# What is the difference between the quantum numbers of the valence electrons of Carbon and the quantum numbers of Carbon? Does the question requires something different than finding the correct sets of quantum numbers of C?

I'm using the diagram below, and I found that all possibilities fit. why are the unselected choices wrong? It's certainly clear that the issue is in the orbitals ml, yet I can't figure out why. • Quantum numbers are a property of electrons, not atoms. Very loosely speaking each electron in each orbital can be identified with one set of quantum numbers (strictly speaking this violates the idea of quantum indistinguishability but let's not go there). So, "quantum numbers of C" doesn't make sense. There are six electrons in a C atom and each of them have their own set of quantum numbers. – orthocresol Dec 19 '16 at 12:14

I think the other two options are wrong because they violate the Hund's rule of maximum multiplicity. The orbitals with l=1 have opposite signs in each of the unselected options, but by the rule, you should have electrons having the same spin sign to have maximum multiplicity.

All the combinations listed are allowed ones for a $p^2$ configuration. (There are 15 combinations in total when Pauli exclusion is added). These combinations produce term symbols $^1\mathrm D$, $^3\mathrm P$ and $^1\mathrm D$, however, you are asked for the ground state which by Hundt's rules is the $^3\mathrm P$.

This has $M_l= m_{l1} + m_{l2} = 1,0,-1$ and spin $M_s= m_{s1}+m_{s2}= 1,0,-1$ which makes 9 combinations.

The $^1\mathrm D$ and $^1\mathrm S$ must have $m_{s1}+m_{s2}= 0$ but S has $m_{l1}+m_{l2}= 0$ and so one combination and D has $m_{l1} + m_{l2} = 2,1,0,-1,2$ making 5 combinations as $M_s=0$ only. (The spin multiplicity superscript on the term symbol is $2M_s+1$).

The first entry in your table has $M_l=1, M_s=0$ so belongs to the singlet state $^1\mathrm D$.

The second entry has $M_l=0, M_s = 1$ so $\rightarrow ^3\mathrm P$.

The third entry has $M_l = 2,M_s=0$ so $\rightarrow ^1\mathrm D$

The forth entry has $M_l = 0,M_s=-1$ so $\rightarrow ^3\mathrm P$.