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CHM 101 CHM 101 Past Questions Objective Question
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Question

What combination of quantum numbers violates the Pauli exclusion principle?
Options
A (n=2, l=1, m=-1, s=+1/2) and (n=2, l=1, m=-1, s=-1/2)
B (n=1, l=0, m=0, s=+1/2) and (n=1, l=0, m=0, s=-1/2)
C (n=3, l=2, m=0, s=-1/2) and (n=3, l=2, m=-2, s=+1/2)
D (n=4, l=3, m=1, s=+1/2) and (n=4, l=2, m=-1, s=-1/2)
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

The Pauli exclusion principle states no two electrons in an atom can have the same set of four quantum numbers. Option D shows different l values (3 vs 2), so they are different orbitals; this does not violate Pauli. The question likely expects identification of a pair that violates, but all given pairs have either different m or different l. Option A and B are valid pairs in the same orbital with opposite spins. None clearly violate Pauli; there may be an error in the options. Option D has different l, so it's allowed.

Hint

Pauli violation occurs when two electrons have identical n, l, m, and s.

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1699 of 1872
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