How many electrons have n = 3, m_l = pm1, and m_s = +1/2?

Oct 31, 2016

For $n = 3$, we are in the 3rd energy level.

• In the $n = 3$ level, we have the values $\textcolor{g r e e n}{l = 0 , 1}$, and $\textcolor{g r e e n}{2}$ available, as $\boldsymbol{{l}_{\max} = n - 1}$ and $n - 1 = 2$.

Therefore, we have the $\boldsymbol{3 s}$, $\boldsymbol{3 p}$, and $\boldsymbol{3 d}$ orbitals available ($l = 0 \leftrightarrow s$, $l = 1 \leftrightarrow p$, $l = 2 \leftrightarrow d$).

• For ${m}_{l} = \pm 1$, we have exactly two specified orbitals (each orbital corresponds to one ${m}_{l}$) for a given $\boldsymbol{l}$. Now we examine each $l$ and see what restriction ${m}_{l} = \pm 1$ places on our set of orbitals.

For $\textcolor{g r e e n}{l = 0}$, ${m}_{l} = \pm 1$ is not valid because it falls outside the range of $l$ (we see that $\pm 1$ is not in the "range" of $0$). So, disregard this for the $3 s$ orbitals, because specifying ${m}_{l} = \pm 1$ has disallowed the $3 s$ orbital, whose ${m}_{l} = \boldsymbol{0}$.

For $\textcolor{g r e e n}{l = 1}$, ${m}_{l} = \pm 1$ is valid, since we would have allowed ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$. So we include two of the $3 p$ orbitals.

For $\textcolor{g r e e n}{l = 2}$, ${m}_{l} = \pm 1$ is also valid, since we would have allowed ${m}_{l} = \left\{- 2 , - 1 , 0 , + 1 , + 2\right\}$. So we include two of the $3 d$ orbitals.

• For ${m}_{s} = + \text{1/2}$, it just says that we specify a single electron with spin "up" of magnitude $\text{1/2}$. (Ordinarily ${m}_{s}$ could have been $\pm \text{1/2}$.)

As a result, our set of orbitals and electron(s) includes:

• Two $3 p$ orbitals (one corresponds to ${m}_{l} = - 1$, the other ${m}_{l} = + 1$)
• Two $3 d$ orbitals (one corresponds to ${m}_{l} = - 1$, the other ${m}_{l} = + 1$)
• One spin-up electron in each (${m}_{s} = + \text{1/2}$).

Therefore, we have an (isolated, theoretical) configuration of $3 {p}^{2} 3 {d}^{2}$, or four electrons.