Draw Lewis structures and MO diagrams for CN+ , CN, and
CN- . Ac...

Question

Draw Lewis structures and MO diagrams for CN+ , CN, and 
CN- . According to the Lewis model, which species is most stable?

Accepted Answer

welcome back everyone in this example. We need to determine the most stable species. Based on our idea, bromide Catalan idea bro. My neutral molecule and our I demand a bromide annan based on our lewis model and based on our molecular orbital theory. So we want to begin with our Lewis model And we're gonna start by drawing the Lewis structure out for each of our ions and molecules. So beginning with our own money bro my caravan we want to calculate total valence electrons before drawing Lewis structures. So we would recall the iodine and bromine on a periodic tables. Both located in group seven a correspond to seven valence electrons. And we also want to recognize that because we have this plus one Catalan charge that means we will lose one electron. So we have minus one electron from our some of all of our valence electrons. And that lease is a total of 13 valence electrons. So that is what we will use to draw lewis structure. We have drawn next to bromine and we are going to start by using our 13 valence electrons to draw in our valence electrons on each of our atoms. So we have 123456 and seven. And then we have 89, 10, 11, 12 and 13 valence electrons. Now we want to go ahead and make our covalin bond connection between eating and grooming because we recognize that these are both nonmetal atoms. So they will share these electrons in a bond here and we want to recall that our formula given from the prompt is a plus one net charge for our caddy on here. And so that means that one of these atoms should have a formal charge of plus one. Recall that our formula to calculate formal charge is that it's equal to our valence electrons for our selected atom subtracted from our non bonding electrons subtracted from our bonding electrons. And so we would recognize that because our aydin adam has 123456 valence electrons drawn in our structure here, it's going to have a formal charge of plus one. Whereas our bro mean we recognized has seven atoms around itself because 1234567. So it's happy with its seven valence electrons. So it has a formal charge of zero. And so we would have a net charge Of Plus one for the entire eye on here. And we should also recall that because iodine is less electro negative than our bromine atom. Based on us recalling our electro negativity trend which increases from left to the upward right of our periodic table. I'm fine being less electro negative than bromine can definitely handle having less electrons than its seven valence electrons. Which is its preference to have. So even though we can draw our positive plus charge on our bromine atom, it would be more favorable to draw our lewis structure with the positive Catalan charge on our idea, less electro negative atom. So moving on to our next lewis structure which is our idea in modern bromide neutral molecule. We're going to calculate total valence electrons by again recalling that iodine and bromine both on our periodic table are located in Group seven A corresponding to seven valence electrons each. We have no net we have a net neutral charge rather so we're going to just total these valence electrons up giving us a total of 14 valence electrons. So we have aydin next to bromine filling in our valence electrons. We have 123456789, 10, 11, 12, 13 and 14. So that leaves us with making our covalin bond between each of our atoms here and that will complete the structure because we would recognize that we have again a net neutral charge given in our formula from the prompt. And because each of our atoms have seven valence electrons, they both have formal charges of zero giving us a net neutral charge for the molecule. So this is the most stable favorable lewis structure here. Moving on to our last given ion for the iodine bromide and ion we want to calculate total valence electrons. So we would recall again, iodine and bromine on our periodic table both in group seven a correspond to seven valence electrons each. But because we recognize that we have that minus one and ion charge, we are going to recall that that means we gain one electron. And so we would add one electron to this. Some of our valence electrons which would give us a total of 15 valence electrons for our Lewis structure. So again, riding next to browning, filling in our valence electrons. We have 123456789, 10, 11, 12, 13, 14. And then We have a 15th valence electron where we would first make our covalin bond to our bromine atom and then realizing that we can remove one of our paired electrons and move it to our bromine atoms so that it now has a total of eight valence electrons around itself to form another covalin bond with our idea adam. And so we would recognize that because bromine now has one more than its valence electron preference of seven electrons. It has a total of 12345678 electrons around itself. That would give us a leftover formal charge of plus one on our or sorry minus one on our bro. Mean adam here because again recall that negative charge means we have extra electrons and in this case we have that one extra electron because we have just a minus one charge. And so this would be capable for bromine to handle because we recall it is more electro negative based on our electro negativity trend on our periodic tables. So it can handle having this negative charge way better than I do I never can because again it is more electro negative. So these are our three lewis structures and we want to determine which is the most stable based on Lewis models. And so we would recall that because we have a minimum charge, we can say minimum net charge. We would therefore have a molecule. Or rather yes, sorry, this is a molecule because it does not have a charge. So we would have therefore a low energy molecule. And so this is going to be the most stable out of the three given species of iodine bromide based on lewis theory. So, for our first part of our answer, we would confirm that Aydin mono bromide is the most stable based on our Lewis theory. So moving on to our next part of our solution here, we need to base off the molecular orbital theory what is the most stable? So we are going to scroll down for more room because we need to draw our molecular orbital diagrams beginning with in the same order with our Dean martin bromide. Carry on, we want to recall that for our molecular orbital diagram, we are only including valence electrons which we recall our outer most electrons. So comparing our two atoms, we would see that idea is lower on the periodic table, being in period five across period five on our periodic tables. And so following Hunt's rule, we are going to begin our molecular orbital diagram by filling in the lowest energy level first for the outermost valence electrons. And that would correspond to our sigma five s bonding molecular orbital which we recall only consist of one orbital where we will fill in our first two electrons. And again let's make note of the fact that above we determined we have a total of 13 valence