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 a**maximum** 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#.

*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#.