array}
\left \{ \begin{array}{ll}
c_{i,\phi_{(1s,H_{A})}} = c_{i,\phi_{(1s,H_{B})}} \\
c_{i,\phi_{(1s,H_{A})}} = -c_{i,\phi_{(1s,H_{B})}} \\
\end{array} \right.
\]](formulas/lcao_mo_9.gif)
An equivalent way of saying this is to say that the MOs of a molecule
are not obtained directly from AOs, but from symmetry-adapted linear
combinations of AOs. These are shown here, for the water molecule, and are
classified by the irrep to which they belong. You should already be
familiar with the names of the irreps for atoms: s, p, d, f, ...; and for
diatomic (or other linear) molecules: 
, , 
, ... You will
learn more about point groups, irreps, symmetry labels, and other
aspects of molecular symmetry in the lecture course with Dr. Norman.
These symmetry-adapted AOs mix, according to the rules given above, to yield MOs for water. Here is a rough diagram showing how the mixing occurs:
Notice that the O 1s orbital does not mix with the others - this is because it is much lower in energy than them. Notice also that the O 2px orbital does not mix with any others - this is because it is the only AO belonging to the b2 irrep.
As already discussed, the exact form of the molecular orbitals for any given molecule cannot be predicted simply from the qualitative rules given above. This is because there are typically lots of symmetry adapted AOs belonging to each irrep, and lots of MOs, and so there are hundreds of coefficients cij. These can be obtained quantitatively by solving the Schrödinger equation. The sort of results obtained can be seen by clicking on the following links to bring up a picture of each of the five lowest MOs of water. The coefficients of each AO for each MO are also given. Note that these links will each open in a new window.
Click to bring up a picture of the corresponding orbital: 1a1 2a1 1b1 3a1 1b2
Overall, because water has ten electrons, these five lowest orbitals are all doubly occupied, giving an electronic configuration which can be written in symbolic form in either of the following ways:
![\[ \Psi\{\mbox{H}_2\mbox{O}\} = \left \{ \begin{array}{lllll}
\psi_{1a_1}(r_1)\psi_{1a_1}(r_2)\psi_{2a_1}(r_3)\psi_{2a_1}(r_4)\ldots\psi_{1b_2}(r_{10}) \\
\mbox{ } \\
\psi_{1a_1}^2\psi_{2a_1}^2\psi_{1b_2}^2\psi_{3a_1}^{2}\psi_{1b_1}^2 \\
\mbox{ } \\
1a_1^22a_1^21b_1^23a_1^21b_2^2 \\ \end{array} \right.
\]](formulas/lcao_mo_10.gif)
Building up the molecular orbitals of large molecules directly from atomic orbitals is hard to do in the same qualitative way we have used above, because there are so many orbitals to consider. In practice, of course, the orbitals are generated by solving the Schrödinger equation. If one is looking for a qualitative understanding of how the molecular orbitals arise, it can sometimes be useful to break up the process of mixing the AOs to form MOs into several steps. For example, one can predict the form of the orbitals of the fragments A and B of a molecule AB, and then concentrate on how these "fragment MOs" mix to form the final MOs. The rules for this mixing procedure between fragment MOs are the same as those we used for mixing AOs
To illustrate this procedure and explain how it helps to understand structure and reactivity, we will analyze two examples chosen from organometallic chemistry.
Fischer Carbene Complexes. Various 16-electron organometallic moieties [M] form "carbene" complexes [M]CR2. An example is the molybdenum species Mo(CO)5=CH2. The structure of this complex is shown here:
This complex has a total of 120 electrons, so there are 60 occupied molecular orbitals. There are 81 occupied atomic orbitals in the 14 atoms making up the molecule so a diagram showing how they all mix would be horrendously complicated. Instead, let us view this molecule as being made up of two fragments, Mo(CO)5 and CH2. We will assume that we already know what the MOs of these fragments look like.
For CH2, we could in fact work them out starting from the atomic
orbitals, because this system is very similar to water, discussed above. In
the singlet state of CH2, the HOMO is similar to the 3a1
orbital of water: it is carbon lone pair, roughly speaking an sp2
hybrid, which is oriented along the Mo-C bond and is thus - loosely speaking -
a orbital. The LUMO is like the 1b2 orbital of water. It is
an empty p orbital orthogonal to the Mo-C bond (a orbital). These
orbitals are shown schematically here:
Click here to bring up a picture of these orbitals derived from an MO computation: HOMO LUMO
The orbitals of the Mo(CO)5 fragment are made up of many more
AOs and are thus very complex. Nevertheless, they can roughly be described as
being a d orbital of symmetry (HOMO), and a -symmetric
orbital made up out of the dz2 orbital, among others (LUMO):
Click here to bring up a picture of these orbitals derived from an MO computation: HOMO LUMO
It is now relatively easy to picture how the bond forms. The frontier
orbitals of the two fragments have matching symmetries (Mo(CO)5 is
sometimes said to be isolobal to CH2), so that the
orbitals mix among themselves, as do the orbitals. The overlap between
the orbitals is greater, so the energy splitting is larger too. This
gives the following rough MO diagram:
The form of three of these orbitals, as derived from an MO computation, can be seen by clicking on the following links: sigma pi pi*.
Note the form of the LUMO - this explains why carbenes react as electrophiles, with nucleophiles adding to the carbon atom. In some respects, the C=Mo double bond behaves like a C=O (carbonyl) double bond!
Palladium Allyl Complexes. These species are formed by the attack of Pd(0) compounds on allyl halides and are important intermediates in the Tsuji-Trost allylic substitution reaction. The structure of the parent compound is shown here (note the plane of symmetry):
Let us break this molecule up into two fragments, Pd(PR3)2 and allyl cation. This way of breaking up the molecule corresponds to the way in which it is formed in organometallic chemistry: by reaction between some or other Pd(0) species Pd(L)2 and an allyl-X derivative (X = acetate, ...). We can now analyse the bonding between the two fragments by looking at some of their MOs. The most important mixing that occurs is between the LUMO of allyl cation, and the HOMO of the Pd fragment. To view these MOs, click here (this link will open in a new window). You can see that both orbitals are antisymmetric with respect to the plane of symmetry - they belong to the A" irrep of the CS point group.
The way the mixing of these orbitals gives rise to the MOs of the complex is shown schematically in the following MO diagram:
To see the HOMO of the complex, click here. Note the strong bonding between palladium and the two terminal carbon atoms. To see the LUMO of the complex, click here. (These links will open in new windows) Note the anti-bonding nature of the orbitals between the Pd and C atoms. Note also the large lobes on the terminal carbon atoms, on the side opposite to palladium. What does this suggest for the stereochemistry of nucleophilic substitution of an allyl acetate by Nu, catalysed by Pd(0)?
The analysis of reactivity in this way is referred to as Frontier Molecular Orbital (FMO) theory, developed among others by Fukui. Using FMO theory is one of the most rigorous ways to derive the Woodward-Hoffmann rules, which you have met in Organic Chemistry lectures and in Dr. Mulholland's Year 2 Lecture Course.
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