A solution of $KMnO_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

A solution of KMnO₄ on reduction yields either a colourless solution or a brown precipitate or a green solutio

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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 reduction of $KMnO_4$ proceeds through distinct colour changes depending on pH: in acidic medium it gives colourless $Mn^{2+}$ , in neutral/weakly alkaline it gives brown $MnO_2$ precipitate, and in strongly alkaline it gives green $MnO_4^{2-}$ . These represent successive stages of manganese reduction from +7 to +2.

The key to understanding this lies in the variable oxidation states of manganese and how pH controls the stability of the intermediate species. Permanganate ion ( $MnO_4^-$ ) is a powerful oxidising agent in all media, but the products differ because the reduction potential and the stability of manganese species change dramatically with pH.

Let me walk through each case systematically.

Acidic medium (pH < 1–2) In strong acid, the reduction goes all the way to $Mn^{2+}$ , which is colourless in dilute solution. The half-reaction is:

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

The $E^\circ$ is +1.51 V, making it the most powerful oxidising condition.

How to carry it out: Add dilute $H_2SO_4$ to the $KMnO_4$ solution, then add a reducing agent like oxalic acid, $FeSO_4$ , or $H_2O_2$ . The purple colour fades to colourless as $Mn^{2+}$ forms.

Neutral or weakly alkaline medium (pH ~7–9) Here the reduction stops at $MnO_2$ , a brown insoluble precipitate. The half-reaction is:

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

Notice that water provides the oxygen, and hydroxide ions are produced — so the solution becomes alkaline as the reaction proceeds.

How to carry it out: Simply add a reducing agent (like $Na_2SO_3$ or $KI$ ) to a neutral $KMnO_4$ solution. No acid or strong base is added. The purple colour turns brown as $MnO_2$ precipitates.

Strongly alkaline medium (pH > 12) In concentrated alkali, the reduction yields the green manganate ion $MnO_4^{2-}$ (oxidation state +6). The half-reaction is:

MnO_4^- + e^- \rightarrow MnO_4^{2-}

This is a one-electron reduction. The green colour is characteristic of $MnO_4^{2-}$ .

How to carry it out: Add excess $KOH$ or $NaOH$ to $KMnO_4$ solution (making it strongly alkaline), then add a mild reducing agent like $KI$ or $Na_2SO_3$ in small amounts. Alternatively, you can heat solid $KMnO_4$ with $KOH$ — but that's a different method. …

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Method: pH-Dependent Reduction of Permanganate

This problem is solved using the Redox Speciation Method — tracking how the oxidation state of manganese changes with pH.

Concept First (Why this happens)

$KMnO_4$ contains manganese in its +7 oxidation state. The reduction product depends on the $H^+$ concentration because:

In acidic medium, $H^+$ ions are available to stabilize lower oxidation states as cations.
In neutral/weakly basic medium, $MnO_2$ (insoluble) forms.
In strongly basic medium, the manganate ion ( $MnO_4^{2-}$ ) is stable.

Steps of the Method

Step 1: Identify the three reduction stages

pH condition	Product	Colour	Mn oxidation state
Acidic	$Mn^{2+}$ (aq)	Colourless	+2
Neutral/weakly basic	$MnO_2$ (s)	Brown precipitate	+4
Strongly basic	$MnO_4^{2-}$ (aq)	Green solution	+6

Step 2: Write the half-reactions for each case

Acidic medium (colourless $Mn^{2+}$ ):

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

Neutral/weakly basic (brown $MnO_2$ ):

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

Strongly basic (green $MnO_4^{2-}$ ):

MnO_4^- + e^- \rightarrow MnO_4^{2-}

Step 3: How to carry out each reduction

| Desired product | Reducing agent | Conditions |

|----------------|----------------|------------| …

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Here is a breakdown of the common mistakes students make on this classic inorganic synthesis question, along with how to avoid each.

