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Regardless of its electron configuration, it must always be paramagnetic when it's a single, neutrally charged atom:
(A) Carbon (B) Nitrogen (C) Oxygen (D) Neon (E) Argon

At first, I eliminated D and E. Then, I tried to find the answer by checking the electron confiurations of the atoms; I saw the phrase "regardless of its electron configuration", but I didn't know what else I can do. Then, I found out that all three atoms (A, B, and C) have some unpaired electrons …

I checked charts with paramagnetic elements filled in; three first pictures on google said different things (check here), but after all the info I have found on the net I understood that C (oxygen) is the most likely answer. I don't understand why … I checked a few videos that said that if an atom has 1 or more unpaired electrons it is paramagnetic; then, A, B, and C are all paramagnetic. A few sites said we need to look not at the elements, but at the compounds (not oxygen, but $\ce{O2}$); the task said, anyway, we are looking at the single, neutrally charged atom.

Please tell if it is oxygen and if yes why it is paramagnetic but other elements are not.

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    $\begingroup$ Answer is nitrogen, because it has an odd number of electrons, and therefore there must be at least one unpaired electron. For carbon and oxygen which have an even number of electrons, it's always possible to pair them up somehow (not in the ground state, though; as your research indicates, all of C, N, and O are paramagnetic in the ground state). I don't like the question's phrasing, but that's what it's going for. $\endgroup$ – orthocresol Aug 20 '17 at 15:37
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Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’.

The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^0$ or any other combination of two of the three equivalent p orbitals half-populated. However, in early organic chemistry courses you typically rapidly learn about the hybridisation approach which changes that electronic configuration to $\mathrm{1s^2\, 2s^1\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^1$ — four instead of two unpaired electrons. Likewise, it is possible to imagine an electronic configuration as follows: $\mathrm{1s^2\, 2s^2\,2p}_x{}^2\, \mathrm{2p}_y{}^0 2\mathrm{p}_z{}^0$. In this case, all electrons are paired and the atom is diamagnetic.

Performing this type of analysis with all the atoms at your disposal — carbon, nitrogen and oxygen — you will be able to succeed in pairing all electrons for carbon and oxygen but not for nitrogen since it has five valence electrons. Five, being an odd number, cannot be mapped pairwise. Thus, a nitrogen atom is always paramagnetic.

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