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.

NCERT Exemplar · Q34

KMnO_4

Andhra Pradesh BieapShort· 2mImportance★★★★★

70% · 92/132 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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:

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

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution
Sphalerite (ZnS) is roasted to ZnO, reduced with coke to impure zinc, then purified by fractional distillation because zinc has a conveniently low boiling point.

Concept and Intuition

Zinc's chief ore, sphalerite (ZnS), is a sulfide ore, so the standard pretreatment step for sulfide ores is roasting (heating strongly in air) — this converts the sulfide to the oxide and drives off $\mathrm{SO_2}$ , giving $\mathrm{ZnO}$ (calcination, by contrast, is used for carbonate/hydroxide ores to drive off $\mathrm{CO_2}$ / $\mathrm{H_2O}$ — not applicable here since sphalerite is a sulfide).

$\mathrm{ZnO}$ is then reduced with coke (carbon) at high temperature to metallic zinc, which distils off as vapour (zinc boils at ~1180 K, well below iron's melting point, so it separates as vapour directly in the retort/furnace). This crude zinc still carries impurities (Pb, Cd, Fe), and because zinc's boiling point is so much lower than these impurities, fractional distillation cleanly separates and purifies it.

Step-by-Step Solution

Sphalerite (ZnS) → froth flotation → concentrated ZnS.

X: roasting converts $\mathrm{ZnS}\rightarrow \mathrm{ZnO} + \mathrm{SO_2}$ .

$\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}$ (reduction with coke) gives impure zinc. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{K_2Cr_2O_7}$
›Reveal solutionSolution
The chromate–dichromate equilibrium is pH-dependent: chromate (yellow) dominates in alkaline/neutral solution, dichromate (orange) dominates on acidification (decreasing pH).

Concept and Intuition

In aqueous solution, chromate and dichromate exist in a pH-sensitive equilibrium:
$2\mathrm{CrO_4^{2-}} + 2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$
Adding acid (H⁺) pushes this equilibrium to the right (Le Chatelier), converting yellow chromate into orange dichromate. This is exactly the industrial step in the dichromate manufacturing process (chromite ore → sodium chromate solution → acidified with $\mathrm{H_2SO_4}$ → sodium dichromate crystallised out).

Step-by-Step Solution

Sodium chromate solution (yellow) is obtained after fusing chromite ore with $\mathrm{Na_2CO_3}$ and roasting/leaching.

To convert it to dichromate, the solution is acidified (typically with $\mathrm{H_2SO_4}$ ), i.e. the pH is decreased. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$Na_2CO_3$
›Reveal solutionSolution
Chromite ore fusion is the standard extraction step for chromium compounds; identifying A as sodium chromate (not dichromate) and B as iron(III) oxide resolves which of the three statements are correct. The answer is (B) II, III only.

Concept and Intuition

Chromite ore ( $FeCr_2O_4$ ) is fused with sodium carbonate in the presence of air (an oxidative fusion). The air oxidises chromium(III) in the ore to chromium(VI), while iron ends up as iron(III) oxide, an insoluble residue. The soluble product is sodium chromate, $Na_2CrO_4$ , which is yellow in colour — this is compound A. Only on acidifying this yellow chromate solution does it convert to the orange dichromate ion ( $Cr_2O_7^{2-}$ ), which does have the Cr–O–Cr bridging linkage; the chromate ion itself is a simple monomeric tetrahedral $CrO_4^{2-}$ ion with no such linkage.

Step-by-Step Solution

Reaction: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$ .

Compound A (yellow solution) = sodium chromate, $Na_2CrO_4$ ; residue B = iron(III) oxide, $Fe_2O_3$ .

