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

Mar 12, 2016

$- \frac{1}{2}$

#### Explanation:

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

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

${m}_{l} = - l , - \left(l - 1\right) , \ldots , - 1 , 0 , 1 , \ldots , \left(l - 1\right) , l$

Finally, the spin quantum number, ${m}_{s}$, can only take two possible values, $\pm \frac{1}{2}$.

Now, the s-orbital can hold a maximum of two electrons. The first electron will have ${m}_{s} = + \frac{1}{2}$ and the second electron would have ${m}_{s} = - \frac{1}{2}$.

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

This means that the fourth quantum number for a $2 {s}^{2}$ electron is

${m}_{s} = - \frac{1}{2} \to$ spin-down

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

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

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

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