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  • During the detection of phosphorus in an organic compound the yellow colour is due to:Option: 1 <img alt="\mathrm{\left( NH _{4}\right)_{3} PO _{4} \

During the detection of phosphorus in an organic compound the yellow colour is due to:

Option: 1

\mathrm{\left( NH _{4}\right)_{3} PO _{4} \cdot 12 MoO _{3}}


Option: 2

\mathrm{F e_{4}\left[F e(C N)_{6}\right]}


Option: 3

\mathrm{NH _{4} I}


Option: 4

\mathrm{N a_{4}\left[ Fe ( CN )_{6}\right]}


Answers (1)

best_answer

As we have learnt,

Test for Phosphorus: The compound is heated with an oxidising agent (sodium  peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate.

Result: A yellow colouration or precipitate of Ammonium Phosphomolybdate indicates the presence of phosphorus.

\mathrm{Na}_{3} \mathrm{PO}_{4}+3 \mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{NaNO}_{3}

\\\mathrm{H}_{3} \mathrm{PO}_{4}+12\left(\mathrm{NH}_{4}\right)_{2} \mathrm{MoO}_{4}+21 \mathrm{HNO}_{3} \longrightarrow\: \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}\: +\: 21 \mathrm{NH}_{4} \mathrm{NO}_{3}\: +\:12\mathrm {H_2O}

Hence, the correct answer is Option (1)

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chirag

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