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The hybridisation of $'Al'$ in $AlCl_4^-$ is

$\begin{array}{1 1}(a)\;sp^2&(b)\;sp\\(c)\;sp^3&(d)\;\text{None of the above}\end{array}$

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Among four bonds between $Al$ and chlorine atoms,three bonds are covalent and one bond is dative bond,central '$Al$' undergoes $sp^3$ hybridisation.
Hence (c) is the correct answer.
answered Mar 4, 2014 by sreemathi.v

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