Question #1b3f0

Jun 20, 2017

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

Explanation:

Start by writing the electron configuration of a beryllium atom

$\text{Be: } 1 {s}^{2} \textcolor{red}{2} {s}^{2}$

As you can see, the fourth electron added to a beryllium atom is located on the second energy level, so

• $n = \textcolor{red}{2}$

The principal quantum number, $n$, tells you the energy level on which an electron resides inside an atom

This electron is located in the $2 s$ subshell, so you can say that you have

• $l = 0$

The angular momentum quantum number, $l$, tells you the subshell in which the electron is located.

Consequently, this electron is located in the $2 s$ orbital, which is described by

• ${m}_{l} = 0$

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

Finally, you know that this is the second electron added to the $2 s$ orbital. As you know, Pauli's Exclusion Principle specifies that each orbital can hold a maximum of $2$ electrons, one having spin-up and the other spin-down.

By convention, we take the first electron added to an orbital to have spin-up and the second electron added to the same orbital to have spin-down.

This means that you have

• ${m}_{s} = - \frac{1}{2}$

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

You can thus say that the fourth electron added to a beryllium atom can be described by the following quantum number set

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

This set describes an electron located on the second energy level, in the 2s subshell, in the 2s orbital, which has spin-down.