Calculation of bond order of N2
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hello friend...!!!
according to the question we should calculate the bond order of N2.
⇒ the total number of electron in N2 is 14 .
⇒ N2 = σ1s2 σ*1s2 σ2s2 σ*2s2 π2px² π2py² σ2pz¹
⇒ BOND ORDER = 1 / 2 ( BMO - ABMO )
here , BMO --------> bonding molecular orbit .
ABMO ----------------> anti - bonding molecular orbit .
therefore for N2.
bond order = 1/2 ( 10 - 4)
= 1/2 (6 )
= 3
therefore the bond order is 3
implies triple bond exist
it is dia - magnetic in nature
-------------------------------------------------------------
hope it hepls...!!
according to the question we should calculate the bond order of N2.
⇒ the total number of electron in N2 is 14 .
⇒ N2 = σ1s2 σ*1s2 σ2s2 σ*2s2 π2px² π2py² σ2pz¹
⇒ BOND ORDER = 1 / 2 ( BMO - ABMO )
here , BMO --------> bonding molecular orbit .
ABMO ----------------> anti - bonding molecular orbit .
therefore for N2.
bond order = 1/2 ( 10 - 4)
= 1/2 (6 )
= 3
therefore the bond order is 3
implies triple bond exist
it is dia - magnetic in nature
-------------------------------------------------------------
hope it hepls...!!
Znoor:
Thanks alot
your welcome
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