# Question 5a170

Oct 5, 2017

Ten electrons.

#### Explanation:

As you know, sulfur, $\text{S}$, is located in period $3$, group $16$ of the Periodic Table. Now, the problem wants you to figure out how many electrons in a neutral sulfur atom can have

$l = 1$

For a sulfur atom, the principal quantum number, $n$, which tells you the energy shell in which an electron is located, can take the possible values

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

The angular momentum quantum number, $l$, tells you the subshell in which the electron is located. The values of the angular momentum quantum number depend on the value of the principal quantum number

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

You know that

$l = 1$

designates the $p$ subshell, which corresponds to the $p$ block in the Periodic Table. Each group located in the $p$ block designates an electron that can occupy the $p$ subshell. Since the $p$ block covers groups $13$ through $18$, you can say that the $p$ subshell can hold a maximum of $6$ electrons.

For sulfur, you can have $l = 1$ for the second and third energy levels, i.e. periods in the Periodic Table.

Notice that for period $2$, all the groups that correspond to the $p$ block are completely filled.

This means that on its second energy level, a sulfur atom has its $2 p$ subshell completely occupied with electrons. Since each orbital can gold a maxium of $2$ electrons, you can say that the $2 p$ subshell contains

3 color(red)(cancel(color(black)("orbitals"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "6 e"^(-)

Now, for period $3$, you have $2$ half-filled and $1$ completely filled orbitals.

The trick here is to realize that according to Hund's Rule, every orbital located in a subshell must be half-filled before any of the orbitals can be completely filled.

This means that for period $3$, you will have $1$ electron added to all $3 p$ orbitals and a single electron added to one of the three $p$ orbitals. This means that for period $3$, i.e. in the $3 p$ subshell, you have

overbrace(1 color(red)(cancel(color(black)("orbital"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))))^(color(blue)("completely filled")) + overbrace(2 color(red)(cancel(color(black)("orbitals"))) * "1 e"^(-)/(1color(red)(cancel(color(black)("orbital")))))^(color(blue)("half-filled")) = "4 e"^(-)

Therefore, the total number of electrons located in the $p$ block, i.e. that are located in the $p$ subshell and therefore have $l = 1$, is equal to

overbrace("6 e"^(-))^(color(blue)("in the 2p subshell")) + overbrace("4 e"^(-))^(color(blue)("in the 3p subshell")) = color(darkgreen)(ul(color(black)("10 e"^(-))))#