# How many electrons in an atom can have n=7 and m_l = +3 ?

Jun 12, 2017

${\text{8 e}}^{-}$

#### Explanation:

The trick here is to focus on the value of the magnetic quantum number, ${m}_{l}$, which tells you the orientation of the orbital in which an electron resides inside an atom.

You know that every value the magnetic quantum number takes represents an orbital. You also know that every orbital can hold a maximum of $2$ electrons.

You can thus say that

${m}_{l} = + 3$

can be shared by a maximum of $2$ electrons per subshell.

Now, we use a total of four quantum numbers to describe the position and spin of an electron inside an atom. As you can see, the value of the magnetic quantum number depends on the value of the angular momentum quantum number, $l$, which, in turn, depends on the value of the principal quantum number, $n$.

More specifically, you know that

${m}_{l} = \left\{- l , - \left(l - 1\right) , \ldots , - 1 , \textcolor{w h i t e}{-} 0 , + 1 , \ldots , \left(l - 1\right) , l\right\}$

and that

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

For $n = 7$, you can have six possible values for the angular momentum quantum number

$l = \left\{0 , 1 , 2 , 3 , 4 , 5 , 6\right\}$

Notice that you can only have ${m}_{l} = + 3$ for values of the angular momentum quantum number that begin with $l = 3$. This means that only $4$ of the $7$ possible subshells can hold orbitals that have ${m}_{l} = + 3$.

$l = \left\{\textcolor{red}{\cancel{\textcolor{b l a c k}{0 , 1 , 2}}} , 3 , 4 , 5 , 6\right\}$

You can thus say that you have

• $l = 3$

${m}_{l} = \left\{- 3 , - 2 , - 1 , 0 , + 1 , + 2 , \textcolor{red}{+ 3}\right\}$

• $l = 4$

${m}_{l} = \left\{- 4 , - 3 , - 2 , - 1 , 0 , + 1 , + 2 , \textcolor{red}{+ 3} , + 4\right\}$

• $l = 5$

${m}_{l} = \left\{- 5 , - 4 , - 3 , - 2 , - 1 , 0 , + 1 , + 2 , \textcolor{red}{+ 3} , + 4 , + 5\right\}$

• $l = 6$

${m}_{l} = \left\{- 6 , - 5 , - 4 , - 3 , - 2 , - 1 , 0 , + 1 , + 2 , \textcolor{red}{+ 3} , + 4 , + 5 , + 6\right\}$

So, you know that for $n = 7$, you can have a total of $4$ orbitals described by ${m}_{l} = + 3$. Since each orbital can hold a maximum of $2$ electrons, you will have

4 color(red)(cancel(color(black)("orbitals"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "8 e"^(-)

Therefore, you know that a maximum of $8$ electrons can share these two quantum numbers

$n = 7 , {m}_{l} = + 3$