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The bond angle of $O-Xe-F$ in $XeOF_4$ is

$\begin{array}{1 1}(a)\;90^{\large\circ}&(b)\; 120^{\large\circ}\\(c)\;72^{\large\circ}&(d)\;90^{\large\circ},120^{\large\circ}\end{array}$

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1 Answer

$O-Xe-F$ bond angle is $90^{\large\circ}$
Hence (a) is the correct answer.
answered Mar 6, 2014 by sreemathi.v
 

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