Sapling #17

(Polar molecules, Non-polar molecules, etc.)

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Valerie Doan 3I
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Sapling #17

Postby Valerie Doan 3I » Sat Nov 28, 2020 12:08 pm

Could someone explain how they found the bond angles and hybridizations for C3H4? Thanks
Screen Shot 2020-11-28 at 12.07.00 PM.png
Screen Shot 2020-11-28 at 12.06.39 PM.png

Stuti Pradhan 2J
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Re: Sapling #17

Postby Stuti Pradhan 2J » Sat Nov 28, 2020 12:13 pm

C3H4 has three resonance structures.

In the first structure given in this image, the carbon on the left has two regions of electron density and is therefore, linear and sp hybridized. This has a 180 degree bond angle. The carbon atom in the center has a triple bond and a single bond, which indicate two regions of electron density (linear) and sp hybridization. This atom also has 180 degree bond angles. The carbon on the right has a single bond to another carbon and 3 single bonds with hydrogens. This indicates four regions of electron density (tetrahedral), sp3 hybridization, and 109.5 degree bond angles.

For the second resonance structure, the carbon atom in the center just has two double bonds, so since it has only two regions of electron density, the central atom is linear and sp hybridized (180 degree bond angle). The two carbons at the end of the chain have a double bond to the central carbon atom and two single bonds to each hydrogen atom, giving them three regions of electron density (trigonal planar), an sp2 hybridization, and 120 degree bond angles.

The third resonance structure has one carbon with four single bonds (tetrahedral), and two carbons with 2 single bonds and a double bond (trigonal planar). This corresponds to sp3 hybridization and sp2 hybridization respectively. However, in this case, since the molecule is cyclic, the bond angle is 60 degrees because the carbons all bond to form a triangle, and to all be the furthest apart from each other, they form an equilateral triangle, which has 60 degree bond angles.

Hope this helps!
Last edited by Stuti Pradhan 2J on Sat Nov 28, 2020 12:32 pm, edited 3 times in total.

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Re: Sapling #17

Postby IsaacLaw1E » Sat Nov 28, 2020 12:16 pm

C3H4 has three structures: triangle, triple bonded, and double bonds.
For hybridizations, just look at the different number of electron densities each carbon can have in the three structures. So, in triangle it has 3 and 4, in triple bonded it has 2 and 4, and in double bonds it has 2 and 3. So the hybridizations that are represented are the ones that represent carbon with 2, 3, or 4 electron densities. These are sp (2 densities) sp2 (3) and sp3 (4).
For bond angles, just look at the VSEPR models. In triangle the carbon ring has 60deg bond angles because it's a triangle, 109.5deg because of the tetrahedral carbon, and 120deg because of the trigonal planar carbon. Only one answer has all three of those. The 180deg bond angle is from the other two structures.

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Re: Sapling #17

Postby arisawaters2D » Sat Nov 28, 2020 5:22 pm

Do we have to know this type of question for the exam?

Ven Chavez 2K
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Re: Sapling #17

Postby Ven Chavez 2K » Sat Nov 28, 2020 5:29 pm

For the hybridization states, the number depends on the number of regions of electron density. Therefore, a Carbon with 2 bonds would have a sp hybrid, a carbon with 3 bonds would be sp2, and a carbon with 4 bonds would be sp3.

For the bond angles, we see the 180 degree angle in the C-C-C linear structure. In the structure where Carbon has three hydrogens and no lone pairs the angles would be 109.5. The triangle structure would be 60 and the carbon with 2 hydrogens and one double bonded carbon would be 120.

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Re: Sapling #17

Postby Sebastian2I » Sat Nov 28, 2020 5:48 pm

Thank you for the explanation! This was all extremely helpful!

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