When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water,

NCERT Exemplar · Q44

Q. When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

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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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This is a classic manganese redox sequence: MnO₂ (A) is oxidised to K₂MnO₄ (B, dark green), which disproportionates to KMnO₄ (C, purple) and MnO₂. KMnO₄ then oxidises KI to KIO₃ (D) in alkaline medium, reforming MnO₂.

The entire problem revolves around the variable oxidation states of manganese — from +4 in MnO₂ to +6 in manganate, then +7 in permanganate, and back down. Each colour change is a direct signal of the oxidation state shift. Let’s walk through it step by step.

Identifying compound A

The starting material is an “oxide of manganese”. The most common stable oxide is manganese dioxide, MnO₂. It is a brown-black solid, widely used in dry cells and as an oxidising agent. When fused with KOH and an oxidising agent (like KNO₃ or air), MnO₂ is oxidised from Mn(IV) to Mn(VI).

So A = MnO₂.
Formation of compound B — the dark green solution

The fusion reaction:

2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \xrightarrow{\Delta} 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}

(KNO₃ can replace O₂ as the oxidiser.)

The product is potassium manganate, K₂MnO₄, which dissolves in water to give a dark green solution. The colour comes from the manganate ion, MnO₄²⁻, where Mn is in the +6 state.

So B = K₂MnO₄.

Disproportionation of B to give purple compound C In neutral or acidic solution, the manganate ion is unstable and disproportionates:

3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}

One Mn(VI) is oxidised to Mn(VII) (purple permanganate) and another is reduced to Mn(IV) (brown MnO₂). The purple colour is unmistakable.

So C = KMnO₄ (or simply the permanganate ion, MnO₄⁻).

Watch out

A common mistake is to think that K₂MnO₄ directly gives KMnO₄ without any byproduct. But the disproportionation always produces MnO₂ as well — that’s why the solution turns muddy if not filtered.

Reaction of C with KI in alkaline medium The problem says: “An alkaline solution of compound C oxidises potassium iodide solution to a compound D and compound A is also formed.” …

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Here is the clear solution method for this problem.

Method: Systematic Deduction via Oxidation States & Colour

This method uses the characteristic colours and oxidation states of manganese compounds, combined with the logic of redox reactions (disproportionation and oxidation), to identify each species step-by-step.

Step 1: Identify Compound (A) from the starting material.

The problem begins with "an oxide of manganese (A)". The most common and stable manganese oxide is manganese dioxide ( $MnO_2$ ).
Result: A = $MnO_2$ (Manganese dioxide, brown-black solid).

Step 2: Identify Compound (B) from the fusion reaction.

Reaction: $MnO_2$ is fused with KOH in the presence of an oxidising agent (like $KNO_3$ or air). This oxidises Mn(IV) to Mn(VI).
Colour clue: The product gives a dark green solution in water. The only common manganese compound with a dark green colour is potassium manganate ( $K_2MnO_4$ ).
Result: B = $K_2MnO_4$ (Potassium manganate, dark green).
Reaction: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$

Step 3: Identify Compound (C) from disproportionation.

Reaction: Compound B ( $K_2MnO_4$ ) disproportionates in neutral or acidic solution. Mn(VI) is unstable and converts to Mn(VII) and Mn(IV).
Colour clue: The product is purple. The only common purple manganese compound is potassium permanganate ( $KMnO_4$ ).
Result: C = $KMnO_4$ (Potassium permanganate, purple).
Disproportionation reaction:
- In neutral solution: $3K_2MnO_4 + 2H_2O \rightarrow 2KMnO_4 + MnO_2 + 4KOH$
- In acidic solution: $2K_2MnO_4 + 2H_2SO_4 \rightarrow 2KMnO_4 + MnO_2 + 2K_2SO_4 + 2H_2O$
- (Note: $MnO_2$ is also formed here, which is compound A). …

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Here are the common mistakes students make on this classic inorganic synthesis problem, along with the correct conceptual approach to avoid them.

