Q.Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
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🔒 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, CuSOX4⋅5HX2O.
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:
Cu+2HX2SOX4(conc⋅)CuSOX4+SOX2+2HX2O
Then you evaporate the solution carefully. Blue crystals of CuSOX4⋅5HX2O 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., 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:
Percentage Yield=Theoretical YieldActual Yield×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 FeClX3 from Fe and ClX2, if you have 1 mol Fe and 2 mol ClX2, Fe is limiting (1:1.5 stoichiometry), so theoretical yield is based on Fe.
2. The Atom Economy Formula: Why It Measures “Greenness”
Atom Economy=Sum of Molecular Masses of All ReactantsMolecular Mass of Desired Product×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 A+BC) means less waste. A low atom economy (e.g., substitution reactions with leaving groups) means more byproducts.
Example: In the synthesis of NaCl from Na and ClX2:
2Na+ClX2→2NaCl
Atom economy = 2×22.99+70.902×58.44×100%=100% — because all atoms end up in the product.
3. The Solubility Product and Precipitation: Why Ksp Controls Synthesis
For a sparingly soluble salt like AgCl:
AgCl(s)AgX+(aq)+ClX−(aq)
Ksp=[AgX+][ClX−]
Why this holds:
- Ksp is an equilibrium constant derived from the law of mass action. It applies only to saturated solutions.
- In synthesis, you use Ksp to predict whether a precipitate will form when mixing solutions. If the ion product Q=[AgX+][ClX−] exceeds Ksp, precipitation occurs.
- The formula is temperature-dependent (because ΔG∘=−RTlnKsp). 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 Ksp 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:
LFSE=(−0.4×nt2g+0.6×neg)Δo
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 t2g (three orbitals) and the higher-energy eg (two orbitals).
- The splitting energy Δo is the energy difference between these sets.
- Electrons fill the t2g orbitals first (Hund’s rule), and each electron in t2g stabilizes the complex by −0.4Δo relative to the barycenter (average energy). Each electron in eg destabilizes by +0.6Δo.
- The formula explains why certain coordination numbers are preferred: for example, [Co(HX2O)X6]X2+ (high-spin d7) has LFSE = −0.8Δo, while [CoClX4]X2− (tetrahedral) has a smaller LFSE — so the octahedral form is more stable. …
The key idea is that potassium dichromate (K2Cr2O7) acts as a strong oxidising agent in acidic medium, where the dichromate ion (Cr2O72−) is reduced to Cr3+ (green). The half-reaction is:
Cr2O72−+14H++6e−→2Cr3++7H2O
Step 1 – Reaction with iodide (I−): Iodide is oxidised to iodine (I2). Balancing electrons (6e⁻ from dichromate, 2e⁻ per I2 molecule) gives:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
Step 2 – Reaction with iron(II) solution (Fe2+): Fe2+ is oxidised to Fe3+ (1e⁻ each). Balancing:
Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+ …
Potassium dichromate in acidic medium acts as a strong oxidising agent because the dichromate ion (Cr2O72−) gets reduced to Cr3+, gaining six electrons. It oxidises iodide to iodine, iron(II) to iron(III), and hydrogen sulphide to sulphur.
Why Potassium Dichromate is an Oxidising Agent
The oxidising power of potassium dichromate (K2Cr2O7) comes from chromium in its +6 oxidation state. In acidic solution, the dichromate ion accepts electrons and gets reduced to the green Cr3+ ion. The half-reaction is:
Cr2O72−+14H++6e−→2Cr3++7H2O
This is a six-electron reduction. The standard reduction potential (E∘=+1.33 V) is high enough to oxidise many common reducing agents. The reaction is strongly favoured in acidic medium — in neutral or alkaline conditions, dichromate converts to chromate (CrO42−), which is a much weaker oxidant.
A common mistake is to forget that the reduction of dichromate consumes 14 H⁺ ions. If the medium is not sufficiently acidic, the reaction slows down or stops. Always write the full ionic equation with H+ and H2O.
Step-by-Step Ionic Equations
1. Reaction with Iodide (I−)
Iodide is oxidised to iodine. Each I− loses one electron, while the dichromate ion accepts six electrons, so six iodide ions are needed to supply those six electrons.
Half-reactions:
- Oxidation: 2I−→I2+2e− (but this gives only 2 electrons; we need 6)
- Multiply by 3: 6I−→3I2+6e−
- Reduction: Cr2O72−+14H++6e−→2Cr3++7H2O
Combined:
Cr2O72−+14H++6I−→2Cr3++3I2+7H2O
The iodine produced gives a brown colour in solution, or a violet colour if extracted into an organic solvent like chloroform.
To balance redox equations quickly: balance atoms other than H and O first, then balance O with H2O, then H with H+, and finally charge with electrons. Then make electrons equal in both halves.
