MCQ
$\ce{(NH_4)_2Cr_2O_7}$ on heating gives :
- A$ N_2O$
- B$NH_3$
- C$NO_2$
- D$N_2$
✓
Answer
On heating, $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ gives, nitrogen gas.
$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}$
$($Orange$) ($Green$)$
But whenever, $\mathrm{NH}_{4} \mathrm{NO}_{2}$ and $\mathrm{NH}_{4} \mathrm{NO}_{3}$ heated, then nitrogen and nitrogen dioxide gas liberated.
$\mathrm{NH}_{4} \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{NH}_{4} \mathrm{NO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}+2 \mathrm{H}_{2} \mathrm{O}$
So, the correct option is $\mathrm{D}$
$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}$
$($Orange$) ($Green$)$
But whenever, $\mathrm{NH}_{4} \mathrm{NO}_{2}$ and $\mathrm{NH}_{4} \mathrm{NO}_{3}$ heated, then nitrogen and nitrogen dioxide gas liberated.
$\mathrm{NH}_{4} \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{NH}_{4} \mathrm{NO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}+2 \mathrm{H}_{2} \mathrm{O}$
So, the correct option is $\mathrm{D}$
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