## How many electrons are in n = 2? What about n = 4, l = 3? What about n = 6, l = 2, m_{l} = -1?

Answer:

Here’s what I got.

Explanation:

Since the question is a bit ambiguous, I will assume that you’re dealing with three distinct sets of quantum numbers.

In addition to this, I will also assume that you’re fairly familiar with quantum numbers, so I won’t go into too much details about what each represents.

The principal quantum number, \(n\) tells you the energy level on which an electron resides. In order to be able to determine how many electrons can share this value of \(n\), you need to determine exactly how many orbitals you have in this energy level.

The number of orbitals you get per energy level can be found using the equation

\(\text { no. of orbitals }=n^{2}\).

Since each orbital can hold amaximum of two electrons, it follows that as many as

\(\text { no. of electrons }=2 n^{2}\).

In this case, the second energy level holds a total of

\(\text { no. of orbitals }=n^{2}=2^{2}=4\).

orbitals. Therefore, a maximum of

\(\text { no. of electrons }=2 \cdot 4=8\).

electrons can share the quantum number n = 2.

\(\text { • } 2^{\text {nd }} \text { set } \rightarrow n=4, l=3\).

This time, you are given both the energy level, n = 4, and the subshell, l = 3, on which the electrons reside.

Now, the subshell is given by the angular momentum quantum number, l, which can take values ranging from 0 to n−1.

- l = 0 → the s-subshell
- l = 1 → the p-subshell
- l = 2 → the d-subshell
- l = 3 → the f-subshell

Now, the number of orbitals you get per subshell is given by the magnetic quantum number, m_{l}, which in this case can be

\(\begin{aligned}

&m_{l}=-l, \ldots,-1,0,1, \ldots,+l \\

&m_{l}=\{-3 ;-2 ;-1 ; 0 ; 1 ; 2 ; 3\}

\end{aligned}\).

So, the f-subshell can hold total of seven orbitals, which means that you have a maximum of

\(\text { no. of electrons }=2 \cdot 7=14\).

electrons that can share these two quantum numbers, n = 4 and l = 3.

\(\text { • } 3^{\mathrm{rd}} \text { set } \rightarrow n=6, l=2, m_{l}=-1\).

This time, you are given the energy level, n = 6, the subshell, l = 2, and the exact orbital, m_{l} = 1, in which the electrons reside.

Since you know the exact orbital, it follows that only two electrons can share these three quantum numbers, one having spin-up, \(m_{s}=+\frac{1}{2}\) and the other having spin-down, \(m_{s}=-\frac{1}{2}\)