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All right, guys, we're doing problem 80 of chapter 9 in chemistry, central science.
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So if we assume the energy level diagrams for homonuclear diatomic ion shown a figure, 9 .43 can be applied to heteronuclear diatomic molecules and ions.
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We predict the bond or magnetic behavior of these molecules.
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So let's go here.
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So we're going to start with co plus.
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So we first have to look at the electron configuration of carbon, which is helium, 2s2, 2p2, and oxygen 2s2, 2p4.
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Now, we're going to have one of our, since we have a positively charged ion, one of our atoms is going to lose an electron.
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In terms of like what we draw on our diagram, it doesn't particularly matter.
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So i chose oxygen is going to lose that atom.
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So now we have our non -bonding, our electrons in our non -bonding orbitals.
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So we're going to put that into our bonding orbitals.
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So we fill up, so we have five electrons in total.
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So we have two in our sigma bond, two in our sigma orbital, two in our pi orbital.
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Oh, and one in our other pi orbital.
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So then, so you can see that we have one unpaired, electron.
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So that means that we are going to be paramagnetic.
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Now we're going to look at our bond order.
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So if we look at our bond order, we have filled up electrons in one, in our sigma orbital, and also in one of our pi orbitals.
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And we are half filled in the other pi orbital.
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So our bond order is going to be somewhere between two and three.
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So let's approximate it and say 2 .5.
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Now, let's look at no.
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So we have our nitrogen with three electrons, and we have our oxygen, and we have our oxygen with four electrons into p orbital.
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And then we have another electron because we have no minus.
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So now we, as i just gave that to my nitrogen.
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So now we take our electrons in our non -bonding orbital.
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We're going to put them in our bonding orbital.
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So we're going to have a fill up our sigma, both our pi.
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Orbitals and then put an electron each of our antipi orbitals.
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So it looks like that, it looks like the bond.
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Well, first of all, our substance is going to be paramagnetic because we have unpaired electrons in our antipy orbital.
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And our substance is going as and our bond order, we have filled out the orbitals for single, double and triple bonds.
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We have two electrons in our anti -pie orbital.
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So these two electrons are going to weaken our bonding.
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So you can approximately you can think of each bond, each electron anti -pi orbital as dropping our bond order down by 0 .5...