Use oxidation state, ligand environment and reaction medium together to explain colours, precipitates and electron-transfer stoichiometry.
JEE Main · JEE Advanced · NEET UG48 min study guide10 indexed sections
BY THE END, YOU CAN
Find metal oxidation states and d-electron counts without confusing ion charge and coordination.
Distinguish redox from acid–base, ligand-exchange and precipitation changes.
Balance permanganate and dichromate reactions for their specified endpoints.
Connect transition-compound colour, magnetic behaviour and acid–base properties to structure.
Solve original titration, disproportionation and potential problems with explicit assumptions.
BEFORE YOU START
Oxidation numbers, half-reaction balancing and mole calculations.
Electron configurations, coordination compounds and simple crystal-field ideas.
Equilibrium quotients and the sign convention for reduction potentials.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A metal symbol is not enough to predict the chemistry
The same transition element can occur in different oxidation states, geometries, ligand sets and phases. A statement about “chromium” can mean metal, an aquated Cr(III) ion, chromate(VI), dichromate(VI) or an insoluble oxide. These species differ in charge, colour, reactivity and coordination. Write the actual formula before assigning a property.
Transition metals have accessible valence states involving ns and (n−1)d electrons, supporting multiple oxidation states. Across the early first-row series, high oxidation states become prominent, reaching Mn(VII) in permanganate. Later members commonly favour lower states in simple aqueous chemistry. Oxygen and fluorine often stabilize high oxidation states, but ligand identity and medium can change the stability ordering.
For a first-row metal ion in the ordinary ionic-counting model, remove 4s electrons before 3d electrons when forming common cations. Fe²⁺ is d⁶, Fe³⁺ d⁵, Cr³⁺ d³, Mn²⁺ d⁵ and Cu²⁺ d⁹. In a coordination compound, first use ligand charges to find the formal metal oxidation state, then find the d count. Coordination number is a separate count of donor atoms.
Oxidation state is formal accounting rather than a measurement of actual atomic charge. Mn is +7 in MnO₄⁻ even though the whole ion has charge −1. Writing a free aqueous Mn⁷⁺ ion would misrepresent the species. High states are often stabilized in covalent oxo groups rather than as bare highly charged aquated cations.
ORIGINAL WORKED EXAMPLE
Oxidation number, d count and ion charge
Determine metal oxidation state and d count for MnO₄⁻, Cr₂O₇²⁻ and [Fe(CN)₆]⁴⁻.
MnO₄⁻: x−8=−1, so Mn is +7; group 7 minus 7 gives d⁰.
Cr₂O₇²⁻: 2x−14=−2, so each Cr is +6; group 6 minus 6 gives d⁰.
For the cyanido complex, x+6(−1)=−4, so Fe is +2; group 8 minus 2 gives d⁶.
The complex has six donor atoms, but coordination 6 did not determine oxidation+2.
Answer Mn(VII) d⁰; Cr(VI) d⁰; Fe(II) d⁶.
02 / UNDERSTAND & APPLY
2. Compound properties: colour, spin, hydration and ligand environment
Partially occupied d orbitals can permit electronic transitions between crystal-field-split levels. Their energies depend on the metal, oxidation state, ligands and geometry. Absorption removes selected wavelengths from white light, producing an observed transmitted or reflected colour. Changing ligands can therefore change colour without changing oxidation state.
A d⁰ or d¹⁰ count rules out an ordinary transition between occupied and empty metal d levels of that simple kind, but it does not guarantee a colourless compound. Purple permanganate and yellow chromate are d⁰ ions whose strong colours arise mainly from charge-transfer transitions. Organic ligands can also absorb visible light. “Coloured means unpaired d electrons” is an invalid universal rule.
Magnetic behaviour depends on the occupied orbital pattern, not simply the total d-electron count. A high-spin aquated Fe(III) d⁵ ion has five unpaired electrons, whereas a sufficiently strong-field octahedral Fe(III) complex can have one. In the spin-only approximation μ≈√[n(n+2)] μB, giving about 5.92 μB and 1.73 μB respectively. Orbital contributions and interactions can make real measurements differ.
