Problem 50 Hard Difficulty

Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) $\mathrm{B}_{2}$ or $\mathrm{B}_{2}+$
(b) $\mathrm{F}_{2}$ or $\mathrm{F}_{2}+$
(c) $\mathrm{O}_{2}$ or $\mathrm{O}_{2}^{2+}$
(d) $\mathrm{C}_{2}+$ or $\mathrm{C}_{2}-$

Answer

a) More is the bond order, more will be the bond strength. Therefore, $\mathrm{B}_{2}$ has the stronger bond than $\mathrm{B}_{2}^{+}$ .
b) More is the bond order, more will be the bond strength. Therefore, $\mathrm{F}_{2}^{+}$ has the stronger bond than $\mathrm{F}_{2}$ .
c) More is the bond order, more will be the bond strength. Therefore, $\mathrm{O}_{2}^{2+}$ has the stronger bond than $\mathrm{O}_{2}$ .
d) More is the bond order, more will be the bond strength. Therefore, $C_{2}^{-}$ has the stronger bond than $C_{2}^{+}$ .

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Chemistry 101

Chemistry

Chapter 8

Advanced Theories of Covalent Bonding

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Video Transcript

So to predict the bon strings, we can look as a pounding order off a compound. The higher number off the bonding order, the stronger the pound is. And for the B two electron configuration, we can get it from textbook table 8.3, which will be Sigma Tau s two. And how'd bonding Sigma Tau s two on pie to pee? Why? Pie to P Z two does for B two on the's compound has abounding order off two plus two minus two over too. Because the 1st 2 comes from the electrons from Sigma Tau s pounding. The second to come from the two electrons from pi pounding on the two being subjected to come from the number of electrons in the anti pounding signal. Tewes divide by two, which should be one. However, if we take one, electrons from this compound forming be true plus the bonding order. Well, change into to steal from the sigma to ass pounding. However, in the pipe bounding there will only be one electrons left. So we have one minus two divided by two, which equals to 0.5. So in this case, the stronger bounding will happen among wheezing. Be too okay for F too. We can predict that its electron configuration is sigma to s two. Untie bonding Sigma to us two sigma to P x two on pie bonding Toupee. Why too easy four on dhe, four electrons in the pie to a P Y onto busy anti bonding. So the bounding order Welby two electrons from Sigma Tau US plus two electrons from Sigma Tau P X plus four electrons from pie to be Y two pz minors to elections from on high bonding Sigma Tau US minus four elections from and abounding pie to be white and two pz and divided by two, which would be one. However, if one additional electron is being removed from f, too forming after, plus the bonding order. So here for the pie and high bonding, one electoral well would be removed from this orbital leaving turning for into three. So the bony order for the F two plus well be tool plus two class four minus two. However, here we minus three because there will be only three elections left in the pie. Um, anti bonding divide by two. You close to 1.5 on 1.5 is hires and five. So we know that after class pounding well be stronger than f to bonding. Okay, so for the end, to all true, it has electron configuration off Sigma Tool s two Sigma and high bonding to us Two sigma to a p x two pie to P Y. Hi Coop. Easy four and then four more electrons in the anti bonding to P Y Oh, okay. Anti bonding Hi to appease e four two Oops, two. So the bond the order well be two plus two plus four minus two minus to divide by two You close to two. Okay. For the 022 plus two electrons is removed so there will be no longer pie and high bonding in the electron configuration, so is upon the order. Instead, while turning to two plus two plus floor things, there's no electron change. We things abounding orbital's. However, only one set off. Two electrons will be removed because, well, there are only two electrons remaining in the anti bonding Sigma Tau s orbital's divide. By two. The number end up being three. So all 22 plus well be stronger, have stronger bonds and old too, because the bond the order is higher. Okay, for question D, we have C to a compound on for zero charge to see to We have configuration off Sigma to s two suit and high bonding Sigma Tau US two hi to P y pie to appease e four. So for this seat who plus where one election is being subjected, we have only three elections left. You see pie bonding on for C two minus. We're one additional. Ah, electron is being added through the compound. We have five electrons in the Oh, sorry, My apology. We still have for electrons in the pie bonding because of maximum occupancy off electrons in pie. It's for, however, for the upcoming always hold level, which is Sigma to Pee X. We'll have one more electrons occupying it. Okay, so the bonding order off. See tool plus well be three plus two minus two. Divide by two, which equals to 1.5 on. For the bonding order off. See to minors. We have two plus It's war plus one minus two, divided by two, which has a final number off 2.5. So in comparison, Seattle miners have much higher bond. The order, then see to plus Soc to minors will have stronger bounding than see to pass.

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