# Question #f307f

Jan 17, 2018

$n = 3 , l = 2 , {m}_{l} = 1 , {m}_{s} = - \frac{1}{2}$

#### Explanation: The first quantum number set given to you is not correct because the spin quantum number, ${m}_{s}$, can only take two possible values.

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

So the quantum number set that has

${m}_{s} = + 1$

cannot be used to describe the location and the spin of an electron inside an atom.

The second quantum number set is also incorrect because the angular momentum quantum number, $l$, cannot take the same value as the principal quantum number, $n$.

Similarly, the third quantum number set is incorrect because the angular momentum quantum number cannot take a negative value.

More specifically, the angular momentum quantum number can take

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

So the quantum number set that has

$n = 1 , l = 1$

is not valid because the value of the angular momentum quantum number is not possible given the value of the principal quantum number.

The set that has

$l = - 3$

is not valid because of the negative value assigned to the angular momentum quantum number.

The fourth quantum number set is correct because all the four quantum numbers take accepted values.

$n = 3 , l = 2 , {m}_{l} = 1 , {m}_{s} = - \frac{1}{2}$

This set describes an electron located in the third energy shell, in the $3 d$ subshell, let's say in the $3 {d}_{x y}$ orbital, that has spin-down.

The value of the magnetic quantum number, ${m}_{l}$, is valid here because this quantum number can take the possible values

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

So for

$l = 2$

you can have

${m}_{l} = 1$

as a valid value for the magnetic quantum number.