The Core Concept (The "Why")

The key is that $KMnO_4$ (manganese in +7 oxidation state) is a powerful oxidising agent. The pH of the solution dictates the final reduction product of manganese because the reduction half-reaction involves $H^+$ ions.

In Acidic Medium ( $pH < 7$ ): $MnO_4^-$ is reduced to the colourless $Mn^{2+}$ ion.
In Neutral/Faintly Alkaline Medium ( $pH \approx 7-9$ ): $MnO_4^-$ is reduced to a brown precipitate of $MnO_2$ .
In Strongly Alkaline Medium ( $pH > 10$ ): $MnO_4^-$ is reduced to a green solution of $MnO_4^{2-}$ (manganate ion).

Common Mistake #1: Confusing the Colour of the Products

The Mistake: Students often mix up which product is formed in which medium. For example, they might say "green solution in acidic medium" or "colourless solution in alkaline medium."

How to Avoid It:

Memorise the "pH-Colour" Triad: Create a simple mental map.
- Acid → Colourless ( $Mn^{2+}$ )
- Neutral → Brown ( $MnO_2$ )
- Alkaline → Green ( $MnO_4^{2-}$ )
Use a Mnemonic: "Acid gives Clear, Neutral gives Brown, Alkaline gives Green." (A-C, N-B, A-G).
Visualise the Ions: Remember that $Mn^{2+}$ is a very pale pink (appears colourless in dilute solution), $MnO_2$ is a solid brown precipitate, and $MnO_4^{2-}$ is a distinct green colour in solution.

Common Mistake #2: Writing the Wrong Half-Reactions

The Mistake: Students write the reduction half-reaction incorrectly, especially forgetting to balance $H^+$ and $H_2O$ or using the wrong number of electrons.

How to Avoid It:

Always Balance by the "ION-ELECTRON" Method: For each medium, write the balanced half-reaction. This is non-negotiable for exam accuracy.

1. Acidic Medium (to $Mn^{2+}$ ):

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

*Note: 5 electrons are gained. The solution becomes colourless.*

**2. Neutral/Faintly Alkaline Medium (to $MnO_2$):**

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

*Note: 3 electrons are gained. The brown precipitate is $MnO_2$.*

**3. Strongly Alkaline Medium (to $MnO_4^{2-}$):**

MnO_4^- + e^- \rightarrow MnO_4^{2-}

*Note: Only 1 electron is gained. The green colour is due to the manganate ion.*

Check the Number of Electrons: The number of electrons gained decreases as the pH increases (5 → 3 → 1). This is a good sanity check.

Common Mistake #3: Forgetting the "How" (The Reagents)

The Mistake: Students can state the products but cannot describe how to carry out the reduction (e.g., what reagent to add).

How to Avoid It:

Learn the Specific Reducing Agents: The question asks "how are they carried out?" You must know the common reagents.
- For Colourless $Mn^{2+}$ (Acidic): Add a reducing agent like oxalic acid ( $H_2C_2O_4$ ) or ferrous sulphate ( $FeSO_4$ ) in the presence of dilute $H_2SO_4$ .
  - Example: $2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
- For Brown $MnO_2$ (Neutral): Add a reducing agent like sodium sulphite ( $Na_2SO_3$ ) or hydrogen peroxide ( $H_2O_2$ ) in neutral or faintly alkaline conditions.
  - Example: $2KMnO_4 + 3Na_2SO_3 + H_2O \rightarrow 2MnO_2 + 3Na_2SO_4 + 2KOH$
- For Green $MnO_4^{2-}$ (Strongly Alkaline): Add a reducing agent like potassium sulphite ( $K_2SO_3$ ) or potassium iodide ( $KI$ ) in a concentrated solution of KOH or NaOH. …

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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. …
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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. …
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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. …
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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. …
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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. …
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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. …
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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. …
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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 …
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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. …
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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. …
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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). …
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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. …
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Q.A solution of KMnO4KMnO_4KMnO4​ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

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Q.A solution of $KMnO_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

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