Question #a5f5a

Sep 7, 2017

Here's how you can do that.

Explanation:

The azimuthal quantum number, which you'll often see as the angular momentum quantum number, $l$, tells you the energy subshell in which an electron is located inside an atom.

In other words, the angular momentum quantum number tells you the shape of the orbital in which the electron is located, but not its specific orientation.

The number of orbitals present in each energy subshell is given by the number of values that the magnetic quantum number, ${m}_{l}$, can take.

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

As you can see the number of orbitals present in a given subshell $l$ is equal to

$\text{no. of orbitals} = 2 l + 1$

Now, you know that each orbital can hold a maximum of $2$ electrons $\to$ think Pauli's Exclusion Principle here.

This means that the maximum number of electrons that can occupy a given subshell $l$ is equal to

$\text{max no. of e"^(-) = 2 xx "no. of orbitals}$

which is equivalent to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{max no. of orbitals} = 2 \cdot \left(2 l + 1\right)}}}$

Let's take, for example, the $p$ subshell, which is denoted by $l = 1$. You have

$\text{no. of orbitals} = 2 \cdot 1 + 1 = 3$

and

${\text{max no. of e}}^{-} = 2 \cdot 3 = 6$

Therefore, you can say that for any energy level that has $n \ge 2$, the $p$ subshell holds a maximum of $6$ electrons located in three $p$ orbitals.