MO Theory: Bond Order Example 1

Jules Bruno
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determine the bond order of the N. O minus. I'll So here remember we're going to set up our molecular orbital diagram and will be based on the less electro negative element nitrogen is less electro negative than oxygen. So we're gonna use the molecular orbital diagram for nitrogen to set that up. We're gonna have here sigma to us, Sigma star to us. Then we're going to have pi to peep And then we're gonna have here σ two p. Then we're gonna have here pi star to pee and then finally we're gonna have up here sigma star to pee. Now if we take a look here nitrogen in Group five A. So it has five valence electrons, oxygen is in Group six A. So it has six valence electrons and then minus one means we've gained an additional electrons. So that's gonna be another electron on top. So this comes out to 12 valence electrons. So we start filling this in. So it be one too 56, 7, 8, 9, 10, 11 and 12. So we fill this in. And remember that the bond order formula. So bond order we're gonna bring it to be oh equals half of your bonding electrons minus your anti bonding electrons. So here we say half of so you're bonding electrons are 2468. So we have eight minus your anti bonding ones are the ones with the stars. So that's going to be two four anti bonding. So that's 8 -4 which is 44 times a half is equal to two. So the bond order here for this particular molecule will be too.