Water ligands around a small highly charged metal ion are polarized, making their O–H protons easier to lose. Aquated Fe(III) is more acidic than aquated Fe(II) in a comparable elementary comparison. Hydrolysis can change pH and lead to precipitated hydrated oxides/hydroxides. Writing Fe³⁺ in aqueous equations is often shorthand for hydrated and possibly hydrolyzed species.
Transition compounds also form complexes with ammonia, halides and multidentate ligands. Ligand exchange, precipitation and electron transfer are different events. For example, replacing water by ammonia around Cu(II) can deepen the blue colour while Cu remains +2. Metal compounds can catalyze reactions through changing coordination and/or oxidation states, but not every catalytic step must be a metal-centred redox step.
ORIGINAL WORKED EXAMPLE
A colour change need not be redox
A blue Cu(II) aquo solution forms a deeper-blue ammine complex after suitable ligand substitution. Has Cu necessarily been reduced?
Neutral H₂O and NH₃ are neutral ligands.
Replacing one neutral donor by another does not itself change the oxidation-state charge balance.
A changed ligand field can change visible absorption energies.
Reduction would require an electron balance showing Cu’s oxidation state decreased, which this substitution alone does not provide.
Answer No. Cu can remain +2 throughout the ligand-exchange colour change.
03 / UNDERSTAND & APPLY
3. Oxidation state also affects acid–base and bonding properties
Within a metal’s oxide family, lower oxidation states often give more basic, more ionic oxides, while higher oxidation states favour acidic covalent oxides or oxoanions. Chromium illustrates the trend: CrO is basic, Cr₂O₃ amphoteric and CrO₃ acidic. This is a comparative pattern, not permission to classify every oxide from oxidation number alone.
Cr₂O₃ or its hydrated Cr(III) hydroxide can react with acid and with sufficiently strong base. A useful simplified hydroxide scheme is Cr(OH)₃+3H⁺→Cr³⁺+3H₂O and Cr(OH)₃+3OH⁻⇌[Cr(OH)₆]³⁻. Actual alkaline hydroxo speciation depends on conditions. Chromium stays +3 in both processes; amphoteric dissolution is not oxidation.
Many aquated transition cations form sparingly soluble hydroxides when pH increases. Cu(OH)₂ is characteristically blue, Fe(OH)₂ is commonly pale green before air oxidation, and Fe(III) hydrated hydroxide/oxide precipitates are brownish. These are useful qualitative observations with composition and ageing caveats. Excess ligand can sometimes dissolve a precipitate by complex formation; excess hydroxide does not dissolve every metal hydroxide equally.
Halides in high oxidation states often show substantial covalent character and hydrolyze. TiCl₄, for example, is a molecular compound that gives hydrated titanium oxide/oxo products and HCl with water; its behaviour differs from simply dissolving an ionic salt. Fluoride can stabilize metal oxidation states inaccessible or less stable with iodide, which is easier to oxidize.
Mixed-valence solids should not be forced into an impossible identical-ion picture. Fe₃O₄ has average Fe oxidation state +8/3 and is conventionally described using one Fe(II) and two Fe(III) per formula unit. This formal distribution helps balance redox, though the detailed electronic and magnetic solid-state structure requires more than isolated-ion arithmetic.
04 / UNDERSTAND & APPLY
4. Chromate and dichromate: a colour change with no electron transfer
Chromate CrO₄²⁻ is yellow and dichromate Cr₂O₇²⁻ orange in their familiar solution comparison. Chromium is +6 in both. Their interconversion couples condensation and acid–base speciation: 2CrO₄²⁻+2H⁺⇌Cr₂O₇²⁻+H₂O. Because the oxidation number does not change, no electrons appear in the balance.
Adding acid favours the dichromate side in this simplified equilibrium, while adding hydroxide removes H⁺ and favours chromate. Total chromium concentration matters as well because two monomeric units combine into one dimer. Other species, including hydrogen chromate, can be important; the two-species equation is a useful classroom model rather than a complete speciation diagram at every concentration and pH.
