Match the reaction of $Cl X_{2}$ in List I with the nitrogen- or chlorine-containing product in List II. List I: (P) cold dilute NaOH, (Q) hot concentrated NaOH, (R) excess $NH X_{3}$ , (S) excess $Cl X_{2}$ with $NH X_{3}$ . List II: (1) $NaClO X_{3}$ , (2) $N X_{2}$ , (3) $NCl X_{3}$ , (4) $NaOCl$ .

Answer: C

Cold dilute NaOH gives

NaOCl

(P-4); hot concentrated NaOH gives

NaClO X_{3}

(Q-1); excess

NH X_{3}

gives

N X_{2}

(R-2); excess

Cl X_{2}

gives

NCl X_{3}

(S-3).