Which of the following is NOT a valid choice of #l# if #n = 4#, #m_l = 2#, and #m_s = +1/2# are already known?

Only #D#.

• The principal quantum number #n = 1, 2, 3, . . . # indicates the energy level.
• The angular momentum quantum number #l = 0, 1, 2, 3, . . . , n-1# indicates the shape of the orbital, i.e. #s,p,d,f, . . . #
• The magnetic quantum number #m_l# is in the set #{-l,-l+1, . . . , 0, . . . , l-1, l}#, and corresponds to each orbital of the same angular momentum #l#.
• The spin quantum number #m_s = pm1/2# is purely a property fermions, namely electrons. All electrons have one of these spins.

For #n = 4# and #m_l = 2#, we have either a #4d# or #4f# orbital, since #l_(max) = n-1 = 3# and the set of #m_l# cannot be greater than what #l# actually is.

As such, #l = 2# or #l = 3# could be correct...

And obviously, any orbital can hold a spin-up electron, so #m_s = +1/2# is fine...

Therefore, #A#, #B#, and #C# are all valid inclusions to describe a quantum state. Of course, you cannot have all three at once. Either you have #A# and #B#, or #A# and #C#.

#bbD# is the only incorrect choice, because clearly, #l# can only take on one value at a time, no matter what value it is.

Maybe not so clearly, the value of #l# limits the maximum #m_l#, and #l# cannot be larger than #n-1#, so #l# can only be #2# or #3#.