Structurally, chromate has a tetrahedral CrO₄ arrangement. Dichromate consists of two such chromium–oxygen coordination tetrahedra sharing one oxygen atom: six terminal oxygens plus one bridge give seven in total. The atoms are not two independent chromate ions simply placed beside each other, and the Cr–O–Cr bridge is not a direct Cr–Cr bond.
Acidifying a solution that also contains a reducing agent can trigger a separate Cr(VI)-to-Cr(III) redox reaction. That is chemically different from the yellow/orange equilibrium. Track the other reactants before interpreting colour: a final greenish Cr(III)-containing solution is not merely “more dichromate”. Chromium(III) colour also varies with ligands; the pure hexaaqua ion need not have the same colour as chloride-containing Cr(III) mixtures.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Coordination-connectivity sketches with schematic projected angles, not quantitative molecular geometry or full Lewis structures. Net ion charges apply to the whole units; bond orders and lone pairs are deliberately omitted. Other protonated species may contribute to real solution speciation.
ORIGINAL WORKED EXAMPLE
Condensation conserves chromium atoms
In an idealized conversion, 0.120 mol chromate units are converted completely to dichromate using the two-species equation. Find dichromate formed, H⁺ consumed and electrons transferred.
Two chromate units form one dichromate, so 0.0600 mol dichromate forms.
The coefficient ratio H⁺/chromate is 2/2, so 0.120 mol H⁺ is consumed.
Chromium is +6 on both sides.
The process transfers zero electrons in this balance.
Answer 0.0600 mol Cr₂O₇²⁻, 0.120 mol H⁺ consumed and no electron transfer.
05 / UNDERSTAND & APPLY
5. Acidic dichromate accepts six electrons per formula unit
Each Cr(VI) centre reduced to Cr(III) accepts three electrons. A dichromate ion contains two chromium atoms, so the total is six electrons per ion. Balancing oxygen with water and hydrogen with H⁺ gives Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O. Always check both atom count and charge: the left charge is −2+14−6=+6, matching two Cr³⁺.
Combining with Fe²⁺→Fe³⁺+e⁻ requires six Fe(II) ions per dichromate. The endpoint chemistry is Fe(II) oxidation to Fe(III), not precipitation of iron metal. The full net equation contains the acid consumed; omitting H⁺ can conceal the fact that changing acidity affects the reaction’s driving force and speciation.
For iodide oxidation, each iodine atom goes from −1 to zero. Six iodides provide six electrons and form three I₂ molecules per dichromate. For a primary alcohol oxidized completely to a carboxylic acid, four electrons per organic molecule are transferred; for a secondary alcohol oxidized to a ketone, two. State the organic endpoint before using an equivalent factor.
Potassium dichromate is often used as a standard oxidant because appropriately pure material is stable enough for analytical use. Its orange colour does not generally make it the same convenient self-indicator as permanganate in Fe(II) titrations; suitable indicators or instrumental endpoints can be needed. These are analytical principles, not a protocol for handling carcinogenic Cr(VI) compounds.
Cr₂O₇²⁻+14H⁺+6Fe²⁺→2Cr³⁺+7H₂O+6Fe³⁺
Every dichromate unit accepts six electrons; each Fe²⁺ supplies one.
ORIGINAL WORKED EXAMPLE
Determine iron from a dichromate amount
A sample containing only Fe(II) as reductant consumes 18.50 mL of 0.02000 mol L⁻¹ dichromate in excess acid. Find moles and mass of Fe using 55.85 g mol⁻¹.
Dichromate amount is 0.01850×0.02000=0.0003700 mol.
Fe(II) amount is six times this: 0.002220 mol.
Iron mass is 0.002220×55.85=0.123987 g.
Round to four significant figures consistent with the given measurements.
Answer 0.002220 mol Fe, or 0.1240 g Fe.
06 / UNDERSTAND & APPLY
6. Permanganate has different electron counts in different media
Permanganate MnO₄⁻ contains Mn(VII) and is intensely purple. In the ordinary sufficiently acidic reduction model, the product is Mn²⁺, usually only faintly coloured in dilute solution. The half-reaction is MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O. One permanganate therefore accepts five electrons in this specified endpoint.
