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Use a simple bonding model to explain the observed bond distances in S2042 2- d(S-S) = 3.0 A d(S-0) = 1.4 A d(0-0) = 3.1 A...

Question

Use a simple bonding model to explain the observed bond distances in S2042 2- d(S-S) = 3.0 A d(S-0) = 1.4 A d(0-0) = 3.1 A

Use a simple bonding model to explain the observed bond distances in S2042 2- d(S-S) = 3.0 A d(S-0) = 1.4 A d(0-0) = 3.1 A



Answers

Use the observed bond lengths to answer each question. (a) Why is bond [1] longer than bond [2]? (b) Why are bonds [3] and [4] equal in length, and shorter than bond [2]?

The next course just said what had the relationship among on hoarders, barmaids and bond energy. Okay, so no high end upon harder the shot of the bond length and the stronger the bond all the higher, the one energy and vice versa. Okay, so he said, according to Monica Lahore bitter Cherry Ward Haider, Bt too all be heat. You bus be expected to exist. Okay, so well, uh, bury him. He's there off element on the periodic table. So considering dice in the second period, the heart emotional configuration all it's just maybe he will be. I want us to trust you. So the villain shell has just to, um, entrance. So I'm going to draw the money. Kalak Orbiter off. The two has obvious just off the burying of mixing together if we have Ah, um this looks like there's okay. So you call this be the 1st 1 who called his p e C column. So this has to and this has got to and this is the two s Mr Stewart's. So you have two electrons in each other to SRD itu's, and, uh, we can put this transcendently kilometers, so we don't want to 34 And this is going to be a Sigma to us bonding. And this is Sigma Asta. Jewish untie bonded. So the bondholder here on harder this number of bonding minus number of anti bonnets and asked you was we have two electrons and the bonding minus two. I'm trying the antibody enterprise I choose. So that's zero Oh, undefined. So whereas when we look at me, he to both be two. Plus it means that we heard him over one electron, so we'll have those. So remember, let Trump on the taipan in Arbiter here for the Bondi on harder in this case will be two minus one to bite you and does there by five. So because they exist upon here, be each, of course can exist. Whereas B E cannot because there exists no bond. So how we expense the beach of prostitutes East and B E chu So three here to us can exist and beat you two cannot just

In this problem. We are asked to do two things. Were asked to compare and contrast bond under bond order, bond length and bond energy. And were asked to discuss a couple of brilliant molecules First, let's go ahead and define each of these three terms bond order is half the difference between the number of electrons in bonding orbital's and the number of electrons in anti bonding orbital's bond energy. It's a value that describes the amount of energy required to break a bond. And of course bond length represents the distance between two nuclei connected connected by a bond. Okay, so those are definitions right here bond order. We find a nice contrasting green here bob order can be considered a bond strength indicator, which means that a higher bond order means a stronger bond bond order is proportional to bond energy. And that should make sense to you. Higher bond energy. Higher bond order. So it means it's proportional and um a lot of the the bond length is reverse. Let me do this, we're gonna write this, the bond order is inversely proportional to bond length, longer bond lower bond energy. So longer bond means lower bond energy. I'm gonna abbreviate energy and lower bond order. And I'm just going to write order for this. So there I have compared to trusted those pretty thoroughly. I think next. Um we are asked to let me read what my question says again here, according to molecular orbital theory, would either of the species that we have listed right here? What either of these exist? And let's go ahead and make a molecular orbital diagram for each one of these. And I am not the world's greatest at these. So there we go. So brilliant to has these electrons and that means by electrons will go on like this, here's my sigma one s or two S and here's my sigma start to s so when I'm doing my bond order on this, it would equal two minus two over to equal zero. And with the bond order of zero that's not stable. It won't exist now for a bond order of here's my brilliance. And in order to have a plus one charge, we're going to lose one of these, which means that here will be what my molecular orbital's look like. Again, here's my two s bonding and my two s anti bonding, and now I'll have to minus one equals a bond order of one half. And yes, this can exist, We're done.

