Explain why does colour of $KMnO_4$ disappear when oxalic acid is added to its solution in acidic medium.

Explain why does colour of KMnO₄ disappear when oxalic acid is added to its solution in acidic medium.

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KMnO_4

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Concept understanding — Inorganic Synthesis

Inorganic Synthesis – What It Really Means

Imagine you want to build a house. You need bricks, cement, steel, and a plan to put them together. Inorganic synthesis is exactly that — but for making chemical compounds that do not contain carbon-hydrogen bonds (the domain of organic chemistry). You take simple starting materials (elements or simple compounds) and, through a controlled chemical reaction, build a more complex inorganic product.

The intuition is simple: you are a chemist-craftsman. You decide what to make, choose the right ingredients, set the right conditions (temperature, pressure, solvent, time), and then isolate the pure product. The "synthesis" part is the entire journey from idea to pure substance.

The Precise Statement

Inorganic synthesis is the branch of chemistry concerned with the design, planning, and execution of chemical reactions to prepare inorganic compounds — including metals, alloys, coordination complexes, main-group compounds, solid-state materials, and nanomaterials — with controlled purity, structure, and properties.

It is not just "mixing chemicals." It involves:

Choosing the correct starting materials (precursors) — often simple salts, oxides, or elements.
Selecting a reaction method — solid-state heating, solution precipitation, electrochemical deposition, sol-gel, hydrothermal, etc.
Controlling reaction conditions — temperature, pressure, pH, concentration, atmosphere (inert gas, air, vacuum).
Purifying the product — recrystallization, distillation, sublimation, chromatography.
Characterising the product — proving you actually made what you intended (X-ray diffraction, spectroscopy, elemental analysis).

A Concrete Example: Making Copper(II) Sulfate Pentahydrate

You want to make the familiar blue crystal, $\ce{CuSO4·5H2O}$ .

Intuition: You have copper metal (a wire) and dilute sulfuric acid. Copper does not react with dilute acid directly — you need an oxidising agent. So you add nitric acid or simply heat copper with concentrated sulfuric acid.

Reaction:

\ce{Cu + 2H2SO4(conc.) -> CuSO4 + SO2 + 2H2O}

Then you evaporate the solution carefully. Blue crystals of $\ce{CuSO4·5H2O}$ appear.

What you did: You synthesised an inorganic compound from elemental copper and an acid. You controlled the concentration, temperature, and evaporation rate. You then filtered and dried the crystals.

Why It Matters

Inorganic synthesis is the foundation of:

Catalysts (e.g., $\ce{Pt}$ on alumina for car exhausts)
Electronic materials (silicon wafers, gallium arsenide for LEDs)
Medicinal compounds (cisplatin for cancer therapy)
Pigments (titanium dioxide white, Prussian blue)
Batteries (lithium cobalt oxide electrodes)

Without inorganic synthesis, modern technology would not exist.

A Common Misconception …

Why this formula?

Inorganic Synthesis: Why the Key Formulae Hold

Inorganic synthesis is the branch of chemistry concerned with the preparation of inorganic compounds — from simple salts to complex coordination compounds, organometallics, and solid-state materials. The key formulae in this field are not arbitrary; they arise from fundamental principles of stoichiometry, thermodynamics, kinetics, and coordination chemistry.

Let’s break down the reasoning behind the most important formulae.

1. The Yield Formula: Why It’s Not Just “Product/Reactant”

The most basic formula in any synthesis is:

\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

Why this holds:

Theoretical yield is calculated from the limiting reagent — the reactant that runs out first. This is based on the law of conservation of mass and the stoichiometric coefficients from the balanced chemical equation.
Actual yield is always less than theoretical because of:
- Side reactions (competing pathways)
- Incomplete reactions (equilibrium limitations)
- Loss during purification (filtration, crystallization, etc.)
The formula is a ratio because yield is a fractional measure of efficiency — it tells you how much of the maximum possible product you actually obtained.

Key insight: The formula works only if you correctly identify the limiting reagent. For example, in the synthesis of $\ce{FeCl3}$ from $\ce{Fe}$ and $\ce{Cl2}$ , if you have 1 mol Fe and 2 mol $\ce{Cl2}$ , Fe is limiting (1:1.5 stoichiometry), so theoretical yield is based on Fe.

2. The Atom Economy Formula: Why It Measures “Greenness”

\text{Atom Economy} = \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Reactants}} \times 100\%

Why this holds:

This formula was introduced by Barry Trost (1991) to quantify how much of the starting materials ends up in the product.
It is not a yield — it’s a theoretical maximum based on the balanced equation. It assumes 100% yield.
The denominator includes all reactants (including solvents if they are consumed, but usually only stoichiometric reagents).
A high atom economy (e.g., 100% for addition reactions like $\ce{A + B -> C}$ ) means less waste. A low atom economy (e.g., substitution reactions with leaving groups) means more byproducts.

