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The $ONO$ bond angle is maximum in

$(a)\;NO_3^{-} \\ (b)\;NO_2^{-} \\(c)\;NO_2 \\(d)\;NO_2^{+} $
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$NO_3^{-}$ has $SP^2$ hybridisation and three resonating structures.
Hence $ONO$ bond angle $=120^{\circ}$
$NO_2^{+}$ has $SP$ hybridisation and N of it contains lone pair of electrons.
Hence shape is linear
with bond angle $=180^{\circ}$
$NO_2$ has one lone pair of electron whereas $NO_2$ has one lone pair of electrons. .
Hence in $NO_2^{-}$, the repulsion on the bond pairs are more and angle is less.
Hence d is the correct answer.
answered Apr 2, 2014 by meena.p

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