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I am having some problems reading character table. I will describe what I understand and thought to be right.

Consider the $\pi$ MOs formed from overlap of p-orbitals in benzene. These two degenerate $\pi$ MOs below have $E_{1g}$ symmetry:

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I don't understand why that is the case. Based on what I have read, if a symmetry operation $g$ maps an orbital to itself, e.g.

enter image description here

then $\chi'(g) = 1$.

Elif $g$ maps the orbital to negative itself (invert the sign of orbital),

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then $\chi'(g) = -1$.

Else, $\chi'(g) = 0$.

For two degenerate orbitals as above, final $\chi(g)$ is the sum of $\chi'(g)$ for two orbitals, where $\chi(g)$ is character of a symmetry operation $g$, that is recorded in the character table.


Applying this, consider the $C_6$ of benzene,

enter image description here

Not mapped to $\pm$ itself, so $\chi'(g) = 0$.

enter image description here

Also not mapped to $\pm$ itself, so $\chi'(g) = 0$.

So $\chi(g) = 0 + 0 = 0$.

But if we look at the character table for $D_{6h}$, for the symmetry species $E_{1g}$, $\chi(C_6) = 1 \neq 0$.

Enlighten me, please, I must have understood something terribly wrong.

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