paramagnetic

[Farcus]

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A) N2^2-

B) O2

C) O2^-

D) N2^2+

E) O2^+

which of the following is not paramagnetic

OK the answer is D. But can someone help me go through this problem for each one with electrons. I don't get understand how to do it when there is a compound. Like O2, there is 2 oxygens so there is 16 electrons so how do I know this isn't unpaired?

 

[Foghorn]

To re-phrase the question, "Which one of these diatomic species exhibits diamagnetic properties i.e. have all electrons paired?"

In order for you to answer the question, you have to know Molecular Orbital Theory (HOMO, LUMO) and use it to figure out how the molecular orbitals are filled. Keep in mind when you do this, the Aufbau Principle still holds.

Here's a hint: Molecular oxygen (O2) exhibits paramagnetism, having 2 unpaired electrons. What happens if an electron is added/removed from its HOMO to create an ionic species? Will the the ionized form still be paramagnetic or will it be diamagnetic?

A similar approach can subsequently be used for molecular nitrogen (N2) and its ionic species derivative.

 

[Farcus]

so when there is a molecule you still count each atom as individuals? and then count if its para or diamagnetic?

 

[tastybiscuit]

I have studied a hella lot and have yet to see a specific discrete type question asking this detailed of knowledge re: MO theory. You should be able to diagram e- configurations but this would require diagramming the configuration and looking at HOMO/LUMO...this is not covered in any MCAT review books. This is really, really low yield stuff. Know configurations and what paramagnetic/diamagnetic means and you would be fine. However, here goes:

Oxygen: sigma, sigma^*, sigma^z are all filled (6 e-), both pi are filled (4e-), and one each in the pi^* orbitals...creating a divalent molecule that is paramagnetic.

Nitrogen^2+: sigma, sigma^@, sigma^z are all filled(6 e-), and one pi is filled(2e-). Thus it is diamagnetic.

To understand this you really need to see the diagrams, understand bond order, etc. It is not required knowledge.