What could the fourth quantum number of a #2s^2# electron be?

As you know, a total of four quantum numbers are used to describe the positions and the spin of an electron in an atom.

![figures.boundless.com]()

In your case, you must decide which possible values can be assigned to the fourth quantum number of an electron that resides in a #2s^2# orbital.

The principal quantum number describes the energy level, or shell, on which the electron resides. In this case, the electron is located on the second energy level, so #n=2#.

The angular momentum quantum number, #l#, describes the subshell in which the electron is located. For an s-orbital, the angular momentum quantum number is equal to #l=0#.

The magnetic quantum number, #m_l#, tells you the specific orbital in which you can find the electron. For an s-orbital, the magnetic quantum number can only take the value #0#, since it depends on the value of #l#

#m_l = -l, -(l-1), ..., -1, 0, 1, ..., (l-1), l#

Finally, the spin quantum number, #m_s#, can only take two possible values, #+-1/2#.

Now, the s-orbital can hold a maximum of two electrons. The first electron will have #m_s = +1/2# and the second electron would have #m_s = -1/2#.

The first electron to occupy the 2s-orbital would be represented using the notation #2s^1#, while the second electron would be represented using the notation #2s^2#.

This means that the fourth quantum number for a #2s^2# electron is

#m_s = -1/2 -># spin-down

This implies that the 2s-orbital already contains an electron that has

#m_s = +1/2 -># spin-up

You can put all this together to find the quantum number set that describes the #2s^2# electron

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