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H How to the pi electrons eldocroth the eysteren atom (e) Acrolein; 2 Furan,...

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H How to the pi electrons eldocroth the eysteren atom (e) Acrolein; 2 Furan,

H How to the pi electrons eldocroth the eysteren atom (e) Acrolein; 2 Furan,



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Which of the following molecular ions have electrons in $\pi$ antibonding orbitals? (a) $\mathrm{O}_{2}^{-} ;$ (b) $\mathrm{O}_{2}^{2-} ;$ (c) $\mathrm{N}_{2}^{2-} ;$ (d) $\mathrm{F}_{2}^{+}$ (e) $\mathrm{N}_{2}^{+} ;$ (f) $\mathrm{O}_{2}^{+} ;(\mathrm{g}) \mathrm{C}_{2}^{2+} ;$ (h) $\mathrm{Br}_{2}^{2+}$

We've got a problem that's telling us to draw molecular liberal diagrams for the following islands. We've got C two to minus into two minus 02 to minus nbr, two to minus. We want to determine which of them has electrons in the pie anti bombing orbital's. So we have two of those and we're going Teoh went to Green Arrow to those or Google's. Those orbital's are right here. So for each of these, we're gonna determine which has electrons and that work. So first things first, we got C two to minus that we need a fair harmony of what Trans we have. Carbon has four. We've got two of them. But we also add two extra electrons for the Tu minus and I on. So we've got eight plus two is 10. So we've got 10 to fill in. So we've got 12 in r two s, 34 and R two s to in our bonding and to our anti bonding. So we take out four and we've got six left to fill in on our our two peoples. So we've got one too. Three one to three. So now we can fill in our orbital's down here. We've got six to use. So correct. Gotta fill in each individual shell first and then pair of them. So we've got four 56 So in this case, we don't have anything in the two pi or the pie to pee Anti bombing Orbital's. So for C two to minus answer to our question is no too small, though. Yeah, I know. Okay, so now we've got our second example. I'm gonna go on ahead and leave the two s shelf full, because we always have that full for each of these examples. So we're race. All of our electrons want to use this diagram a couple times because it is a pain to draw every time. So you have copies to use of this. Feel free to do that. What caused the video after each example. So now we've got our second example into two minus. So nitrogen has five bales like tones. We have two of each year to night organs and then we've got plus too far to minus, and I on So 10 plus two is 12 to fill in what we've used for for our two s so minus for the exhaust with eight to fill in our peoples. So we thought one, 23 for by 67 eight. So now we can fill in our shells without one Sirio three for 567 eight. Now, for this one, we can see we just put two and our pies hooping. Anti bonding Orbital city Answer your question. Which of them has electrons in the high anti bond orbital's? We would say into two minus is a Yes. We do have electrons enough. So pause it here from seat up race, these electrons use this again Empty or two p shell. Now we can do our third example. We've got oxygen, this farm, so go back through. We've got oxygen. Oxygen has six in its Vaillant show and we have two of them and then we have two extra electrons. So 12 plus two is 14 to work with. We've used for in our s Orbital's over here. We got 10 left work with, so we're gonna put those in here 123 for five. So that's half. And that one too three four five scenario got tend to work with. So about one, 23456789 So now we see we've got four in our pie to be anti bonding Orbital's. So the answer to this question for two to minus is also yes. Now we've got our fourth example which we do have to change up each of these numbers because, bro mean does not it's Vaillant. Shell does not account for two p into us. It's Vaillant. Shell is on the fourth home. So we've got to change all of these twos these and or erase each of these electrons. Just skip this. If you don't watch me race all these, I promise there's nothing information just going to I could leave These electrons just need to change the number of Rs shell. So over here, we've got four s for us. 80 there. Sigma Forest Anti bonding signal for us. I got four p for he the draw a little correctly. So all of these twos just changed it. Four because we've gone down to the fourth grow instead of the second. Okay, so race these last few electrons arts and now we're ready to work with broom Almost almost ready. So now without bro mean to Su minus about seven valence electrons to bro means. And then we add to our our two months in I on about 14 plus 2 16 electrons to work with we've used for in our four s shells used for were loved. Well, it's OK, so 16 minus 4 12 So we'll fill in 12 about 123456 12345 and six that we filled up our s and r P shells that these in our sectionals too. 3456789 10 11 12 We used all 12 and we can see that we again have filled up our pie for p anti bonding orders. So rpai anti bonding orbital's we do have electrons will be are two to minus is what an easy way to look at This is we said no for our first apartment because we have Let's see, we had four times two's eight plus two is 10. So since we had 10 there, we weren't able to put any in our four p anti bonding levels. But from end to and on, we increase that amount of electrons, so we were able to say yes for nitrogen. We would also say yes for oxygen and grooming just because we ADM or electrons to them. So we had 10 and then we had nitrogen was five times to use 10 so that we had 12 and then for ox training. At 14 we had 16 for brewing. So since that was the case, we knew that we were going toe have electrons in those orbital's for nitrogen oxygen broke. So the answer to whether we have electrons in high anti bonding orbital is yes, for N two to minus 02 to minus and br to to.

This question wants you to draw to resident structures for the Form eight ion and then identify where the molecular orbital would be the pie molecular orbital in order to account for the D localized electrons suggest. In the two resonant structures Format has the Formula HCO 2-. So we have a total of 18 valence electrons When we use all 18. We end up with this as our lewis structure, but carbon does not have an octet, so one of the oxygen's needs to share. This could be one of the resonant structures, or we could have put a double bond here for a second resonant structure. Therefore, there must be a pie molecular orbital that spans both oxygen's and the central carbon in order to account for the D localized electrons.

This is the answer to Chapter 17. Problem number 31 fromthe Smith Organic chemistry textbook. This problem asks us how many pie electrons are contained in each molecule. Ah, and so, um, we need thio. Basically. Just count the pie bonds and remember that each pi bond eyes made of two electrons. And so, um, for a we have five pi bonds on so five pi bonds is gonna be 10 electrons for B. Ah, we have, um, three pie bonds. It's gonna be six electrons, but then we have an unparalled electron as well. So be is gonna be seven electrons, uh, for C. We have, um, 12345 pi bonds. So that's gonna be 10 electrons offer D. We have 357 pi bonds for a total of 14 pie electrons. Ah, and then any? We have six pie bonds for a total of 12 pie electrons. Okay, so that's it. Um, the only riel outlier here is is part B where we have that un paired electron that we have to account for. Otherwise, we can just town the number of double bonds and each of these and multiply by two on then for being. We re do the same thing and just add one for that un parent electron that we have on DSO. That's the answer to Chapter 17 problem number 31.

I know the question is the number of electrons in the cyclo built up by and I am So we are going to discuss how many die Bahn is present in the cycle. You can buy in, I'll in my own societal Right now. The structure is not this one. Yeah, This structure is this in this structure, how many five? Always present? And this structure? There is 2 5 this present, right? This is not goodbye. So we can see that protect also is the first he is the character of some. I hope you are working in concert. Okay, here's


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