In a common neutral or mildly alkaline reduction model, the product is brown MnO₂(s), with Mn(IV). The balanced alkaline-form equation is MnO₄⁻+2H₂O+3e⁻→MnO₂+4OH⁻. It accepts three electrons. A solid precipitate is part of this reaction, so the visual observation differs from acidic decolourization.
Under appropriately strongly alkaline conditions and a suitable reducing partner, permanganate can be reduced to green manganate, MnO₄²⁻. Mn changes +7 to +6, giving MnO₄⁻+e⁻→MnO₄²⁻. This is a one-electron endpoint. The medium and reductant must support it; pH alone does not guarantee that every reduction stops at manganate.
The equivalent factor of KMnO₄ is consequently reaction-dependent: 5 for Mn²⁺, 3 for MnO₂ and 1 for manganate. A quoted “normality of permanganate” is incomplete without the endpoint convention. Molarity is amount per volume independent of which redox reaction is later chosen.
In a suitable acidic titration, a slight excess of permanganate produces a persistent faint colour, providing a self-indicating endpoint. Other coloured species, side reactions and slow kinetics can complicate interpretation. A suitable acid medium avoids introducing an independently oxidizable anion or a competing oxidant; hydrochloric and nitric acids are therefore not universally interchangeable with the usual sulfuric-acid medium.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
These are specified reaction models, not a full Pourbaix diagram. pH alone does not guarantee the endpoint independently of the reductant and other conditions. All half-reactions conserve atoms and charge.
ORIGINAL WORKED EXAMPLE
One solution, three formal oxidizing capacities
A 25.00 mL portion of 0.01600 mol L⁻¹ MnO₄⁻ is reduced completely to each stated endpoint in separate conceptual reactions. Find electron amounts.
Permanganate amount is 0.02500×0.01600=0.0004000 mol.
To Mn²⁺, multiply by 5 to obtain 0.002000 mol electrons.
To MnO₂, multiply by 3 to obtain 0.001200 mol electrons.
To manganate, multiply by 1 to obtain 0.0004000 mol electrons.
Answer Electron capacities are 0.002000, 0.001200 and 0.0004000 mol respectively.
07 / UNDERSTAND & APPLY
7. Pair the oxidant with the actual reducing half-reaction
Fe²⁺ loses one electron per ion, so acid permanganate requires five Fe²⁺ per MnO₄⁻. Oxalate C₂O₄²⁻ instead loses two electrons per ion when both carbons become CO₂: C₂O₄²⁻→2CO₂+2e⁻. Matching five and two electrons gives the least common total of ten, with two permanganates per five oxalates.
The oxalate reaction can show an induction period and acceleration as Mn²⁺ accumulates; the product assists the reaction pathway, an example of autocatalytic behaviour. Favourable redox thermodynamics does not guarantee a rapid initial reaction. A stoichiometry question that supplies complete reaction can still use the balanced equation, while an experimental interpretation must consider rate.
Hydrogen peroxide acts as a reductant toward sufficiently acidic permanganate, being oxidized to O₂: H₂O₂→O₂+2H⁺+2e⁻. Combining yields 2MnO₄⁻+5H₂O₂+6H⁺→2Mn²⁺+5O₂+8H₂O. The acid coefficient is six after cancellation, not sixteen from copying the permanganate half-reaction twice and ignoring protons produced by peroxide oxidation.
Peroxide can act as either oxidant or reductant with other partners. Its role must be established from oxygen oxidation-state changes and the coupled reaction, not memorized as “always an oxidant”. Likewise, using a gas volume to quantify peroxide requires specified temperature, pressure, water-vapour treatment and exclusive gas-producing chemistry.
2MnO₄⁻+5C₂O₄²⁻+16H⁺→2Mn²⁺+10CO₂+8H₂O
The oxalate-to-carbon-dioxide endpoint supplies two electrons per oxalate.
