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In which of the following changes there is no change in hybridisation of the central atom?
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The correct answer is:
$\mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}$
The total number of electron pairs (bond pairs + lone pairs) is the same for $\mathrm{NH}_3$ and $\mathrm{NH}_4^{+}$. Thus, their hybridization is also the same that is $\mathrm{sp}^3$.
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