Group Theory Topics

Symmetry Operations for Phosphine

First we look at two pictures showing the molecule Phosphine along with its axis of rotation (orange) and three planes of reflection:

top down view–easier to see planes of reflection, harder to see the position of the free electrons relative to the rest of the molecule
side view–easier to see the free electrons relative to the rest of the molecule, harder to see the planes of reflection

The molecule is positioned so that the axis of rotation coincides with the z-axis. The hydrogen that is labeled “A” is on the x-axis. Looking down on the molecule, a positive rotation moves in the counterclockwise direction.

The Symmetry Operations are:

  • E – Identity
  • C_3 (+) -Rotation of +120 degrees
  • C_3 (-) -Rotation of -120 degrees
  • \sigma_{v} (a) -Reflection across the plane that bisects hydrogen A
  • \sigma_{v} (b) -Reflection across the plane that bisects hydrogen B
  • \sigma_{v} (c) -Reflection across the plane that bisects hydrogen C

We can build a Cayley Table with these operations:

\begin{bmatrix} C_{3v} & E & C_3 (+) & C_3 (-) & \sigma_{v} (a) & \sigma_{v} (b) & \sigma_{v} (c) \\ E & E & C_3 (+) & C_3 (-) & \sigma_{v} (a) & \sigma_{v} (b) & \sigma_{v} (c) \\ C_3 (+) & C_3 (+) & C_3 (-) & E & \sigma_{v} (c) & \sigma_{v} (a) & \sigma_{v} (b) \\ C_3 (-) & C_3 (-) & E & C_3 (+) & \sigma_{v} (b) & \sigma_{v} (c) & \sigma_{v} (a) \\ \sigma_{v} (a) & \sigma_{v} (a) & \sigma_{v} (b) & \sigma_{v} (c) & E & C_3 (+) & C_3 (-) \\ \sigma_{v} (b) & \sigma_{v} (b) & \sigma_{v} (c) & \sigma_{v} (a) & C_3 (-) & E & C_3 (+) \\ \sigma_{v} (c) & \sigma_{v} (c) & \sigma_{v} (a) & \sigma_{v} (b) & C_3 (+) & C_3 (-) & E \\ \end{bmatrix}

The matrices that correspond to these Symmetry Operations are shown below:

Identity – E - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Rotation (+) = C_3 (+) = \begin{bmatrix} cos(120) & sin(120) & 0 \\ - sin(120) & cos(120) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - \dfrac {1} {2} & - \dfrac { \sqrt(3)} {2} & 0 \\ \dfrac {+ \sqrt(3)}{2} & - \dfrac {1} {2} & 0 \\ 0 & 0 & 1 \end{bmatrix}

Rotation (-) = C_3 (-) = \begin{bmatrix} cos(- 120) & sin(- 120) & 0 \\ - sin(- 120) & cos(- 120) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - \dfrac {1} {2} & + \dfrac {\sqrt(3)} {2} & 0 \\ - \dfrac {\sqrt(3)}{2} & - \dfrac {1} {2} & 0 \\ 0 & 0 & 1 \end{bmatrix}

Reflection (a) = \sigma_{v} (a) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Reflection (b) = \sigma_{v} (b) = \begin{bmatrix} + \dfrac {1} {2} & - \dfrac {\sqrt(3)} {2} & 0 \\ - \dfrac {\sqrt(3)}{2} & - \dfrac {1} {2} & 0 \\ 0 & 0 & 1 \end{bmatrix}

Reflection (c) = C_3 (b) = \begin{bmatrix} + \dfrac {1} {2} & + \dfrac {\sqrt(3)} {2} & 0 \\ + \dfrac {\sqrt(3)}{2} & - \dfrac {1} {2} & 0 \\ 0 & 0 & 1 \end{bmatrix}

Symmetry Operations for Water

We will repeat this work, working with water. We won’t show the two lone pairs of electrons–it will be your homework to model them on a computer and show that the symmetry operations discussed below move them appropriately.

This time there is only one rotation, since “rotate +180” accomplishes the same thing as “rotate -180”.

Rotation = C_2 = \begin{bmatrix} cos(180) & sin(180) & 0 \\ - sin(180) & cos(180) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} - 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Also, “rotate +180” is its own inverse:

(rotate +180)(rotate +180) = E

There are two planes of reflection:

Reflection (xz) = \sigma_{xz}

Reflection (yz) = \sigma_{yz}

\begin{bmatrix} C_{2v} & E & C_2 & \sigma_{xz} & \sigma_{yz} \\ E & E & C_2 & \sigma_{xz} & \sigma_{yz} \\ C_2 & C_2 & E & \sigma_{yz} & \sigma_{xz} \\ \sigma_{xz} & \sigma_{xz} & \sigma_{yz} & E & C_2 \\ \sigma_{yz} & \sigma_{yz} & \sigma_{xz} & C_2 & E \end{bmatrix}

Matrices for the Symmetry Operations for Water

\begin{bmatrix} - 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} - 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} +1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

There are two planes of reflection:

Reflection (xz) = \sigma_{xz} = \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}

Reflection (yz) = \sigma_{yz} =\begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}

We can build a Cayley Table with the above matrices and everything in this table will correspond to the Cayley Table showing the multiplication of the Symmetry Operations. This will prove that, for these matrices and the Symmetry Operations, one is a Representation of the Other.

\begin{bmatrix} & E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & C_2 = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \sigma_v(xz) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \sigma_v(yz) == \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ C_2 = = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ \sigma_v(xz) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ \sigma_v(yz) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{bmatrix}
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