Carbon monoxide is produced by incomplete combustion of
fossil f...

Question

Carbon monoxide is produced by incomplete combustion of 
fossil fuels. 
(c) What is the bond order of CO? Does this match the 
bond order predicted by the electron-dot structure?

Accepted Answer

Hello everyone in this field. We're dealing with the molecule nitric acid with a formula of N. O. So the question is being asked that if we calculate the bond order using the mo diagram format versus just using the lewis dot structure, is this bond order going to be the same? So first we'll just go ahead and drop the mo diagram and they will calculate the bond order for that and then join up the lewis dot structure and predicting the bond order of that as well and seeing if they do match up or if they're different. So having the M. O diagram then of course we have increasing energy from bottom to top here. We're dealing with nitrogen was whenever Adams auctioned as the other and in the middle will be our molecule which is just an Oh so of course we're dealing with our s orbital's and p orbital's in this case because we're over to the right of our periodic table. So then we have our to us orbital's central will have our top end bottom. So the bottom is going to be our sigma orbital and top will be our antibiotic sigma star. Then we'll just have this interactions going on here. Alright now for our pure metals on the move this over to the right a little bit. Alright so over to the right we have our p overall so 12 and three again, same thing to the left then we have 12 and three. So these need to be in the same energy level. So that's just like that. Two people. All right, so then we have our again bonding and anti bonding. So bonding we have starting with sigma and then we have our pie orbital's simply from the top of our anti bonding. We have to for our pie star and one orbital for our sigma star. So, I'll just go ahead and extend this here, giving us a little bit more space here. Alright, again, we have of course our interactions represent that with a dotted line. All right, so, imagine it's in group five A. So we have five valence electrons. So we'll add to the left of our M O diagram here. So, we have 1234 and five over to the right, we're talking about oxygen, oxygen is in group 68. So, I have six valence electrons. So that'll be 12345 and six. So, here, because we're dealing with the molecule we have the total number of electrons. So, again, we have five from our nitrogen and six from our auction five plus six is 11. To go ahead and add in 11 electrons. So we have 123456789, 10 and 11. Alright, so first we'll go ahead and calculate for our bond order. I'll just now move all of this over to the left. Alright, so bond order. This equation is going to be equal to half times the sum of all the electrons in our bonding orbital's minus electrons in our anti bonding. All right, so looking at the mo diagram here, the number of electrons in the bonding orbital. So bonding orbital or any without the star. So just these and these. So for our bonding we'll have 246 and eight. So eight electrons there, we put it into the formula the electrons in our anti bonding orbital czar anyone with the star. So this one here and this one here, which is just two and three. Alright, so eight minus three is of course 55 divided by two is equal to 2.5. So, the bonding orbital or bond order using the molecular orbital diagram is equal to 2.5. And take a look at the low start structure. Let's go ahead and draw the lewis dot structure Actually we're doing color blue now. So of course we need to go ahead and first going ahead to calculate the total number of valence electrons. So, even though we are determined that over to the left, when the mo diagram, we'll do it again just for protocol's sake. So for nitrogen is in Group 58. So that provides five electrons here auctions in Group 68. So provides six bands electrons taking some of those two. That will equal to 11 electrons now for nitrogen and auction we don't have a central atom because we only are dealing with two atoms present in this molecule. So auction usually likes to have two electrons. So we'll do one here and because we have used total four again, we'll add the two electrons to auction because auction likes to have two bonds as well as to lowest um or two lone pairs. So the remaining three electrons will be a lone pair on the nitrogen and just one singular extra electrons on our nitrogen. So then of course we can see that this is going to have a lewis structure or a resonance structure. So again we have our nitrogen, oxygen. So nitrogen then likes to have usually three bonds and one lone pair, so have a triple bond and then a long pair on the nitrogen. So remaining three electrons will go of course go to the right, so we have a long pair and then one extra. So we can see here that of our lower structures are resonant structures over to the left, we have a double bond double bond over to the right, we have a triple bond. So calculating the bond order, then we'll just do the average. So we'll do this in the green color like we did above the bond order now is well we have a double bond plus a number of two plus three for the triple bond or define that by two to get the average. So two plus three is equal to +55 divided by two is equal to 2.5. So you see your hair that using the lower structure method or lewis dot structure method, we get the bond order of 2.5 from our final answer is that yes, the bond orders are simplified as B. O is the same. So regardless of whichever method we use, we get the same exact bond order, and that's going to be my final answer for this problem.

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