# In a ground-state "P" atom in the gas phase, how many electrons have the quantum numbers n = 3, l = 1, m_l = –1 ?

Dec 16, 2017

One electron.

#### Explanation:

The key here is the electron configuration of a neutral atom of phosphorus, which looks like this

$\text{P: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{3}$

Now, the $p$ subshell contains a total of $3$ orbitals. According to Hund's Rule, every orbital present in a given subshell must be half filled before any one of the orbitals can be completely filled.

This means that all the three $p$ orbitals present in the $p$ subshell must contain one electron, i.e. be half-filled, before any one of these orbitals can contain two electrons, i.e. be completely filled.

Notice that phosphorus has $3$ electrons in the $3 p$ subshell. This tells you that each $3 p$ orbital will hold a single electron.

Now, you know that the principal quantum number, $n$, tells you the energy shell in which an electron is located inside an atom.

$n = 3 \to$ the third energy shell

The angular momentum quantum number, $l$, describes the energy shell in which the electron is located.

You know that you have

• $l = 0 \to$ the $s$ subshell
• $l = 1 \to$ the $p$ subshell
• $l = 2 \to$ the $d$ subshell
$\vdots$

and so on. In your case, you're dealing with the $p$ subshell.

Finally, the magnetic quantum number, ${m}_{l}$, tells you the orientation of the orbital in which the electron is located. For the $p$ subshell, you have

$l = 1 \implies {m}_{l} = \left\{- 1 , 0 , + 1\right\}$

${m}_{l} = - 1$
which refers to one of the three orbitals present in the $3 p$ subshell. Since all three $3 p$ orbitals contain a single electron, you can say that the incomplete quantum number set
$n = 3 , l = 1 , {m}_{l} = - 1$
describes one electron present in the third energy shell, in the $3 p$ subshell, in one of the three $3 p$ orbitals--let's say $3 {p}_{y}$.