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Hybridisation and geometry of $XeO_4$ is

$\begin{array}{1 1}(a)\;\text{Tetrahedral}-sp^3\\(b)\;\text{Square planar}-dsp^2\\(c)\;\text{Pyramidal}-sp^3\\(d)\;\text{Pyramidal}-sp^2\end{array}$

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1 Answer

According to VSEPR theory the hybridisation is $sp^3$ and geometry is tetrahedral.
Hence (a) is the correct answer.
answered Mar 7, 2014 by sreemathi.v
 

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