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

Nov 17, 2015

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.

• ${1}^{\text{st}}$ set $\to n = 2$

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

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

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

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

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 $n = 2$.

• ${2}^{\text{nd}}$ set $\to n = 4 , l = 3$

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 $n - 1$.

• $l = 0 \to$ the s-subshell
• $l = 1 \to$ the p-subshell
• $l = 2 \to$ the d-subshell
• $l = 3 \to$ 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

${m}_{l} = - l , \ldots , - 1 , 0 , 1 , \ldots , + l$

m_l = {-3; -2; -1; 0; 1; 2; 3}

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 $l = 3$.

• ${3}^{\text{rd}}$ set $\to n = 6 , l = 2 , {m}_{l} = - 1$

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, ${m}_{s} = - \frac{1}{2}$.