Midterm Practice with Unpaired Electrons

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005115864
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Joined: Fri Sep 28, 2018 12:15 am

Midterm Practice with Unpaired Electrons

Postby 005115864 » Mon Nov 05, 2018 7:29 am

Hi all, I have a practice problem that I need help on. The question is: For the following molecules how many unpaired electrons are there for each molecule?

O2

and

C2^-


So for this, I don't know why they gave us the molecules in diatomic form but I treated it as a normal atom and did a a drawn out electron configuration for each molecule. For oxygen, it would be a 2p^4, so drawing that out would result in 2 unpaired electrons and that is correct


However, for C2^-, I did the same and assumed it would now be a 2p^3 so I thought it would be 3 unpaired electrons since electrons like to fill all the subshells with one electron before doubling in electrons because of electron-electron repulsions BUT that answer is not correct. The correct answer indicated is 1. Why is this? Can anyone explain.

Ana Pedreros
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

Re: Midterm Practice with Unpaired Electrons

Postby Ana Pedreros » Mon Nov 05, 2018 11:30 am

Im pretty sure you are right if you are analyzing these molecules as atoms, however, if they are molecules, C2 would have to have a quadruple bond and the negative charge would indicate an extra unpaired electron.

Cody Do 2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: Midterm Practice with Unpaired Electrons

Postby Cody Do 2F » Mon Nov 05, 2018 11:38 am

First, count the amount of electrons present! One carbon has 4 valence electrons, so two carbons will give us 8 electrons. As the C2 has a negative charge associated with it, add an extra electron. Thus, there is a total of 9 electrons.

When drawing the Lewis Structure, draw the two Carbons with a triple bond between them. This triple bond will use 6 of the 9 electrons, leaving us with 3 electrons left over. You're right that electrons would fill separate subshells first with one electron before doubling up, but carbon only has 4 outer subshells (2s, 2px, 2py, 2pz) and six electrons (from the triple bond) are already in play. That means that, prior to adding in the 3 electrons we still have, each carbon has a configuration of 2s2, 2px2, 2py1, 2pz1. Thus, with the remaining 3 electrons, you will add 2 electrons to one carbon to complete its 2py and 2pz orbitals. The last electron will fill the other carbon's 2py orbital, leaving the 2pz the only orbital with one electron unpaired.


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