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The hybridisation of Sn in $SnCl_6^{2-}$ is

$\begin{array}{1 1}(a)\;sp^3d^3&(b)\; sp^3d^2\\(c)\;dsp^2&(d)\;sp^3d\end{array}$

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$SnCl_6^{2-}$ is in octahedral shape with $sp^3d^2$ hybridisation.
Hence (b) is the correct answer.
answered Mar 5, 2014 by sreemathi.v
 

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