quantum mechanics - Ground state of Beryllium (${rm Be}$)

Why is the ground state of Beryllium (${\rm Be}$) with electronic configuration $[{\rm He}]2s^2$ is $^1S_0$ and not $^3S_1$? The state $^3S_1$ has higher spin multiplicity.

Answer

There is no such state. A triplet $^3S$ spin state would be exchange-symmetric on the spin sector, which would require an antisymmetric state on the orbital sector, and this is impossible on a $2s^2$ orbital configuration, since both electrons are in the same state and that is intrinsically symmetric.

(On the other hand, a $^3S_1$ state is perfectly possible, say, in a $1s\: 2s$ configuration. But it will be extremely hard to find a real atomic system with a ground state of that type, as the higher energy of the $2s$ (or other such orbital) means that flipping to a $^1S$ configuration by dropping to both electrons on the lower orbital will be energetically favourable. And, indeed, a short scan using the Mathematica curated ElementData shows that no real atoms have a $^3S_1$ ground state.)

The Hund maximum-multiplicity rule does provide the ground state (except in the cases where it breaks), but you need to maximize over the set of states that do exist ;-).