# What are the sets of quantum numbers that describe the last three electrons of N?

Dec 22, 2017

Here's what I got.

#### Explanation:

For starters, you know that nitrogen is located in the second period of the Periodic Table, which implies that its outermost electrons are located in the second energy shell.

Moreover, the fact that nitrogen is located in group $15$ tells you that its outermost energy shell contains a total of $5$ electrons.

Now, the second energy shell is comprised of the $s$ subshell and the $p$ subshell. The $s$ subshell contains a single orbital, the $s$ orbital, while the $p$ subshell contains three orbitals, the ${p}_{z}$, ${p}_{y}$, and ${p}_{x}$ orbitals.

The $s$ orbital is lower in energy than the three $p$ orbitals, you can say that $2$ out of the $5$ electrons located in the second energy shell of an atom of nitrogen will be added to the $2 s$ orbital.

This means that the remaining $3$ electrons will be added to the $2 p$ orbitals.

Now, Hund's Rule states that all the orbitals present in a given energy subshell must be half-filled before any one of those orbitals can be completely filled.

Since you have $3$ electrons and $3$ orbitals, you can say that each orbital will hold a single electron. So, you can say that the quantum number sets that will describe these electrons will be

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

This set describes an electron that is located in the second energy shell, in the $2 p$ subshell, in the $2 {p}_{y}$ orbital and that has spin-up

$n = 2 , l = 1 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

This set describes an electron that is located in the second energy shell, in the $2 p$ subshell, in the $2 {p}_{z}$ orbital and that has spin-up

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

This set describes an electron that is located in the second energy shell, in the $2 p$ subshell, in the $2 {p}_{x}$ orbital and that has spin-up

By convention, an electron added to an empty orbital is assigned spin-up, which is why all three electrons have the spin quantum number, ${m}_{s}$, equal to $+ \frac{1}{2}$.

Notice that the principal quantum number, $n$, and the angular momentum quantum number, $l$, are equal for all three electrons, which is what you would expect to see for three electrons located in the same energy shell, i.e. $n = 2$, and in the same energy subshell, i.e. $l = 1$.