Pawan Prajapati
Last Activity: 4 Years ago
Answer :
The bond order of a bond is half the difference between the number of bonding and antibonding electrons.
BO =½(B–A)
The C−C sigma bonds
Each C−Cσ bond is a localized bond. It has 2 bonding electrons and 0 nonbonding electrons.
σ BO = ½(B–A)=½(2–0)=1
The C−C pi bonds
Benzene has 6 molecular π orbitals.
Of these, three are bonding and three are anti-bonding. The six π electrons go into the three bonding orbitals.
π BO = ½(B–A)=½(6–0)=3
This is the π bond order for 6 C−C bonds.
For one C−Cπ bond, BO = 3/6=0.5.
For a single C−C bond in benzene, the total BO = σ+π=1+0.5=1.5.
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