Mistake 1: Misidentifying Compound (A) — The Starting Oxide

The Mistake: Students often guess $\text{MnO}_2$ (manganese dioxide) for (A) because it is the most common oxide of manganese. While $\text{MnO}_2$ can be used, the problem specifies the oxide is fused with KOH in the presence of an oxidising agent. This implies (A) is in a lower oxidation state so it can be oxidised to the dark green compound.
Why it’s wrong: If (A) were $\text{MnO}_2$ (Mn in +4 state), fusing it with KOH and an oxidant (like air or $\text{KNO}_3$ ) would still work, but the problem’s wording “an oxide of manganese (A) is fused…” often points to the most common lower oxide that undergoes this specific fusion.
The Correct Approach: Recall the fusion of $\text{MnO}_2$ with KOH and an oxidant gives potassium manganate ( $\text{K}_2\text{MnO}_4$ ). However, the problem’s phrasing “an oxide of manganese (A)” is ambiguous. The most common exam answer is that (A) is $\text{MnO}_2$ (manganese dioxide). The fusion reaction is:

2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \xrightarrow{\Delta} 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}

**Avoid this mistake by:** Remembering that the fusion of any manganese oxide (usually $\text{MnO}_2$) with KOH in air/oxidant yields the manganate ion ($\text{MnO}_4^{2-}$), which is dark green.

Mistake 2: Confusing Compound (B) — The Dark Green Solution

The Mistake: Thinking the dark green solution is potassium permanganate ( $\text{KMnO}_4$ ), which is purple.
Why it’s wrong: Permanganate ( $\text{MnO}_4^-$ ) is purple. The dark green colour is characteristic of the manganate ion ( $\text{MnO}_4^{2-}$ ).
The Correct Approach: Compound (B) is potassium manganate ( $\text{K}_2\text{MnO}_4$ ). The colour comes from the $\text{MnO}_4^{2-}$ $MnO_{4}^{2 -}$ ion. Remember the colour code:
- Green → Manganate ( $\text{MnO}_4^{2-}$ , +6 oxidation state)
- Purple → Permanganate ( $\text{MnO}_4^-$ , +7 oxidation state)

Mistake 3: Forgetting the Disproportionation Condition

The Mistake: Stating that compound (B) disproportionates in any solution, or writing the wrong products.
Why it’s wrong: The disproportionation of manganate is pH-dependent. In strongly alkaline solution, $\text{MnO}_4^{2-}$ is stable. It only disproportionates when the solution is made neutral or acidic.
The Correct Approach: The reaction is:

3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}

- **Purple product:** $\text{MnO}_4^-$ (compound C)
- **Brown precipitate:** $\text{MnO}_2$ (which is actually compound A again!)
**Avoid this mistake by:** Always linking disproportionation of manganate to the addition of acid or $\text{CO}_2$ (which makes the solution acidic via carbonic acid).

Mistake 4: Misidentifying Compound (D) — The Product of Iodide Oxidation

The Mistake: Saying compound (D) is $\text{I}_2$ (iodine), by analogy with permanganate's reaction with iodide in acidic medium.
Why it’s wrong: The question specifies an alkaline solution of compound C. In alkaline medium, iodide is oxidised all the way to iodate ( $\text{IO}_3^-$ ) — any $\text{I}_2$ formed does not survive and is itself further oxidised by the excess $\text{OH}^-$ /permanganate present.
The Correct Approach: The reaction of alkaline $\text{KMnO}_4$ (compound C) with KI: $2\text{KMnO}_4 + \text{KI} + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + \text{KIO}_3 + 2\text{KOH}$ …

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

TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$(\mathrm{CH}_3)_3\mathrm{CH} \xrightarrow{\mathrm{KMnO}_4} X \xrightarrow{\mathrm{Cu}\ 573\mathrm{K}} Y$
›Reveal solutionSolution
The reaction sequence oxidises isobutane to tert-butyl alcohol, then dehydrates it to isobutylene; Y has two sp³ and two sp² carbons, so the answer is (C).

The key is to track the carbon skeleton through two transformations. The starting molecule is isobutane, $(\mathrm{CH}_3)_3\mathrm{CH}$ . The first step uses hot $\mathrm{KMnO}_4$ , a strong oxidising agent. The second step uses copper at 573 K, a classic dehydration condition for alcohols.

Identify the starting compound and the first reaction.

Isobutane has a tertiary carbon (the CH group) and three methyl groups. Hot $\mathrm{KMnO}_4$ oxidises alkanes at the most substituted carbon, converting a C–H bond to a C–OH bond. So the product X is tert-butyl alcohol: $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{OH}$ .

Why this works: The tertiary C–H is the weakest, so it’s attacked first; permanganate inserts oxygen.

Determine the second reaction.

Passing an alcohol over copper at 573 K (about 300 °C) causes dehydration. For a tertiary alcohol like tert-butyl alcohol, elimination is very easy — it loses water to form an alkene. The only possible alkene here is isobutylene (2-methylpropene): $(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2$ .

