Why do most character tables (e.g. $C_{3h}$ but not $C_{2h}$) include an $(x^2 + y^2)$ term?

Is it an abbreviated form for $d_{z^2}$, applying only where $3d$ (and higher) orbitals might be involved, or just a generic binomial term with no specific relevance?

The only clue I can find is on this page where it lists the full expression for the angular wave function for $3d_{z^2}$ as

$$Y_{3d_{z^2}} = √(5/4){2z^2-(x^2 + y^2)}/r^2 × (1/4\pi)^{1/2} $$

where it also comments that "the $3d_{2z^2-x^2-y^2}$ orbital is abbreviated to $3d_{z^2}$ for simplicity".

Thoughts appreciated.

  • $\begingroup$ Thanks to initial editor of the question, for showing me how to improve super- and sub-scripting $\endgroup$ – iSeeker Jun 17 at 14:42

The functions on the right hand side of the table are the 'basis functions' for the irreducible representation, which are the rows of character, say $A_g$ or $B_g$ in C$_{2h}$. The basis functions have the same symmetry properties as the atomic orbitals with the same name (but other functions in $x, x^2, x^3$ etc and similarly for $y,z$, could be used).

To determine where an $xy$ function goes, look at a $d_{xy}$ orbital shape and subject it to each of the symmetry operations in turn,$E, C_{2}, i$ etc depending on point group. (Assume z is the principal axis). Count +1 if the orbital (or function) is unchanged and -1 if not. So in C$_{2h}$, a $C_2$ operation leaves $xy$ unchanged as does $\sigma_h$ and $i$ so $xy$ is a 'basis' for A$_g$. The $xz$ transforms as B$_g$; a $C_2$ operation gives -1, as does $\sigma_h$, but $i$ produces +1.

(note in cases such as $z^2 -(x^2+y^2)/r^2$ the $x^2+y^2$ part is totally symmetric in the $xy$ plane and does not change anything so can be ignored; think of the donut around d$_{z^2}$.)

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