Compound A reacts with $\mathrm{NH}_{4} \mathrm{Cl}$ and forms a compound B . Compound B reacts with $\mathrm{H}_{2} \mathrm{O}$ and excess of $\mathrm{CO}_{2}$ to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound $\mathrm{A}$ . $\mathrm{B}$ and C , are respectively.

(C) $\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{NH}_{3}, \mathrm{NH}_{4} \mathrm{HCO}_{3}$

(A) $\mathrm{CaCl}_{2}, \mathrm{NH}_{3}, \mathrm{NH}_{4} \mathrm{HCO}_{3}$

(B) $\mathrm{CaCl}_{2}, \mathrm{NH}_{4}{ }^{+},(\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}$

(D) $\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{NH}_{4}{ }^{+},(\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}$

s-Block Elements — JEE Main · 2023 · 25 Jan 2023 · Shift 1 Question with Solution

Compound A reacts with

NH_{4} Cl

and forms a compound B . Compound B reacts with

H_{2} O

and excess of

CO_{2}

to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound

A

B

and C , are respectively.

(A) $CaCl_{2}, NH_{3}, NH_{4} HCO_{3}$
(B) $CaCl_{2}, NH_{4}^{+}, (NH_{4})_{2} CO_{3}$
(C) $Ca (OH)_{2}, NH_{3}, NH_{4} HCO_{3}$ ✓ Correct
(D) $Ca (OH)_{2}, NH_{4}^{+}, (NH_{4})_{2} CO_{3}$

Answer: (C)

Solution

Warming slaked lime with ammonium chloride liberates ammonia,

Ca (OH) X_{2} + 2 NH X_{4} Cl CaCl X_{2} + 2 NH X_{3} + 2 H X_{2} O

, so the first substance is calcium hydroxide and the gas obtained is ammonia. This ammonia, in water with excess carbon dioxide, forms ammonium hydrogen carbonate,

NH X_{3} + H X_{2} O + CO X_{2} NH X_{4} HCO X_{3}

. Passing this through saturated sodium chloride solution throws down the sparingly soluble sodium hydrogen carbonate,

NH X_{4} HCO X_{3} + NaCl NaHCO X_{3} + NH X_{4} Cl

. The three compounds are calcium hydroxide, ammonia and ammonium hydrogen carbonate.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · s-Block Elements

Solution

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