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The hybridisation of $'Xe'$ atom in $XeF_4$ molecule is

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

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Square planar $(XeF_4)$ with $sp^3d^2$ hybridisation.
Hence (d) is the correct answer.
answered Mar 7, 2014 by sreemathi.v
 

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