ORIGINAL WORKED EXAMPLE
Oxalate standardization as a mole-ratio problem
A solution containing 0.001500 mol oxalate consumes 24.00 mL of permanganate in sufficient acid, with complete exclusive reaction. Find permanganate molarity.
The MnO₄⁻/oxalate ratio is 2/5.
Permanganate consumed is 0.001500×2/5=0.0006000 mol.
Divide by 0.02400 L.
No five-electron factor is applied a second time after using the balanced mole ratio.
Answer 0.02500 mol L⁻¹ MnO₄⁻.
08 / UNDERSTAND & APPLY
8. Disproportionation, comproportionation and selected metal couples
Manganate(VI) can disproportionate when conditions no longer stabilize it sufficiently: 3MnO₄²⁻+4H⁺→2MnO₄⁻+MnO₂+2H₂O. Two manganese atoms rise from +6 to +7 and one falls from +6 to +4; the two electrons lost equal the two gained. Strong alkalinity favours manganate stability relative to this acid-driven representation.
Cu(I) disproportionates in the elementary uncomplexed aqueous model: 2Cu⁺→Cu²⁺+Cu(s). But ligands and precipitation can stabilize Cu(I), as in insoluble CuI or suitable complexes. A general statement that Cu(I) can never exist is therefore false. Compare the actual species and medium rather than only the elemental symbol.
Cr²⁺ is a useful example of a reducing aqueous ion because oxidation to Cr³⁺ can be favourable with appropriate oxidants. Mn³⁺ can be strongly oxidizing toward suitable partners, while high-spin Mn²⁺ has a particularly stable d⁵ configuration in many simple environments. These electron-configuration patterns are helpful context, but reduction potentials include hydration, bonding and entropy as well as d-shell counts.
Comproportionation is the reverse pattern: two oxidation states react to form an intermediate state. Fe(s)+2Fe³⁺→3Fe²⁺ provides a clear balance when the stated conditions support it. One Fe atom rises 0→+2 and two fall+3→+2. The mere existence of adjacent oxidation states does not guarantee disproportionation or comproportionation; the relevant free-energy change decides.
ORIGINAL WORKED EXAMPLE
Balance manganate electron bookkeeping
For complete disproportionation of 0.0900 mol manganate by the stated acidic equation, find permanganate and MnO₂ formed.
Three manganates give two permanganates and one MnO₂.
Permanganate formed is 0.0900×2/3=0.0600 mol.
MnO₂ formed is 0.0900×1/3=0.0300 mol.
The two +6→+7 oxidations release two electrons for every one +6→+4 reduction accepting two.
Answer 0.0600 mol MnO₄⁻ and 0.0300 mol MnO₂.
09 / UNDERSTAND & APPLY
9. Potentials predict tendency under specified conditions
For a cell written in its forward direction, E°cell=E°reduction at the cathode−E°reduction for the anode couple. Multiplying a half-reaction to balance electrons does not multiply its electrode potential. Free energy is extensive, ΔG°=−nFE°, so adding free-energy changes is the correct route when combining reduction stages.
A positive E°cell predicts a thermodynamically favourable standard-state forward reaction, but does not guarantee fast visible reaction. Passivation, ligand-substitution barriers and electron-transfer kinetics can delay it. Under nonstandard conditions, the Nernst quotient can change the potential and even the favourable direction.
For acidic permanganate reduction, the quotient contains a(H⁺) to the eighth power in the denominator: Q=a(Mn²⁺)/[a(MnO₄⁻)a(H⁺)⁸], with pure liquid water omitted in the usual dilute model. At 298 K and fixed manganese-species activity ratio, E falls by approximately (0.05916×8/5)=0.0947 V for each unit increase in pH, as long as that half-reaction remains the correct description.
Changing pH enough can change the stable manganese product to MnO₂ or manganate, so extrapolating that same straight line across all pH values is invalid. Complexation and precipitation can also alter free-ion activities and conditional potentials. A colour, precipitate or ligand change can therefore influence redox indirectly even if the initial step itself transfers no electrons.
E=E°−(0.05916/n)log₁₀Q at 298 K
Use a dimensionless activity quotient and the electron number of the written reaction.
