Four possible electron configurations for a nitrogen atom are shown below, but only one schematic represents the correct configuration for a nitrogen atom in its ground state. Which configurations violate the Pauli exclusion principle?

Video transcript

Hello everyone in this problem. We are given several electric configurations. So we have four here A through, we were seeing if any of these valley, the poly exclusion principle, let's have a little refresher about what the pole exclusion principle states. What says that? No two electrons can have the same quantum numbers. So let's go ahead and write this out. So I'll say over here baby Deep Polly says no same quantum numbers. So that means the end we have our L have R, M. L. And we have our M. S. Values. So usually in this type of problem, the most common mistake or violation for the pol exclusion principle is when these have the same M. S number. So let's put that over here. So M. S. Is spin number. So that's if The spin number is going to be positive 1/2 Or -1/2. So no electron can have both positives are both negatives. So that usually looks something like if ever to be positive will both be pointing up if it's down or if it's negative half it'll be pointing downwards like So. So let's take a look here for the first row, do we see any of the same pointed arrows going the same direction in one box here. See this is okay, this is good, this is good, this is good and this is good. So this is fine. This does not violate the Pauli exclusion principle. What about B. So this is okay, this is okay and this is okay. Although this should not be there because we need to fill up the lower energy orbital's first. This is not what the poly exclusion principle states, so this although it may not be correct, it will not violate the Pauli exclusion principle to be exact, but this is still okay, so that's not an answer. That's why I put an X. As for C. Let's see here, we have this being okay, the two S. Is okay. And what about this 1? And I said if we have to same M. S. Numbers, then we will have two arrows pointing upwards, or two arrows pointing downwards. As a matter of fact does so C. Is going to be our answer for this problem.