${\mathrm{NH}}_3+{\mathrm{O}}_2\underset{700-800°\mathrm{C}}{\overset{\mathrm{Pt}/\mathrm{Rh} \mathrm{gauge}}{\to }}\underset{\underset{\left[\mathrm{Z}\right]}{ \downarrow {\mathrm{O}}_2}}{[\mathrm{X}}{]}_{\mathrm{gas}}+\underset{\mathrm{neutral} \mathrm{and} {\mathrm{sp}}^3 \mathrm{hybridised}}{{\left[\mathrm{Y}\right]}_{\mathrm{gas}}}+\mathrm{Heat}$

Choose the correct option.

Overall reaction is:

$$\begin{gathered} 4{\mathrm{NH}}_3 + 5{\mathrm{O}}_2\to \underset{\left(\mathrm{X}\right)}{4\mathrm{NO}}\left(\mathrm{g}\right) +\underset{\mathrm{Y}}{6{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}} \\ \downarrow {\mathrm{O}}_2 \\ \underset{\left(\mathrm{Z}\right)}{{\mathrm{NO}}_2\uparrow } \end{gathered}$$

NO₂ has sp² hybridization and has a bent shape. Due to presence of lone pair of electrons, it has $\cong 134°$ bond angle. It is deep-orange color gas and is paramagnetic because of presence of an unpaired electron.