hybridization question

[Contach]

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can someone explain to me how to figure out the hybridization of atoms.
it is easy for Carbons: CH3 is sp3, CH2=CH2 is sp2 etc.
however, when dealing with things like..: NH3, or NO2+, what do you do?
thanks..

 

[PreDent2009]

It's rather simple. All you do is count the number of electron domains around the atom you are trying to find the hybridization of and use
(s,p,p,p) as a template. So for a CH3 attached to a CH2, the CH3 carbon has 4 bonds around it, so therefore you use an s,p,p, and p, hence sp3. Now for NH3, there is a lone pair on top of the nitrogen, that counts as one electron domain. So N has 4 e-domains and is also sp3. Double and triple bonds count as one electron domain so something like
H-C-Triple Bond-C-CH3, the two C directly attached to the triple bond only have 2 edomains, so therefore they are sp hybridized.

You have to be able to draw correct lewis structures to make sure that you have the correct number of lone pairs, then count the e-domains and the number you count is how many you take for the (s,p,p,p) template. Hope this helps.

can someone explain to me how to figure out the hybridization of atoms.
it is easy for Carbons: CH3 is sp3, CH2=CH2 is sp2 etc.
however, when dealing with things like..: NH3, or NO2+, what do you do?
thanks..

 

[Contach]

Wow! thanks so much. I knew it was something simple like this.. Just didn't know which method to take (counting e- domains, or looking at shape)

Thanks

It's rather simple. All you do is count the number of electron domains around the atom you are trying to find the hybridization of and use
(s,p,p,p) as a template. So for a CH3 attached to a CH2, the CH3 carbon has 4 bonds around it, so therefore you use an s,p,p, and p, hence sp3. Now for NH3, there is a lone pair on top of the nitrogen, that counts as one electron domain. So N has 4 e-domains and is also sp3. Double and triple bonds count as one electron domain so something like
H-C-Triple Bond-C-CH3, the two C directly attached to the triple bond only have 2 edomains, so therefore they are sp hybridized.

You have to be able to draw correct lewis structures to make sure that you have the correct number of lone pairs, then count the e-domains and the number you count is how many you take for the (s,p,p,p) template. Hope this helps.

 

[Contach]

i just read on these forums that the oxygen in furan is sp2.. that is inconsistent with what you told me about counting electron domains and going s,p,p,p.... now i'm confused T_T

 

[Doc Numbskull]

That's because the lone pair on oxygen in furan in a conjugated system with the 2 C=C bonds next to it. In cyclic systems like that of furan, determining hybridization becomes a little more tricky. You have to realize which way the lone pairs on the heteroatom are orientated in relation to the double bonds. If they are parallel, then they are conjugated and the Oxygen (or Nitrogen) is sp2.

 

[PreDent2009]

My way works with all inorganic structures as well as organic structures as you are not dealing with aromatic systems. This goes more into the ability to take lone pair e- and put them into the pi system to make the compound aromatic. All other cases work. Don't worry I wouldn't steer you wrong. Just like anything with organic chemistry, there are always exceptions.

i just read on these forums that the oxygen in furan is sp2.. that is inconsistent with what you told me about counting electron domains and going s,p,p,p.... now i'm confused T_T

 

[artanis]

2 things I encountered on my tests that more than 50% of the class got wrong

1) O=C=O , the carbon is SP hydridized.
2) resonance structures need to be considered when determining the hybridization.