electrons for our caddy on here. So filling in our first two electrons with opposite spins. We then want to move up in energy to our sigma anti bonding five S molecular orbital which again is only one orbital where we fill in our next two electrons with opposite spins moving up higher in energy. We recall we have our sigma five p bonding molecular orbital which consists of one orbital where we have our next two opposite spin electrons moving up higher in energy. We have our pi five p bonding molecular orbital and correction not pi five P. But actually we have our and sorry overthinking that. So yes pi five P bonding molecular orbital is higher up in energy next however, recall that this consists of a total of two orbital's where we would recall our poly exclusion principle. So we fill in our electrons as follows. We have six electrons filled in so far. So we then have seven eight and then 10. We're sorry, 7, 8, 9. And then we have 10. That leaves us with three more electrons to fill in. So recall that higher up in energy. We have our pie anti bonding 5p molecular orbital which also consists of two orbital's total. So let's make some more room here. So filling in our last three electrons we have according to our poly exclusion principle 11, 12 and then 13. And then even though we are done filling in our valence electrons here recall that our highest energy level orbital is also going to be up next the sigma anti bonding five P molecular orbital which is just one orbital. Now let's keep going and draw out our molecular orbital diagram for our next molecule which is our idea bromide neutral molecule, recall that above. We stated that that has a total of 14 valence electrons. So beginning with our lowest energy level orbital which again according to Dean's position on the periodic table being in period five, we would recall that that is going to be again our sigma five S molecular orbital with just one single orbital filling in our first two electrons moving up higher in energy as before. We have our sigma anti bonding five S molecular orbital which consists of one orbital filling in our opposite spin next to electrons than going up higher in energy. We have our sigma five p bonding molecular orbital which only consists of two electrons moving up in energy. We have our pi five p bonding molecular orbital which consists of a total of two orbital's where we fill in our next four electrons. So according to the Pauli exclusion principle 123 and four. Moving up higher in energy, we have our pie anti bonding 5P molecular orbital which consists of a total of two orbital's we can fill in our last four electrons here. So according to the Pauli exclusion principle we have 123 and four. And to make sure we have enough electrons 2468, 10, 12, 14 of our valence electrons filled in. And just so it's clear our highest energy level molecular orbital is going to be again our sigma anti bonding five P molecular orbital, which again we leave empty in this case. So moving on to our last molecular orbital diagram for aydin, mono, bromide and ion we have, as we stated above in our calculations, a total of 15 valence electrons. So drawing out our molecular orbital diagram, we begin with the lowest orbital energy to fill in first, which is, we recall again our sigma five s molecular orbital. Again that is going to contain our first two valence electrons of opposite spins. Moving up higher in energy. We have our sigma anti bonding five S molecular orbital where we fill in our next two electrons of opposite spins. Then after that we have our sigma five p bonding molecular orbital which is just one orbital where we fill in our next two electrons of opposite spins. Higher up in energy, we have our pi five p bonding molecular orbital which consists of two orbital's where we fill in our next four electrons of opposite spin. So 123 and four. Moving up higher in energy. We have our anti bonding pi five p molecular orbital which consists of again two electrons. We fill in our next four electrons according to Pauli exclusion principle. So 123 and four. And higher up in energy, we have our sigma anti bonding five p molecular orbital where we fill in our 15th electron. It can go in any spin direction. I'm just gonna go with an upward spin. So these three are our molecular orbital diagrams and our last step is to calculate bond order based on each of these diagrams. So beginning with our calculation of bond order for our and actually we'll use it as a subscript. So we'll say the bond order for our idea. Mono bromide cat ion. We're going to recall that our formula for bond order is equal to one half which is multiplied by the difference between our bonding electrons subtracted from our anti bonding electrons. And so following this formula, we have one half multiplied by our bonding electrons which above we can see is all the molecular orbital's without the asterix. So that's 2468 of our bonding electrons filled in for the Catalan here. So that's eight subtracted from our anti bonding electrons, the molecular orbital with the aspect. So that would be of our total anti bonding molecular orbital electrons filled in. So that's eight minus five. We have eight minus five times one half which should give us a result of three halfs which we understand is equal to a decimal value of 1.5. Moving on to our next bond order calculation for iodine bromide neutral molecule, we follow the formula one half multiplied by our bonding electrons. So going back to our diagram we see that we have a total of 2468 bonding electrons again. So feeling that in our formula we have eight subtracted from our anti bonding electrons located in our molecular orbital with the asterisk, we have 246 anti bonding electrons. So we feel that in our formula eight minus six times one half is going to give us a bond order of one. And sorry this is the bond order calculation for I demand a bromide. Moving on to our last calculation of bond order, we have our iodine bromide and ion bond order calculation, we follow the formula one half, multiplied by our bonding valence electrons according to our diagram. Those are the ones with the without the asterix. So that's 2468 total bonding valence electrons plugging that in our formula and subtracting that from our non bonding valence electrons according to our diagram. Those are the ones with the asterix, So that's 246 and seven and that's eight minus seven for our last calculation times one half. This is going to give us a bond order equal to a value of one half, which we understand is a decimal value of 10.5. And we are going to recognize that because we have the highest bond order associated with our idea, bromide, Cagayan. We would say that therefore this is the most stable according to our molecular orbital theory and so this would be our second part of our final answer. So our idea monochrome acadian is the highest bond order and so it's the most stable according to molecular orbital theory. So everything highlighted in yellow in our solution represents our final answers to complete this example. So I hope that everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.

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