# Question #a47b4

Oct 5, 2017

Two electrons.

#### Explanation:

As you know, the principal quantum number, $n$, and the angular momentum quantum number, $l$, can only take positive values, so you only have a handful of possibilities for

$n + l = 2$

The angular momentum quantum number depends on the principal quantum number as follows

$l = 0 , 1 , \ldots , n - 1$

You know that zinc, $\text{Zn}$, is located in the fourth period of the Periodic Table. This implies that the electrons located in an atom of zinc will have

$n = \left\{1 , 2 , 3 , 4\right\}$

This means that you can have

• $n + l = 2 \text{ "-> " } \textcolor{red}{\cancel{\textcolor{b l a c k}{n = 1 \implies l = 1}}}$

Not possible because $l$ cannot be equal to $n$.

• $n + l = 2 \text{ " -> " " n = 2 implies l = 0" " " } \textcolor{g r e e n}{\sqrt{}}$

These electrons will be located on the second energy level, in the $s$ subshell.

• $n + l = 2 \text{ "-> " } \textcolor{red}{\cancel{\textcolor{b l a c k}{n = 3 \implies l = - 1}}}$

Not possible because $l$ cannot have a negative value.

• $n + l = 2 \text{ "-> " } \textcolor{red}{\cancel{\textcolor{b l a c k}{n = 4 \implies l = - 2}}}$

Not possible because $l$ cannot have a negative value.

Now, the number of values that the magnetic quantum number, ${m}_{l}$, can take will give you the number of orbitals located in each subshell.

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

You will have

$l = 0 \implies {m}_{l} = 0$

A single orbital is located in the $s$ subshell.

As you know, each orbital can hold a maximum of $2$ electrons of opposite spins.

This implies that the total number of electrons that can have $n + l = 2$ in a zinc atom is

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{orbital"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = color(darkgreen)(ul(color(black)("2 e}}^{-}}}}$

Both electrons will be located in the $2 s$ subshell, in the $2 s$ orbital, and have opposite spins. The complete sets of quantum numbers for these electrons will be

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

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