Heating of phosphate salt to about $50-60^{\circ} \mathrm{C}$ with concentrated $\mathrm{HNO}_{3}$ and…

Heating of phosphate salt to about $50-60^{\circ} \mathrm{C}$ with concentrated $\mathrm{HNO}_{3}$ and ammonium molybdate produces
  1. yellow precipitates of $\left(\mathrm{NH}_{4}ight)_{3}\left[\mathrm{P}\left(\mathrm{Mo}_{3} \mathrm{O}_{10}ight)_{4}ight]$
  2. red precipitates of $\left(\mathrm{NH}_{4}ight)_{3} \mathrm{PO}_{4} \cdot 10 \mathrm{MoO}_{3}$
  3. gree precipitates of $\left(\mathrm{NH}_{4}ight)_{3} \mathrm{PO}_{4} \cdot 6 \mathrm{MoO}_{3}$
  4. blue precipitates of $\mathrm{MO}_{3}\left(\mathrm{PO}_{4}ight)_{2}$

Solution

When ammonium molybdate is treated with a solution containing phosphate a canary yellow crystalline precipitate of ammonium phosphomolybdate (ammonium dodecamolybdatophosphate), $\left(\mathrm{NH}_{4}ight)_{3}\left[\mathrm{P} \mathrm{Mo}_{12} \mathrm{O}_{40}ight]$ or $\left(\mathrm{NH}_{4}ight)_{3}\left[\mathrm{P}\left(\mathrm{Mo}_{3} \mathrm{O}_{10}ight)_{4}ight]$ is formed.
In the compound formed, the $\mathrm{Mo}_{3} \mathrm{O}_{10}$ group replaces each oxygen atom in phosphate.
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Asked in: JEE-TOPICTESTS-CHEMISTRY

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