Average of high-spin states: $n$ electrons in MO with
degenerate $n_{ir}$.

$$a = \frac{n_{ir}\bigl(4k(k+l-1)+l(l-1)\bigr)}{(n_{ir}-1)n^2}$$

$$b = \frac{2 n_{ir}\bigl(2k(k+l-1)+l(l-1)\bigr)}{(n_{ir}-1)n^2}$$

where: $k = \max(0,n-n_{ir})$ , $l = n-2k = 2S$   (spin)

This covers most of the cases given above. A CSF results only if $n = \{1, (n_{ir}-1),$ $n_{ir},$ $(n_{ir}+1),$ $(2n_{ir}-1)\}$ since there is a single high-spin CSF in these cases.

The last equations for $a$ and $b$ can be rewritten in many ways, the probably most concise form is

$$a = \frac{n(n-2)+2S}{(n-2f)n}$$

$$b = \frac{n(n-2)+(2S)^2}{(n-2f)n}\,.$$

This applies to shells with one electron, one hole, the high-spin couplings of half-filled shells and those with one electron more ore less. For $d^2$, $d^3$, $d^7$, and $d^8$ it represents the (weighted) average of high-spin cases: $^3$F + $^3$P for $d^2$,$d^8$, $^4$F + $^4$P for $d^3$, $d^7$.