What is the difference between the quantum numbers of the valence electrons of Carbon and the quantum numbers of Carbon?

[![enter image description here](https://i.stack.imgur.com/dPCon.png)](https://i.stack.imgur.com/dPCon.png)


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.  

[![enter image description here](https://i.stack.imgur.com/PJGwY.png)](https://i.stack.imgur.com/PJGwY.png) 




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.


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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 $\rightarrow ^3\mathrm P$.


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