How would you find out the number of electrons from just the 4 quantum numbers?

Nov 17, 2015

Here's what I would suggest.

Explanation:

If a complete set of quantum numbers is given to you, then you can say for a fact that you're only dealing with one electron.

As you know, quantum numbers are used to describe the location and spin of an electron that surrounds an atom's nucleus.

More specifically, the four quantum numbers will tell you

• the energy level on which the electron resides - given by $n$, the principal quantum number
• the subshell in which the electron resides - given by $l$, the angular momentum quantum number
• the exact orbital in which you can find the electron - given by, ${m}_{l}$, the magnetic quantum number
• the spin of the electron - given by ${m}_{s}$, the spin quantum number

Now, each complete set of quantum numbers can only describe one electron. This means that if you're given something like this

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

then you know for a fact that only one electron can have that set of quantum numbers.

Now, let's assume that you are not given all four quantum numbers. These are some rules to help you figure out how many electrons can share an incomplete set of quantum numbers

• You are given the value of $n$

If this is the case, then use the fact that the number of electrons you get per energy level is equal to

$\textcolor{b l u e}{\text{no. of electrons} = 2 {n}^{2}}$

• You are given $n$ and $l$

This time, you know the energy level and the subshell in which the electrons can be found. To determine how many electrons can share these quantum numbers, use ${m}_{l}$, which can take values from $- l$ to $l$.

Once you know how many orbitals you have per subshell, multiply that value by $2$ to get the number of electrons

$\textcolor{b l u e}{\text{no. of electrons" = 2 xx "no. of subshells}}$

• You are ginve $n$, $l$, and ${m}_{l}$

Now you know the energy level, subshell, and specific orbital in which the electrons can reside.

Since the only possible values for ${m}_{s}$ are $\pm \frac{1}{2}$, it follows that every orbital can hold a maximum of two electrons.