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Draw the structure ofa compound that meets the following criteria:Molecular (ormula: CsxHx ( Two Sp" hybridized carbon atoms (Clearly label these on the struct...

Question

Draw the structure ofa compound that meets the following criteria:Molecular (ormula: CsxHx ( Two Sp" hybridized carbon atoms (Clearly label these on the structure) Six sp' hybridized carbon atoms (Clearly label these on the structure) (Clearly label these un the structure) Four primary (1 8) carbon atoms carbon atoms (Clearly label these on the structure) One secondary (20 (Clearly label these on the structure) One tertiary (39 carbon atom

Draw the structure ofa compound that meets the following criteria: Molecular (ormula: CsxHx ( Two Sp" hybridized carbon atoms (Clearly label these on the structure) Six sp' hybridized carbon atoms (Clearly label these on the structure) (Clearly label these un the structure) Four primary (1 8) carbon atoms carbon atoms (Clearly label these on the structure) One secondary (20 (Clearly label these on the structure) One tertiary (39 carbon atom



Answers

Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has a. three $s p^{3}$ hybridized carbons. b. one $s p^{3}$ hybridized carbon and two $s p^{2}$ hybridized carbons. c. two sp^' hybridized carbons and two sp hybridized carbons.

They were asked essentially to depict the chemical structures of different compounds. So essentially in our first case we have three sp three hybridized carbons, which means that each carbon atom has to be connected basically to four basically different Different, has basically four bonds, basically two different atoms. Well in the second case we have uh two S. P two and one sp three. So sp two basically for carbons would have to be connected to three, three different atoms. Swale, for sp three you have to be connected to four atoms. So in the third case we have to sp And to Sp three. So in the first case the simplest example would be the structure of propane And so propane is H 3, 2 and CH three. So each carbons connected essentially to for four different atoms in this case. So for a part two, you have Essentially to Sp two And one sp three. And the simplest example would be some type of an al keen with a double bond. So we also need an sp three, so we have to add some form of a metal group here, then we can add hydrogen. So the method group is always going to be sp three offer Elkins Elkins carbons or sp two. So we basically see that we have two SB two and one sp three hybridized carbon. In our third case, we have to S. P three and to sp and one of the only examples we have in this case is essentially a central AL kind, which has a triple bond. Then we have Essentially Metal Groups Here Associate three H 3 C. And these are basically carbons that are part of the al Qaeda or sp hybridized. While these metal groups are always going to be sp three hybridized and that's it.

Yeah, let's take a look at this acid Tate Ill. So this is a rudimentary acid Tate ion, just a c o. Negative. And what can happen is well, first, we know that this oxygen has two double bonds. This oxygen iss are two lone pairs and this auction as three loan players. So you know that this oxygen does that speak to hybridized. And we know that this carbon makes three different bonds, is also expected Hybridized this oxygen looks like it's SP three item hybridized. However, what it can do is donate these electrons two residents and we know that if those electrons can engage in residence, they exist in a P orbital, which means that it's in fact, not SB three, but SB two because SP two allows for that free P orbital. So SP two plus p orbital equals residents and that's key. And we need three of them adjacent so that it will involve in resonance. And what we get is, well, something that looks entirely similar to the first structure, where now the negative charges on the top oxygen and of course it can go backwards and go that way. Well, it either way that makes it that the all three of these Adams, the carbon theocracy gin and the other oxygen our SP two hybridized. Now, if we were to select one of these structures, let's go with this one. So we have carbon or so Let's draw the lone pairs on these or at the bonding pairs on these. So when we have a bond or a double bond specifically, well, let's start with a single blow. Single bond is a signal bond, and a Sigma bond is an overlap of SP to Orbital's. In this case, SP two Orbital's on That is a signal bond, and the same can be done with our our group. It's It's an overlap to sharing of those electrons, and the same can be done with this Carbonell carbon. So one of those bonds is a Sigma Bond, s P two. However, there is a pipe bond between this carbon and this oxygen. And what that looks like is, well, r p orbital going to be something like this, and it's not quite a direct overlap. It's more of like an attraction of distant of through distance, but like it za weak poll. But it's not a direct overlap. And that's rpai bond in R P orbital and this oxygen also has. It's lone electron here or lone pair, as a matter of fact, giving it a negative charge because charges exist in the P orbital. No matter what. If there is a charge it exists in the P orbital, there's no way can exist in a hybridized orbital and it has its normal electron pairs, which are in SP hybridize. That should be red, but but yes, so the overlap. The only overlapping orbital's are the SP two orbital's. Those were the only overlapping orbital's to form bonds. The A P orbital doesn't technically overlap to form a bond, it's more of an attraction, but the main overlap is from the SP to hybridize orbital.

So here we have a molecular structure to consider. So we have CH three C. H. Double bond C. H C H two carbon double bond, oxygen, CH two C. O. H. We can count the number of carbon, says we have 1234567 So carter album carbon atom 1 46. They are sp three hybridized and carbon atom 235 and seven R. Sp two hybridized. So the given compound can exist in two different is Emma's, which will draw here. So we can have the sets. So that looks as follows carbon single bond, carbon double bond, carbon hydrogen can just fill in all about hydrogen is here and then from MR bond we have carbon carbon, carbon carbon oh H double lung oxygen To hide regions, double bond oxygen To hide regions. Alternatively, we can have the trans. So now CH three carbon double bond carbon hydrogen, hydrogen, carbon, carbon double bond oxygen, carbon, carbon double bond, oxygen, oxygen, hydrogen. So just recap we can't have the cIS and trans eczema because of the presence of this double bond that imposes restricted rotation.

So we were just looking are bonding. Orbital's in a few different structures. So I've drawn the mall upon screen to save us time. And so we're gonna be running through what orbital's we are working with filling with electrons as well as the hybridization off those orbital's. So, with our first example, where we've got CEO Seo too. So we have a carbon I'll present. So that means that we have a double bond attached to our central carbon. So therefore we know that to be sp two hybridized while we then have bonding angles of about 120 degrees in our plane, a structure, another orbital's that we are using all the two s two into p two, where we fill up our to peel, but with two electrons. So next we can look at our square based pyramid structure with our central bro. Mean so we are filling up to the forest to and then we have five electrons in our four p orbital where the hybridization is SP three d two. Where are bonding angles are about 90 degrees between those equatorial florins. So two more examples here we've got a linear structure with a noble gas in the center, so we essentially have three equatorial lone pairs. Where are hybridization is s P three d was trying to use some of those de orbital's at this point where we are filling up to our five s 25 p six orbits also were completely felling that five p orbital. Lastly, we've got our eye a dean molecule where we have three iodine atoms involved, and again with feeling up to all five us to five p six. And that is because we have a negative charge on the I A dean. And so we again have SP three d hybridization.


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