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.

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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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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