Example: In the synthesis of $\ce{NaCl}$ from $\ce{Na}$ and $\ce{Cl2}$ :

$2\ce{Na} + \ce{Cl2} \rightarrow 2\ce{NaCl}$

Atom economy = $\frac{2 \times 58.44}{2 \times 22.99 + 70.90} \times 100\% = 100\%$ — because all atoms end up in the product.

3. The Solubility Product and Precipitation: Why $K_{sp}$ Controls Synthesis

For a sparingly soluble salt like $\ce{AgCl}$ :

\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}

K_{sp} = [\ce{Ag+}][\ce{Cl-}]

Why this holds:

$K_{sp}$ is an equilibrium constant derived from the law of mass action. It applies only to saturated solutions.
In synthesis, you use $K_{sp}$ to predict whether a precipitate will form when mixing solutions. If the ion product $Q = [\ce{Ag+}][\ce{Cl-}]$ exceeds $K_{sp}$ , precipitation occurs.
The formula is temperature-dependent (because $\Delta G^\circ = -RT \ln K_{sp}$ ). So you must control temperature to control precipitation.

Reasoning: The equilibrium constant arises from the balance between the lattice energy (holding the solid together) and the hydration energy (stabilizing ions in solution). A very small $K_{sp}$ means the solid is very stable — useful for gravimetric synthesis.

4. The Coordination Number and Ligand Field Stabilization Energy (LFSE)

For an octahedral complex, the LFSE is:

\text{LFSE} = \left( -0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g} \right) \Delta_o

Why this holds:

This formula comes from crystal field theory (CFT). In an octahedral field, the five $d$ orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals).
The splitting energy $\Delta_o$ is the energy difference between these sets.
Electrons fill the $t_{2g}$ orbitals first (Hund’s rule), and each electron in $t_{2g}$ stabilizes the complex by $-0.4\Delta_o$ relative to the barycenter (average energy). Each electron in $e_g$ destabilizes by $+0.6\Delta_o$ .
The formula explains why certain coordination numbers are preferred: for example, $\ce{[Co(H2O)6]^{2+}}$ (high-spin $d^7$ ) has LFSE = $-0.8\Delta_o$ , while $\ce{[CoCl4]^{2-}}$ (tetrahedral) has a smaller LFSE — so the octahedral form is more stable. …

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The purple colour of $\text{KMnO}_4$ disappears because $\text{MnO}_4^-$ (Mn in +7 oxidation state) is reduced to colourless $\text{Mn}^{2+}$ by oxalic acid in acidic medium — the reaction is a classic redox titration where the deep purple permanganate ion acts as a self-indicator.

Why This Happens: The Core Idea

The colour of $\text{KMnO}_4$ comes from the permanganate ion, $\text{MnO}_4^-$ , where manganese is in its highest oxidation state, +7. This ion absorbs light in the visible region (green-yellow) due to charge transfer transitions, giving the solution its characteristic deep purple colour.

When you add oxalic acid ( $\text{H}_2\text{C}_2\text{O}_4$ ) in acidic medium, a redox reaction occurs. The permanganate ion is a powerful oxidising agent, and oxalic acid is a reducing agent. The Mn in $\text{MnO}_4^-$ gets reduced from +7 to +2, forming the nearly colourless $\text{Mn}^{2+}$ ion. The oxalic acid gets oxidised to carbon dioxide and water.

The key point: the product $\text{Mn}^{2+}$ does not absorb visible light strongly — it appears pale pink only in very concentrated solutions, but in typical lab conditions it looks colourless. So as the reaction proceeds, the purple colour fades and eventually disappears completely.

Watch out

A common mistake is to think the colour disappears because permanganate "reacts" without specifying the change in oxidation state. The colour change is directly linked to the change in electronic structure of Mn — from $\text{MnO}_4^-$ (d⁰, intense charge transfer) to $\text{Mn}^{2+}$ (d⁵, weak d-d transitions). It is not simply "dilution" or "neutralisation".