Right here. I have some ah, molecular orbiters already drawn. And what I'm gonna do is that I'm gonna use, for example, be to so I'm gonna use the being would be to neutral molecule. And I'm gonna use B two plants, and I'm gonna use B two minus. And then we are going to write them in order off decreased inborn energy and decreasing Ah, bond length. So let's first do the neutral one. So you know that Boren has, ah, three valence electrons. So if this is born to So this is one boron DC's another born so one to born to. Now I'm gonna place those electrons into the atomic orbital first. So 12 and three, one to and three Remember that your atomic all videos need to have exactly the same amount off electrons as you're our molecular overdose. A new always begin with the lowest energy. So I have one, 234 on my SS. So I have one to three four on my s is and then I have one, and to for might be so DC's one and to perfect. So what I'm gonna do is that I'm gonna change the cooler off beat doing here just to show you in the same diagram. Different colors. Okay, how be to B two plus on B two negative change. And what I'm gonna do is that I'm going to change the colors of this too, so we can distinguish. Okay, so for B to close, we're gonna use the green and for B to negative, we're gonna use the blue. Okay, Before I move on, I'm gonna find the born order for this. So the born order for B two equals. Remember, it's bonding in a drugs minus anti bonding electrons divided by two. Because we have to. Or but I was too Orbital's and so one. So I have one too. Three m four electrons on my bond in, or videos. I have two electrons on my and they wanted or were also four minus two is too divided by two D. C. S one. So they want order right here for B two east one. Now, if you look in here if you're looking here, this is telling me that there be two is losing one electron. So that means that these electoral right here, he's going to be lost now, once that electron is lost, remember, If you lose or gain electrons, you're born. Order is going to change. So let's find the bone order for this. So we're gonna find a bone or there, and we're gonna do exactly the same thing. Now we have one to we right, one to three electrons minus two. Anti bonding divided by two. So these is zero Boyne five from perfect. And now what we're gonna do is we're gonna do exactly the same for be too negative. And we're gonna use the blowing here now for being too negative is telling me that this molecule is gaining and no other one. So this is gonna be thes warning here on. We're going to do exactly the same. Born or there equals 12 345 five born in miners to untie bonding divided by two. So five minus two is 33 Divided by two is 1.5. Now that we have the born order Okay, we can make the determination and we can write them in order off decreasing bond length. Now, remember, the higher your born order the low where you boned link and the higher the bone order, the higher you belong energy. So if we're gonna put this in order off, decreasing Bonin's, we're gonna have be to minus. Then we're gonna have be to, and then we're gonna have be to Plus now, if we're gonna what did them in order off, decreasing. And actually DC's decreasing born energy decreasing born and energy. And now we're gonna do the decreasing bond needs. Okay, so right here we will have been the opposite. Be to laws be to know, troll, and then be too minus

So in this problem, we are estimating the bond length between two atoms. Um, say Adams A and Adam B. Um, and we want to calculate what the bond length is between these two atoms. Generally, the rule of thumb is we can first go to Table 10.2 and see if the table lists out the bond length for these particular atoms. However, if it doesn't, one way to estimate the bond length is to look at the bond of Adam A to Adam A and the Bond ling of Adam B to Adam B. Bond length. And we can estimate, um, use these values estimate. So we'll take one half of the A to a bond length plus half of the B two b bond ling, and this will give us and estimate of the A B bond length. Um, and we're gonna use this principle throughout this problem. Okay, so in our first problem, we want to look at the bond length between I cl um from table 10 point to the bond length of this particular pair is not listed. However, we do know that the bond length of an I I bond is 266 Pekka meters and the bond length of CL two cl is 199 PICO leaders Following this rule of thumb, we can then calculate the bond length of ice cl one half of 266 Pekka meters plus one half of 199 Pekka meters. Um and this gives us an average bond length of 232.5 centimeters. Yeah, and so that's what we can estimate this bond as in our next problem, we're looking at the bond length between O and C L. Similarly, we can find from table 10 point to that. The 00 Bond length is 145 Pekka meters and we have the CLC. The CL cl bond ling is 199 pedometers. We can follow this equation as well one half of the 00 bond length and then we average it with the cl cl bond link and we get 172 kilometers. Okay, moving on to the third part of this problem, we can follow the same general, um, trend an equation and we get we're looking at the CF bond. So from table 10.2, we have a carbon carbon bond is 100 and 54 Pekka meters and the flooring flooring bond is 143 Pekka meters. We can average these together and we'll get an average bond line of 148.5 Pekka meters. Lastly, we're looking here at a carbon, um, roaming bond, um, from table 10.2 and also from the previous problem. We know that the carbon carbon bond is 154 Pekka meters and a bro Ming roaming bond is 228 Pekka meters. Um, we can again estimate this bond length by averaging these two together to get a bond leg of 191 pedometers.


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