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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!