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

Dec 3, 2016

A conceptual approach is to simply count electrons in a bond and treat each bonding 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 $: \text{N"-="N} :$ 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 ${\text{NO}}_{3}^{-}$, a conceptual approach is to count the number of electrons in the bond and see how many bonds it is distributed across. So, in ${\text{NO}}_{3}^{-}$, which has one double bond in its resonance structure, has $2$ electrons in its $\pi$ bond, distributed across three $\text{N"-"O}$ bonds.

That means its $\boldsymbol{\pi}$-bond order is simply $\frac{1}{2} \cdot \left(\text{2 pi electrons")/("3 N"-"O bonds") = color(blue)("0.333}\right)$, making the bond order for each $\text{N"-"O}$ bond overall be:

${\text{BO" = "BO"_sigma + "BO}}_{\pi} = 1 + 0.333 = 1.333$.

Therefore, ${\text{NO}}_{3}^{-}$ on average actually has three "$\boldsymbol{\text{1.333}}$" 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

${\text{O}}_{2}$ actually has two singly-occupied ${\pi}^{\text{*}}$ 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 ${\text{O}}_{2}^{+}$? From the discussion above, we may expect $1.5$, but it's NOT $1.5$, even though ${\text{O}}_{2}^{+}$ has one less valence electron. What is it actually?

You may realize that we would have removed one electron from an ${\pi}^{\text{*}}$ 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 ${\text{O}}_{2}^{+}$ is $\boldsymbol{2.5}$, stronger than ${\text{O}}_{2}$!