Which hybridization scheme allows the central atom to form
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Question

Which hybridization scheme allows the central atom to form 
more than four bonds? 
sp3, sp3d, sp2

Accepted Answer

welcome back everyone in this example, we need to identify the hybridization that conform exactly six bonds. So beginning with choice A. Were given a hybridization of sp two. We should recall that for our molecular geometry which corresponds to our shape of our molecule based on only bonded electrons. We would have a molecular geometry that is tribunal planner. And so when we imagine our central atom of our molecule will just say x. We would have a total of three bonds to our central molecule in the following shape. So we would count three bonds. And because that is not six bonds, we would rule out choice A. Moving onto choice B. We have sp three hybridization which we should recall corresponds to a tetrahedron molecular geometry where we can imagine our central atom with a total of four bonds. So we have one out of the top of our central atom. And then we have three more where one of the three at the bottom half is a wedge coming out of the plane of our paper towards ourselves. And then our third bond is here. So this gives us a total of four bonds. And because that is not six, we can rule out choice be moving on to choice C. We have sp three D two hybridization which we should recall corresponds to an Octa he'd rel molecular geometry where we can imagine our central atom with three bonds coming from the top half of the molecule and to wedge bonds coming towards the or making of the bottom half of our molecule which come out of the plane of our page here. And then we just have a third regular bond here at the bottom. So this gives us a total of six bonds which is definitely exactly what the prompt asks for. So we can confirm that C. Is a great answer, choice. But let's move on and consider choice D. Where were given sp three D. Three hybridization. And we should recall that this corresponds to pentagonal by pyramidal molecular geometry which we should recall has the shape where we have our central atom in the center where a wedge bond is coming out of our plane of our paper from the top half of our molecule. And let's make that a better which we have two more regular bonds making up the top half of our molecule here we have a dashed bond coming towards the bottom half of our molecule which we should recall. Dash bonds are away from the plane of the paper of our view here. So they're away from us that is surrounded by two other regular bonds. And sorry about that. So for this portion of our molecule and then we have another bond extending towards the center of our central atom here. And this is what makes up our pentagonal by pyramidal molecular geometry here. So we can count a total of one, bonds here for this shape. And so to complete this example, we're going to rule out also choice D. Meaning that the only correct choice that we can confirm with exactly six bonds corresponding to the hybridization, given sp three D two is choice C. So C is our final answer. To complete this example. I hope that everything that I reviewed was clear, but if you have any questions, please leave them down below and I will see everyone in the next practice video.

Which hybridization scheme allows the central atom to form more than four bonds? sp3, sp3d, sp2

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