Problem

Which of the period 2 homonuclear diatomic molecu…

03:32

Problem 43 Easy Difficulty

Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) $\mathrm{H}$ and $\mathrm{H}_{2}$
(b) $\mathrm{N}$ and $\mathrm{N}_{2}$
(c) $\mathrm{O}$ and $\mathrm{O}_{2}$
(d) $\mathrm{C}$ and $\mathrm{C}_{2}$
(e) $\mathrm{B}$ and $\mathrm{B}_{2}$

Answer

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Chemistry 101

Chemistry

Chapter 8

Advanced Theories of Covalent Bonding

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Video Transcript

So here we're just gonna be looking out memos, theory. And we've got quite a few Die Atomics to get through so fast it will start off with boron. And now we know that boron has three valence electrons per atom. So for our electron counting for boron, we simply have three times to, which gives us six. However, four of those are in the to s minus four. Now we're filling our to Pete with two electrons. So here they post go in or sigma bonding orbital. And that is because the pie orbital's are of a higher energy. And so it is less energy costly not to pair the electrons rather than to put the electron Singley into the segment on GPI orbital. So is boron. We have compared in the Sigma orbital. So now if we move onto carbon, carbon has four electrons. So that is four times two. That gives us eight, and then we take with taken away for again because I'm not drawing out my to eths. And that gives us four electrons to fill. So again we have 1234 Now we are approaching the pie bonding Orbital's here on. We Singley fell rpai orbital's because they are degenerate because they are both derived from the same sorts of P orbital interaction. So here we have a completely occupied Sigma orbital at the bottom and then we have our Singley occupied high orbital's, both of which are bonding. So now if we want to move on to nitrogen, nitrogen is group five. Multiply that by two. That gives us 10 electrons. But remember for what electrons are in that two s orbital. So we're subtracting for which leaves us with six electrons to populate R two p orbital with so a fully occupied or sigma bonding. Then we singularly occupied both of our pie bonding orbital's and then we go back on doubly occupy so you can see him that nitrogen die. Nitrogen has a bond order off three and that is because we have occupied all of our bonding Orbital's. So now if we move onto a few more Die Atomics here, we've got to f to an e too so fastly to look oxygen. So remember that we have four like drugs in or two s orbital. So with oxygen are six Valent electrons multiplied by two gives us 12 and again we're taking away for as we have accounted for those in our two ass orbital. And so we are filling eight electrons into this diagram. 1234 five, 67 eight. So, as you can see here, we have started to populate or anti bonding orbital's. Ah, Now we denote on anti bonding orbital using the ass tricks. So the bond order off 02 is too. Whereas the bonding order of end to is three, so are bonding on this starting to decrease because we're populating those anti bonding orbital's. So now we can look at F two so f two. They have seven electrons each because they are in Group seven. There were halogen multiplied by two gives us 14 take away, for we're left with tan electrons to fill so we can go ahead and throw 10 electrons into our orbital's want to three, 456789 10. Now we're putting even more electrons into the anti bonding orbital. So are bonding. Order here is one. We're destabilizing the molecule by having electron density in those anti bonding orbital's. So then, if you were to look a neon to Neon is Group eight. So then we're multiplying that by to give us 16. And then again, you need to account for your two p electrons where we're taking away for. And then that leaves us with 12 electrons to fill where we have already filled up 10. So he simply fell or anti bonding sigma. And now you can see that our bond order is zero and the diatonic of neon is not viable. It probably won't exist, and that is because neon already has it full up tap so it doesn't need to bond in order to fulfill any sort of shall configuration that might want to, and so is very un reactive for that reason. So, as you can see, moving from left to right of the table, we increasingly populate our bonding and they're not anti bonding orbital's, and we see a stabilization as we fill our bonding and then a destabilization as we then start to populate or anti bonding

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