Question

Question #1b3f0

Answer 1

`n=2, l=0, m_l = 0, m_s = -1/2`

Explanation:

Start by writing the electron configuration of a beryllium atom

`"Be: " 1s^2 color(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 = color(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 `2s` 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 `2s` 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 `2s` 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 = -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=color(red)(2), l=0, m_l = 0, m_s = -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.