JEE Main · 2020 · Shift-IImediumPB12-105

In the following reactions, products (A) and (B), respectively, are: {NaOH + Cl_2 (A) + {side products}}{(hot and…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

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

NaOH+Cl2(A)+side products(hot and conc.)\mathrm{NaOH + Cl_2 \rightarrow (A) + \text{side products}}\quad\text{(hot and conc.)}

Ca(OH)2+Cl2(B)+side products(dry)\mathrm{Ca(OH)_2 + Cl_2 \rightarrow (B) + \text{side products}}\quad\text{(dry)}

Options
  1. a

    NaClO3\mathrm{NaClO_3} and Ca(OCl)2\mathrm{Ca(OCl)_2}

  2. b

    NaClO3\mathrm{NaClO_3} and Ca(ClO3)2\mathrm{Ca(ClO_3)_2}

  3. c

    NaOCl\mathrm{NaOCl} and Ca(OCl)2\mathrm{Ca(OCl)_2}

  4. d

    NaOCl\mathrm{NaOCl} and Ca(ClO3)2\mathrm{Ca(ClO_3)_2}

Correct Answera

NaClO3\mathrm{NaClO_3} and Ca(OCl)2\mathrm{Ca(OCl)_2}

Detailed Solution

Reaction 1: Cl2\mathrm{Cl_2} + hot concentrated NaOH: 3Cl2+6NaOHhot,conc5NaCl+NaClO3+3H2O\mathrm{3Cl_2 + 6NaOH \xrightarrow{hot,\,conc} 5NaCl + NaClO_3 + 3H_2O} A = NaClO3\mathrm{NaClO_3} (sodium chlorate)

Reaction 2: Cl2\mathrm{Cl_2} + dry Ca(OH)2\mathrm{Ca(OH)_2}: 2Ca(OH)2+2Cl2dryCa(OCl)2+CaCl2+2H2O\mathrm{2Ca(OH)_2 + 2Cl_2 \xrightarrow{dry} Ca(OCl)_2 + CaCl_2 + 2H_2O} B = Ca(OCl)2\mathrm{Ca(OCl)_2} (calcium hypochlorite, bleaching powder component)

Note: With cold dilute NaOH, A would be NaOCl.

Answer: Option (1) — NaClO3\mathrm{NaClO_3} and Ca(OCl)2\mathrm{Ca(OCl)_2}

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