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QUESTION

# What is bond order? How we can determine it? Please basics. Thank you so much.

A conceptual approach is to simply count electrons in a bond and treat each valence electron as half a bond order.

This works for many cases, except for when the highest-energy electron is in an antibonding molecular orbital.

SIMPLE CASE

For example, the bond order of ##:"N"-="N":## fairly straightforward because it's a triple bond, and each bonding valence electron contributes half a bond order.

So:

##"BO"_"triple" = "BO"_sigma + 2"BO"_pi = 1/2 xx ("2 electrons") + 2(1/2 xx ("2 electrons")) = 3## for the bond order, as we should expect, since bond order tells you the "degree" of bonding.

MULTI-ATOM CASE

Or, in a more complicated example, like ##"NO"_3^(-)##, a conceptual approach is to count the number of electrons in the bond and see how many bonds it is distributed across.

So, in ##"NO"_3^(-)##, which has one double bond in its structure, has ##2## electrons in its ##pi## bond, distributed across three ##"N"-"O"## bonds.

That means its ##bb(pi)##-bond order is simply ##1/2*("2 pi electrons")/("3 N"-"O bonds") = color(blue)("0.333")##, making the bond order for each ##"N"-"O"## bond overall be:

##"BO" = "BO"_sigma + "BO"_pi = 1 + 0.333 = 1.333##.

Therefore, ##"NO"_3^(-)## on average actually has three "##bb("1.333")##" bonds overall (instead of one double bond and two single bonds), meaning it is one third of the way between a single bond and a double bond.

EXCEPTION EXAMPLE: O2

##"O"_2## actually has two singly-occupied ##pi^"*"## antibonding orbitals.

If we were to calculate its bond order, we would get ##2## normally, corresponding to the ##:stackrel(..)("O")=stackrel(..)"O":## Lewis structure.

But what if we wanted the bond order for ##"O"_2^(+)##? From the discussion above, we may expect ##1.5##, but it's NOT ##1.5##, even though ##"O"_2^(+)## has one less valence electron. What is it actually?

You may realize that we would have removed one electron from an ##pi^"*"## antibonding molecular orbital. That means we've removed half a bond order corresponding to antibonding character, which is the same as adding half a bond order corresponding to bonding character.

So, by removing an antibonding electron, we've done the equivalent of adding a bonding electron.

In other words, we've decreased a bond-weakening factor, thereby increasing the bonding ability of the molecule.

Therefore, the actual bond order of ##"O"_2^(+)## is ##bb(2.5)##, stronger than ##"O"_2##!