MO Theory: Bond Order Example 2

Jules Bruno
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using Ml theory and bond order determine the number of bonds connecting the nitrogen atoms within the end to two minus ion. Alright, so remember we said that if you have a single bond, it's one. If it's double bond it's too here though, we don't have an actual visualization of the bond between the two nitrogen. So we're gonna have to use our M. O. Diagram since we're dealing with nitrogen. Remember that would be what sigma to us. Sigma star to us, pi two P sigma to pete pi star to pee and then finally sigma star to pee here, nitrogen group five days what has five valence electrons and there are two nitrogen. So that's five times two which is 10. And then we have an additional two electrons so add to two, this is 12. So all we gotta do now is fill in those electrons 12, aN:aN:000NaN 3456789, 10, 11. And then 12. And then remember here bond order which we're gonna abbreviate his B. O. Equals half of your bonding electrons. So you're bonding electrons would be too 468. So you have eight bonding electrons minus your anti bonding electrons, the ones with the star. So too four. So we do 8 -4, which is four times a half gives me too. So the bond order between the two nitrogen would be too and if we wanted, if I wanted to show you what this would look like it looks like this. This is what the ion actually looks like. And we can clearly see that there's a double bond between the nitrogen. So this visualization is definitely reinforced by the fact that we found a bond order of two. So that means there's a double bond between the two nitrogen, right? So we'd say that determine the number of bonds connecting the two nitrogen would say that they're connected by a double bond.