Reaction of BCl$_3$ with NH$_4$Cl at 140 °C produces compound P. Further, P reacts with NaBH$_4$ to give a colorless liquid Q. The reaction of Q with H$_2$O at 100 °C produces compound R and a diatomic gas S. Among the following, the CORRECT statement is
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Borazine (B$_3$N$_3$H$_6$) is called inorganic benzene. It is formed by heating BCl$_3$ and NH$_4$Cl and reacts with NaBH$_4$ to give borane derivatives.
Step 1: Formation of compound P.
Boron trichloride reacts with ammonium chloride at high temperature (140°C) to form **borazine (B$_3$N$_3$H$_6$)**, also called inorganic benzene.
\[
3 \text{BCl}_3 + 3 \text{NH}_4\text{Cl} \rightarrow \text{B}_3\text{N}_3\text{H}_6 + 9 \text{HCl}
\]
Step 2: Reaction with NaBH$_4$.
Borazine reacts with sodium borohydride to form a colorless liquid borane derivative (Q), typically **borazine hydride**. Step 3: Hydrolysis of Q.
On hydrolysis, borazine hydride produces **boric acid** and hydrogen gas.
\[
\text{Q} + \text{H}_2\text{O} \rightarrow \text{R (boric acid)} + \text{H}_2
\]
Step 4: Conclusion.
Hence, compound P is B$_3$N$_3$H$_6$ (borazine).
