Which of the following hybridisation is correct for transition state during hydrolysis of $S i C l_{4}$ ?

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${sp}^{3} d$

${sp}^{3}$

${sp}^{3} d^{2}$

${sp}^{2}$

answer is C.

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Detailed Solution

$s p^{3} d$ is correct because $S i C l_{4}$ has a vacant orbital that can easily accommodate electrons from water, and the intermediate product is $s p^{3}$ hybridized with trigonal bipyramidal shape.

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