# Question 64fc6

Jun 12, 2017

${\text{3 e}}^{-}$

#### Explanation:

The idea here is that each individual orbital can hold a maximum of $2$ electrons, one having spin-up, or ${m}_{2} = + \frac{1}{2}$, and the other having spin-down, or ${m}_{s} = - \frac{1}{2}$.

This implies that in order to figure out how many electrons that are located in the $2 p$ subshell can have ${m}_{2} = - \frac{1}{2}$, you must determine how many orbitals are present in this subshell.

As you know, the number of orbitals present in a given subshell $l$ is given by the number of values that the magnetic quantum number, ${m}_{l}$, can take.

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

In your case, you are dealing with a $p$ subshell, so you should know that $l = 1$ since you have

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

and so on. This means that any $p$ subshell, including the $2 p$ subshell, which is simply the $p$ subshell located on the second energy level, will have

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

So, three values for ${m}_{l}$ give you $3$ orbitals for any $p$ subshell.

Now, you know that you get a maximum of $2$ electrons per orbital, but that only $1$ electron can have ${m}_{2} = - \frac{1}{2}$ in an orbital--the other electron located in the same orbital must have ${m}_{2} = + \frac{1}{2}$.

This means that you have

3 color(red)(cancel(color(black)("2p orbitals"))) * ("1 e"^(-)color(white)(.)"with m"_s = -1/2)/(1color(red)(cancel(color(black)("2p orbital")))) = "3 e"^(-)# ${\text{with m}}_{s} = - \frac{1}{2}$