# Question #6cfda

Dec 5, 2017

$n = 2 , l = 1 , {m}_{l} = - 1 , {m}_{s} = + \frac{1}{2}$

#### Explanation:

For starters, you know that nitrogen's electronic configuration looks like this

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

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

This basically means that the first $3$ electrons added to the $2 p$ subshell will be added to different orbitals. Now, you're dealing with the $\textcolor{b l u e}{2} p$ subshell, so you know that the principal quantum number, $n$, will be equal to $\textcolor{b l u e}{2}$.

$n = 2 \to$ the second energy shell

For a $p$ subshell, the angular momentum quantum number, $l$, which denotes the energy subshell in which an electron resides, is equal to $1$.

$l = 1 \to$ the $p$ subshell

In this case, the magnetic quantum number, ${m}_{l}$, can take three possible values, each representing an orbital located in the $p$ subshell.

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

Finally, the spin quantum number, ${m}_{s}$, which describes the spin of the electron, can take two possible values

${m}_{s} = \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

By convention, electrons added to an empty orbital are assigned spin-up, so

${m}_{s} = + \frac{1}{2}$

This means that one electron located in the $2 p$ subshell of a neutral atom of nitrogen will have the following quantum number set

$n = 2 , l = 1 , {m}_{l} = - 1 , {m}_{s} = + \frac{1}{2}$

The other two electrons will have quantum number sets that differ only in the value of the magnetic quantum number.