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The correct order of bond angles of $CH_4, NH_3$ and $H_2O$ are

$\begin{array}{1 1}(a)\;CH_4 < NH_3 < H_2O&(b)\;CH_4 < H_2O < NH_3\\(c)\;CH_4 > NH_3 > H_2O&(d)\;CH_4 > NH_3 = H_2O\end{array}$

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$ CH_4 > NH_3 > H_2O$
$CH_4(109.50), NH_3(107.50), H_2O(105.50) $
According to VSEPR theory, more the number of lone pairs on a central atom , the greater is the contraction caused in the angle between the bondpairs.
answered Dec 9, 2013 by sreemathi.v

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