# Question #21ca6

##### 1 Answer
Dec 3, 2017

Here's what I got.

#### 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 know that the $p$ subshell contains a total of $3$ orbitals, each capable of holding a maximum of two electrons as stipulated by Pauli's Exclusion Principle.

As you can see, for a neutral atom of nitrogen, the $2 p$ subshell contains a total of $3$ electrons. According to Hund's Rule, which states that for a given subshell, every orbital must be half-filled before any one of the orbitals present in said subshell can be completely filled.

This means that a neutral atom of nitrogen has $3$ unpaired electrons, all $3$ located in the $2 p$ subshell. In this case, the principal quantum number, which gives you the energy shell in which the electrons reside, is equal to $2$ for all three electrons.

$n = 2 \to$ designates the second energy shell

The angular momentum quantum number, which gives you the identity of the subshell in which an electron is located, is equal to $1$ for all three electrons.

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

The magnetic quantum number, which tells you the orientation of the orbital in which an electron is located, can take the following values for a $p$ subshell

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

Since all three orbitals located in the $2 p$ subshell contain an electron, you can assign a value to each electron present in this subshell.

Finally, the spin quantum number, which tells you the spin of the electron, can take one of two possible values

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

By convention, the electrons added to an empty orbital are assigned a spin-up, so you can say that all three electrons will have

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

This means that the complete quantum number sets for the three electrons are going to look like this

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

• $n = 2 , l = 1 , {m}_{l} = \textcolor{w h i t e}{-} 0 , {m}_{s} = + \frac{1}{2}$

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

Assuming that you want the sum of all $12$ quantum numbers, you will end up with

$\sum \left(n , m , {m}_{l} , {m}_{s}\right) = 3 \times 2 + 3 \times 1 + \left[\textcolor{red}{\cancel{\textcolor{b l a c k}{- 1}}} + 0 + \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{- 1}}}\right)\right] + 3 \times \frac{1}{2}$

$\sum \left(n , m , {m}_{l} , {m}_{s}\right) = \frac{21}{2}$