(C) The reaction between excess $PCl_{5}$ and concentrated $H_{2}SO_{4}$ is as follows:$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3}

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(C) The reaction between excess $PCl_{5}$ and concentrated $H_{2}SO_{4}$ is as follows:$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3} + 2HCl$In this reaction,$PCl_{5}$ acts as a chlorinating agent and replaces the hydroxyl groups of $H_{2}SO_{4}$ with chlorine atoms to form sulphuryl chloride $(SO_{2}Cl_{2})$.