Question

Circle all of the allowed sets of quantum numbers for an electron in a hydrogen atom?

n=6, l=3, ml =-2, ms=+1/2
n=2, l=1, ml =0, ms=-1/2
n=2, l=2, ml =1, ms=+1/2
n=3, l=0, ml =1/2, ms=-1/2
n=3, l=1, ml =0, ms=0
n=0, l=0, ml =0, ms=-1/2

Answer 1

Recall the quantum numbers:

The three main quantum numbers describe the energy level, shape, and projection of the orbitals onto the xyz axes. Bonus: there is a fourth which describes the spin of the electron(s) in the orbital.

• `n` is the principle quantum number which describes the energy level. `n >= 1` and is in the set of integers. That is, `n = 1,2, . . . , N`, for some finite `N` (only one of those numbers at a time).
• `l` is the orbital angular momentum quantum number which describes the shape of the orbital. `l >= 0` and is an integer. `l = 0, 1, 2, ..., n-1` for `s, p, d, ...` orbitals, respectively (only one of those numbers at a time).
• `m` (more specifically, `m_l`) is the magnetic quantum number, which describes the projection of the orbital in the `0,pm1, pm2, . . . , pml` directions.

It takes on every possible value of `l` and `0` in integer steps (not just the value of `l`, but every integer from `-l` to `0` to `+l`).

That is, `m_l = {0,pm1,pm2, . . . , pml}`. Note that `m_l` tells you how many subshells there are for a given orbital type (`s,p,d, . . .`).

• `m_s` (see why we need to say `m_l`?) is the electron spin quantum number, which describes the spin of the electron. It is only `pm1/2`, independent of the other quantum numbers.

(1)

With `n = 6` and `l = 3`, we are looking at a `6f` orbital, since `l = 3` corresponds to an `f` orbital.

For `l = 3`, we really have `m_l = {0,pm1,pm2,pm3}`, so `m_l = -2` is allowed. Lastly, `m_s = +1/2` is allowed.

This is really saying that there is a spin-up electron in the `6f` orbital with an orbital angular momentum of `-2` (the second orbital from the left, for a `-3,-2,-1,0,+1,+2,+3` order).

(2)

With `n = 2` and `l = 1`, we are looking at a `2p` orbital, since `l = 1` corresponds to an `p` orbital.

For `l = 1`, we really have `m_l = {0,pm1}`, so `m_l = 0` is allowed. Lastly, `m_s = -1/2` is allowed.

This is really saying that there is a spin-down electron in the `2p` orbital with an orbital angular momentum of `0` (the middle orbital for a `-1,0,+1` order).

(3)

With `n = 2` and `l = 2`, we are looking at a so-called "`2d`" orbital, which doesn't exist.

For `l = 0,1,2, . . . , n-1`, we walk through `0 harr s, 1 harr p, 2 harr d` and find that `l = 2 harr d`.

For `d` orbitals, we note that `n >= 3` for `d` orbitals, but `n = 2` is impossible if `l = 2`.

Thus, this is an impossible configuration.

(4)

With `n = 3` and `l = 0`, we are looking at a `3s` orbital.

For `l = 0`, `m_l = {0}`. So, `m_l = 1/2` is impossible.

Thus, this is an impossible configuration.

(5)

With `n = 3` and `l = 1`, we are looking at a `3p` orbital.

For `l = 1`, `m_l = {0,pm1}`. So, `m_l = 0` is allowed. However, only `m_s = pm1/2` is allowed. So, `m_s = 0` is forbidden.

Thus, this is an impossible configuration.

(6)

With `n = 0`, consider that `n = 1,2,3, . . . , N`, for some finite `N`. But of course, `0 < 1`. It's outside the range of possible values of `n`.

Thus, this is an impossible orbital.