# For the ground state of tin, why is it not possible to have a triplet D state?

I have been looking at electron configuration and understand the use of hund's rules, the Aufbau principle and the Pauli exclusion principle but am having difficulty with a question that has come up in a text book I am reading.

It says to find the ground state of tin (Z=50).I have calculated and checked that this leaves two remaining electrons in the 5p shell and that you ignore the filled shells as they contribute nothing towards calculations.

Iv worked through it and using the rules specified above and found that the lowest energy available (ground state) points to a triplet D state, but in the book it simply says that this is not possible due to the Pauli exclusion principle and doesn't explain further.

I am not sure how this relates to the quantum numbers and am wondering if it is due to the m_l quantum number??

Thank you

• Just to clarify, I had calculated that S=1 for the highest spin and that L=2 is the highest value you can obtain for total l but realize this is not possible and require L=1, but do not understand the maths behind it to prove this result (but understand the concept). Commented May 27, 2017 at 15:36

(I'm not sure what is a triplet $D$ state since $D\to L=2$ would have $5$ states but this is not essential to my answer.)
By Hund's rule, the $S=1$ state for two aligned valence spins is symmetric under permutations of the spin, meaning the spatial part of the wave function must be antisymmetric.
In the coupling of the $\ell=1$ states, the symmetric part contains $L=2$ and $L=0$ and the antisymmetric part contains only $L=1$. Thus, the spatial wave function must have $L=1$.
Possibly the most direct way to check the symmetry under permutation is to look at the Clebsch-Gordan coefficient $C_{1m_1;1m_2}^{LM}$ which couples your two $\ell=1$ states to angular momentum $L$. Permuting the particles gives $C_{1m_2;1m_1}^{LM}=(-1)^{2-L}C_{1m_1;1m_2}^{LM}$, showing that the odd $L$ is antisymmetric under permutation.
Alternatively, one can establish there are a total of 6 permutation symmetric states: three of the type $$\vert 1m_1\rangle \vert 1m_2\rangle + \vert 1m_2\rangle \vert 1m_1\rangle$$ for $(m_1,m_2)=(-1,0),(-1,1),(0,1)$ and three more of the type $$\vert 1m_1\rangle \vert 1m_1\rangle$$ with $m_1=m_2=(-1,-1),(0,0),(1,1)$. This leaves 3 antisymmetric states, which can only have angular momentum $L=1$ since $(\ell=1)\otimes(\ell=1)= L=2\oplus L=1\oplus L=0$ and neither the $L=2$ nor the $L=0$ have $3$ states.