# Question a5f60

Mar 13, 2017

Here's what I got.

#### Explanation:

As you know, we can use a set of four quantum numbers to describe the location and spin of an electron in an atom. Your goal here is to figure out the maximum number of electrons that can share the incomplete sets of quantum numbers given to you.

Keep in mind that every complete set of quantum number is unique and can only belong to a single electron. In other words, no two electrons can have the same values for all four quantum numbers.

Before moving on, it's worth noting that

$\left\{\begin{matrix}n = \text{energy level" \\ l = "subshell" \\ m_l = "orbital" \\ m_s = "spin}\end{matrix}\right.$

Also, keep in mind that each individual orbital can hold a maximum of $2$ electrons, as described by Pauli's Exclusion Principle.

So, let's start looking at the incomplete sets given to you.

$\textcolor{w h i t e}{a}$

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = 3 , {m}_{l} = - 2}}}$

In this case, you know that the electrons are located on the third energy level, i.e. in the third energy shell, because they have $n = 3$.

Now, ${m}_{l} = - 2$ can only correspond to an orbital located in the $l$ subshell because

$n = 3 \implies l = \left\{0 , 1 , 2\right\}$

but only $l = 2$ has

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

This means that the maximum number of electrons that can share these two quantum numbers is equal to $2$. These electrons are located on the third energy level, in the $3 d$ subshell, in one of the five $3 d$ orbitals.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{n = 3 , {m}_{l} = - 2 \implies \text{2 e"^(-)color(white)(.)"max}}}}$

$\textcolor{w h i t e}{a}$

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = 4 , l = 3}}}$

This time, you know that you're working on the fourth energy level because $n = 4$.

In this case, the angular momentum quantum number can take the values

• $l = 0 \to$ the s subshell
• $l = 1 \to$ the p subshell
• $l = 2 \to$ the d subshell
• $l = 3 \to$ the f subshell

This means that you must figure out the maximum number of electrons that can occupy the $f$ subshell. In this case, the magnetic quantum number can take the values

${m}_{l} = \left(- 3 , - 2 , - 1 , 0 , 1 , 2. 3\right\}$

Since each value of ${m}_{l}$ corresponds to an individual orbital, you can say that the $f$ subshell holds a total of $7$ orbitals.

This means that the maximum number of electrons that can share these two quantum numbers is equal to

7 color(red)(cancel(color(black)("orbitals"))) * ("2 e"^(-)color(white)(.)"max")/(1color(red)(cancel(color(black)("orbital")))) = "14 e"^(-)color(white)(.)"max"#

Therefore, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{n = 4 , l = 3 \implies \text{14 e"^(-)color(white)(.)"max}}}}$

$\textcolor{w h i t e}{a}$

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = 5 , l = 3 , {m}_{l} = 2}}}$

In this case, you know the energy level, the subshell, and the orbital, so you should be able to say without too much trouble that a maximum of $2$ electrons can share these quantum numbers.

These electrons are located on the fifth energy level, in the $5 f$ subshell, in one of the seven $5 f$ orbitals.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{n = 5 , l = 3 , {m}_{l} = 2 \implies \text{2 e"^(-)color(white)(.)"max}}}}$

$\textcolor{w h i t e}{a}$

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = 4 , l = 1 , {m}_{l} = 0}}}$

Once again, you are given the energy level, subshell, and orbital, so you can conclude that a maximum of $2$ electrons can share these quantum numbers.

These electrons are located on the fourth energy level, in the $4 p$ subshell, in one of the three $4 p$ orbitals.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{n = 4 , l = 1 , {m}_{l} = 0 \implies \text{2 e"^(-)color(white)(.)"max}}}}$