# Hund's rules for Li, contradiction?

Hund's rules basically state that we sould maxiumise $S$ and for a given value of $S$ maximise $L$. So let us take the case of Li. Li has 3 electrons, clearly the first 2 are in the $n=1$ state, with $S=0$ and $L=0$. For our remaining electron it must go into one of the $n=2$ states. No matter which state we put it in $S=1$, so we now use Hund's rule to maximise the value of $L$, which would indicate that we put it into the $p$ state giving $L=1$ rather then $s$ state giving $L=0$. But it does go into the $s$ state, which appears to violate Hund's rule. Why does it do this?

(p.s. my notation may be slightly out)

• The $2\mathrm{s}$ orbital is lower in energy than the $2\mathrm{p}$ orbital. Their energy difference is larger than the energy you might gain by maximising $L$. Thus $2\mathrm{s}$ gets populated. – Philipp May 3 '16 at 12:50
• Now is a good time to learn that many chemistry 'rules' are, at best, general advice with the caveat that you have to pay attention to all the 'outliers'... – Jon Custer May 3 '16 at 14:42

For example, with a $\mathrm{2p^2}$ configuration (e.g. carbon), the allowed term symbols are $^1D$, $^3P$, and $^1S$. From Hund's first rule, you can predict that the most stable one out of these three is the $^3P$ term.
You cannot compare these three terms with a term arising from a different electronic configuration, e.g. a $\mathrm{2p3p}$ configuration, which could give rise to a $^3D$ term which would be "more stable" on the basis of Hund's second rule.
Just to show you why it doesn't make sense: with your lithium example, if you really wanted to, then you might as well promote the 2s electron to a 3d orbital, since that would give you a $^2D$ state. Higher $L$, wonderful! Why not go further and promote it to a 4f orbital, or a 5g orbital? We might as well ionise it...