# Question #a3028

Dec 23, 2016

Here's what I got.

#### Explanation:

As you know, we can use four quantum numbers to describe the location and spin of an electron in an atom.

These four quantum numbers describe

• $n \to$ the energy shell in which the electron is located
• $l \to$ the subshell in which the electron resides
• ${m}_{l} \to$ the exact orbital that holds the electron
• ${m}_{s} \to$ the spin of the electron

As you can see, the exact orbital that holds the electron is given by the magnetic quantum number, ${m}_{l}$. In other words, you can only know the identity of the orbital, i.e. its orientation, if you know the value of ${m}_{l}$.

One exception to this rule is actually your first example

$n = 1 , l = 0$

For the first energy shell, which is described by $n = 1$, you can only have one value for $l$. Consequently, you can only have one value for ${m}_{l}$, since

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

Therefore, you can say for sure that an electron that has $n = 1$ and $l = 0$ will reside in the $1 s$ orbital, since that's the only value that can exist for ${m}_{l}$.

For your second set

$n = 3 , l = 1$

you must keep in mind that you have

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

and so on. This means that this set describes an electron located in the third energy shell, since $n = 3$, and also in the $p$ subshell, since $l = 1$.

However, you have $3$ possible values for ${m}_{l}$ here

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

This means that you can't say for sure which exact orbital holds this electron. The same can be said for the third set

$n = 4 , l = 2$

since the $d$ subshell can have

${m}_{l} = \left\{- 2 , - 1 , 0 , 1 , 2\right\}$

On the other hand, the fourth set is not valid because it doesn't follow the rule

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

For $n = 1$ you can only have $l = 0$, which is why the set is said to be invalid.