How many electrons in an atom can have #n + l =6# ?

As you know, we use four quantum numbers to describe the location and spin of an electron in an atom.

![figures.boundless.com]()

In your case, you must find the number of electrons that can have

#n + l = 6" " " "color(orange)("(*)")#

The angular momentum quantum number, #l#, which tells you the subshell in which the electron is located, depends on the value of the principal quantum number, #n#, which tells the energy level on which the electron resides, as given by

#l <= n-1#

Right from the start, you can tell that the first energy level that can hold electrons that satisfy condition #color(orange)("(*)")# corresponds to #n=4#, since for

#n = 3 implies l = (0, 1, 2}#

you can't have a pair that matches the given condition

• #3 + 0 < 6" "color(red)(xx)#

• #3 + 1 < 6" "color(red)(xx)#

• #3 + 2 < 6 " "color(red)(xx)#

The same is true, of course, for #n=1# and #n=2#.

Now, for #n=4# you have

#n = 4 implies l = {0, 1, 2, 3}#

#4 + 2 = 6" "color(green)(sqrt())#

For the angular momentum quantum number, you have

• #l=0 -># the s-subshell
• #l=1 -># the p-subshell
• #l=2 -># the d-subshell

The d-subshell holds a total of #5# orbitals as given by the magnetic quantum number, #m_l#, which in this case can take the values

#m_l = {-2, -1, 0, 1, 2} -># for the d-subshell

Now, each orbital can hold a maximum of two electrons, which means that a total of

#5 color(red)(cancel(color(black)("d-orbitals"))) * "2 e"^(-1)/(1color(red)(cancel(color(black)("orbital")))) = "10 e"^(-)#

can have #n=4# and #l=2#. Now move on to #n=5#, for which

#n=5 implies l = {0, 1, 2, 3, 4}#

This time, you can have

#5 + 1 = 6" "color(green)(sqrt())#

The p-subshell holds a total of #3# orbitals as given by #m_l#

#m_l = {-1,0,1} -># for the p-subshell

Each of those orbitals can hold #2# electrons, so

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

can have #n=5# and #l=1#. Finally, move on to #n=6#, for which

#n=6 implies l = {0,1,2,3,4,5}#

This time, you have

#6 + 0 = 6" "color(green)(sqrt())#

The s-subshell holds a single orbital, since

#m_l = 0 -># for the s-subshell

This means that only

#1 color(red)(cancel(color(black)("s-orbital"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "2 e"^(-)#

can have #n=6# and #l=0#.

Therefore, the total number of electrons that can have #n+l = 6# will be

#overbrace("10 e"^(-))^(color(blue)(4 + 2 = 6)) + overbrace("6 e"^(-))^(color(darkgreen)(5 + 1 = 6)) + overbrace("2 e"^(-))^(color(purple)(6 + 0 = 6)) = "18 e"^(-)#