Resonance structures and hybridization | Organic chemistry | Khan Academy

Resonance structures, hybridization of orbitals, and localized and delocalized electron density. Created by Jay.
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Organic Chemistry on Khan Academy: Carbon can form covalent bonds with itself and other elements to create a mind-boggling array of structures. In organic chemistry, we will learn about the reactions chemists use to synthesize crazy carbon based structures, as well as the analytical methods to characterize them. We will also think about how those reactions are occurring on a molecular level with reaction mechanisms. Simply put, organic chemistry is like building with molecular Legos. Let's make some beautiful organic molecules!
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[@necktiekakmalith8036]

you're the best 12 minutes video make to understand resonance and hybridization than two hours lecture I don't understand what the hell my lecturer was doing

[@sarahholland5980]

thank you so much for explaining that last example!!! my professor kept mentioning how that lone pair can't participate in pi bonding because the orbitals are orthogonal to each other and I just did not understand what that meant. I can now see in your example that they are orthogonal to each other because the lone pair is contributing to sp2 hybridization because the unhybridized p is already being used by the pi bond. Thank you!

[@noecazaresjr7284]

Truly a godsend. Educating people of the world with youtube access.

[@Mariam-qd8of]

Loved it! Thank you

[@glendasworld4125]

How does the know the formal charge right away? How does he know if the lone país are localized or not? How does he know where to move the electrons?

[@TheToxicMegacolon]

excellent video!

[@favourafolabi9589]

Thank you!

[@Bigolho97]

In the last example, why can't N change to sp1 and form2 double bounds? In the other examples we had sp3 structures which changed to sp2 so that a double bound could be formed

[@czarwu3359]

That is impossible on the right hand side because there is still a hydrogen bond to the carbon on the right of nitrogen, so the 2 pi bonds plus 2 sigma bonds will actually break the octet rule of the carbon.

[@bamikuma7684]

Thank you very much Khan team. My prof might as well have been speaking Spanish before I happened upon this video!

[@mollysolomon6191]

How do you tell whether a lone pair is localized or delocalized? Is if if there is a double bond(P orbital) near?

[@du0377]

it must be adjacent to SP2 hybridized atoms.

[@RiddhiThorat-rf4ew]

One lone pair is in conjugation that is delocalised And (Not in conjugation) then other lone pair is localised

[@mahekvanjani1561]

Can this happen in phenol like in oxygen hybridisation would seem sp3 but actually it should be sp2 ??

[@natej0556]

On the last example, why can't we move change the lone pair into a pi bond and change the pi bond into a lone pair at the same time(and move the bottom pi bond to the right)

[@rishikeshlatthe8560]

How we came to know that the electrons are delocalized or not

[@InnerMittenSignal]

God's musical chairs...

[@InnerMittenSignal]

expelled no intelligence allowed

[@catseye4332]

kzbin.info/www/bejne/jGaogayjmLyjY8k

[@harrrgh944]

You can't say "amines" without "Am I?..."

[@LordChundy]

Can you explain why there is a resonance structure in the first example? When I calculate the formal charge for nitrogen and oxygen they are both 0 in the first structure. Where as in the second structure nitrogen and oxygens formal charge is not 0. Therefor would the first structure not be the favoured structure?

[@matthewbk]

+Alex Watkins One of the reasons why is that there's a pattern. When you have a LP adjacent to a double bond (in this case the O), there is always resonance. As long as the overall charge of the resonance structure is the same as the original, that is as "favored" as the other one---the more resonance structures there is, the more stable the molecule. In this case, the original structure has no formal charges on any of the atoms. In its resonance structure, the oxygen atom gains a negative charge while the nitrogen atom gains a positive charge thus making the overall charge of the amide 0.

[@Badgerbahalwan]

@ ~.40 in this video, it states that Sp3 is trigonal pyramidal. I thought it was Tetrahedral?🧐

[@Badgerbahalwan]

is it due to it being a more planar structure?? i.e. no wedges and no dashes

[@skarloey2808]

ik you probably got it by now but it’s because there’s a lone pair.

[@taranjitkaur1854]

In pink, why does carbon have a -1 formal charge? I thought its formal charge was -2.

[@du0377]

Carbon has 4 valence electrons, here there is one lone pair, 2 carbon-carbon single bonds(each counts for 1), and the carbon hydrogen bond(1). So we get 4-5= -1. Thus the formal charge is -1

[@shazwannaim8543]

the question I had now is why delocalization happened? I hope my question is not stupid

[@gadrri1167]

Delocalization allows stability..In simple terms, sharing the negative charge of e = less strain on any individual atom