When XO is fused with an alkali metal hydroxide in presence of an oxidizing agent 2 such as KNO , a dark green product is formed which disproportionate in acidic 3 solution to afford a dark purple solution. X is

The correct answer is A. Complete step-by-step solution with explanation available on PadhoLikhoJEE.

JEE Main 2018 D and F Block Elements — Fused Alkali Metal Hydroxide Presence Oxidizing

Quick Summary

Concept Tested: Manganese Chemistry (Oxidation States & Colors) in Alkaline Medium.

Chapter: D and F Block Elements.

Difficulty: ★☆☆☆☆

Time Required: 1-2 Minutes.

Key Formula: $$MnO + 2KOH + KNO3 \rightarrow K_2MnO_4 + KNO_2 + H_2O$$

Answer: (A) Mn

Reasoning: Manganese forms potassium manganate (K₂MnO₄), which is dark green. In acidic medium, it disproportionates to form potassium permanganate (KMnO₄), which is dark purple.

The Question

When XO is fused with an alkali metal hydroxide (KOH) in the presence of an oxidizing agent (KNO₃), a dark green product is formed. This product disproportionates in acidic solution to give a dark purple solution. Identify X from the given options:

(A) Mn

(B) Cr

(C) V

(D) Ti

Quick Answer

The correct answer is (A) Mn.

Manganese(II) oxide (MnO) is oxidized in the alkaline fusion with KNO₃ to form potassium manganate (K₂MnO₄), which is dark green. Upon acidification, K₂MnO₄ undergoes disproportionation to form potassium permanganate (KMnO₄), which is dark purple. Chromium and Vanadium do not produce a green intermediate in this reaction.

Why Other Options Are Incorrect

(B) Cr (Chromium): When chromium(II) oxide (CrO) undergoes alkaline fusion, it forms potassium chromate (K₂CrO₄). Unlike manganese, chromate is yellow, not dark green. Furthermore, its acidification yields orange dichromate (K₂Cr₂O₇), not purple.

(C) V (Vanadium): Vanadium(II) oxide (VO) forms potassium orthovanadate (K₃VO₄) in this reaction. This compound is typically colorless or light yellow, lacking the dark green color characteristic of manganese compounds. Acidification yields vanadium pentoxide (V₂O₅), which is orange, not purple.

(D) Ti (Titanium): Titanium(II) oxide (TiO) reacts to form potassium titanate (K₂TiO₃), which is colorless. Acidification yields titanium dioxide (TiO₂), which is white. Neither step produces the observed dark green or dark purple colors.

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Understanding the Concept

This problem tests the ability to identify transition metal oxides based on the color of their oxyanions in different oxidation states. The key concept is the disproportionation of manganate (MnO₄^2-) in acidic media.

Manganese exhibits multiple oxidation states (+2, +4, +6, +7). The green color is specific to the +6 state (manganate ion, MnO₄^2-), while the purple color is specific to the +7 state (permanganate ion, MnO₄^–). The stability of these species depends heavily on the pH of the solution.

Detailed Step-by-Step Solution

Step 1: Formation of the Dark Green Product (Alkaline Fusion)

Manganese(II) oxide (MnO) is oxidized in the presence of an alkali metal hydroxide (KOH) and an oxidizing agent (KNO₃). The oxidizing agent (nitrate, NO₃^–) facilitates the conversion of Mn²⁺ to Mn⁶⁺.

Reaction:

$$MnO + 2KOH + KNO_3 \rightarrow K_2MnO_4 + KNO_2 + H_2O$$

Analysis: Manganese changes its oxidation state from +2 (in MnO) to +6 (in K₂MnO₄). The product, potassium manganate, is dark green.

Step 2: Disproportionation in Acidic Medium

The dark green K₂MnO₄ is unstable in acidic conditions. It undergoes disproportionation, where the same element (Mn) is both reduced and oxidized.

