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

Question

Statement-I: An element ‘X’ of P-block forms a hydride H–X, which has the longest bond length, then element ‘X’ will have the shortest covalent radius.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.

Answer 1

To solve this problem, we must identify the compounds A, B, and C involved in the reactions described. Let's break down the steps and reactions:

Compound A reacts with $NH_4Cl$ to form compound B.
- Given the options, and typical reactions, compound A is likely $Ca(OH)_2$ (calcium hydroxide). Its reaction with ammonium chloride ($NH_4Cl$) is:
- $Ca(OH)_2 + 2NH_4Cl \rightarrow CaCl_2 + 2NH_3(g) + 2H_2O$
Compound B reacts with $H_2O$ and excess of $CO_2$ to form compound C.
- Ammonia reacts with water and carbon dioxide to form ammonium bicarbonate ($NH_4HCO_3$).
- $NH_3 + H_2O + CO_2 \rightarrow NH_4HCO_3$
Passing compound C through saturated $NaCl$ solution forms sodium hydrogen carbonate.
- $NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$

Based on the above reactions, the compounds are:

Compound A: $Ca(OH)_2$ (Calcium hydroxide)
Compound B: $NH_3$ (Ammonia)
Compound C: $NH_4HCO_3$ (Ammonium bicarbonate)

Hence, the correct answer is the third option: $Ca(OH)_2, NH_3, NH_4HCO_3$.

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

The Correct Option is C

Solution and Explanation

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