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Topic: bond order (Read 16195 times)
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ixi
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bond order
«
on:
May 06, 2008, 10:57:48 PM »
Could someone please explain to me how this works?
I know the formula is 1/2 (total bonding electrons - total antibonding electrons) but I'm not sure how to use this formula.
I found the configuration for O2 to be
(1 sigma g)^2 (1 sigma u *)^2 (2 sigma g)^2 (2 sigma u *)^2 (3 sigma g)^2 (1 pi u)^2 (1 pi u)^2 (1 pi g*) (1 pi g*)
How would I find the bond order for O2?
Thanks
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tamim83
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Re: bond order
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Reply #1 on:
May 08, 2008, 01:53:14 PM »
This may help, only consider the valence electrons (i.e.-ignore the MOs made from the 1s orbitals-they are core electrons) So there are 8 bonding electrons and 4 antibonding electrons. Now plug it into the formula and you'll have your answer.
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ixi
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Re: bond order
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Reply #2 on:
May 09, 2008, 08:01:37 PM »
thanks
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bond order