chemistry help please?

Question

what is a possible set of quantum numbers for the 2 highest energy electrons in a 82Pb atom? (n,l,ml,ms)

Answer 1

Pb has the electronic structure: [Xe].4f14.5d10.6s2.6p2, where [Xe] represents the electronic configuration of the element xenon (that part of the electronic structure of Pb is not relevant to this question).

The highest energy electrons are the ones in 6p orbital, so we know immediately that n = 6, l = 1 (because they are in a p orbital). The possible values of ml are +1, 0 or -1, and the possible values of ms are +1/2 and -1/2.

We know from the Pauli exclusion principle that no two electrons can have the same set of quantum numbers, and we also know from Hund's rule that every orbital in a subshell is occupied by a single electron before any of the orbitals are doubly occupied, and that alll electrons in singly occupied orbitals have the same spin.

That means that the possible pairs of ground-state quantum numbers for the 6p electrons in Pb are:

(6,1,+1,+1/2) (6,1,0,+1/2)
(6,1,+1,+1/2) (6,1,-1,+1/2)
(6,1,0,+1/2) (6,1,-1,+1/2)
(6,1,+1,-1/2) (6,1,0,-1/2)
(6,1,+1,-1/2) (6,1,-1,-1/2)
(6,1,0,-1/2) (6,1,-1,-1/2)

plus the additional 6 pairs that correspond to interchanging the two electrons.