Since the question is a little bit ambiguous, I will certainly assume the you"re taking care of three distinctive sets that quantum numbers.

In enhancement to this, i will additionally assume that you"re fairly familiar v quantum numbers, so ns won"t get in too lot details around what each represents.

*

#1^"st"# set # -> n=2#

The principal quantum number, #n#, speak you the power level on which an electron resides. In order to be able to determine how countless electrons have the right to share this value of #n#, you need to determine precisely how countless orbitals you have actually in this energy level.

The variety of orbitals you gain per energy level deserve to be uncovered using the equation

#color(blue)("no. Of orbitals" = n^2)#

Since every orbital have the right to hold amaximum of 2 electrons, it complies with that as countless as

#color(blue)("no. Of electrons" = 2n^2)#

In this case, the 2nd energy level stop a full of

#"no. Of orbitals" = n^2 = 2^2 = 4#

orbitals. Therefore, a best of

#"no. Of electrons" = 2 * 4 = 8#

electrons have the right to share the quantum number #n=2#.

#2^"nd"# set #-> n=4, l=3#

This time, you are given both the energy level, #n=4#, and also the subshell, #l=3#, on i beg your pardon the electrons reside.

Now, the subshell is offered by the angular momentum quantum number, #l#, which have the right to take values ranging from #0# come #n-1#.

#l=0 -># the s-subshell#l=1 -># the p-subshell#l=2 -># the d-subshell#l=3 -># the f-subshell

Now, the variety of orbitals you gain per subshell is given by the magnetic quantum number, #m_l#, i m sorry in this instance can be

#m_l = -l, ..., -1, 0, 1, ..., +l#

#m_l = -3; -2; -1; 0; 1; 2; 3#

So, the f-subshell can hold complete of seven orbitals, which method that you have a maximum of

#"no. Of electrons" = 2 * 7 = 14#

electrons that have the right to share these two quantum numbers, #n=4# and #l=3#.

#3^"rd"# set #-> n=6, l=2, m_l = -1#

This time, friend are provided the power level, #n=6#, the subshell, #l=2#, and the exact orbital, #m_l = 1#, in i beg your pardon the electron reside.


You are watching: N=4 l=1 how many electrons


See more: How Many Ounces Of Grated Parmesan In A Cup, How Do You Measure Cheese

Since you recognize the specific orbital, it adheres to that only two electrons deserve to share these three quantum numbers, one having spin-up, #m_s = +1/2#, and the other having spin-down, #m_s = -1/2#.