# Question #d02f6

Oct 7, 2017

Here's what I got.

#### Explanation:

For starters, you know that the energy shell in which an electron is located inside an atom is given by the principal quantum number, $n$.

$n = 4$

Now, the energy subshell in which an electron is located is given by the angular momentum quantum number, $l$.

The $d$ subshell, which is the subshell that contains the $d$ orbitals, is given by

$l = 2$

The orientation of an orbital located in a given subshell $l$ is given by the magnetic quantum number, ${m}_{l}$.

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

The $d$ subshell contains a total of $5$ orbitals, each described by a different value of the magnetic quantum number

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

Finally, the spin quantum number, ${m}_{s}$, tells you the spin of an electron inside an orbital.

${m}_{s} = \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

You can now write the quantum number sets that can describe an electron located in the $\text{4th}$ energy shell, in the $4 d$ subshell.

I'll give you several examples to go by

• $n = 4 , l = 2 , {m}_{l} = - 2 , {m}_{s} = + \frac{1}{2}$
• $n = 4 , l = 2 , {m}_{l} = - 2 , {m}_{s} = - \frac{1}{2}$
• $n = 4 , l = 2 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$
• $n = 4 , l = 2 , {m}_{l} = + 1 , {m}_{s} = + \frac{1}{2}$
• $n = 4 , l = 2 , {m}_{l} = + 2 , {m}_{s} = - \frac{1}{2}$