Chemical Bonding And Molecular Structure Ques 26
26. Which one of the following arrangements represents the increasing bond orders of the given species?
[1999]
(a) $\mathrm{NO}^{+}<\mathrm{NO}<\mathrm{NO}^{-}<\mathrm{O}_2^{-}$
(b) $\mathrm{O}_2^{-}<\mathrm{NO}^{-}<\mathrm{NO}<\mathrm{NO}^{+}$
(c) $\mathrm{NO}^{-}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{NO}^{+}$
(d) $\mathrm{NO}<\mathrm{NO}^{+}<\mathrm{O}_2^{-}<\mathrm{NO}^{-}$
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Answer:
Correct Answer: 26.(b)
Solution: (b) $\mathrm{NO}^{+}=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_x^2$ $\pi 2 p_y^2=\pi 2 p_z^2$
Bond order of $\mathrm{NO}^{+}=\frac{1}{2}\left(\mathrm{~N} _ {\mathrm{b}}-\mathrm{N}_ {\mathrm{a}}\right)$
$ =\frac{1}{2}(10-4)=\frac{1}{2} \times 6=3 $
Similarly, Bond order of $\mathrm{NO}=\frac{1}{2}(10-5)$ $ -\frac{1}{2}(5)=2.5 $
Bond order of $\mathrm{NO}^{-}=\frac{1}{2}(10-6)=\frac{1}{2}(4)=2$
Bond order of $\mathrm{O}_2^{-}=\frac{1}{2}(10-7)=\frac{1}{2}(3)=1.5$
By above calculation, we get
Decreasing bond order
$ \mathrm{NO}^{+}>\mathrm{NO}>\mathrm{NO}^{-}>\mathrm{O}_2^{-} $
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