Question

A) n=1 l=0 ml=0 ms=1/2
B) n=3 l=1 ml=-1 ms=1/2
C) n=2 l=0 ml=0 ms=-1/2
D) n=2 l=1 ml=0 ms=1/2
E) n=3 l=2 ml=0 ms=1/2
F) n=2 l=1 ml=1 ms=-1/2




Which of the above sets of quantum numbers could possibly describe an electron in the ground-state configurations of sulfur

Answer 1

16S = 1s2 2s2 2p6 3s2 3p4
What is it about this that you don't understand? Here are the rules.
n can be any whole number starting with 1.
l can be any whole number less than n-1.
ml can be -ell to +ell including 0.
ms can be either +1/2 or -1/2.

Answer 2

What I don't get is how do i determine the quantum numbers that match up with S's electron configuration

Answer 3

OK. Let's look at the two 1s electrons.
Remember those configurations give you n and l.
So for 1s2 n must be 1 and s electrons have ell = 0 and the two electrons of 1s2 can have +1/2 and other one -1/2
Therefore, one electron has
n = 1, l = 0 mL = 0 (see the rules;; i.e., since l = 0 then mL can be -l to +l incl 0 and in this case 0 is the only choice), then ms = +1/2
The other one of the pair is
n = 1, l = 0, ml = 0 and ms = -1/2.

For the 2s2.
n = 2, s means l = 0, ml = 0 and ms = +1/2 and =1/2.
You go through the electron progression just as I've done for the 1s2 and 2s2 but you cover the others. Remember the code: s means l = 0, p means l = 1, d means l = 2 and f means l = 3
I'll be glad to check your work if you want to post it.
A appears to be a choice. E can't be a choice because for l to be 2 we must have a d electrons and there aren't any d electrons in Sulfur. I'll be happy to answer any further questions. This can be very confusing; that's why you need to follow the rules.