Fall 2012 Q6A [ENDORSED]

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Armo_Derbarsegian_3K
Posts: 35
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Fall 2012 Q6A

Hello,

For this problem I don't understand how you find the hybridizations.

Do we assume the carbon will just make two hydrogen bonds? Or is it possible it may form a double bond? Would we have to calculate FC and figure that out, or can we assume it will just get 2 H bonds as problem only states that H and lone pairs are omitted?

Thanks.

Chem_Mod
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Re: Fall 2012 Q6A  [ENDORSED]

The problem description states that there are "lone pairs omitted" and "...some of the hydrogen atoms attached to carbon atoms are omitted."

Thus, it is helpful to write in the missing H atoms on the ring and the lone pairs on the N and both O atoms.

If you do this, the hybridization (based on the number of regions of electron density) should become apparent.

Krishil_Gandhi_1D
Posts: 27
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Re: Fall 2012 Q6A

I still don't understand this question. Even after writing in the hydrogen atoms, I do not understand hoe the carbon atoms have a sp^3 hybridization. From my point of view, I can only see 3 regions of electron density around the carbon atoms in the ring structure. Thus, I keep getting sp^2. Is there a lone pair somewhere that I am missing? Thanks

Armo_Derbarsegian_3K
Posts: 35
Joined: Sat Jul 09, 2016 3:00 am

Re: Fall 2012 Q6A

In the ring, the carbons each make two bonds, either to another carbon or to the nitrogen. Since the problem states hydrogens have been omitted and we know carbon likes to make 4 bonds, we would add in two hydrogens to each carbon to fulfill their octet. Therefore, each carbon actually makes 4 bonds. 4 bonds will give us an sp3 hybridization.

Hope that helps.

Fengting Liang 1F
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Fall 2012 Q6A

It actually has four areas of electron density because each carbon is attached to two carbons (or one carbon and the nitrogen), as well as two hydrogens.
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