Question

identify a set of quantum numbers for the last electron added to the sulfur atom.
$n$, $l$, $m_l$, $m_s$

Explanation:

Step1: Find sulfur's atomic number

Sulfur (S) has an atomic number of 16, so it has 16 electrons.

Step2: Write electron configuration

Electron configuration: $1s^22s^22p^63s^23p^4$
The last electron is in the 3p subshell.

Step3: Assign principal quantum number ($n$)

For the 3p subshell, $n=3$ (the shell number).

Step4: Assign azimuthal quantum number ($l$)

For p subshells, $l=1$ (s=0, p=1, d=2, f=3).

Step5: Assign magnetic quantum number ($m_l$)

p subshell has 3 orbitals with $m_l = -1, 0, +1$. The 4th electron in 3p fills the second orbital, so $m_l=0$ (or $m_l=+1/-1$ are also valid as orbitals are degenerate, but 0 is a standard choice for the 4th p electron).

Step6: Assign spin quantum number ($m_s$)

The electron is the second in its orbital, so spin is $-\frac{1}{2}$.

Answer:

$3, 1, 0, -\frac{1}{2}$
(Note: $m_l$ can also be $+1$ or $-1$, and $m_s$ could be $+\frac{1}{2}$ if considering the first electron in a new p orbital, but the above is a standard valid set.)