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Chemistry Question on Molecular Orbital Theory

The correct order of bond dissociation energy among N2,O2,O2−N_2, O_2, O_2^- is shown in which of the following arrangements ?

A

N2>O2−>O2N_{2}> O_{2}^{-}>O_{2}

B

O2−>O2>N2O^{-}_{2}> O_{2}>N_{2}

C

N2>O2>O2−N_{2}> O_{2}>O_{2}^{-}

D

O2>O2−>N2O_{2}> O_{2}^{-}>N_{2}

Answer

N2>O2>O2−N_{2}> O_{2}>O_{2}^{-}

Explanation

Solution

Bond energy ∝\propto Bond order bondorder :-

N2=Nb=10,Na=4N_{2} =Nb=10, Na=4
B.O.=(N2)=10−42=3B.O.=\left(N_{2}\right)=\frac{10-4}{2}=3
O2=Nb=10,Na=6O_{2}=Nb=10, Na=6
B.O(o2)=10−62=2B.O_{\left(o_2\right)}=\frac{10-6}{2}=2
O2−=Nb=10,Na=7O_{2}^{-}=Nb=10, Na=7
B.O.(o2)=10−72=32=1.5B.O._{\left(o_2\right)}=\frac{10-7}{2}=\frac{3}{2}=1.5

Hence the order of B.O.

N2> O2> O2−N_{2}>\, O_{2} >\, O_{2}^{-}