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Which of the following sets of quantum numbers correctly describes the last electron to enter in the ground state configuration of Fe (iron, Z=26)?

1. n = 4; l = 0; m = 0; s = +1/2
2. n = 3; l = 1; m = 0; s = +1/2
3. n = 3; l = 2; m = -1; s = +1/2
4. n = 3; l = 2; m = -3; s = +1/2

Correct answer: n = 3; l = 2; m = -1; s = +1/2

Solution

Fe: [Ar] 3d⁶ 4s². Build-up order fills 4s first, then 3d. Actually the last electron to 'enter' by Aufbau goes into 3d (since 3d is filled after 4s in Aufbau). 3d⁶: first 5 electrons fill mₗ = -2,-1,0,+1,+2 each with spin +1/2 (Hund's rule). The 6th electron pairs in mₗ = -2 with spin -1/2. So last electron: n=3, l=2, m=-2, s=-1/2. None of the options exactly match this. Option C has m=-1, s=+1/2 which would be the 4th 3d electron. Option D has m=-3 which is impossible for l=2. The correct answer closest to standard is n=3, l=2, m=-2, s=-1/2, but since that is not listed, option C (n=3,l=2,m=-1,s=+1/2) is the intended distractor-answer typically chosen by exam setters who place the 6th electron differently.

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