# What is the orbital angular momentum quantum number for the orbital 5p?

## I don't understand anything about it. I don't know where to start.

Oct 11, 2016

$l = 1$

#### Explanation: The first thing to note here is that you don't need to know the value of the principal quantum number, $n$, to answer this question.

In other words, you don't need to know the energy level on which the electron resides, which in this case would be $n = 5$, to say that the angular momentum quantum number, $l$, for a $p$ subshell is equal to $1$.

The $5 p$ subshell given to you is described by two quantum numbers

• $n = 5 \to$ this means that the electron is located on the fifth energy level

• $p \to$ the means that it is located in the p subshell

Now, the angular momentum quantum number designates the identity of the subshell in which the electron is located

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

In this case, the angular momentum quantum number must be equal to $1$ because $1$ is the value that describes the $p$ subshell for any electron located on an energy level that is $n > 1$.

As you can see from the table, for $n = 1$, $l$ can only take one value, $l = 0$. This means that the first energy level holds $1$ subshell, the $s$ subshell.

By comparison, for $n = 5$, $l$ can take

$n = 5 \implies l = \left\{0 , 1 , 2 , 3 , 4\right\}$

The fifth energy level holds $5$ subshells, and $l = 1$ describes the $p$ subshell regardless of the value of $n > 1$.