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The molecular shapes of $SF_4,CF_4$ and $XeF_4$ are

$\begin{array}{1 1}(a)\;\text{the same with 2,0 and 1 lone pairs of electrons on the central atoms,respectively}\\ (b)\;\text{the same with 1,1 and 1 lone pair of electrons on the central atoms,respectively} \\(c)\;\text{different with 0,1 and 2 lone pair of electrons on the central atoms,respectively}\\(d)\;\text{different with 1,0 and 2 lone pair of electrons on the central atoms,respectively}\end{array} $

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$SF_4$ (hybridisation =$sp^3d$,geometry=trigonal bipyramidal with one equatorial position occupied by 1 lone pair),$CF_4$(hybridisation =$sp^3$,geometry=tetrahedral,no lone pair),$XeF_4$(hybridisation =$sp^3d^2$,geometry=square planar,two lone pairs.
Hence (d) is the correct option.
answered Mar 26, 2014 by sreemathi.v
 

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