An oxide of nitrogen is reddish brown and paramagnetic at room temperature but it decolourises and also…
- Pure $\mathrm{NO}_{2}$
- Pure $\mathrm{N}_{2} \mathrm{O}_{4}$
- equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}$ and $\mathrm{NO}_{2}$
- $\mathrm{N}_{2} \mathrm{O}_{5}$
Solution
i.e. can be prepared by heating heavy metal nitrate of $\mathrm{Pb}\left(\mathrm{NO}_{3}ight)_{2}$. It is a red brown gas but on cooling below $273 \mathrm{~K}$ converts to colourless liquid and retains as dimer. In $\mathrm{NO}_{2}$ molecules $17$ valence $\mathrm{e}^{-}$are there ($12$ that of $\mathrm{O}$ & $5$ that of $\mathrm{N}$ ) i.e. one $\mathrm{e}^{-}$unpaired results to paramagnetic nature. So this is an odd $\mathrm{e}^{-}$molecule on being dimer with even $\mathrm{e}^{-}$becomes stable & loses paramagnetic characteristics.


Asked in: JEE-TOPICTESTS-CHEMISTRY
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