MHT CET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)
Concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with \(\mathrm{PCl}_{5}\) to produce
- A \(\mathrm{HClO}_{2}\)
- B \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\)
- C \(\mathrm{SOCl}_{2}\)
- D \(\mathrm{HClO}_{4}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\)
Step-by-step Solution
Detailed explanation
When excess of \(\mathrm{Pa}_{5}\) reacts with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\), it gives suphuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) as product
\(2 PCl _5+ H _2 SO _4\) (conc) \(\longrightarrow \underset{\substack{\text { (sulphurye) } \\ \text { chloride) }}}{ SO _2 Cl _2}\) \(+~2 POCl _3+2 HCl\)
\(2 PCl _5+ H _2 SO _4\) (conc) \(\longrightarrow \underset{\substack{\text { (sulphurye) } \\ \text { chloride) }}}{ SO _2 Cl _2}\) \(+~2 POCl _3+2 HCl\)
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