An organic compound containing phosphorus on | vedclass

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Question

(B) $1$. When an organic compound containing phosphorus is fused with sodium peroxide $(Na_2O_2)$,the phosphorus is oxidized to phosphate ions,forming

Vedclass · Accepted Answer

$Na_3PO_4, (NH_4)_3PO_4 \cdot 12MoO_3$

Vedclass · Answer

(B) $1$. When an organic compound containing phosphorus is fused with sodium peroxide $(Na_2O_2)$,the phosphorus is oxidized to phosphate ions,forming sodium phosphate $(Na_3PO_4)$ as compound '$X$'.$2$. When '$X$' $(Na_3PO_4)$ is boiled with concentrated nitric acid $(HNO_3)$ and then treated with ammonium molybdate reagent,it forms a yellow precipitate of ammonium phosphomolybdate,which is $(NH_4)_3PO_4 \cdot 12MoO_3$ (or $(NH_4)_3[P(Mo_3O_{10})_4]$),identified as '$Y$'.$3$. Therefore,'$X$' is $Na_3PO_4$ and '$Y$' is $(NH_4)_3PO_4 \cdot 12MoO_3$.

List-$I$ (Test/Reagents/Observation)	List-$II$ (Species detected)
$(a)$ Lassaigne's Test	$(i)$ Carbon
$(b)$ $Cu(II)$ oxide	$(ii)$ Sulphur
$(c)$ Silver nitrate	$(iii)$ $N, S, P,$ and halogen
$(d)$ The sodium fusion extract gives black precipitate with acetic acid and lead acetate	$(iv)$ Halogen specifically

An organic compound containing phosphorus on oxidation with $Na_2O_2$ gives a compound '$X$'. This '$X$' when boiled with $HNO_3$ followed by treatment with ammonium molybdate reagent gives a yellow precipitate '$Y$'. '$X$' and '$Y$' respectively are:

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