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The hybridisation and shape of the $[Pt(Cl)_4]^{2-}$ is

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

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The hybridisation and shape of the $[Pt(Cl)_4]^{2-}$ is $dsp^2$ and Square planar.
Hence (d) is the correct answer.
answered Dec 6, 2013 by sreemathi.v
 

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