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# The change in hybridisation of the Al atom in the reaction of $AlCl_3+Cl^-\rightarrow AlCl_4^-$

$\begin{array}{1 1}(a)\;sp^2\;to\;sp^3\;hybridisation\\(b)\; sp^3\;to\;dsp^2\;hybridisation\\(c)\;dsp^2\;to\;sp^3\;hybridisation\\(d)\;\text{there is no change}\end{array}$

$AlCl_3+Cl^-\rightarrow AlCl_4^-$
$AlCl_3\Rightarrow sp^3$(Trigonal planar)
$AlCl_4^-\Rightarrow sp^3$(Tetrahydral)
Hence (d) is the correct answer.