Question #5a170

1 Answer
Oct 5, 2017

Answer:

Ten electrons.

Explanation:

As you know, sulfur, #"S"#, is located in period #3#, group #16# of the Periodic Table.

http://www.knowledgedoor.com/2/elements_handbook/sulfur.html

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.

https://saylordotorg.github.io/text_introductory-chemistry/s12-04-electronic-structure-and-the-p.html

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"^(-))))#