MCQ
Pure $'N_2'$ obtain by
- A$N{H_4}N{O_2}\,\xrightarrow{\Delta }$
- B$N{H_4}N{O_3}\,\xrightarrow{\Delta }$
- C$(NH_4)_2Cr_2O_7\,\xrightarrow{\Delta }$
- ✓$NaN_3\,\xrightarrow{\Delta }$
$2 \mathrm{NaN}_{3} \stackrel{573 \mathrm{K}}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{N}_{2}$
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Assertion $(A):$ $\mathrm{H}_2 \mathrm{Te}$ is more acidic than $\mathrm{H}_2 \mathrm{~S}$.
Reason $(R):$ The bond dissociation enthalpy of $\mathrm{H}_2 \mathrm{Te}$ is less than that of $\mathrm{H}_2 \mathrm{~S}$.
In the context of the above statements, choose the correct answer from the following options:
$Benzene$ $\xrightarrow{HCHO+HCl}\,A$$\xrightarrow{AgCN}\,B$



$\mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_2, \mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{NH}_3, \mathrm{HCl}, \mathrm{CHCl}_3, \mathrm{BF}_3$