# Question #5afee

Sep 17, 2016

Two electrons.

#### Explanation:

The thing to remember about quantum numbers is that each electron located in an atom is described by an unique set of four quantum numbers

In your case, the problem provides you with values for three out of the four possible quantum numbers.

Right from the start, you should be able to look at this information and say that these three quantum numbers can be shared by a maximum of $2$ electrons.

Here's why.

The principal quantum number, $n$, tells you the energy level on which the electron is located. For your electrons, $n = 3$, which is the same as saying that the electrons reside on the third energy level.

The angular momentum quantum number, $l$, tells you the subshell in which the electron resides. The $l = 2$ value for the angular momentum quantum number designates a d subshell.

The magnetic quantum number, ${m}_{l}$, tells you the exact orbital in which the electron resides. The d subshell contains a total of $5$ d orbitals designated by

${m}_{l} = \left(- 2 , - 1 , 0 , 1 , 2\right\}$

The value ${m}_{l} = 1$ designates one of the $5$ d orbitals present in the d subshell, let's say $3 {d}_{y z}$.

Now, each orbital can hold a maximum of $2$ electrons of opposite spin, as given by the Pauli Exclusion Principle.

This means that the three quantum numbers given to you can be shared by a maximum of $2$ electrons, one having spin-up, for which the spin quantum number, ${m}_{s}$ is equal to $\frac{1}{2}$, and the other having spin-down, for which ${m}_{s}$ is equal to $- \frac{1}{2}$.

Therefore, you can say that two complete quantum number sets can be written here

$n = 3 , l = 2 , {m}_{l} = 1 , {m}_{s} = + \frac{1}{2}$

This set describes an electron located on the third energy level, in the d subshell, in the $3 {d}_{y z}$ orbital, having spin-up

$n = 3 , l = 2 , {m}_{l} = 1 , {m}_{s} = - \frac{1}{2}$

This set describes an electron located on the third energy level, in the d subshell, in the $3 {d}_{y z}$ orbital, having spin-down