2. Reaction with Iron(II) Solution (Fe2+)
Iron(II) is oxidised to iron(III). Each Fe2+ loses one electron. Since dichromate accepts six electrons, we need six Fe2+ ions.
Half-reactions:
- Oxidation: Fe2+→Fe3++e− (multiply by 6)
- 6Fe2+→6Fe3++6e−
- Reduction: Cr2O72−+14H++6e−→2Cr3++7H2O
Combined:
Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2O
This is a classic titration reaction used to estimate iron(II) in solution. The colour change from orange (dichromate) to green (Cr³⁺) marks the endpoint.
In the lab, this reaction is often done in the presence of dilute H2SO4. Hydrochloric acid is avoided because chloride ions can also be oxidised by dichromate, interfering with the result.
3. Reaction with Hydrogen Sulphide (H2S) …
Method: Oxidation–Reduction (Redox) Half-Reaction Method
This method breaks the overall reaction into two half-reactions — oxidation and reduction — then balances atoms and charge step-by-step.
Steps:
- Identify the oxidising agent (gets reduced) and the reducing agent (gets oxidised).
- Write the half-reaction for reduction of dichromate.
- Write the half-reaction for oxidation of the given species.
- Balance atoms other than H and O.
- Balance oxygen by adding H2O.
- Balance hydrogen by adding H+ (acidic medium).
- Balance charge by adding electrons (e−).
- Multiply half-reactions so electrons cancel.
- Add the half-reactions and simplify.
Oxidising Action of Potassium Dichromate (K2Cr2O7)
In acidic medium, dichromate ion (Cr2O72−) is a strong oxidising agent. It gets reduced to green Cr3+:
Cr2O72−+14H++6e−→2Cr3++7H2O
This is the reduction half-reaction used in all three cases below.
(i) Reaction with Iodide (I−)
Oxidation half-reaction:
2I−→I2+2e−
Reduction half-reaction (from above):
Cr2O72−+14H++6e−→2Cr3++7H2O
Balance electrons: Multiply oxidation half by 3:
6I−→3I2+6e−
Add both half-reactions:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
Ionic equation:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
(ii) Reaction with Iron(II) Solution (Fe2+)
Oxidation half-reaction:
Fe2+→Fe3++e−
Reduction half-reaction:
Cr2O72−+14H++6e−→2Cr3++7H2O
Balance electrons: Multiply oxidation half by 6:
6Fe2+→6Fe3++6e−
Add both half-reactions:
Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+
Ionic equation: …
Here is a breakdown of the common mistakes students make when answering this question, along with the correct reasoning and exam-ready solutions.
The Core Concept (Why Students Slip Up)
The key to this question is stoichiometry and medium. Potassium dichromate (K2Cr2O7) is a strong oxidising agent only in acidic medium. In neutral or basic medium, its oxidising power is much weaker. Students often forget to specify the medium or balance the half-reactions incorrectly.
Common Mistake #1: Forgetting the Acidic Medium
The Mistake: Writing the reaction without mentioning H+ ions or writing the product as Cr3+ without balancing the oxygen with water and hydrogen ions.
Why it happens: Students memorise the half-reaction as:
Cr2O72−→2Cr3+
but forget that this is not balanced for charge or atoms.
How to Avoid:
Always write the balanced half-reaction in acidic medium:
Cr2O72−+14H++6e−→2Cr3++7H2O
- Check: Left side: 2 Cr, 7 O, 14 H, charge = -2 + 14 = +12. Right side: 2 Cr, 7 O, 14 H, charge = +6. The 6 electrons balance the charge.
- Exam tip: If the question says "acidified potassium dichromate", you must include H+ in the ionic equation.
Common Mistake #2: Incorrect Stoichiometry with Iodide (I−)
The Mistake: Writing the product as I2 but getting the mole ratio wrong (e.g., 1:1 instead of 1:6).
Why it happens: Students forget that each Cr2O72− gains 6 electrons, while each I− loses 1 electron to form 21I2.
How to Avoid:
- Step 1: Write the oxidation half-reaction:
2I−→I2+2e−
- Step 2: Multiply by 3 to match the 6 electrons gained by dichromate:
6I−→3I2+6e−
- Step 3: Combine with the reduction half-reaction:
Cr2O72−+14H++6e−→2Cr3++7H2O
- Final balanced equation:
Cr2O72−+14H++6I−→2Cr3++3I2+7H2O
Key result: 1 mole K2Cr2O7 oxidises 6 moles of I−.
Common Mistake #3: Confusing Iron(II) with Iron(III) Products
The Mistake: Writing Fe2+→Fe3+ but forgetting to balance the charge or writing the wrong number of electrons.
Why it happens: Students think it's a simple 1-electron transfer but forget to account for the 6 electrons from dichromate.
