# Advanced Inorganic Chemistry/Td Molecular Orbitals

## Td Point Group

The Td symmetry group is a one of the point groups considered to have higher symmetry. Molecules in the Td group, such as methane (CH4) have four C3 rotational axes.

The character table for the Td point group is shown below:

Td E 8C3 3C2 6S4 6σd
A1 1 1 1 1 1 x2+y2+z2
A2 1 1 1 −1 −1
E 2 −1 2 0 0 (2z2x2y2, x2y2)
T1 3 0 −1 1 −1 (Rx, Ry, Rz)
T2 3 0 −1 −1 1 (x, y, z) (xy, xz, yz)

## Constructing Molecular Orbitals

### Determining Reducible Representations

Figure 1. Bond vectors in a tetrahedral (Td) molecule.

The first step to constructing the molecular orbitals (MOs) is to determine the reducible representations for the σ and π bond vectors, denoted ${\displaystyle \Gamma _{\sigma }}$  and ${\displaystyle \Gamma _{\pi }}$ , respectively. The character corresponding to each symmetry operation can be determined by tracking how the bond vectors transform. Each vector that is left unmoved will contribute +1, each vector that is shifted to a new position will contribute 0, and each vector that undergoes a sign reversal will contribute −1. The reducible representations ${\displaystyle \Gamma _{\sigma }}$  and ${\displaystyle \Gamma _{\pi }}$  are listed below:

Td E 8C3 3C2 6S4 6σd
${\displaystyle \Gamma _{\sigma }}$  4 1 0 0 2
${\displaystyle \Gamma _{\pi }}$  8 −1 0 0 0

### Reducing the Reducible Representations

The reducible representations obtained from the previous section can be written as linear combinations of the irreducible representations shown in the character table. The contribution of each irreducible representation to the reducible representation is given by the formula

${\displaystyle n_{i}={\frac {1}{h}}\sum N\chi _{R}\chi _{I}}$

where ${\displaystyle n_{i}}$  is the coefficient of the ith reducible representation, ${\displaystyle h}$  is the order of the point group, ${\displaystyle N}$  is the number of symmetry operations in the class, ${\displaystyle \chi _{R}}$  is the character of the reducible representation corresponding to the class, ${\displaystyle \chi _{I}}$  is the character of the irreducible representation corresponding to the class, and the summation is taken over all classes of symmetry operations. Applying this method to the ${\displaystyle \Gamma _{\sigma }}$  and ${\displaystyle \Gamma _{\pi }}$  representations gives

${\displaystyle \Gamma _{\sigma }=A_{1}+T_{2}}$

${\displaystyle \Gamma _{\pi }=E+T_{1}+T_{2}}$

### Determining SALCs of Pendant Ligands

The Symmetry Adapted Linear Combinations (SALCs) can be found using the projection operator technique. For example, suppose we using the following labelling scheme for the ligand groups in Figure 1: (1) top-left-front, (2) top-right-back, (3) bottom-right-front, (4) bottom-left-back. Then the normalized SALCs for the s orbitals of the ligand groups are

${\displaystyle SALC(A_{1})={\frac {1}{2}}(s_{1}+s_{2}+s_{3}+s_{4})}$

${\displaystyle SALC(T_{2})={\frac {1}{2}}(s_{1}-s_{2}-s_{3}+s_{4})}$

${\displaystyle SALC(T_{2})={\frac {1}{2}}(s_{1}-s_{2}+s_{3}-s_{4})}$

${\displaystyle SALC(T_{2})={\frac {1}{2}}(s_{1}+s_{2}-s_{3}-s_{4})}$

The purpose of using the projection operator method is to be able to draw the SALCs.

### Drawing the Molecular Orbital Diagram

Structure of methane molecule.

Consider CH4 as a specific example of a molecule within the Td symmetry group. Before drawing the MO diagram, the valence atomic orbitals on the central atom and their symmetries must be noted. The symmetry of the orbitals can be determined by determining how they transform under the symmetry operations or by simply looking at the right most columns on the character table. The valence orbitals of the carbon atom include 2s (A1 symmetry) and 2px, 2py, 2pz (T2 symmetry).

In CH4, the only potentially bonding orbitals of the ligand groups are the 1s orbitals in the hydrogen atom. The irreducible representations contributing to the symmetry of the ligand group SALCs determines the AOs of the central atom with which they can interact. As found previously, the SALCs of the hydrogen atoms have A1 + T2 symmetry, so they can interact with all of the valence orbitals on the carbon atom to form bonding and antibonding MOs. Bonding MOs are positive linear combinations of the central atom AOs and ligand group SALCs, while antibonding MOs are negative linear combinations. The number of bonding and antibonding MOs created must be equal to the number of AOs and ligand group orbitals used to create them.

Although in CH4, all AOs and SALCs formed of bonding and antibonding MOs, this is not always the case. For example, if a central atom AO has symmetry that is absent from the ligand group SALCs, it will be unable to participate in the formation of bonding and antibonding MOs. In such cases, non-bonding MOs are formed.

The relative energy of the MOs are influenced by a few trends. For MOs formed with similar AOs, the following tend to be true:

• Bonding MOs have lower energy than antibonding MOs. Non-bonding MOs tend to have energy intermediate between bonding and antibonding MOs.
• π bonding MOs have higher energy than σ bonding MOs.
• The energy of MOs increases with the number of nodes.
• The totally symmetric σ bonding MO has the lowest energy among σ bonding MOs.

Illustrations of CH4 MOs can be found in Pfennig's text, Principles of Inorganic Chemistry, pp. 300.