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The maximum $'S'$ character is in

$\begin{array}{1 1}(a)\;BF_4^-&(b)\;XeO_3\\(c)\;NH_4^+&(d)\;\text{equal}\end{array}$

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$BF_4^-,XeO_3$ and $NH_4^+$ all are with $Sp^3$ hybridisation .
Hence $'S'$ character is same in all the molecules.
Hence (d) is the correct answer.
answered Mar 5, 2014 by sreemathi.v
 

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