Why this works: Tertiary alcohols dehydrate via an E1 mechanism; the carbocation intermediate is stable, and the most substituted alkene (Saytzeff product) forms.

Count the hybridisation in Y (isobutylene).

Draw the structure:
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$\mathrm{CuCO_3.Cu(OH)_2}$
›Reveal solutionSolution
Siderite is an iron carbonate ore. Its chemical formula is $\mathrm{FeCO_3}$ , which corresponds to option (B).

The question asks for the composition of siderite ore. This is a straightforward recall from metallurgy — the study of how metals are extracted from their ores. Each ore has a distinct chemical identity, and siderite is one of the common iron ores.

The key idea: ores are naturally occurring minerals from which metals are extracted. Iron ores include hematite ( $\mathrm{Fe_2O_3}$ ), magnetite ( $\mathrm{Fe_3O_4}$ ), limonite ( $\mathrm{Fe_2O_3 \cdot 3H_2O}$ ), and siderite. The name "siderite" comes from the Greek word for iron, sideros, which hints at its composition.

Let’s go through the options one by one.

Option (A): $\mathrm{CuCO_3 \cdot Cu(OH)_2}$

This is malachite, a copper ore. The dot here indicates a hydrated or basic carbonate — copper carbonate combined with copper hydroxide. Not siderite.

Option (B): $\mathrm{FeCO_3}$

This is iron(II) carbonate. Siderite is indeed ferrous carbonate. It forms in sedimentary environments and is an important but less common iron ore. The iron is in the +2 oxidation state.

Option (C): $\mathrm{CuFeS_2}$ …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$_2$
›Reveal solutionSolution
The key idea is that alkali metals form different oxides depending on their size and the oxygen supply: larger metals (like Cs) form superoxides, while smaller metals (like Na) form peroxides. The correct option is (C).

The relevant concept here is the trend in oxide formation among alkali metals. As you go down Group 1, the metal cation becomes larger and less polarizing. This stabilizes larger, more complex oxygen anions:

Small Li forms only the normal oxide (O²⁻).

Na forms the peroxide (O₂²⁻) under normal conditions.

K, Rb, and Cs form superoxides (O₂⁻) because the large cation stabilizes the big superoxide ion. Also, excess vs. limited oxygen doesn’t change the product for Cs — it always forms the superoxide.

Let’s work through it step by step:

Identify the product for Cs with oxygen

Cesium is the largest alkali metal. Its cation (Cs⁺) is very large and has low charge density, which stabilizes the large superoxide ion (O₂⁻).

With excess O₂: Cs + O₂ → CsO₂ (superoxide).

With limited O₂: Cs still forms CsO₂ because the superoxide is the most stable oxide for Cs. So X = superoxide.

Identify the product for Na with oxygen

Sodium is smaller than Cs. Its cation (Na⁺) has a higher charge density, which stabilizes the peroxide ion (O₂²⁻) better than the superoxide.

Na + O₂ → Na₂O₂ (peroxide). So Y = peroxide.

Match with the options …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$Q. Match the following List-I (Ore name) A. Calamine B. Copper pyrites C. Zincite D. Malachite List-II (Type of ore) I. Oxide II. Sulphide III. Carbonate-hydroxide IV. Carbonate The correct answer is (A) A - II, B - I, C - IV, D - III (B) A - IV, B - III, C - I, D - II (C) A - IV, B - II, C - I, D - III (D) A - III, B - II, C - IV, D - I$
›Reveal solutionSolution
This is a matching problem linking ore names to their chemical type. The key is knowing the chemical composition of each ore: Calamine is zinc carbonate, Copper pyrites is a sulphide, Zincite is zinc oxide, and Malachite is a basic copper carbonate (carbonate-hydroxide). The correct match is A–IV, B–II, C–I, D–III, which corresponds to option (C).

The question tests your memory of common ores and their classification by chemical nature — oxide, sulphide, carbonate, or carbonate-hydroxide. In metallurgy, ores are grouped by the anion present: oxides (O²⁻), sulphides (S²⁻), carbonates (CO₃²⁻), and sometimes mixed salts like basic carbonates (carbonate-hydroxide). Getting this right is just a matter of recalling the formula of each ore.

Let’s go through each one.

Calamine – This is an ore of zinc. Its chemical formula is ZnCO₃. That’s a pure carbonate — no hydroxide group. So it belongs to List-II: IV (Carbonate).

Copper pyrites – Also called chalcopyrite, this is the most common copper ore. Its formula is CuFeS₂. The presence of S²⁻ makes it a sulphide. So it matches List-II: II (Sulphide).

