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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 81a

Determine the electron configurations for CN+, CN, and CN-. (a) Which species has the strongest C¬N bond?

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hey everyone in this example, we need to identify which of the given compounds. Either the carbon fluoride caddy on carbon florida as a neutral atom or the carbon florid. An ion has the strongest bond. So what we're going to recall is our concept of bond order and we can think of calculating bond order by taking one half, multiplied by our number of electrons in our bonding molecular orbital and then subtracted from our number of electrons in our anti bonding molecular orbital. Now we want to think of bond order because we would recall also that the highest bond order value corresponds to the strongest bond for a given molecule. So what we want to do is first look at the molecular orbital diagram for the carbon fluoride neutral atom. We want to begin by calculating its total valence electrons. So we're going to find carbon on the periodic table and see that it's in Group four A which corresponds to four valence electrons. So we would see that we have one carbon atom multiplied by four electrons. And then we're going to add to this our valence electrons for the flooring atom. So we would have here one flooring adam and we see on our periodic tables that florian is in Group seven a corresponding to seven valence electrons. So we would have seven plus four. And this gives us a total of 11 valence electrons. And so we also should recognize that on our periodic tables. Carbon and flooring are in the second period of the periodic table. So we're going to begin our molecular orbital diagram at the sigma to us energy level here which is our bonding molecular orbital and we're going to place our first two electrons here. Moving up in energy. We're going to go to the anti bonding sigma two s molecular orbital where we're going to fill in our next two electrons. Further up in energy, we're going to move up to the sigma two p bonding molecular orbital where we fill in another two electrons. So so far we have six electrons filled in of our 11. So we have five more electrons to fill in. So moving up in energy, we would go to the pi two p bonding molecular orbital where we have two orbital's and we're going to fill in 123 and four electrons here, pairing them up last and then we just have one more electron to fill in which will fill in the anti bonding pi two P molecular orbital. We still have to orbital's but we're only going to place one electron it and you can place it in any direction spin of your choice. So this is our molecular orbital diagram for our carbon fluoride neutral atom and now we can go ahead and calculate the bond order for the carbon fluoride molecule. And so what we're going to do is take one half and multiply this by our electrons in the bonding molecular orbital. So here we would count two. And remember the bonding molecular orbital czar the orbital's molecular orbital without asterix. So we have two electrons here, we have four. We have 68. So that gives us eight electrons in the bonding molecular orbital. And let's just make this neater and then minus our electrons. And the anti bonding molecular orbital which are molecular orbital with the asterix. So we would have two and then three. So that's eight minus three anti bonding electrons. So we would have eight minus three, that gives us five. And so that would leave us with one half times five. And we would have a bond order equal to five halfs for our carbon fluoride neutral atom. And so calculating now the bond order for the carbon fluoride an ion. We should recognize that because this is an anti in that means that we gain an extra electron. So for the carbon florida, an ion we gain one electron in the higher energy orbital, which is our pi two p anti bonding molecular orbital. So we'll represent that in purple here where we're gaining this extra electron. And so now we have actually 12 electrons. So this is going to change our bond order. So what we would calculate is one half we still have eight electrons in the bonding molecular orbital. But now we're going to subtract from our number of electrons in the anti bonding molecular orbital where we still have to here, but now we have another two here. So we would say we have four electrons in the anti bonding molecular orbital. And now we're going to have a bond order equal to eight minus four, which is four and then one half times four, Which would give us a value of two for the bond order. And so lastly we have the bond order to calculate for our sorry, carbon fluoride cat eye on where we would recall that a positive charge means we lose one electron for our carbon fluoride cat ion. And so this means we would go ahead and we're not going to consider the electron added for the an ion. So we're just going to remove this electron here which was originally there for our neutral atom of carbon fluoride. And so this means that now our bond order is going to be one half, we still have eight electrons in the bonding molecular orbital. However, now our electrons in the anti bonding molecular orbital will now only be two. So we would have two electrons since we removed one from the highest energy level orbital which is our pi two p anti bonding molecular orbital. And now we have a bond order of eight minus two which is six and then one half times two or one half times six rather is going to give us a value of three for our bond order. So we can say that our carbon fluoride cat ion has the highest bond order Equal to three. And so therefore the carbon fluoride Catalon has the strongest bond between the carbon and fluoride atom. And so this here strongest bond for carbon fluoride as a cat ion would be our final answer. So I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.