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CBSE - XI Science - Chemistry - Chemical Bonding and Molecular Structure

could you please explain question no. 40 in the exercise of the chapter 'chemical bonding and the molecular structure'?

Asked by PRAJITHA P 12th August 2012, 8:39 PM
Answered by Expert


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(Nb-Na)

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

= 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 = Nb and the number of anti-bonding electrons = 4 = Na.

Bond order=1/2(Nb-Na)

 = 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of can be written as:


Nb = 8

Na = 3

Bond order of

= 2.5

Thus, the bond order of is 2.5.

The electronic configuration of ion will be:


Nb = 8

Na = 5

Bond order of =

= 1.5

Thus, the bond order of ion is 1.5.

Answered by Expert 14th August 2012, 12:36 PM

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