Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Spiffy Girl on April 20, 2008, 11:06:00 PM
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I don't understand how to get the bond order by looking at a Lewis Structure. I know that in the MO diagram it's 1/2(# of electrons in bonding MO - #of electrons in antibonding MO)
In my practice midterm a question asks "Use Lewis Structure to predict the bond order for a nitrogen-oxygen bond in the nitrite ion, NO2-."
I drew two resonance structures but can't figure out why 3/2 is the bond order. Is there a formula?
Here are the resonance structures.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Ffacultyfp.salisbury.edu%2Fdfrieck%2Fhtdocs%2F121%2FLewis%2FLEWIS.htg%2F4.gif&hash=68bd02ed02d9b7e21456b0b259b8253ef776ebf7)
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Two keys here to answering the question.
Your question, with one word slightly emphasized:
"Use Lewis Structure to predict the bond order for a nitrogen-oxygen bond in the nitrite ion, NO2-."
A, one of them.
Now, you have two diagrams. The electrons can switch place, as you diagrammed. When you focus on only one N-O bond, you will always have 1 bond. Now some of the time, you will have an additional bond, and at other times you will not.
Because you have two equally stable resonance structures, you can mathematically average them and say half the time it is between one N-O bond and the other half of the time between the other. Or, 1 + 1/2 = 3/2.
It might also be helpful to see if you look at it another common way of being represented:
O....N....O
That is, one full bond to each oxygen, and spread the resonance bond out. Now, only when they are of equal stability can you easily do this.
If you had three equal resonance forms, you would have a 1/3, etc.
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Same thing, but for a resonance I'm remembering it as:
# of shared electrons pairs / # of bonds = bond order
In the case above, on the left, there are two pairs on the left, one pair on the right, so 3 shared pairs. There are two bonds. So 3/2 is the result.