How to Avoid:
- Oxidation half-reaction:
Fe2+→Fe3++e−
- Multiply by 6:
6Fe2+→6Fe3++6e−
- Combine:
Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2O
Key result: 1 mole K2Cr2O7 oxidises 6 moles of Fe2+.
Common Mistake #4: Wrong Product for H2S Oxidation
The Mistake: Writing H2S→S but forgetting that H2S is a gas and S is a solid (precipitate). Or writing SO2 or SO42− as the product.
Why it happens: Students over-oxidise H2S because they think "strong oxidiser" means all the way to sulfate. But in acidic medium, H2S is typically oxidised only to elemental sulfur (S), not to SO2 or SO42−.
How to Avoid:
- Oxidation half-reaction:
H2S→S+2H++2e−
- Multiply by 3 to match 6 electrons:
3H2S→3S+6H++6e−
- Combine:
Cr2O72−+14H++3H2S→2Cr3++3S+7H2O+6H+
- Simplify H+: 14H+−6H+=8H+ on left side.
- Final: …
Showing the 12 most recent of 23 on this concept.
- CBSE 2025Set ANNUAL1 markQ.How will you prepare K2MnO4 from pyrolusite? (Give chemical equation only)
›Reveal solutionSolution
Fusion of pyrolusite (MnO2) with KOH in the presence of an oxidising agent (air/O2 or KNO3) gives potassium manganate.
Pyrolusite (MnO2) is fused with KOH in presence of air (or an oxidising agent like KNO3):
2MnO2+4KOH+O2fuse2K2MnO4+2H2O
…
- CBSE 2025Set ANNUAL1 markQ.How will you prepare Potassium dichromate from Sodium dichromate? (Give chemical equation only)
›Reveal solutionSolution
KCl is added to a solution of sodium dichromate; the less soluble potassium dichromate crystallises out.
Sodium dichromate solution is treated with potassium chloride:
Na2Cr2O7+2KCl→K2Cr2O7+2NaCl
…
- CBSE 2024Set 56/2/11 markMCQQ.When MnO2 is fused with KOH in air, it gives : (A) KMnO4 (B) K2MnO4 (C) Mn2O7 (D) Mn2O3
›Reveal solutionSolution
Fusing MnO2 with KOH in air oxidises Mn(IV) to Mn(VI), forming the green manganate ion MnO42−. The product is potassium manganate, K2MnO4, option (B).
This is a classic example of an oxidation reaction in a fused alkaline medium. The key is to track the oxidation state of manganese and the role of the environment.
Why this approach works: In solid-state or fused-salt reactions, the strong alkaline medium (KOH) and the oxidising power of atmospheric oxygen work together. MnO2 is already a common starting material for manganese chemistry. When you fuse it with KOH, you create a melt rich in OH− ions. Air (O2) acts as the oxidising agent, pulling electrons away from manganese. The Mn(IV) in MnO2 cannot stay at +4 in such a strongly oxidising, basic melt — it gets pushed to a higher stable state. The +6 state (manganate) is particularly stable in alkaline conditions, while the +7 state (permanganate) requires even stronger oxidising conditions or a different workup.
Let’s walk through the reasoning step by step.
-
Identify the starting oxidation state. In MnO2, oxygen is −2 (usual for oxides). Let the Mn oxidation state be x. Then x+2(−2)=0, so x=+4. Manganese is in the +4 oxidation state.
-
Recognise the reaction conditions. “Fused with KOH in air” means:
- High temperature (fusion) — the mixture is molten.
- Strongly basic medium — excess KOH provides OH− ions.
- Presence of atmospheric oxygen (O2) — a good oxidising agent.
-
Predict the likely product. In alkaline conditions, manganese can exist in several oxidation states. The +6 state, as the manganate ion MnO42−, is well-known and stable in basic solution. The +7 state, as permanganate MnO4−, is more stable in acidic or neutral conditions. Here, the basic melt favours the manganate. Also, O2 is a moderately strong oxidiser — it can take Mn from +4 to +6, but not easily to +7 (that usually requires a stronger oxidant like KNO3 or KClO3).
-
Write the balanced chemical equation. The reaction is:
2MnO2+4KOH+O2→2K2MnO4+2H2O
Check: Mn goes from +4 to +6 (loss of 2 electrons per Mn). O2 goes from 0 to −2 (gain of 4 electrons per O2). Two Mn atoms lose 4 electrons total, exactly balancing the gain by one O2 molecule. The KOH provides the potassium ions and the oxygen for the water. …
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- CBSE 2024Set ANNUAL1 markMCQQ.The chemical formula of chromite ore is -(a) MnO2(b) Na2Cr2O4(c) FeCr2O4(d) Na2CrO4
›Reveal solutionSolution
Chromite ore, the main source of chromium, has the formula FeCr2O4 (iron(II) chromite, a mixed oxide of iron and chromium).
