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Speclfy thckoxal clcctroa grorcreries zbout thc zlomns babelkd #-4 Unshured ckcuon patrs aflect local For cach Jom cntcr onc oftkx followang #trabedral squarr plxna...

Question

Speclfy thckoxal clcctroa grorcreries zbout thc zlomns babelkd #-4 Unshured ckcuon patrs aflect local For cach Jom cntcr onc oftkx followang #trabedral squarr plxnar trigonal planar, trlgena nnltul Varar

Speclfy thckoxal clcctroa grorcreries zbout thc zlomns babelkd #-4 Unshured ckcuon patrs aflect local For cach Jom cntcr onc oftkx followang #trabedral squarr plxnar trigonal planar, trlgena nnltul Varar



Answers

The gcomerrical shapc of spred hybriclization is (a) Trigonal bipyramidal (b) Lincar (c) "letrahedral (d) Square planar

In this problem, in one side, I will write the molecule lamb and in other side I will write the safe. Just look at it carefully so we all know that CLF three head distorted the shape. So it is corrected statement in X C F two in except to the same page linear. So it is also correct. An ex c e o f four had a square planner sape, so it is also correct. So according to the given option option A C. D uh correct. Benson of some A C. D are correct option for this problem. I hope you understand the solution of this problem.

So the hybridization is associated with the following electron geometries can be determined by thinking about how many um groups. So how many bonds or how many electron pairs here on the central atom. So starting with Tribunal planner, that means that we have a central atom And it has three bonds. That means that it is sp two hybridized linear. But it looks something like this. And we know that that's sp hybridized. This central atom would have to bonds Tetra. He'd rel means that it has four bonds. So it is sp three hybridized Octavia federal. If we draw the planes we have orbital above, below and then four on this plane. So I think there's 123456 bonds will be involved. That would be sp three D. To hybridized orbital's. And lastly tribunal by pure middle again, if we draw the axis we would have won two on the top and bottom and then 123 orbital's on that plane. So that'll be sp three D. Hybridized.

Arcos chinese. The medical shape of geometrical people sp three hybridized and the three D. I. Agreed. I guess in is We need to identify geometrical safe off sp 300 days. Who The correct answer for the SPT geometrical shape of sp three hybridization is diagonal by pyramidal. It is I gurnal by pyramidal diagonal by kilometers so we can draw the structure also. This is fresh. This one is second, this is first, this one second. This one is heard when his fourth. This one is 50 and this one is it's Here. The angle will be 120° If the angle is 90°. So this is tribunal by paramedics. So option A. Is current.

Fresher off our core content at the very top I've discussed in detail in previous podcasts. I won't do so here just to save time. So we're gonna be drawing out some molecular geometries. So our fast one, we have C two h two. So this is all kind. I know this because off the number of protons hydrogen atoms that are bonded to my carbons when they have 100 you not some bonded. Then we have a triple bond between our carbons. But now, looking to my next structure, we have a double bond present because we have four carbons. So each carbon four protons so each carbon must be able to accommodate to each so that we have a double bond. And then my last example c two h six. This is just on. I'll key Al Cane where we just have a single bond. So it my fast structure I've drawn up is SP three hybridized. So that is mania. My second struck Dr drawn his SP to hybridize. So we have the geometry of tribunal playing art centers about our carbons are now this final structure I've drawn where we have SP three hybrid. I centers about our carbons. We have bonding uncles about 109.5 degrees. Because they have the geometry of Tetra, he drools. So we're going to come back to this very same material Podcast 72 where we will look out the bonding angles on each of these German trees.


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