Statement I: "A contains Cr–O–Cr linkage." The chromate ion $CrO_4^{2-}$ is a single tetrahedral unit with no bridging oxygen between two chromium atoms — that bridging (Cr–O–Cr) only appears in the dichromate ion $Cr_2O_7^{2-}$ , formed later on acidification. So statement I is false. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$Q. Wrought iron is prepared from cast iron in a reverberatory furnace. The substance commonly used to line the furnace and the chemical process involved in it are respectively (A) Magnetite, reduction (B) Magnetite, oxidation (C) Haematite, oxidation (D) Haematite, reduction$
›Reveal solutionSolution
The puddling process converts cast iron to wrought iron in a haematite-lined reverberatory furnace, where the haematite oxidises the impurities out of the molten iron.

Concept and Intuition

Cast iron contains several % of carbon plus other impurities (Si, S, P, Mn) that make it brittle. Wrought iron is the purest commercial form of iron (almost no carbon). To go from cast iron to wrought iron, impurities must be removed by oxidation — this is done in a reverberatory furnace whose lining ( $Fe_2O_3$ , haematite) itself acts as an oxidising agent, converting impurities to their oxides which float off as slag, in the classic "puddling" process.

Step-by-Step Solution

Molten cast iron is melted in a reverberatory furnace lined with haematite ( $Fe_2O_3$ ).

Haematite oxidises carbon (to CO/CO $_2$ ), silicon (to $SiO_2$ ), sulphur, phosphorus and manganese in the melt.

These oxidised impurities combine with the furnace lining/flux to form a fusible slag, which is skimmed off. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$Q. The method by which very pure nitrogen can be obtained is (A) Thermal decomposition of ammonium dichromate (B) Thermal decomposition of barium azide (C) Reaction of aqueous solution of ammonium chloride with sodium nitrite (D) Thermal decomposition of ammonium nitrate$
›Reveal solutionSolution
The routine lab preparation of N $_2$ (from NH $_4$ Cl + NaNO $_2$ ) carries trace NO/HNO $_3$ impurities; genuinely pure, dry nitrogen is instead obtained via thermal decomposition of an azide such as barium (or sodium) azide.

Concept and Intuition

Several routes give nitrogen gas, but not all give it in a pure state. The everyday laboratory method (heating an aqueous mixture of ammonium chloride and sodium nitrite) is convenient but contaminated by small amounts of NO and HNO $_3$ as side products. To get genuinely pure nitrogen, chemists instead thermally decompose an ionic azide, which cleanly releases only N $_2$ gas (plus the metal), with no such side products.

Step-by-Step Solution

$NH_4Cl(aq) + NaNO_2(aq) \xrightarrow{\Delta} N_2 + NaCl + 2H_2O$ — this is the standard lab method but yields nitrogen contaminated with traces of NO and HNO $_3$ .

To obtain very pure, dry $N_2$ , azides are thermally decomposed instead: $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2\uparrow$ (similarly $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$ ).

This decomposition gives clean N $_2$ gas without the nitrogen-oxide contaminants of the aqueous nitrite method. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$K_2CO_3$
›Reveal solutionSolution
The classic limitation of the Solvay process is that it cannot be used to make $K_2CO_3$ , because $KHCO_3$ is too soluble to precipitate — this makes statement (A) the incorrect one.

Concept and Intuition

The Solvay (ammonia–soda) process manufactures sodium carbonate by passing $CO_2$ and $NH_3$ into brine, precipitating relatively insoluble $NaHCO_3$ , which is then calcined to $Na_2CO_3$ . The entire process depends on $NaHCO_3$ being sufficiently insoluble to crystallize out. The analogous potassium salt, $KHCO_3$ , is much more soluble in water and does not precipitate under the same conditions — so the Solvay process simply cannot be adapted to prepare $K_2CO_3$ , which is why potassium carbonate is manufactured by a different industrial route.

Step-by-Step Solution

Check (A): 'K2CO3 can be prepared by solvay process' — false, as explained above (textbook exception of the Solvay process).

