Arrange $H_{2} O, N H_{3}, C H_{4}, S F_{6}, B F_{3}, C O_{2}$ in order of increasing
bond angle.

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Solution

$S F_{6} < H_{2} O < N H_{3} < C H_{4} < {C O}_{2}$
$S h a p e \Rightarrow$ Octahedral bend pyramidal tertahdral linear
$B o n d A n g l e \Rightarrow$ $(90^{o}) ({104.5}^{o}) ({106.5}^{o}) (109.28) (180^{o})$

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