Since the question is a little ambiguous, I will assume that you"re dealing with *three distinctive sets* the quantum numbers.

In enhancement to this, ns will likewise assume that you"re relatively familiar with quantum numbers, so ns won"t get in too lot details about what every 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 plenty of electrons deserve to share this value of #n#, you must determine exactly how many **orbitals** you have actually in this energy level.

The variety of orbitals you acquire *per energy level* can be uncovered using the equation

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

Since every orbital can hold a**maximum** of two electrons, it follows that as numerous as

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

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

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

orbitals. Therefore, a best of

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

electrons deserve to share the quantum number #n=2#.

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

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

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

*the s-subshell*#l=1 ->#

*the p-subshell*#l=2 ->#

*the d-subshell*#l=3 ->#

*the f-subshell*

Now, the number of **orbitals** you get *per subshell* is provided by the *magnetic quantum number*, #m_l#, which in this case 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 total of **seven** orbitals, which method that you have a preferably of

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

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

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

This time, girlfriend are offered the power level, #n=6#, the subshell, #l=2#, and the **exact orbital**, #m_l = 1#, in i m sorry the electrons reside.

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Since you recognize the precise orbital, it follows that just **two electrons** have the right to share these 3 quantum numbers, one having spin-up, #m_s = +1/2#, and also the other having spin-down, #m_s = -1/2#.