Question

Question #5a170

Answer 1

Ten electrons.

Explanation:

As you know, sulfur, `"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 = {1, 2, 3}`

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 = {0, 1, ..., n - 1}`

You know that

`l = 1`

designates the `p` subshell, which corresponds to the `p` block in the Periodic Table.

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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 `2p` subshell completely occupied with electrons. Since each orbital can gold a maxium of `2` electrons, you can say that the `2p` 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 `3p` orbitals and a single electron added to one of the three `p` orbitals.

![http://kaffee.50webs.com/Science/activities/Chem/Activity.Electron.Configuration.html]()

This means that for period `3`, i.e. in the `3p` 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"^(-))))`