Step-by-Step Breakdown

Identify the species and their roles

In acidic medium, $\text{KMnO}_4$ dissociates to give $\text{K}^+$ and $\text{MnO}_4^-$ . Oxalic acid exists mostly as $\text{H}_2\text{C}_2\text{O}_4$ (or its conjugate base $\text{HC}_2\text{O}_4^-$ depending on pH). The medium is acidified, typically with dilute $\text{H}_2\text{SO}_4$ , to provide $\text{H}^+$ ions.
Write the half-reactions

The reduction half-reaction for permanganate in acidic medium is:

\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

The oxidation half-reaction for oxalic acid is:

\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^-

Balance the overall redox equation To balance electrons, multiply the reduction half by 2 and the oxidation half by 5:

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{H}_2\text{C}_2\text{O}_4 \rightarrow 10\text{CO}_2 + 10\text{H}^+ + 10e^-

Adding them gives:

2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}

2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}

Track the colour change …

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Method: Redox Titration (Permanganate vs. Oxalic Acid)

This is a classic redox reaction where potassium permanganate ( $KMnO_4$ ) acts as an oxidising agent and oxalic acid ( $H_2C_2O_4$ ) acts as a reducing agent in acidic medium.

Step-by-step reasoning

Identify the colour change

$KMnO_4$ in solution is deep purple due to the presence of the permanganate ion ( $MnO_4^-$ ).
Recognise the reaction conditions

The reaction is carried out in acidic medium (usually dilute $H_2SO_4$ ). The acid provides $H^+$ ions necessary for the reduction of $MnO_4^-$ .
Write the half-reactions
- Reduction half (permanganate ion is reduced):

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

 Here, $Mn$ goes from **+7** oxidation state (purple) to **+2** (colourless/pale pink in very dilute solutions).

Oxidation half (oxalic acid is oxidised):

H_2C_2O_4 \rightarrow 2CO_2 + 2H^+ + 2e^-

 Carbon goes from **+3** in oxalic acid to **+4** in $CO_2$.

4. Combine the half-reactions

Multiply the oxidation half by 5 and the reduction half by 2 to balance electrons:

2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O

Explain the colour disappearance …

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Let’s break it down.

Why the colour disappears (the correct concept)

$KMnO_4$ in acidic medium is a deep purple because of the charge transfer from oxygen to manganese in the $\text{MnO}_4^-$ ion.

When oxalic acid ( $\text{H}_2\text{C}_2\text{O}_4$ ) is added, a redox reaction occurs:

$KMnO_4$ is reduced from $\text{Mn}^{+7}$ (purple) to $\text{Mn}^{2+}$ (colourless/pale pink — essentially colourless in dilute solution).
Oxalic acid is oxidised to $\text{CO}_2$ and water.

The balanced equation (in acidic medium):

2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + 8\text{H}_2\text{O}

The colour disappears because $\text{Mn}^{2+}$ does not absorb visible light strongly — the purple colour is gone.

Common mistakes students make

1. ✗ Saying “oxalic acid bleaches” or “decolourises” without mentioning redox

Why it’s wrong: Bleaching implies a physical or non-redox process (like adsorption). This is a chemical reduction.

✓ How to avoid: Always say: “Oxalic acid reduces $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ , and the purple colour is lost because $\text{Mn}^{2+}$ is colourless.”

2. ✗ Forgetting to specify “acidic medium”

Why it’s wrong: In neutral or alkaline medium, $KMnO_4$ reduces to $\text{MnO}_2$ (brown) or $\text{MnO}_4^{2-}$ (green) — the colour does not disappear.

✓ How to avoid: Always mention: “In acidic medium, the reduction product is $\text{Mn}^{2+}$ (colourless).”

3. ✗ Writing the wrong oxidation state change

Common errors:

Saying Mn goes from +7 to +4 (that’s neutral/alkaline medium).
Saying Mn goes from +7 to +6 (that’s alkaline).

✓ How to avoid: Memorise the three conditions:

Medium	Product	Colour change
Acidic	$\text{Mn}^{2+}$ (+2)	Purple → colourless
Neutral/weakly alkaline	$\text{MnO}_2$ (+4)	Purple → brown
Strongly alkaline	$\text{MnO}_4^{2-}$ (+6)	Purple → green

4. ✗ Confusing the role of $\text{H}_2\text{SO}_4$

Some students think $\text{H}_2\text{SO}_4$ is a catalyst or is consumed.

✓ How to avoid: $\text{H}_2\text{SO}_4$ provides the acidic medium (H⁺ ions) needed for the reduction to $\text{Mn}^{2+}$ . It is not consumed — it’s a reactant that gets regenerated.

5. ✗ Forgetting to balance the equation or write the ionic equation

Exams often ask for the ionic equation:

$2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$ …

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Q.Explain why does colour of KMnO4KMnO_4KMnO4​ disappear when oxalic acid is added to its solution in acidic medium.

Why This Happens: The Core Idea

Step-by-Step Breakdown

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Q.Explain why does colour of $KMnO_4$ disappear when oxalic acid is added to its solution in acidic medium.

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