10 / UNDERSTAND & APPLY
10. Solved practice and an electron audit
First name the endpoint and medium. Then balance electrons, confirm mass and charge, and only afterwards convert moles to measured quantities. All questions are original conceptual exercises, not laboratory instructions.
ORIGINAL PRACTICE · 1
Does CrO₄²⁻→Cr₂O₇²⁻ by acidification change the chromium oxidation state?
Reveal answer & reasoning
No; Cr remains +6.
The two-species balance uses H⁺ and water but no electrons. Condensation and acid–base speciation account for the change.
ORIGINAL PRACTICE · 2
How many Fe²⁺ ions can one MnO₄⁻ oxidize when the manganese product is Mn²⁺?
Reveal answer & reasoning
Five.
Mn(+7) gains five electrons to become +2; each Fe²⁺ loses one to become Fe³⁺.
ORIGINAL PRACTICE · 3
How many electrons does one dichromate ion accept on complete reduction to Cr³⁺?
Reveal answer & reasoning
Six.
Each of its two Cr atoms changes+6→+3, accepting three electrons. Counting only one chromium gives an incorrect factor of three.
ORIGINAL PRACTICE · 4
Why can purple MnO₄⁻ be coloured even though Mn(VII) is d⁰?
Reveal answer & reasoning
Charge-transfer absorption can produce colour.
The d⁰ count excludes ordinary metal-centred d–d excitation but does not exclude ligand-to-metal charge transfer or other electronic transitions.
ORIGINAL PRACTICE · 5
What is the formal Fe(II):Fe(III) ratio in stoichiometric Fe₃O₄?
Reveal answer & reasoning
1:2.
Four O²⁻ require total metal charge+8. One Fe²⁺ plus two Fe³⁺ gives+8, whereas three identical integer-valent Fe ions would not.
ORIGINAL PRACTICE · 6
A 0.0100 mol amount of H₂O₂ is oxidized exclusively to O₂ by acidic permanganate. Find MnO₄⁻ required.
Reveal answer & reasoning
0.00400 mol.
The balanced ratio MnO₄⁻/H₂O₂ is 2/5, so 0.0100×2/5=0.00400 mol.
ORIGINAL PRACTICE · 7
For a hypothetical standard cell with cathode reduction potential+1.20 V and anode-couple reduction potential+0.35 V, find E°cell.
Reveal answer & reasoning
+0.85 V.
Subtract the reduction potentials:1.20−0.35. Do not multiply either potential by electron-balancing coefficients.
ORIGINAL PRACTICE · 8
Using the stated permanganate/Mn²⁺ Nernst model at 298 K, how does E change when pH rises by 2.00 while the manganese activity ratio remains fixed?
Reveal answer & reasoning
It decreases by approximately 0.189 V.
The per-pH-unit change is−0.05916×8/5=−0.094656 V. Multiplying by 2.00 gives−0.189312 V, provided the same species and endpoint remain valid.
ORIGINAL PRACTICE · 9
Is redissolving Cr(OH)₃ in strong base necessarily oxidation to Cr(VI)?
Reveal answer & reasoning
No.
Hydroxo-complex formation can retain Cr(III), for example [Cr(OH)₆]³⁻. A separate oxidant and electron-transfer balance are required to form chromate.
ORIGINAL PRACTICE · 10
Can Cu(I) exist despite the simple aqueous disproportionation equation?
Reveal answer & reasoning
Yes, when suitable ligands or solids stabilize it.
The uncomplexed-aqueous model does not cover insoluble CuI or all coordination environments. Species-specific thermodynamics matters.
KEEP THE BIG IDEAS
Quick revision
Always specify species, medium and redox endpoint.
Colour changes can reflect ligand exchange or acid–base equilibrium without electron transfer.
Permanganate accepts 5,3 or 1 electrons in the three specified endpoint models.
Dichromate accepts six electrons when both chromium atoms become Cr(III).
Oxide basicity and high-state covalency are qualified trends.
Potentials are intensive; free energy and electron count must be combined consistently.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.