Two oxides of nitrogen, NO and NO_2, react to | vedclass

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(B) $NO + NO_2 \xrightarrow{253 \ K} N_2O_3 \ (X)$$N_2O_3 + H_2O \longrightarrow 2HNO_2 \ (Y)$The anion of $Y$ $(HNO_2)$ is $NO_2^-$.In $NO_2^-$, the

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triangular planar

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(B) $NO + NO_2 \xrightarrow{253 \ K} N_2O_3 \ (X)$$N_2O_3 + H_2O \longrightarrow 2HNO_2 \ (Y)$The anion of $Y$ $(HNO_2)$ is $NO_2^-$.In $NO_2^-$, the nitrogen atom is $sp^2$ hybridized with one lone pair, resulting in a bent or angular shape. However, among the given options, the geometry of the electron domain is triangular planar. Given the options provided, the intended answer is triangular planar.

Two oxides of nitrogen, $NO$ and $NO_2$, react together at $253 \ K$ to form a compound of nitrogen $X$. $X$ reacts with water to yield another compound of nitrogen $Y$. The shape of the anion of $Y$ is:

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