Zincite – Another zinc ore, but this one is ZnO. That’s straightforward — an oxide. So it goes with List-II: I (Oxide). …
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TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
$_5$
›Reveal solutionSolution
The compound CrO $_5$ contains peroxide linkages, which means not all oxygen atoms have an oxidation state of -2. Accounting for these, the oxidation state of Cr is $\boxed{+6}$ .

The oxidation state of an element in a compound represents the hypothetical charge it would have if all bonds were ionic. While there are general rules for assigning oxidation states, it's crucial to understand that these rules have exceptions, especially when dealing with peroxides or superoxides.

For oxygen, the most common oxidation state is -2. However, in peroxides (containing an O-O bond), each oxygen atom has an oxidation state of -1. In superoxides, it's -1/2. When calculating oxidation states, if the standard assumption (e.g., O = -2) leads to an impossibly high or low oxidation state for the central atom, it's a strong indicator that an exception to the general rules is present, and the compound's structure needs to be considered.

Let's determine the oxidation state of Cr in CrO $_5$ :

Initial Calculation (and why it's incorrect):

If we assume all five oxygen atoms have an oxidation state of -2 (the most common state for oxygen), then for the neutral compound CrO $_5$ :

Let the oxidation state of Cr be $x$ .

$x + 5 \times (-2) = 0$

$x - 10 = 0$

$x = +10$

This result, +10, is impossible for chromium. Chromium is a d-block element in Group 6, and its maximum possible oxidation state is +6 (corresponding to the total number of valence electrons it can lose). An oxidation state of +10 would imply the loss of more electrons than it possesses, which is chemically unfeasible. This tells us that our initial assumption about all oxygen atoms being -2 is incorrect.

Understanding the Structure of CrO $_5$ :

The compound CrO $_5$ has a characteristic "butterfly" structure. This structure reveals the presence of peroxide linkages.

It consists of:

One oxygen atom double-bonded to chromium (a normal oxide oxygen).

Two peroxide linkages, each containing two oxygen atoms. This means there are four oxygen atoms involved in O-O bonds. So, out of the five oxygen atoms, one is a normal oxide, and four are peroxide oxygens.

Assigning Correct Oxidation States to Oxygen:

The oxygen atom that is double-bonded to chromium (Cr=O) has an oxidation state of -2. …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$Q. The reagent used in the determination of the chemical oxygen demand (COD) of water sample is (A) Potassium permanganate (B) Sulphuric acid (C) Acidified Potassium dichromate (D) Potassium dichromate$
›Reveal solutionSolution
The chemical oxygen demand (COD) of a water sample is determined by oxidizing organic and inorganic matter using a strong oxidizing agent in an acidic medium. The primary reagent used for this purpose is acidified potassium dichromate.

The Chemical Oxygen Demand (COD) test is a crucial measure of water quality, indicating the amount of oxygen required to chemically oxidize all oxidizable organic and inorganic substances present in a water sample. This test provides a rapid estimate of the pollution load in wastewater.

The core idea behind COD determination is to subject the water sample to a powerful chemical oxidation process. A strong oxidizing agent is used to break down the complex organic molecules into simpler, stable inorganic end products like carbon dioxide and water. The amount of oxidant consumed is then measured, which is directly proportional to the oxygen equivalent of the oxidizable matter.

Here's why a specific reagent system is chosen:

Need for a Strong Oxidizing Agent: To ensure nearly complete oxidation of a wide range of organic compounds, a very strong oxidizing agent is required.

Potassium dichromate ( $K_2Cr_2O_7$ ) is an excellent choice because it is a powerful oxidant, stable, and can oxidize almost all types of organic compounds (except for a few aromatic hydrocarbons and pyridine) to carbon dioxide and water. Its oxidation efficiency is high and consistent.

Potassium permanganate ( $KMnO_4$ ), while also an oxidant, is generally less effective than dichromate for COD determination. Its oxidizing power can vary with pH, and it does not oxidize all organic compounds as completely as dichromate, leading to lower and less reliable COD values. It is more commonly used in other specific titrations or for measuring oxygen consumed in milder conditions (e.g., Permanganate Value).

Role of Acidification: The oxidizing power of potassium dichromate is significantly enhanced in an acidic environment.

Sulphuric acid ( $H_2SO_4$ ) is added to provide the necessary acidic medium. In this acidic environment, the dichromate ion ( $Cr_2O_7^{2-}$ ) is converted to chromic acid, which is a much stronger oxidizing agent. The reaction proceeds as follows: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ …
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