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Molecular Aocen General Chemistry CHEM II1AFexperimen V09167019chemican name/lormulaValcncn enctronshrbridizatlon (lor each central atom)Acetylene/C,H,i0 VSEPR skct...

Question

Molecular Aocen General Chemistry CHEM II1AFexperimen V09167019chemican name/lormulaValcncn enctronshrbridizatlon (lor each central atom)Acetylene/C,H,i0 VSEPR skctchLewris Structuresketchmolecuize geometry eachcentralatomi2lectron klometny loreach centra atom)chemical name/(ormulavalence electronthvbridization Mor &ach cenua Jompotassium chloride/Lewuis StructureVSEPR sketchve sketchmolecular geometry (for t each central atom)electron geometry (for each central atom)126

Molecular Aocen General Chemistry CHEM II1AF experimen V09167019 chemican name/lormula Valcncn enctrons hrbridizatlon (lor each central atom) Acetylene/C,H, i0 VSEPR skctch Lewris Structure sketch molecuize geometry eachcentralatomi 2lectron klometny loreach centra atom) chemical name/(ormula valence electront hvbridization Mor &ach cenua Jom potassium chloride/ Lewuis Structure VSEPR sketch ve sketch molecular geometry (for t each central atom) electron geometry (for each central atom) 126



Answers

Determine the molecular geometry about each interior atom and sketch each molecule.
\begin{equation}\begin{array}{l}{\text { a. } \mathrm{C}_{2} \mathrm{H}_{2} \text { (skeletal structure HCCH) }} \\ {\text { b. } \mathrm{C}_{2} \mathrm{H}_{4} \text { (skeletal structure } \mathrm{H}_{2} \mathrm{CCH}_{2} )} \\ {\text { c. } \mathrm{C}_{2} \mathrm{H}_{6} \text { (skeletal structure } \mathrm{H}_{3} \mathrm{CCH}_{3} )}\end{array}\end{equation}

So here we're just gonna be looking at electron geometries. So this is the arrangement of our electoral groups. So if lone pairs of electrons in other words electrons that are not bonded to the atoms are located on our molecule, then this will affect the molecular geometry. So here we're gonna be determining the electron are molecular geometry for a few different examples we have so we can start off withdrawing some of the mountain. It's my faster army of C H 30 h. So we have a structure that is toucher. He drill Andi, we use all off our valence electrons on carbon So we have no lone pairs. So we have a electron on molecular geometry of the Tetra hydro. The next example is an ISA, so we can have the electron geometry off a touch. Wahid RL However, it has the molecular German tree of a bank Confirmation on again for my last example, we have hydrogen peroxide. So again we have not touched a He'd roll electron geometry with four lack drawn groups around each oxygen atom. However, we have a bank confirmation in reality

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.

The first compound is a Salford with two chlorine. Therefore so for will be the central Adam have to. Chlorine is coming off of it and then it will have to compared Alright, trance and just are the rest of my impaired period of electrons. And the geometry of this one will be bent. Since there's on Lee, two things connecting so for the next example we have is P I three. So phosphorus will be the central Adam and then it will have are dying coming off of it in three places and then it will have one set of impaired electrons and the geometry of this one will be tribunal criminal. The reason for that is because it will have three substitue INTs in one, um pair, pair of electrons. The next example we have a seal to with an oxygen so the oxygen will be our central Adam. Since it's like to bonds, both of the CIA will be bonded to it. So just like our first example, it has to substitute wince and then two pairs of, um paired electrons. And so again, this geometry will be that a next example is in H two with a Corinne. So nitrogen will be our central Adam and then to the too hard Children will be connected to it as well. A so chlorine. And then I'll have one set of unparallel our trunks. And so, just like our second example, we have we had, it will be tribunal perimeter all because it had three Substitute Wint's in one hair of electrons that are not paired with anything. Our next example. We have a sigh, which will be arson. True, Adam. And then I'll have three chlorine za round it and then one bro Ming. So since it has four substitute Wint's, the geometry of this compound will be tetra. He'd roll. Then for last example. We have oxygen, nitrogen and chlorine, so nitrogen will be or central compound and they don't have a chlorine connected to it and then will be double bonded to an oxygen. Since an oxygen likes to have two bonds and arches, one likes to have three bonds, so this will give nitrogen one pair of impaired electrons and this compound is also spent. Find that out, get free all

This problem wants us to determine the molecular geometry around each central Adam and then sketch it. And so the first is end, too. Uh, any dye atomic? So any element with two of them is always going to be linear. And so the sketch will look like this with a triple bond between the nitrogen Sze and each of them having a lone pair. B is end to H two. Now this is going to be bent, and it will look like this. And with the double bond and then bent for the two ages and then each end still has a loan pair. So the ends always have to have a lone pair. And so that's why this central bond can't be a triple bond. And so then, if you look at each end, you can see it's bent compared to the other end and to the H. Now let her see is end to H four. This is going to be Trigon a ll Pura Middle, And how will show that is with N and each of them has a loan pair and then we have our ages and so each and is connected to one other end and two ages. And because of that loan pair, each of them is Trigon all pyramidal


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