Chemical Bonding and Molecular Structure - Result Question 97
####99. Decreasing order of stability of $O_2, O_2^{-}, O_2^{+}$and $O_2^{2-}$ is :
[2015 RS]
(a) $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$
(b) $O_2^{2-}>O_2^{-}>O_2>O_2^{+}$
(c) $O_2>O_2^{+}>O_2^{2-}>O_2^{-}$
(d) $O_2^{-}>O_2^{2-}>O_2^{+}>O_2$
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Solution:
- (a) According to molecular orbital theory as bond order decreases stability of the molecule decreases
Bond order $=\frac{1}{2}(N_b-N_a)$
Bond order for $O_2^{+}=\frac{1}{2}(10-5)=2.5$
Bond order for $O_2=\frac{1}{2}(10-6)=2$
Bond order for $O_2^{-}=\frac{1}{2}(10-7)=1.5$
Bond order for $O_2^{2-}=\frac{1}{2}(10-8)=1.0$
hence the correct order is
$O_2^{+}>O_2>O_2^{-}>O_2^{2-}$
