A translucent white waxy solid (A) on heating in an inert atmosphere is converted to its allotropic form (B). Allotrope (A) on reaction with very dilute aqueous $KOH$ liberates a highly poisonous gas (C) having rotten fish smell. After reaction with excess of chlorine it forms (D) which after hydrolysis gives compound (E). Identify compounds (A) to (E).
Answer
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Hint- Phosphorus comes in two forms, red and white. Red phosphorus is innocuous yet the white structure is exceptionally toxic and responds viciously with oxygen, so it must be put away submerged or paraffin to stop it suddenly combusting.
Complete answer:
Translucent waxy solid is actually white phosphorus ${P_4}$. We can say so because-
${P_4}$ on heating results into red phosphorus in inert atmosphere
$ \Rightarrow {P_4}\left( s \right)\xrightarrow[{gas}]{{Heat}}{P_4}\left( s \right)$
Since allotrope (A) on boiling with dilute aqueous $KOH$ gives a highly poisonous gas (C) having rotten fish smell, therefore, it is confirmed that allotrope (A) is white phosphorus and the poisonous gas (C) is phosphine $\left( {P{H_3}} \right)$.
$ \Rightarrow {P_4}\left( s \right) + 3KOH\left( {aq} \right) + 3{H_2}O\left( l \right)\xrightarrow{{Heat}}3K{H_2}P{O_2}\left( {aq} \right) + P{H_3}\left( g \right)$
Since phosphine (C) reacts with excess of chlorine to form a compound (D) which after hydrolysis results into compound (E), therefore, (D) should be phosphorus pentachloride $\left( {PC{l_5}} \right)$ and (E) must be phosphoric acid $\left( {{H_3}P{O_4}} \right)$.
$
\Rightarrow P{H_2}\left( g \right) + 4C{l_2}\left( g \right) \to PC{l_5}\left( s \right) + 3HCl\left( g \right) \\
\\
\Rightarrow PC{l_5}\left( s \right) + 4{H_2}O\left( l \right) \to {H_3}P{O_4}\left( {aq} \right) + 5HCl\left( {aq} \right) \\
$
Thus, we have the following compounds-
A is white phosphorus
B is red phosphorus
C is phosphine $P{H_3}$
D is phosphorus pentachloride $PC{l_5}$
E is phosphoric acid ${H_3}P{O_4}$
Note: Phosphoric acid, otherwise called orthophosphoric acid or phosphoric(V) acid, is a feeble acid with the chemical formula ${H_3}P{O_4}$. It is typically experienced as a colourless syrup of 85% focus in water. The unadulterated compound is a colourless solid.
Complete answer:
Translucent waxy solid is actually white phosphorus ${P_4}$. We can say so because-
${P_4}$ on heating results into red phosphorus in inert atmosphere
$ \Rightarrow {P_4}\left( s \right)\xrightarrow[{gas}]{{Heat}}{P_4}\left( s \right)$
Since allotrope (A) on boiling with dilute aqueous $KOH$ gives a highly poisonous gas (C) having rotten fish smell, therefore, it is confirmed that allotrope (A) is white phosphorus and the poisonous gas (C) is phosphine $\left( {P{H_3}} \right)$.
$ \Rightarrow {P_4}\left( s \right) + 3KOH\left( {aq} \right) + 3{H_2}O\left( l \right)\xrightarrow{{Heat}}3K{H_2}P{O_2}\left( {aq} \right) + P{H_3}\left( g \right)$
Since phosphine (C) reacts with excess of chlorine to form a compound (D) which after hydrolysis results into compound (E), therefore, (D) should be phosphorus pentachloride $\left( {PC{l_5}} \right)$ and (E) must be phosphoric acid $\left( {{H_3}P{O_4}} \right)$.
$
\Rightarrow P{H_2}\left( g \right) + 4C{l_2}\left( g \right) \to PC{l_5}\left( s \right) + 3HCl\left( g \right) \\
\\
\Rightarrow PC{l_5}\left( s \right) + 4{H_2}O\left( l \right) \to {H_3}P{O_4}\left( {aq} \right) + 5HCl\left( {aq} \right) \\
$
Thus, we have the following compounds-
A is white phosphorus
B is red phosphorus
C is phosphine $P{H_3}$
D is phosphorus pentachloride $PC{l_5}$
E is phosphoric acid ${H_3}P{O_4}$
Note: Phosphoric acid, otherwise called orthophosphoric acid or phosphoric(V) acid, is a feeble acid with the chemical formula ${H_3}P{O_4}$. It is typically experienced as a colourless syrup of 85% focus in water. The unadulterated compound is a colourless solid.
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