Animated Solution for Chemistry - d and f-Block Elements: On treating a compound with warm dil. $H_{2}S{O}_4$, gas $X$ is evolved, which turns $K_{2}C{r}_2{O}_7$ paper acidified with dil. $H_{2}S{O}_4$ to a green compound $Y$. $X$ and $Y$ respectively are

The journey to solving this problem begins in the qualitative analysis laboratory. Imagine you are holding a test tube containing an unknown salt. You add a few drops of warm, dilute sulphuric acid ($H_{2}S{O}_4$), and immediately, a gas begins to bubble out. We are tasked with identifying this mystery gas, labeled as $X$, and the subsequent compound $Y$ it forms when tested.

To identify gas $X$, we use a classic chemical test: bringing a filter paper soaked in acidified potassium dichromate ($K_{2}C{r}_2{O}_7$) near the mouth of the test tube. The problem states that the orange paper turns green.

This color change is a massive clue! Potassium dichromate is a well-known, powerful oxidizing agent. If it is changing color, it is undergoing a chemical transformation—specifically, it is being reduced. Therefore, the gas $X$ that caused this change must be a reducing agent.

Identifying Gas X

Looking at our options, gas $X$ is either sulphur dioxide ($S{O}_2$) or sulphur trioxide ($S{O}_3$). To determine which one is the reducing agent, we must examine the oxidation states of sulphur in both molecules.

In $S{O}_3$, the oxidation state of sulphur is $+6$. Since sulphur belongs to Group 16, it has 6 valence electrons, making $+6$ its maximum possible oxidation state. Because it cannot lose any more electrons, $S{O}_3$ cannot be oxidized further and thus cannot act as a reducing agent.

On the other hand, in $S{O}_2$, sulphur is in a $+4$ oxidation state. It has the capacity to lose two more electrons to reach the $+6$ state. Therefore, $S{O}_2$ can easily be oxidized, making it an excellent reducing agent.

This confirms that our mystery gas $X$ is $S{O}_2$.

Identifying Compound Y

Now that we know $X$ is $S{O}_2$, let's trace what happens to the potassium dichromate. In the $K_{2}C{r}_2{O}_7$ molecule, chromium is in a $+6$ oxidation state ($C{r}^{+6}$), which is responsible for its bright orange color.

When $S{O}_2$ reacts with it, $C{r}^{+6}$ gains electrons (gets reduced) and drops to a $+3$ oxidation state ($C{r}^{+3}$). In aqueous solutions, $C{r}^{+3}$ ions exhibit a characteristic green color.

But what is the exact chemical formula of this green compound $Y$? The reaction takes place in a medium acidified with dilute $H_{2}S{O}_4$. This means the solution is rich in sulphate ions ($S{O}_4^{2-}$). The newly formed $C{r}^{+3}$ ions will naturally bond with these sulphate ions.

By crossing their valencies (Chromium is $+3$, Sulphate is $-2$), we get the formula for Chromium(III) sulphate: $C{r}_2(S{O}_4{)}_3$.

The overall balanced redox reaction is:

Thus, the green compound $Y$ is $C{r}_2(S{O}_4{)}_3$. Matching this with our options, we find that the correct answer is option (c).