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4.40     What is meant by the term bond order? Calculate the bond order of :

                N_{2}

                O_{2},

                O_{2}^{+}

                O_{2}^{-}

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Bond order (B.O.) is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals of a molecule.

Bond\ Order = \frac{1}{2}(N_{b}-N_{a})

Where N_{b}\ and N_{a} are the number of electrons occupying bonding orbitals and the number occupying the antibonding orbitals respectively.

 So, bond order for different molecules are:

N_{2} : The electronic configuration is (\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_{x})^2(\pi2p_{y})^2(\sigma2p_{z})^2

Where, the number of bonding electrons N_{b} =10  and number of antibonding electrons, N_{a} =4

So, Bond order of nitrogen molecule = \frac{1}{2}(10-4) = 3

 

O_{2} : The electronic configuration is (\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^1\equiv \pi^*2p_{y}^1)

Where, the number of bonding electrons N_{b} =10  and number of antibonding electrons, N_{a} =6

So, Bond order of nitrogen molecule = \frac{1}{2}(10-6) = 2

 

O_{2}^{+} : The electronic configuration is (\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^1)

Where, the number of bonding electrons N_{b} =8  and number of antibonding electrons, N_{a} =3

So, Bond order of O_{2}^{+} molecule = \frac{1}{2}(8-3) = 2.5

O_{2}^{-} : The electronic configuration is (\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_{z})^2(\pi2p_{x}^2\equiv \pi2p_{y}^2)(\pi^*2p_{x}^2 \equiv \pi^*2p_{y}^1 )

Where, the number of bonding electrons N_{b} =8  and number of antibonding electrons, N_{a} =5

So, Bond order of O_{2}^{-} molecule = \frac{1}{2}(8-5) = 1.5

                                                

Posted by

Divya Prakash Singh

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