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JEE Advance - Chemistry (2013 - Paper 2 Offline - No. 18)

The reaction of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T.
The reaction of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T.
The reaction of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T.
R, S and T respectively, are
SO2Cl2, PCl5 and H3PO4
SO2Cl2, PCl3 and H3PO3
SOCl2, PCl3 and H3PO2
SOCl2, PCl5 and H3PO4

Explanation

The reactions involved are

$$S{O_2} + C{l_2}\buildrel {Charcoal} \over \longrightarrow \mathop {S{O_2}C{l_2}}\limits_{(R)} $$

$$\mathop {10S{O_2}C{l_2}}\limits_{(R)} + {P_4} \to \mathop {4PC{l_5}}\limits_{(S)} + 10S{O_2}$$

$$\mathop {PC{l_5}}\limits_{(S)} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{(T)} + 5HCl$$

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