When PCl5 reacts with sulphuric acid, sulphuryl chloride (SO2Cl2) is formed as the final product. This shows that sulphuric acid:

Question

When PCl5 reacts with sulphuric acid, sulphuryl chloride (SO2Cl2) is formed as the final product. This shows that sulphuric acid: (A) has two hydroxyl groups in its structure (B) is a derivative of sulphur dioxide (C) is a dibasic acid (D) has greater affinity for water

Tardigrade Admin · Accepted Answer

oversetO mathop overset|| mathop undersetO mathop underset|| mathopPCl5+HO-S-OH xrightarrow Cl- undersetO mathop underset|| mathop oversetO mathop overset|| mathopS -Cl+POCl3+H2O PCl5 attacks -OH group and replace it by -Cl group. Hence, reaction of PCl5 with H2SO4 shows the presence of 2 -OH groups in H2SO4 .

Q. When $PC l_{5}$ reacts with sulphuric acid, sulphuryl chloride $(S O_{2} C l_{2})$ is formed as the final product. This shows that sulphuric acid:

has two hydroxyl groups in its structure

is a derivative of sulphur dioxide

is a dibasic acid

has greater affinity for water

$O ∣∣ PC l_{5} + H O - S - O H ∣∣ O$ $Cl - O ∣∣ S ∣∣ O - Cl + POC l_{3} + H_{2} O$ $PC l_{5}$ attacks $- O H$ group and replace it by $- Cl$ group. Hence, reaction of $PC l_{5}$ with $H_{2} S O_{4}$ shows the presence of 2 $- O H$ groups in $H_{2} S O_{4}$ .

Q. When PCl5​ reacts with sulphuric acid, sulphuryl chloride (SO2​Cl2​) is formed as the final product. This shows that sulphuric acid: