# Question #6812b

Aug 1, 2017

Here's why that is the case.

#### Explanation:

For starters, you know that beryllium is located in period $2$, group $2$ of the Periodic Table of Elements.

Now, we can use four quantum numbers to describe the location and spin of an electron inside an atom.

The principal quantum number, $n$, tells you the energy level on which the electron is located. The value of the principal quantum number is given by the period in which the element is located.

In this case, beryllium is located in period $2$, so its outermost electrons will be located on the second energy level.

This means that you have

$n = 2 \to$ the second energy level

The angular momentum quantum number, $l$, tells you the subshell in which the electron resides. The value of the angular momentum quantum number is given by the block in which the element is located in the Periodic Table.

Group $1$ and group $2$ elements, together with helium, make up the $s$ block. This tells you that these elements have their outermost electrons located in the $s$ subshell.

Since you know that you have

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

and so on, you can say that the fourth electron of beryllium will have

$l = 0 \to$ the $s$ subshell

The magnetic quantum number, ${m}_{l}$, tells you the orbital in which the electron is located.

Since the $s$ subshell can hold a single orbital, the $s$ orbital, you can say that you have

${m}_{l} = 0 \to$ the s orbital

Finally, the spin quantum number, ${m}_{s}$, tells you the spin of the electron.

Now, each orbital can hold a maximum of $2$ electrons of opposite spins--one electron having spin-up and the other spin-down, as stated by Pauli's Exclusion Principle.

• ${m}_{s} = + \frac{1}{2} \to$ the electron has spin-up
• ${m}_{s} = - \frac{1}{2} \to$ the electron has spin-down

By convention, we take the first electron to be added to an orbital, i.e. the electron added to an empty orbital, to have spin-up and the second electron to be added to the same orbital, i.e. the electron added to the half-filled orbital, to have spin-down.

In your case, the third electron in a beryllium atom is added to the empty $2$ orbital, so it will have spin-up, and the fourth electron is added to the half-filled $2 s$ orbital, so it will have spin-down.

The electron configuration of beryllium shows why that is the case

$\text{Be: } 1 {s}^{2} 2 {s}^{1} \to$ the third electron is added to the empty $2 s$ orbital

$\text{Be: } 1 {s}^{2} 2 {s}^{2} \to$ the fourth electron is added to the half-filled $2 s$ orbital

This means that you have

${m}_{2} = - \frac{1}{2}$

Therefore, the full set of quantum numbers that we can use to describe the fourth electron in a beryllium atom is

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

The electron is located on the second energy level, in the $2 s$ subshell, in the $2 s$ orbital, and has spin-up.