The bond order of a molecule is given by

Question

The bond order of a molecule is given by (A) the difference between the number of electrons in bonding and antibonding orbitals (B) total number of electrons in bonding and antibonding orbitals (C) twice the difference between the number of electrons in bonding and antibonding orbitals (D) half the difference between number of electrons in bonding and antibonding orbitals

Tardigrade Admin · Accepted Answer

Bond order ( B.O. )=(1/2)(Nb-Na)

Q. The bond order of a molecule is given by

the difference between the number of electrons in bonding and antibonding orbitals

total number of electrons in bonding and antibonding orbitals

twice the difference between the number of electrons in bonding and antibonding orbitals

half the difference between number of electrons in bonding and antibonding orbitals

Bond order $(B . O .) = \frac{1}{2} (N_{b} - N_{a})$

Q. The bond order of a molecule is given by

the difference between the number of electrons in bonding and antibonding orbitals

total number of electrons in bonding and antibonding orbitals

twice the difference between the number of electrons in bonding and antibonding orbitals

half the difference between number of electrons in bonding and antibonding orbitals

Bond order (B.O.)=21​(Nb​−Na​)

Bond order $(B . O .) = \frac{1}{2} (N_{b} - N_{a})$