Animated Solution for Chemistry - s and p-Block Elements: Reaction of an inorganic sulphite X with dilute ${\text{H}}_2{\text{SO}}_4$ generates compound Y. Reaction of Y with NaOH gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z respectively, are

The beauty of inorganic chemistry lies in its interconnectedness. It's like a grand puzzle where every compound is a piece, and the reactions are the interlocking edges. In this problem, we are presented with a fascinating sequence of transformations starting from an unknown inorganic sulphite, which we'll call $X$.

Our mission is to decode this sequence and unveil the true identities of compounds $Y$ and $Z$. Let's embark on this chemical detective journey!

The Acid Test

The problem begins by stating that an inorganic sulphite $X$ reacts with dilute sulfuric acid (${\text{H}}_2{\text{SO}}_4$) to generate a new compound $Y$.

If you recall your qualitative analysis laboratory sessions, this is a classic, textbook test for the sulphite radical (${\text{SO}}_3^{2-}$). When a salt containing a sulphite ion is treated with a dilute acid, a displacement reaction occurs. The stronger acid (sulfuric acid) displaces the weaker sulfurous acid (${\text{H}}_2{\text{SO}}_3$) from its salt.

However, sulfurous acid is highly unstable. It immediately decomposes into water and sulfur dioxide gas.

This pungent, suffocating gas that turns acidified potassium dichromate paper green is none other than sulfur dioxide. Therefore, we have successfully identified our first unknown: Compound $Y$ is ${\text{SO}}_2$.

The Neutralization Dance

Now, the plot thickens. The problem states that when compound $Y$ (${\text{SO}}_2$) reacts with sodium hydroxide ($\text{NaOH}$), it gives back our original compound $X$.

To understand this, we must look at the chemical nature of sulfur dioxide. ${\text{SO}}_2$ is an oxide of a non-metal (sulfur), which makes it an acidic oxide. In fact, it is the acid anhydride of sulfurous acid.

When an acidic oxide meets a strong base like sodium hydroxide, a classic neutralization reaction takes place, yielding a salt and water.

The salt formed in this reaction is sodium sulphite. Since this reaction gives us back compound $X$, we can confidently deduce that our starting inorganic sulphite, Compound $X$, is ${\text{Na}}_2{\text{SO}}_3$.

The Acid Salt Formation

We are now at the final stage of our puzzle. The problem reveals that compound $X$ (${\text{Na}}_2{\text{SO}}_3$) reacts with compound $Y$ (${\text{SO}}_2$) and water to afford compound $Z$.

Let's visualize this mixture: we have sodium sulphite, sulfur dioxide, and water. We already know that ${\text{SO}}_2$ and water combine to form sulfurous acid (${\text{H}}_2{\text{SO}}_3$). So, essentially, we are reacting a normal salt (${\text{Na}}_2{\text{SO}}_3$) with more of its parent acid.

When a normal salt reacts with an excess of the acid, it forms an acid salt. The sodium ions are partially replaced by hydrogen ions.

The resulting compound is sodium hydrogen sulphite, commonly known as sodium bisulphite. Thus, the final piece of our puzzle falls into place: Compound $Z$ is ${\text{NaHSO}}_3$.

By carefully following the breadcrumbs left by the chemical reactions, we have unraveled the entire sequence.

We started with sodium sulphite ($X$), treated it with acid to release sulfur dioxide ($Y$), reacted the gas with a base to get the sulphite back, and finally passed excess gas through the sulphite solution to obtain sodium bisulphite ($Z$).

Comparing our findings with the given options, we see that $Y$ is ${\text{SO}}_2$ and $Z$ is ${\text{NaHSO}}_3$. This perfectly aligns with Option (c).

Chemistry is not just about memorizing equations; it's about understanding the fundamental behaviors of molecules. Keep exploring these fascinating chemical narratives!