Question

Question #a3028

Answer 1

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.

![figures.boundless.com]()

These four quantum numbers describe

• `n->` the energy shell in which the electron is located
• `l->` the subshell in which the electron resides
• `m_l ->` the exact orbital that holds the electron
• `m_s ->` 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 = {-l, ..., -1, 0, 1, ..., l}`

Therefore, you can say for sure that an electron that has `n=1` and `l=0` will reside in the `1s` 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 ->` the s subshell
• `l=1 ->` the p subshell
• `l=2 ->` the d subshell
• `l=3 ->` 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 = {-1, 0, 1}`

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 = {-2, -1, 0, 1, 2}`

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

`l = {0, 1, ..., (n-1)}`

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