# Question #77f2d

Sep 19, 2017

Here's what I got.

#### Explanation:

For starters, the coefficient added in front of the letter tells you that you're dealing with an electron that is located in the third energy shell.

This implies that the principal quantum number, $n$, is equal to $3$.

$n = 3$

The letter that follows the coefficient tells you the name of the subshell in which the electron is located. The name of the subshell is denoted by the angular momentum quantum number, $l$.

For a $d$ subshell, the angular momentum quantum number is equal to $2$.

$l = 2$

Next, the number of orbitals present in a given subshell is denoted by the magnetic quantum number, ${m}_{l}$, which, for a $d$ subshell, can take the possible values

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

This means that a $d$ subshell, in your case, the $3 d$ subshell, can hold a total of five $3 d$ orbitals. If you exclude the value of the spin quantum number, ${m}_{s}$, you can write a total of $5$ incomplete sets of quantum numbers for this electron.

• $n = 3 , l = 2 , {m}_{l} = - 2$
• $n = 3 , l = 2 , {m}_{l} = - 1$
• $n = 3 , l = 2 , {m}_{l} = 0$
• $n = 3 , l = 2 , {m}_{l} = + 1$
• $n = 3 , l = 2 , {m}_{l} = + 2$