In the scheme given below, X and Y, respectively, are
(A) CrO4^2- and Br2
(B) MnO4^2- and Cl2
(C) Mn

Q: Why did we assume the metal is Manganese?

The clue lies in the 'coloured species in solution' formed upon strong oxidation. Permanganate ($\text{MnO}_4^-$) is a classic, intensely purple species formed by oxidizing $\text{Mn}^{2+}$.

Q: What is the role of $\text{PbO}_2$ here?

$\text{PbO}_2$ acts as a powerful oxidizing agent in the acidic medium, capable of oxidizing $\text{Mn}^{2+}$ to its highest oxidation state, $\text{Mn}^{7+}$ in $\text{MnO}_4^-$.

Q: Can we use any other oxidizing agent instead of $\text{PbO}_2$?

Yes, other strong oxidizing agents like $\text{NaBiO}_3$ (sodium bismuthate) or $\text{Pb}_3\text{O}_4$ in acidic medium can also oxidize $\text{Mn}^{2+}$ to $\text{MnO}_4^-$.

Q: Why does KI-starch paper turn blue?

Chlorine gas ($\text{Cl}_2$) oxidizes the iodide ions ($\text{I}^-$) in KI to elemental iodine ($\text{I}_2$). The liberated iodine then forms a deep blue-black inclusion complex with starch.

Q: What is $\text{MnO(OH)}_2$?

It is a hydrated form of manganese dioxide, often written as $\text{MnO}_2 \cdot \text{H}_2\text{O}$. It acts as an oxidizing agent, similar to $\text{MnO}_2$, oxidizing chloride to chlorine gas in the presence of concentrated sulfuric acid.

Question

In the scheme given below, X and Y, respectively, are

Options:

(A) $	ext{CrO}_4^{2-}$ and $	ext{Br}_2$

(B) $	ext{MnO}_4^{2-}$ and $	ext{Cl}_2$

(C) $	ext{MnO}_4^{-}$ and $	ext{Cl}_2$

(D) $	ext{MnSO}_4$ and $	ext{HOCl}$

Sigma Percentile · Accepted Answer

### Unraveling the Mystery of the Metal Halide

Imagine you are a detective in a chemistry lab, and you are handed a mysterious metal halide. Your mission is to identify the unknown species **X** and **Y** produced through a series of reactions. Let's break down the clues step by step.

### The Initial Precipitation

The first clue is that our metal halide reacts with aqueous sodium hydroxide ($	ext{NaOH}$) to form a white precipitate **P** and a filtrate **Q**.

The fact that **P** later forms a highly colored species strongly hints at transition metals, specifically manganese. If we assume the metal halide is manganese chloride ($	ext{MnCl}_2$), reacting it with $	ext{NaOH}$ yields a white precipitate of manganese(II) hydroxide.

$$ 	ext{MnCl}_2 + 2	ext{NaOH} ightarrow 	ext{Mn(OH)}_2 \downarrow + 2	ext{NaCl} $$

So, we have identified **P** as $	ext{Mn(OH)}_2$, and the filtrate **Q** contains sodium chloride ($	ext{NaCl}$).

### The Powerful Oxidation

Now look at the second step. The white precipitate **P**, $	ext{Mn(OH)}_2$, is heated with lead dioxide ($	ext{PbO}_2$) and aqueous sulfuric acid. Lead dioxide is a powerful oxidizing agent.

It oxidizes the manganese from a $+2$ state all the way up to a $+7$ state, forming the permanganate ion, $	ext{MnO}_4^-$. This ion is famous for its intense purple color in solution.

$$ 	ext{Mn(OH)}_2 + 	ext{PbO}_2 + 	ext{H}^+ \xrightarrow{\Delta} 	ext{MnO}_4^- + 	ext{Pb}^{2+} + 	ext{H}_2	ext{O} $$

Therefore, the colored species **X** is $	ext{MnO}_4^-$.

### The Halide Test

Let's shift our focus to the filtrate **Q**, which contains chloride ions ($	ext{Cl}^-$). It is warmed with $	ext{MnO(OH)}_2$ and concentrated sulfuric acid. This is a classic laboratory test for chloride.

Under these acidic and oxidizing conditions, the chloride ions are oxidized to produce chlorine gas ($	ext{Cl}_2$).

$$ 2	ext{Cl}^- + 	ext{MnO(OH)}_2 + 4	ext{H}^+ \xrightarrow{	ext{warm}} 	ext{Mn}^{2+} + 	ext{Cl}_2 \uparrow + 3	ext{H}_2	ext{O} $$

So, the species **Y** is $	ext{Cl}_2$.

### The Confirmation

To confirm this, the problem states that **Y** gives a blue coloration with KI-starch paper. Chlorine gas oxidizes iodide ions ($	ext{I}^-$) from KI into free iodine ($	ext{I}_2$).

$$ 	ext{Cl}_2 + 2	ext{I}^- ightarrow 2	ext{Cl}^- + 	ext{I}_2 $$

This iodine then reacts with starch to form a deep blue complex. This perfectly confirms our identification of **Y**.

So, let's summarize our findings. **X** is the permanganate ion, $	ext{MnO}_4^-$, and **Y** is chlorine gas, $	ext{Cl}_2$. Looking at our options, this matches perfectly with option (C).

Animated Solution for Chemistry - Salt Analysis: In the scheme given below, X and Y, respectively, are

Select Answer:

Visualized Solution

Initial Precipitation of P

Oxidation to form X

Oxidation to form Y

Confirmation with KI-Starch

The Sigma Insight: Salt Analysis

Unraveling the Mystery of the Metal Halide

The Initial Precipitation

The Powerful Oxidation

The Halide Test

The Confirmation

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