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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 102a

Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:

(a) Predict the bond angles around each of the carbon atoms and sketch the molecule.

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Welcome back everyone. We have buddha 13 dying CH two ch ch ch two is a planer molecule whose carbon carbon bond links are as follows, sketch the molecule and predict the bond angles surrounding each carbon atom. So let's begin by numbering each of our carbon. Starting from left to right are left, most carbon is carbon one. We have the central carbon as carbon to the third central carbon as carbon three and the right most carbon as carbon number four. Notice that at carbon number one we have two sigma bonds being our carbon hydrogen single bonds and then we have one pi bond being our carbon carbon double bond. And this together counts as three domains, 3 electron domains. Looking at carbon number two, we can also observe the same pattern where we have two sigma bonds, one to carbon and one hydrogen and one pi bond to carbon. So again three electron domains. And we would also say that at carbon number three we have the same pattern because it's bonded in the same way as carbon number two. Lastly at carbon number four We have σ bond or sorry, we have two sigma bonds, two hydrogen and also one pi bond, two carbon. So all of our carbon atoms all have three electron domains. And this therefore tells us that we have sp two hybridization for each of these carbons. And this tells us that our electron geometry is therefore going to be tribunal planer. We should recall recall that tribunal planer geometry means we have bond angles That are at 120°. And so for the sketch of our structure, we have our four carbon atoms each bonded to a hydrogen atom. And we have this tribunal planner geometry. We have one of our carbon carbon bonds being a sigma bond and to being pi bonds. And this would be the sketch of our structure of Buta 13 dying noticed that we've specifically drawn in our bond angles at 120 degrees so we can fill that in to just for more detail. And so our sketch, as our final answer boxed here in yellow is going to most closely match the correct answer in the multiple choice being choice B. So I hope that this made sense. And let us know if you have any questions.
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Textbook Question

Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:

(b) From left to right, what is the hybridization of each carbon atom in butadiene?

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Textbook Question

Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:

(c) The middle C¬C bond length in butadiene (1.48 Å) is a little shorter than the average C¬C single bond length (1.54 Å). Does this imply that the middle C¬C bond in butadiene is weaker or stronger than the average C¬C single bond?

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Textbook Question

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero.

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