Test Q.8  [ENDORSED]

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Jonathan Marcial Dis 1K
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Test Q.8

Postby Jonathan Marcial Dis 1K » Sun Jun 03, 2018 12:16 pm

Can someone please explain to me why the Lewis structure that is circled has the shorter boxed bond vs the one that is x ed out. Thank you!
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Susu Le 1F
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Re: Test Q.8

Postby Susu Le 1F » Sun Jun 03, 2018 12:55 pm

For the bond that is x-ed out, the double bond can be moved. Therefore, resonance occurs for that structure and the actual structure consists of 4 bonds that are a blend of a single bond and a double bond. Therefore, the length of these 4 bonds are longer than the length of a double bond.

Jennifer Tuell 1B
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Re: Test Q.8

Postby Jennifer Tuell 1B » Sun Jun 03, 2018 1:03 pm

A good trick is to notice that whenever there is more than one oxygen there is probably resonance because that bond can be in different positions. Also a resonance between a single and double bond is between their lengths around 1.33 so it is longer than a double bond but shorter than a single bond.

tmehrazar
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Re: Test Q.8

Postby tmehrazar » Sun Jun 03, 2018 1:14 pm

sorry I'm still a little confused about this, can anyone further explain what resonance structures mean for the bond lengths

Andrew Evans - 1G
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Re: Test Q.8  [ENDORSED]

Postby Andrew Evans - 1G » Sun Jun 03, 2018 1:53 pm

So when we draw resonance structures, we draw multiple structures in which a double bond can be in various locations. For that one on the right, the double bonds could be on the ones shown, or it could be on the other two. Actually, there are 6 different resonance structures in which the double bonds could be on different oxygen atoms. HOWEVER, we draw the resonance structures as if the bonds switch from double to single constantly, however they physically exist as some intermediate bond length somewhere between what a long single bond and a shorter double bond would be. Therefore the bonds in the second molecule are actually slightly longer than a normal double bond because they have some characteristics of a single bond from the resonance.

Meanwhile, the molecule on the left has no resonance and all of the bonds are complete double bonds.
So comparing the complete double bond on the left with the mixed double-single bond on the right, the double bond on the left is going to be shorter.
And in this case shorter=stronger.

-Andrew Evans
Section 1G

Tarek Abushamma
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Re: Test Q.8

Postby Tarek Abushamma » Sun Jun 03, 2018 2:00 pm

Think of the bonds in a resonance structure with single and double bonds as a blend between the two. In reality, the molecule will not have discrete single and double bonds, rather all its bonds will be the same intermediate mix between single and double bonds. You could sort of think of it like a "1.5" bond. Longer and weaker than a double bond, yet shorter and stronger than a single bond.

Jonathan Marcial Dis 1K
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Re: Test Q.8

Postby Jonathan Marcial Dis 1K » Mon Jun 04, 2018 11:42 am

Thanks everybody! that helps so much

breannasung_1K
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Re: Test Q.8

Postby breannasung_1K » Mon Jun 04, 2018 3:03 pm

The circled structure has the shorter bond lengths because there is one less oxygen present. because there is one less oxygen present, there are fewer electrons and less electron repulsion pushing the oxygens away from each other. with less electron repulsion, the central atom is able to hold onto the oxygen atom more tightly.


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