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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 102b

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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Welcome back everyone. The carbon carbon bond lengths for pro pine C three H four and alkaline are illustrated in the following figure. What is the hybridization of each carbon atom in pro pine from left to right. So we're going to begin with our left most carbon here and we'll write a note being one for this. Carbon being carbon number one, we'll have this as carbon number two and this carbon as carbon number three. So beginning with carbon one note that it has a bond to hydrogen hydrogen, which is not shown but is implied because hydrogen is written right next to heart. Carbon and note that it has a Triple Covalin Bond with Carbon # two. So we can say that carbon one has two electron domains again from the single bond, the carbon hydrogen single bond and our carbon carbon triple bond. So these represent our two electron domains. And because we have two electron domains for carbon number one, we can say therefore carbon number one is sp hybridized, that is its hybridization. So now moving on to carbon number two and for less confusion, I'll change the colors here. So let's keep them all the same color going from left to right. We're going to go to carbon number two. Next. So now we're focusing on carbon number two and its electron domains notice that it has a carbon carbon triple bond with carbon number one and a carbon carbon single bond with carbon number three and I wrote the three in the wrong position. So carbon number three is here. So sorry for that confusion. So we can say that we have also two electron domains for carbon number two, the first being the carbon carbon triple bond, that is our first domain and our second domain being our carbon carbon single bond or sigma bond. And so therefore we would consider carbon number two as also being sp hybridized. So so far we have our first two answers for carbon one and two going from left to right now for our right most carbon carbon number three, we want to notice that it has a single carbon carbon, we're sorry, a single carbon carbon bond with carbon number two and three implied single bonds to hydrogen written by the hydrogen next to carbon three. And so this means that carbon three has a total of four electron domains which we can define by again, the first one being the carbon carbon sigma bond, than our second one being our carbon hydrogen implied single bond and the fourth the 3rd and 4th domains also being our carbon hydrogen single bonds. And so therefore we can say that carbon number three, the right most carbon is going to be sp three hybridized whenever you have four electron domains that corresponds to sp three hybridization. And this would be our third answer. And so all of our answers highlighted in yellow correspond to choice A and the multiple choice as the correct answer. So I hope that this made sense. And let us know if you have any questions
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In ozone, O3, the two oxygen atoms on the ends of the molecule are equivalent to one another. (d) How many electrons are delocalized in the p system of ozone?

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Textbook Question

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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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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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. (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? Would you expect the molecule to be planar for both Lewis structures?

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