During detection of $P$ , we get a canary yellow precipitate on the addition of ammonium molybdate. The number of atom(s) in $s p^{3}$ hybridisation from the $15^{t h}$ group is

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Solution

To check the presence of phosphorus in an organic compound , the organic compound is first treated with sodium peroxide, such that all the phosphorus gets converted to sodium phosphate.
The aqueous extract is then heated with concentrated nitric acid and ammonium molybdate. A canary yellow precipitate of ammonium phosphomolybdate is obtained. The formation of this yellow precipitate confirms the presence of phosphorus.

$N a_{3} P O_{4} + 3 H N O_{3} ⟶ H_{3} P O_{4} + 3 N a N O_{3}$
$H_{3} P O_{4} + 12 (N H_{4})_{2} M o O_{4} + 21 H N O_{3} ⟶ (N H_{4})_{3} P O_{4} .12 M o O_{3} y e l l o w p p t + 21 N H_{4} N O_{3} + 12 H_{2} O$

The yellow precipitate formed here is
$(N H_{4})_{3} P O_{4} .12 M o O_{3}$ which has 3 $N$ and 1 $P$ atom in $s p^{3}$ hybrid state.

Thus the number of atoms in $s p^{3}$ hybridisation from the $15^{t h}$ group is 4.

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