Question

Hi! I had a question on chemistry.

The last valence electron in sulfur has a spin of
For this question I assumed the spin was -1 as the last valance electron is down spinning, and m is -l to l. In this case sulfur is 3p meaning l is 1. The correct answer is -1/2. Can someone explain this to me? Is the spin not m? Is m not in relation to l?

Answer 1

The electron spin quantum number is
`m_(s)` and does not depend on another quantum number. It is +1/2 (↑) and -1/2 (↓). For the sulfur, the spin is -1/2.

Step-by-step explanation:

The quantum numbers for atoms are:

- n: the principal quantum number = 1, 2, 3, 4, 5, 6

- l: the orbital angular momentum quantum number = 0, 1, 2, 3, ... n-1

-
`m_(l)`: the magnetic quantum number = -l, .. -1, 0, 1, .. l

-
`m_(s)`: the electron spin quantum number = +1/2 (↑), -1/2 (↓). This quantum number does not depend on another quantum number.

The sulfur, Z = 16 has the following electronic configuration:

1s² 2s² 2p⁶ 3s² 3p⁴ → [Ne] 3s² 3p⁴

Let's now see the quantum numbers for the orbital 3p⁴ (the last electron filling orbital) for the sulfur:

- n = 3

- l = 1 (since 0 = s, 1 = p, 2 = d, 3 = f)

-
`m_(l)` = -1, since the last electron filled in the orbital p (the fourth electron) is in ml = -1:

↑↓ ↑ ↑

n = 3
`p_(x)`
`p_(y)`
`p_(z)`

ml = -1 0 +1

-
`m_(s)` = -1/2. This is because the last electron filled is in ml = -1, in a down position ↓.

Therefore, the electron spin quantum number is
`m_(s)`, and does not depend on another quantum number. It is +1/2 when the electron is ↑ and -1/2 when the electron is ↓. Thus, for the sulfur the last electron filled in the orbital is ↓, so the spin is -1/2.

I hope it helps you!