Bond Order

Asked by Ishaan Kumar | 18th Feb, 2011, 08:10: PM

Expert Answer:

Dear Student
 
bond order = ½ (bonding electrons − antibonding electrons)
 
BO of N2 = 1/2(8-2)= 3

O2: KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)1(p*2py)1
 
Bond Order = 1/2(8-4)= 2


O2+: KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)2 (p*2py)
 
Bond Order = 1/2(8-3) = 2.5

O2-: KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)2 (p*2py)1
 
 Bond Order = 1/2 (8-5) =1.5

 
We hope that clarifies your query.
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Topperlearning

Answered by  | 18th Feb, 2011, 09:07: PM

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