Question

Explain $PC{l_3}$ on hydrolysis gives fumes of:
A.${H_3}P{O_3} + HCl$
B.${H_3}P{O_4} + HCl$
C.${H_3}P{O_2}$ and ${H_3}P{O_3}$
D.${H_3}P{O_2} + HCl$

Answer 1

In order to solve this question we will see the complete reaction. Firstly the lone pair of oxygen of water will attack in phosphorus as a result the chlorine ion will be removed from $PC{l_3}$ then similarly the second and like this all the chlorine ions will be removed and we will get the final answer.

Complete step-by-step answer: For solving this question we will see the full reaction step by step so first of all on hydrolysis of $PC{l_3}$ the lone pair of the oxygen atom of water will attack on phosphorus then the bond between the phosphorus and chorine will be break and the hydroxyl ion will attach with the phosphorus and one mole of hydrochloric acid will be formed.
According to reaction:
$PC{l_3} + {H_2}O \to HO - PC{l_2} + HCl$
Now in second step the lone pair of water will attack on the main product of this reaction and again the chlorine ion will be separated from this:
According to the reaction:
$HO - PC{l_2} + {H_2}O \to {\left( {HO} \right)_2}P - Cl + HCl$
Now as in above two steps all the chlorine ions are removed so similarly this chlorine ion will also be removed from here and finally we will get all the three hydroxyl ions with phosphorus and again one molecule of hydrochloric acid will be formed.
According to reaction:
${\left( {HO} \right)_2}P - Cl + {H_2}O \to P{\left( {OH} \right)_3} + HCl$
Now by combining all the steps we will get the three mole of hydrochloric acid along with $P{\left( {OH} \right)_3}$

So the correct answer will be ${H_3}P{O_3} + HCl$ and the correct option will be A.

Note: While solving these types of questions let me tell you that the product is not formed simultaneously, water will attack in the reactant then the intermediate will be formed, then as the bond between oxygen and phosphorus strengths the bond between phosphorus and chlorine weakens.