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In the reaction $2 \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{4}^{+}+\mathrm{PCl}_{6}^{-}$, the change in hybridisation is from
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The correct answer is:
$s p^{3} d$ to $s p^{3}$ and $s p^{3} d^{2}$
$2 \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{4}^{+}+\mathrm{PCl}_{6}^{-}$
$\uparrow s p^{3} d \quad \uparrow s p^{3} \quad \uparrow s p^{3} d^{2}$
$\uparrow s p^{3} d \quad \uparrow s p^{3} \quad \uparrow s p^{3} d^{2}$
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