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Topic: Bond order in Molecular Orbital Theory.  (Read 8978 times)

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Offline gggggggggg

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Bond order in Molecular Orbital Theory.
« on: January 09, 2010, 10:41:59 AM »
I have these three species

O2
Bond order: 2
Unpaired Electrons : 2

O2+
Bond order: 2.5
Unpaired Electrons: 1


O2-
Bond order : 1.5
Unpaired Electrons :1

Question 1 : Which out of these 3 species has the strongest bond?

My answer: Am i right to assume that the higher the bond order the greater the stability and thus the stronger the bond?

Question 2 : What if all the bond orders were the same? How would i then determine the strongest bond?

Offline stewie griffin

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Re: Bond order in Molecular Orbital Theory.
« Reply #1 on: January 09, 2010, 10:53:04 AM »
Larger bond order does indeed mean stronger bond.
I'm not sure when you'd see a question with molecules all having the same bond order. I mean if the molecules did have the same bond order, they couldn't be a series analogous to what you've got here... there would have to be different atoms present amongst the molecules.

Offline gggggggggg

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Re: Bond order in Molecular Orbital Theory.
« Reply #2 on: January 09, 2010, 10:56:53 AM »
Thanks for the help m8. :)

Offline Ligander

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Re: Bond order in Molecular Orbital Theory.
« Reply #3 on: January 17, 2010, 07:52:02 PM »
Larger bond order does indeed mean stronger bond.
Can you give any example, please?

Offline stewie griffin

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Re: Bond order in Molecular Orbital Theory.
« Reply #4 on: January 17, 2010, 10:13:22 PM »
Ligander:
I'm confused about why you need an example.  ??? The original post is an example of larger bond order corresponding with stronger bond....

Offline Ligander

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Re: Bond order in Molecular Orbital Theory.
« Reply #5 on: January 18, 2010, 12:33:30 AM »
Ligander:
I'm confused about why you need an example.  ??? The original post is an example of larger bond order corresponding with stronger bond....
Sorry, I've read does not indeed :D

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