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A translucent white waxy solid (A) on heating in an inert atmosphere is converted inot its allotropic form (B). Allotrope (A) onreaction with very dilute aqueous KOH liberates a highly poisonous gas (C) having rotten fish smell. With excess of chlorine (C) forms (D) which hydrolyses to compound (E). Identify (A) to (E).

Answer» <html><body><p></p>Solution :(i) Since a <a href="https://interviewquestions.tuteehub.com/tag/translucent-1426181" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSLUCENT">TRANSLUCENT</a> white waxy <a href="https://interviewquestions.tuteehub.com/tag/solid-1216587" style="font-weight:bold;" target="_blank" title="Click to know more about SOLID">SOLID</a> (A), on heating in an inert atmosphere, is converted into its allotrope (B), therefore, (A) is white or yellow phosphorus while (B) is red phosphorus. <br/> `underset("White phosphorus")(P_(4)(s))overset("Heat, inert gas")tounderset("Red phosphorus")(P_(4)(s))` <br/> (ii) 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 `(PH_(3))` <br/> `underset("White phosphine(A)")(P_(4)(s))+3KOH(aq)+3H_(2)O(l)overset("Heat")tounderset("Pot. hypophosphite")(3KH_(2)PO_(2)(aq))+underset("Phosphine (C)")(PH_(3)(g))` <br/> (iii) Since phosphine (C) reacts with <a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a> of <a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a> to form a compound (D) which upon hydrolysis gives compound (E), therefore, (D) must be phsophorus pentachloride `(PCl_(5))` and (E) must be phosphoric acid `(H_(3)PO_(4))`. <br/> `underset("Phosphine (C)")(PH_(2)(g))+underset(("Excess"))(4Cl_(2)(g))tounderset("Phosphorus pentachloride (D)")(PCl_(5)(s))+3HCl(g)` <br/> `underset((D))(PCl_(5)(s))+4H_(2)O(l)tounderset("Phosphoric acid (E)")(H_(3)PO_(4)(aq))+5HCl(aq)` <br/> Thus, (A) is white phosphorus, (B) is red phosphorus, (C) is phosphine `(PH_(3))`. (D) is phosphorus pentachloirde `(PCl_(5))` and (E) is phosphoric acid `(H_(3)PO_(4))`</body></html>


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