# 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

May 11, 2016

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. $n \ge 1$ 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. $l \ge 0$ and is an integer. $l = 0 , 1 , 2 , \ldots , n - 1$ for $s , p , d , \ldots$ 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 $0 , \pm 1 , \pm 2 , . . . , \pm l$ 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, ${m}_{l} = \left\{0 , \pm 1 , \pm 2 , . . . , \pm l\right\}$. 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 $\pm \frac{1}{2}$, independent of the other quantum numbers.

(1)

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

For $l = 3$, we really have ${m}_{l} = \left\{0 , \pm 1 , \pm 2 , \pm 3\right\}$, so ${m}_{l} = - 2$ is allowed. Lastly, ${m}_{s} = + \frac{1}{2}$ is allowed.

This is really saying that there is a spin-up electron in the $6 f$ 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 $2 p$ orbital, since $l = 1$ corresponds to an $p$ orbital.

For $l = 1$, we really have ${m}_{l} = \left\{0 , \pm 1\right\}$, so ${m}_{l} = 0$ is allowed. Lastly, ${m}_{s} = - \frac{1}{2}$ is allowed.

This is really saying that there is a spin-down electron in the $2 p$ 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 "$2 d$" orbital, which doesn't exist.

For $l = 0 , 1 , 2 , . . . , n - 1$, we walk through $0 \leftrightarrow s , 1 \leftrightarrow p , 2 \leftrightarrow d$ and find that $l = 2 \leftrightarrow d$.

For $d$ orbitals, we note that $n \ge 3$ 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 $3 s$ orbital.

For $l = 0$, ${m}_{l} = \left\{0\right\}$. So, ${m}_{l} = \frac{1}{2}$ is impossible.

Thus, this is an impossible configuration.

(5)

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

For $l = 1$, ${m}_{l} = \left\{0 , \pm 1\right\}$. So, ${m}_{l} = 0$ is allowed. However, only ${m}_{s} = \pm \frac{1}{2}$ 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.