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

Dec 24, 2015

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

#### Explanation:

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

Rubidium, $\text{Rb}$, 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 $\text{Rb}$ atom must account for a total of $37$ electrons.

$\text{Rb: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} \textcolor{red}{5} {s}^{1}$

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 $n = \textcolor{red}{5}$.

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

• $l = 0 \to$ s-subshell
• $l = 1 \to$ p-subshell
• $l = 2 \to$ 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

• ${m}_{s} = + \frac{1}{2} \to$ spin-up electron
• ${m}_{s} = - \frac{1}{2} \to$ spin-down electron

For you electron, you can select ${m}_{s} = + \frac{1}{2}$. The complete set of quantum numbers that describes this particular electron will be

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

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