Question #1b3f0

1 Answer
Jun 20, 2017

Answer:

#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.