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What is Meant by the Term Bond Order? Calculate the Bond Order Of: N2, O2, O+2+And O−2 - Chemistry

What is meant by the term bond order? Calculate the bond order of: N2, O2, `O_2^+`and `O_2^-`

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Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If Na is equal to the number of electrons in an anti-bonding orbital, then Nb is equal to the number of electrons in a bonding orbital.

Bond order =  `1/2 (N_b-N_a)`

If Nb > Na, then the molecule is said be stable. However, if Nb ≤ Na, then the molecule is considered to be unstable.

Bond order of N2 can be calculated from its electronic configuration as:


Number of bonding electrons, Nb = 10

Number of anti-bonding electrons, Na = 4

Bond order of nitrogen molecule  = `1/2(10 - 4)`

 = 3

There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:


Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nand the number of anti-bonding electrons = 4 = Na.

Bond order ` = 1/2(N_b - N_a)`

= 1/2 (8-4)

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of `O_2^+` can be written as:


Nb = 8

Na = 3

Bond order of `O_2^+`  = 1/2 (8 -3)

= 2.5

Thus, the bond order of `O_2^+` is  2.5.

The electronic configuration of `O_2^-` ion will be:


Nb = 8

Na = 5

Bond order of `O_2^-` = 1/2 (8-5)

= 1.5

Thus, the bond order of `O_2^-` ion is 1.5.

Concept: Bond Parameters - Bond Order
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NCERT Class 11 Chemistry Textbook
Chapter 4 Chemical Bonding and Molecular Structure
Q 40 | Page 131
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