I'm trying to understand the examples on Hund's rules on Wikipedia, but I have a problem. Wikipedia says that for Silicon the possible combinations of quantum numbers are $1D$, $3P$ and $1S$ (first example). I understand why $S$ can only have a spin multiplicity of 1 ($m_\ell$ must equal 0, so the electrons must have opposite spin), and $P$ of 3 ($m_\ell$ can be $-1,0,1$ so electrons can have the same spin).

However, I don't understand why the spin multiplicity for $D$ is 1: $\ell=2$, so $m_\ell=-2,-1,0,1,2$. I understand that 2 and -2 are not possible as it would require the two electrons to have the same spin in the same state, but why can't there be one electron with $m_\ell=0$ and one with $m_\ell=1$ with the same spin? This would make a spin multiplicity of 3, not 1.

  • $\begingroup$ Note that one is considering here Silicon only in configuration $1s^2 2s^2 2p^6 3s^2 3p^2$. There are no d-electrons in this configuration, and no need to consider them explicitly. See the difference between term symbol and symmetry of of an orbital. Here is exactly the problem solved: youtube.com/watch?v=Cl2JWWoi6T8 $\endgroup$ – Mikael Kuisma Feb 20 '16 at 15:57

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