Question

How would you write a full set of quantum numbers for the outermost electron in an Rb atom?

Answer 1

`n=5, l=0, m_l = 0, m_s = +1/2`

Explanation:

Start by writing the complete electron configuration fir a neutral rubidium atom.

Rubidium, `"Rb"`, is located in period 5, group 1 of the periodic table, and has an atomic number equal to `37`. This means that the electron configuration of a neutral `"Rb"` atom must account for a total of `37` electrons.

`"Rb: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 color(red)(5)s^1`

So, the outermost electron in a rubidium atom is located on the fifth energy level, in an s-orbital.

As you know, four quantum numbers are used to describe the position and spin of an electron inside an atom.

The principal quantum number, `n`, tells you on which energy level the electron resides. In this case, the outermost electron is located on the fifth energy level, and so `n = color(red)(5)`.

The angular momentum quantum number, `l`, describes the subshell in which the electron resides. For `l`, you have

• `l=0 ->` s-subshell
• `l=1 ->` p-subshell
• `l=2 ->` d-subshell

and so on. Since your electron is located in the s-subshell, you will have `l=0`.

The magnetic quantum number, `m_l`, describes the exact orbital in which you can find the electron. For an s-subshell electron, `m_l` can only take one value, `m_l = 0`, since this subshell can only hold one orbital.

Finally, the spin quantum number, `m_s`, describes the spin of the electron. Here you have

• `m_s = +1/2 ->` spin-up electron
• `m_s = -1/2 ->` spin-down electron

For you electron, you can select `m_s = +1/2`. The complete set of quantum numbers that describes this particular electron will be

`n=5, l=0, m_l =0, m_s = +1/2`

The electron is located on the fifth energy level, in the 5s-subshell, in the 5s-orbital, and has spin-up.