Check (B): $CaCl_2$ is indeed a byproduct of the Solvay process ( $CaCO_3 \rightarrow CaO+CO_2$ ; $CaO+2NH_4Cl \rightarrow CaCl_2+2NH_3+H_2O$ ) — true.

Check (C): $Na_2CO_3$ solution is basic because $CO_3^{2-}$ , being the conjugate base of a weak acid ( $HCO_3^-$ ), hydrolyses to give $OH^-$ — true. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{BF_3}$
›Reveal solutionSolution
This tests the industrial vs laboratory preparation of diborane; the industrial route uses $\mathrm{BF_3}$ + NaH at 450 K.

Concept and Intuition

Diborane, $\mathrm{B_2H_6}$ , can be made by more than one route, but exam questions on boron hydrides specifically distinguish the laboratory-scale method from the industrial-scale method because only one is economical at bulk production.

Step-by-Step Solution

Laboratory method: $4\mathrm{BF_3} + 3\mathrm{LiAlH_4} \rightarrow 2\mathrm{B_2H_6} + 3\mathrm{LiF} + 3\mathrm{AlF_3}$ , carried out in diethyl ether — this is a small-scale/lab preparation, not an industrial one.

Industrial method: sodium hydride is reacted with $\mathrm{BF_3}$ at 450 K: $2\mathrm{BF_3} + 6\mathrm{NaH} \xrightarrow{450\,K} \mathrm{B_2H_6} + 6\mathrm{NaF}$ . This is the route used on an industrial scale because NaH is far cheaper and easier to handle in bulk than $\mathrm{LiAlH_4}$ .

Oxidation of $\mathrm{NaBH_4}$ with $\mathrm{I_2}$ gives diborane too, but it is used as a convenient small-scale/lab method, not the industrial process. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{NaBH_4 + I_2 \longrightarrow A + NaI + H_2}$
›Reveal solutionSolution
This is a standard boron-hydride synthesis sequence: sodium borohydride with iodine gives diborane, which then reacts with lithium hydride to give lithium borohydride.

Concept and Intuition

Diborane ( $B_2H_6$ ) is classically prepared in the laboratory by the reaction of sodium borohydride with iodine. It also reacts with electron-rich hydrides like LiH to form borohydride salts by accepting a hydride ion.

Step-by-Step Solution

$NaBH_4+I_2\rightarrow A+NaI+H_2$ : this is the known laboratory preparation of diborane, so $A=B_2H_6$ : $2NaBH_4+I_2\rightarrow B_2H_6+2NaI+H_2$ .

$A+LiH\rightarrow B$ : diborane reacts with lithium hydride (a hydride donor) to form lithium borohydride: $B_2H_6+2LiH\rightarrow2LiBH_4$ , so $B=LiBH_4$ .

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$SO_2$
›Reveal solutionSolution
The Contact Process sources its $SO_2$ feedstock from burning sulfur and roasting iron pyrites ( $FeS_2$ ) — the standard raw materials taught for industrial $H_2SO_4$ manufacture.

Concept and Intuition

The Contact Process converts $SO_2$ to $SO_3$ (catalytically, over $V_2O_5$ ) and then to $H_2SO_4$ . The very first step is generating $SO_2$ itself, which industrially comes from two main sources:

Burning elemental sulfur in air/oxygen: $S + O_2 \rightarrow SO_2$ .

Roasting (oxidative roasting) of iron pyrites (fool's gold, $FeS_2$ ): $4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$ .

Other sulfur-containing species like $FeS$ , $H_2S$ , or $Na_2S$ are not the standard industrial feedstocks for this process (in fact roasting $FeS$ is not the conventional pyrites-roasting reaction taught, and $H_2S$ / $Na_2S$ are not the textbook sources).

Step-by-Step Solution

Recall that the Contact Process needs a supply of $SO_2$ as its starting material. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$\mathrm{HNO_3}$
›Reveal solutionSolution
The standard lab preparation of $N_2$ from $NH_4Cl$ and $NaNO_2$ gives a gas contaminated with NO and $HNO_3$ ; this is purified by passing through acidified potassium dichromate solution, i.e. $H_2SO_4$ (aq) containing $K_2Cr_2O_7$ — option (B).

