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'?
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
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.
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
Thus, the bond order of is 2.5.
The electronic configuration of ion will be:
Nb = 8
Na = 5
Bond order of =
Thus, the bond order of ion is 1.5.
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