In the following reactions, products(A) and (B) respectively are $$\begin{aligned} & \underset{(\text { Hot and } \mathrm{Conc})}{\mathrm{NaOH}}+\mathrm{Cl}_{2} \longrightarrow(\mathrm{~A})+\text { side products } \\ & \underset{(\text { Dry })}{\mathrm{Ca}(\mathrm{OH})_{2}} \mathrm{Cl}_{2} \longrightarrow(\mathrm{~B})+\text { side products } \end{aligned}$$

(A) $\mathrm{NaClO}_{3}$ and $\mathrm{Ca}(\mathrm{OCl})_{2}$

(B) NaOCl and $\mathrm{Ca}(\mathrm{OCl})_{2}$

(C) NaOCl and $\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}$

(D) $\mathrm{NaClO}_{3}$ and $\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · p-Block Elements

In the following reactions, products(A) and (B) respectively are

(Hot and Conc) NaOH + Cl_{2} ⟶ (A) + side products (Dry) Ca (OH)_{2} Cl_{2} ⟶ (B) + side products

(A) $NaClO_{3}$ and $Ca (OCl)_{2}$ ✓ Correct
(B)NaOCl and $Ca (OCl)_{2}$
(C)NaOCl and $Ca (ClO_{3})_{2}$
(D) $NaClO_{3}$ and $Ca (ClO_{3})_{2}$

Answer: (A)

Solution

Hot, concentrated

NaOH

makes chlorine disproportionate all the way to chlorate,

6 NaOH + 3 Cl X_{2} 5 NaCl + NaClO X_{3} + 3 H X_{2} O

, so A is

NaClO X_{3}

. Dry slaked lime with chlorine gives bleaching powder,

2 Ca (OH) X_{2} + 2 Cl X_{2} Ca (OCl) X_{2} + CaCl X_{2} + 2 H X_{2} O

, so B is

Ca (OCl) X_{2}

. Thus A is

NaClO X_{3}

and B is

Ca (OCl) X_{2}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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