What is the maximum number of electrons in an atom that can have the following set of quantum numbers? n = 3, ml = 2, ms = +1/2

Five. Even if you also specified #l = 2#, it would not make a difference.

A short review of quantum numbers:

• #n = 1, 2, 3, . . . # is the principal quantum number, indicating on which energy level the electron resides.
• #l = 0,1, 2, . . . , n-1# is the angular momentum quantum number, which specifies the subshell in a given energy level #n# that the electron is in. It describes the shape of the orbital, and #0 harr s#, #1 harr p#, #2 harr d#, #3 harr f#, etc.
• #m_l = {-l,-l+1, . . . , 0, . . . , +l-1, +l}# corresponds to each orientation of orbitals in a given subshell, and is the magnetic quantum number.
• #m_s = pm1/2# is the spin quantum number of fermions, which is the class of particle an electron belongs to.

Since you have specified all but #l#, we rely on the restriction of #m_l# by #l# to determine how many electrons this can correspond to. Here is the logic you could follow:

1. #n = 3# permits up to #l = 2# only because #l_max = n-1#, so only #3s# (#l = 0#), #3p# (#l = 1#), and #3d# (#l = 2#) subshells exist.
2. #m_l# must lie in the range of #{-l, . . . , +l}#, so if #m_l = 2#, it must lie in the range #{-l, . . . , -2, -1, 0, +1, +2, . . . , +l}#.
   But since #l_max = n-1 = 2#, we know that #|m_(l,max)| = 2#.
3. #m_s = +1/2# requires that the electron be spin-up. From the Pauli Exclusion Principle, no two electrons can be in the same quantum state, so no two electrons in the same orbital can be spin-up at the same time.

Therefore, only five such electrons exist, and they must be in the five #3d# orbitals:

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To be clear, this CANNOT happen:

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