Which of the following sets of quantum numbers is not allowed?

Solution

Chemistry ma’am · 3 steps

Check the chain n, then l, then m: l must stop at n − 1, so with n = 3 the largest l is 2.

1. The rules. For a given n, l can be 0 to n-1. For a given l, mₗ runs from -l to +l, and s is +1/2 or -1/2.

   Option	n, l	Allowed l for this n	mₗ in range?	Verdict
   (a)	n=3, l=2	0, 1, 2	0 lies in -2 to +2	allowed (a 3d orbital)
   (b)	n=3, l=2	0, 1, 2	-2 lies in -2 to +2	allowed (a 3d orbital)
   (c)	n=3, l=3	0, 1, 2	check fails before this	not allowed: l = n
   (d)	n=3, l=0	0, 1, 2	0 is the only value	allowed (the 3s orbital)

2. No 3f orbital. l = 3 would be an f subshell. The f subshell first appears at n = 4, so a 3f orbital does not exist.
3. So. Option (c) is the set that is not allowed. Its mₗ = -3 and s = -1/2 look fine on their own, but l = 3 is impossible when n = 3.

The idea to keep. Test quantum numbers in order: is l ≤ n-1? Is |mₗ| ≤ l? Is s = ±1/2? The first failed test makes the whole set invalid, however reasonable the rest looks.

Answer from the NTA final answer key. Solution by PrepWiser.