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

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

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

The principal quantum number, #n#, speak you the power level on i beg your pardon an electron resides. In bespeak to be able to determine how countless electrons can share this value of #n#, you should determine precisely how many orbitals you have actually in this power level.

The variety of orbitals you obtain per energy level deserve to be discovered using the equation

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

Since each orbital have the right to hold amaximum of 2 electrons, it adheres to that as many as

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

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

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

orbitals. Therefore, a maximum of

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

electrons can share the quantum number #n=2#.

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

This time, friend are provided both the energy level, #n=4#, and also the subshell, #l=3#, on which the electron reside.

Now, the subshell is given 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 get per subshell is offered by the magnetic quantum number, #m_l#, i beg your pardon 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 deserve to hold full of seven orbitals, which way that you have actually a preferably of

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

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

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

This time, you are given the power level, #n=6#, the subshell, #l=2#, and also the exact orbital, #m_l = 1#, in i beg your pardon the electrons reside.

You are watching: N=4 l=2 how many electrons

See more: How Many Grams Does A Paperclip Weigh ? Jumbo Paper Clip

Since you know the precise orbital, it follows that only two electrons can share these 3 quantum numbers, one having spin-up, #m_s = +1/2#, and also the other having actually spin-down, #m_s = -1/2#.