Determine the electron configurations for CN+, CN, and
CN-. (b) ...

Question

Determine the electron configurations for CN+, CN, and 
CN-. (b) Which species, if any, has unpaired electrons?

Accepted Answer

hey everyone in this example, we're given the species below. We need to determine which have unpaid electrons based on their electron configurations. So beginning with choice A were given a cat ion of cobalt oxide with a plus one charge. We should recall that to calculate our electron configuration. We need to start out with finding the total valence electrons for this ion. And so beginning with our carbon atom we recognize on the periodic table. Carbon is located in group four A. So we would take this one carbon atom and multiply it by the four valence electrons. Because we recall that group numbers correspond to the number of valence electrons for our atom Added to this. We would recognize for oxygen that it's located in group six a. So we would take this one carbon atom and multiply it by six valence electrons. Then moving on in our configuration or calculation for our electron configuration, we would see that this plus charge here corresponds to an an acadian charge, meaning that we will lose one electron according to that charge. And so this will give us our total valence electrons equal to We would have six plus four which would give us 10 but then losing that one electron from Arkady in charge would leave us with nine electrons total. And so what we're going to do for our configuration is draw out a molecular orbital configuration. And so to begin our molecular orbital configuration, we're just going to include our nine valence electrons. And because we recognize that carbon and oxygen are in the second period of our periodic table. We're going to start out with the two sigma or the two s sigma bonding molecular orbital. So we would fill in for two electrons here to move up in our configuration. And next we're going to go to the anti bonding to us sigma to us anti bonding molecular orbital where we would also fill in for two electrons because we want them to cancel out our two electrons in the bonding to us sigma two S molecular orbital. So moving up in our configuration, we're then going to get to our pie two P bonding molecular orbital where we're going to fill in a total of four electrons and then right now we can count that we filled in 468 electrons. We need to fill in nine total. So we're going to move up in our configuration and go to our sigma two p bonding molecular orbital where we're only going to fill in just one electron. And so based on this one electron only being filled in here, we can say that therefore we have one um paired electron. And so this would be a great answer choice to consider for this example. So moving onto choice B we're going to look at a neutral atom of cobalt oxide and because this is a neutral atom We can say that this is going to have 10 electrons total or 10 valence electrons total because we wouldn't have lost that one as we did in example, a where we had that plus one caddy on charge. So now we can go ahead and just go straight into our molecular orbital configuration. Beginning with our bonding sigma two s molecular orbital where we're going to fill in for two electrons to cancel those out. We're going to fill in our sigma anti bonding to s molecular orbital with two electrons. Then we're going to go to our pie to p bonding molecular orbital where we will fill in for four electrons. And right now we can count a total of 10 electrons. We need we're sorry, a total of eight electrons because we have 468, we need 10. So we're going to go up to our sigma two p bonding molecular orbital where we will fill in those last two electrons to get to 10 electrons. And so based on this configuration, we can say therefore we have zero um paired electrons And so this would not be a good answer choice. So we can rule out choice b Now we have just troisi to consider. So we're gonna do that below here where we're given the cobalt oxide and ion. So first beginning with our total valence electrons because this is an anti in charge and we recall that an ions mean we would gain that charge of electrons, we can say we have 11, we'll say that we have an an eye on And so we have 11 electrons total. And so to write out our configuration, we're going to begin with the sigma two s bonding molecular orbital filling in with two electrons moving up. We would have the sigma asterix two s anti bonding molecular orbital filled in with two electrons to cancel the ones out in the bonding molecular orbital. Moving on up, we would have the pie to p bonding molecular orbital filled in with four electrons. We would have then the sigma two p bonding molecular orbital filled in with two electrons. And now we need 11 electrons right now we have 10 electrons. So we just need one more filled in. So we would move up and we would have the and I'm just going to make more space actually. Mhm. So we would move up in our configuration and we would have the pie anti bonding to p molecular orbital where we would only fill in one electron and so based on this configuration, we can say therefore we have we'll use red here to keep things consistent. one unpaid electron here, which was in our sigma or pi two P rather anti bonding molecular orbital over here. And so choice C would be also a good answer choice to complete this example. And so what we can confirm is that according to the given answer choices choice D which states that options A and C. Are the ions given that have unpaid electrons based on their configuration as we've proved below on the right hand side. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

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