Concept and Intuition

The reaction $NH_4Cl(aq) + NaNO_2(aq) \xrightarrow{\Delta} N_2(g) + NaCl(aq) + 2H_2O(l)$ is the classic laboratory route to dinitrogen gas. However, due to side reactions during the heating, the evolved $N_2$ carries small amounts of oxides of nitrogen (NO) and nitric acid vapor ( $HNO_3$ ) as impurities. Since $N_2$ itself is chemically very inert (strong $N\equiv N$ triple bond), it passes through an oxidizing acidic scrubbing solution unaffected, while the reactive NO/ $HNO_3$ impurities get oxidized/absorbed and stripped out. Acidified potassium dichromate ( $K_2Cr_2O_7$ in $H_2SO_4$ ) is precisely such an oxidizing scrubber and is the standard purification step taught for this preparation.

Step-by-Step Solution

Recall the preparation: heating aqueous $NH_4Cl$ and $NaNO_2$ gives $N_2$ gas, along with trace NO and $HNO_3$ as side-product impurities.

To purify, the gas stream is bubbled through a scrubbing solution that will react with/absorb the impurities but not the inert $N_2$ .

Acidified $K_2Cr_2O_7$ (an oxidizing agent) reacts with and removes the NO/ $HNO_3$ impurities, letting pure $N_2$ pass through. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
$(NH_4)_2BeF_4$
›Reveal solutionSolution
This tests a specific preparation method of beryllium fluoride; $BeF_2$ comes from thermal decomposition of $(NH_4)_2BeF_4$ .

Concept and Intuition

Beryllium, being the smallest and most electronegative of the alkaline earth metals, forms compounds through routes distinct from its heavier congeners; $BeF_2$ 's standard preparation exploits the thermal instability of the ammonium fluoroberyllate salt.

Step-by-Step Solution

$BeF_2$ is prepared by first forming the complex ammonium salt $(NH_4)_2BeF_4$ (ammonium tetrafluoroberyllate).

On heating, this salt decomposes:

$(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F$

This is the standard textbook preparation of anhydrous $BeF_2$ , distinguishing Be from the other Group 2 metals (whose fluorides are prepared differently, e.g. via direct reaction with $F_2$ /HF). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
$\mathrm{NH_4OH}$
›Reveal solutionSolution
The laboratory preparation of pure $N_2$ is by heating a mixture of ammonium chloride and sodium nitrite. Answer: (B).

Concept and Intuition

Ammonium salts contain $N$ in the $-3$ oxidation state, while nitrites contain $N$ in the $+3$ state. When heated together in solution, these comproportionate: the ammonium ion is oxidised and the nitrite ion is reduced, both converging to $N_2$ (oxidation state $0$ ), which escapes as a gas along with water. This is the standard lab-scale route to pure, dry nitrogen (avoiding the traces of oxides of nitrogen/other impurities that atmospheric nitrogen or other industrial routes carry).

Step-by-Step Solution

Write the mixture: aqueous $NH_4Cl$ + aqueous $NaNO_2$ .

On gentle heating: $NH_4Cl + NaNO_2 \xrightarrow{\Delta} NaCl + 2H_2O + N_2\uparrow$ .

Check the redox: $N$ in $NH_4^+$ is $-3 \to 0$ (oxidised, loses $3e^-$ ); $N$ in $NO_2^-$ is $+3 \to 0$ (reduced, gains $3e^-$ ) — electrons balance, confirming this comproportionation is valid. …
Start your 14-day free trial to see the full solution →

←Previous questionQ33 — Although +3 oxidation state is the characteristic oxidation state of l…→Next questionQ35 — When orange solution containing Cr₂O₇²⁻ ion is treated with an alkali,…

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

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days