Chromite crystallises in the spinel structure, in which Fe2+ ions occupy tetrahedral holes and Cr3+ ions occupy octahedral holes of a close-packed oxide lattice, giving the overall formula FeCr2O4 (equivalently FeO.Cr2O3). …
- CBSE 2023Set ANNUAL1 markMCQQ.Process of commercial production of nitric acid is(a) Haber process(b) Ostwald's process(c) Contact process(d) Deacon's process
›Reveal solutionSolution
Ostwald's process is named specifically for industrial nitric-acid manufacture, distinguishing it from Haber's (ammonia), Contact (sulphuric acid) and Deacon's (chlorine) processes.
In Ostwald's process, ammonia is catalytically oxidised over a Pt-Rh catalyst to nitric oxide, which is further oxidised to NO2 and then absorbed in water to give nitric acid:
4NH3 + 5O2 --(Pt/Rh, 500 K, 9 bar)--> 4NO + 6H2O …
- CBSE 2022Set M1 markQ.Name the method used for concentration of sulphide ore.
›Reveal solutionSolution
Sulphide ores are concentrated by the froth flotation process.
The froth flotation process is used to concentrate sulphide ores. The powdered ore is mixed with water and a collector/frother (e.g. pine oil); air is blown through. The sulphide ore particles are preferentially wetted by the oil and rise with t …
- CBSE 2022Set ANNUAL1 markMCQQ.Zone refining is used for obtaining ultra pure sample of(a) copper(b) sodium(c) germanium(d) zinc
›Reveal solutionSolution
Zone refining purifies a metal based on the difference in solubility of impurities in the molten vs solid state of the metal.
In zone refining, a mobile induction heater melts a narrow zone of an impure metal rod at one end and moves slowly to the other end. Impurities are more soluble in the molten zone than in the solid, so they get swept along with the moving molten zone and concentrate at one end, which is then cut off. This te …
- CBSE 2020Set ANNUAL1 markQ.Iron scraps are advisable and advantageous than zinc scraps for reducing the low grade copper ores. Why?
›Reveal solutionSolution
Iron and zinc both lie above copper in the reactivity series and can reduce Cu2+, but iron scrap is far cheaper and more abundant, so it is the economical choice.
Concept. In hydrometallurgy of copper, a low-grade ore is leached and the copper in solution is displaced by a more reactive metal:
Cu2+(aq)+M→Cu+M2+(aq)
where M must lie above copper in the activity series.
Reason. Both Fe and Zn are more reactive than Cu, so either can reduce Cu2+ to Cu:
Cu2++Fe→Cu+Fe2+ …
- CBSE 2020Set ANNUAL1 markQ.Complete the reaction XeF₆ + H₂O ⟶ ? + 2HF .
›Reveal solutionSolution
One molecule of water partially hydrolyses XeF6 to XeOF4, liberating 2HF.
Concept. Xenon hexafluoride is readily hydrolysed. The extent of hydrolysis depends on the amount of water. With a limited amount (1 mole of water), only partial hydrolysis occurs.
Reaction (partial hydrolysis).
XeF6+H2O→XeOF4+2HF
Here one O atom replaces two F atoms, and the two displaced F combine with the two H of water to give 2HF.
…
- CBSE 2019Set ANNUAL1 markQ.What is the role of depressant (NaCN) in Froth-Flotation method?
›Reveal solutionSolution
NaCN selectively prevents ZnS from being wetted by the collector oil (by forming a complex on its surface), so ZnS sinks while PbS floats — separating a mixed Pb–Zn sulphide ore.
Concept: Froth flotation concentrates sulphide ores: pine-oil collectors make the mineral surface hydrophobic so it rises with the froth. When two sulphides are present, a depressant is used to keep one down.
…
- CBSE 2019Set ANNUAL1 markMCQQ.Which of the following noble gases is abundant in air?(i) He(ii) Ne(iii) Ar(iv) Kr
›Reveal solutionSolution
Argon is the most abundant noble gas in air.
Dry air contains about 0.93% argon by volume, whereas neon, helium and krypton are present only in trace amounts (of the order of parts per million). Hen …
- CBSE 2019Set ANNUAL1 markMCQQ.Which one is the ore of copper?(i) Haematite(ii) Chalcopyrite(iii) Dolomite(iv) Bauxite
›Reveal solutionSolution
Chalcopyrite (CuFeS2) is the ore of copper.
An ore is a mineral from which a metal is extracted profitably. Chalcopyrite (copper pyrites), CuFeS2, is the principal ore of copper. Haematite (Fe2O3) is an iron ore, dolomite (CaCO3·MgCO3) is a c …
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