Reaction:

$$3K_2MnO_4 + 4H^+ \rightarrow 2KMnO_4 + MnO_2 + 2H_2O + 4K^+$$

Analysis: Manganese changes from +6 in manganate to +7 in permanganate (reduction) and to +4 in manganese dioxide (oxidation). The resulting purple compound is potassium permanganate (KMnO₄).

Step 3: Elimination of Other Elements

We compare the behavior of the other options:

Chromium (Cr): Forms K₂CrO₄ (yellow) which turns to K₂Cr₂O₇ (orange) in acid.
Vanadium (V): Forms K₃VO₄ (colorless) which turns to V₂O₅ (orange) in acid.
Titanium (Ti): Forms K₂TiO₃ (colorless) which turns to TiO₂ (white) in acid.

Only Manganese fits the description of dark green (intermediate) and dark purple (final) products.

Final Answer

✔ X = Mn

Essential Formulas for This Topic

1. Oxidation in Alkaline Medium:$$MnO + 2OH^- + NO_3^- \rightarrow MnO_4^{2-} + NO_2^- + H_2O$$

2. Disproportionation of Manganate in Acid:$$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$

Common Mistakes to Avoid

Mistake 1: Confusing the Color of Chromate with Manganate

Wrong Thinking: “The green compound is potassium chromate (K₂CrO₄).”

Correction: Chromate is yellow. The dark green compound is specifically potassium manganate (K₂MnO₄).

Mistake 2: Misidentifying the Purple Product

Wrong Thinking: “The purple solution is dichromate.”

Correction: Dichromate (K₂Cr₂O₇) is orange. The purple solution is potassium permanganate (KMnO₄).

Mistake 3: Ignoring Oxidation State Changes

Wrong Thinking: Assuming all transition metals behave similarly in alkaline fusion.

Correction: Track the oxidation numbers. Manganese goes from +2 (in MnO) to +6 (in K₂MnO₄) and finally to +7 (in KMnO₄). This multi-step oxidation is unique to Mn in this context.

Key Concept Summary

Manganese exhibits oxidation states of +2, +4, +6, and +7.
K₂MnO₄ (Manganate) is dark green.
MnO₂ (Manganese Dioxide) is brown/black.
KMnO₄ (Permanganate) is dark purple.
Manganate disproportionates in acidic medium.

Golden Rule: In the Mn family, the green color indicates the +6 oxidation state, while the purple color indicates the +7 oxidation state.

Frequently Asked Questions

Q: Why is KNO₃ used as the oxidizing agent in alkaline fusion?

A: KNO₃ provides the nitrate ion (NO₃^–), which is a strong oxidizing agent in alkaline conditions. It facilitates the conversion of lower oxidation state metal oxides to higher ones (e.g., Mn²⁺ to Mn⁶⁺).

Q: Is K₂MnO₄ stable in water?

A: No. In neutral or acidic water, K₂MnO₄ disproportionates to form MnO₂ (brown precipitate) and MnO₄^– (purple color). It is only stable in strongly alkaline solutions.

Q: What happens when chromate is acidified?

A: When K₂CrO₄ (yellow) is acidified, it forms K₂Cr₂O₇ (orange) in acidic conditions, not a purple solution.

Q: Can Titanium form a green compound under these conditions?

A: No. Titanium chemistry is generally limited to lower oxidation states (+2, +3, +4) in this context. It does not form a stable green oxyanion like manganese does.

Prerequisites to Solve This Question

Understanding of Oxidation States (how to calculate them).
Knowledge of Disproportionation reactions.
Color identification of common transition metal oxyanions (Chromate, Manganate, Permanganate).

After Solving This, You Can:

✔ Identify manganese oxides based on color.
✔ Predict the products of alkaline fusion reactions.
✔ Distinguish between chromate and manganate chemistry.

Study Tips for This Topic

Memorize the color of oxyanions for the first row transition metals to solve such questions quickly. The Mn family is the most common source of these color-based questions in JEE exams.

Difficulty Rating & Exam Frequency

Difficulty: Easy

JEE Main Frequency: Moderate

Importance: High for Inorganic Chemistry basics.