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The relationship between the dissociation energy of $N_2$ and $N_2^+$ is

$\begin{array}{1 1}(a)\;\text{dissociation energy of }N_2\text{=dissociation energy of }N_2^+\\(b)\;\text{dissociation energy of }N_2\text{ can either be lower or higher than the dissociation energy of }N_2^+\\(c)\;\text{dissociation energy of }N_2\text{> dissociation energy of }N_2^+\\(d)\;\text{dissociation energy of }N_2^+\text{> dissociation energy of }N_2\end{array} $

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Bond order of $N_2=3$,Bond order of $N_2^+=2\large\frac{1}{2}$
Higher the bond order,higher is the dissociation energy.
Hence dissociation energy of $N_2 > $dissociation energy of $N_2^+$
Hence (c) is the correct answer.
answered Mar 26, 2014 by sreemathi.v
 

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