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Transition metal chemistry looks like a table to memorise. Then a question asks why Mn3+ is a strong oxidising agent while Fe3+ is not, or why Cu+ falls apart in water, and the table does not help. This post gives you four ideas that explain most of the pattern, with worked numbers from NCERT data and the mistakes that cost marks.
What does the syllabus ask about oxidation states?
The JEE (Main) 2026 syllabus lists oxidation states among the trends of first-row transition elements, together with K2Cr2O7 and KMnO4. The NEET (UG)-2026 syllabus has the same unit, and the JEE (Advanced) 2026 brochure asks for "Oxidation states and their stability; standard electrode potentials". All the data below come from the NCERT Class 12 chapter The d- and f-Block Elements. Check the syllabus for your own exam year.
Why do transition metals show so many oxidation states?
NCERT's explanation is that the d orbitals are only partly filled, so electrons can be removed one at a time. That is why the states of vanadium run +2, +3, +4, +5, differing by one, while p-block elements usually change by two.
Manganese shows every state from +2 to +7. At the left end there are too few electrons to lose (scandium shows only +3); at the right end there are too many d electrons, so zinc shows only +2. Up to manganese, the highest stable state equals the total of 4s and 3d electrons: TiO2 (+4), VO2+ (+5), CrO4²⁻ (+6) and MnO4⁻ (+7). After manganese the higher states become much less stable.
How do you calculate an oxidation state without slipping?
Write one equation: the oxidation states add up to the charge on the species.
KMnO4: +1 + x + 4(−2) = 0, so x = +7.
K2Cr2O7: 2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6.
K4[Fe(CN)6]: 4(+1) + x + 6(−1) = 0, so x = +2. Treat cyanide as one unit with charge −1.
Ni(CO)4: CO is neutral, so nickel is 0. NCERT notes that ligands like CO allow these low states.
Fe3O4: 3x = 8, so the average is +8/3. A fraction means the atoms are in different states: one Fe(II) and two Fe(III).
Why are Mn2+ and Fe3+ so stable?
Both have the 3d5 configuration: five d electrons, one in each orbital. NCERT's Table 4.2 shows the cost of disturbing it. The third ionisation enthalpy of manganese, which takes Mn2+ (d5) to Mn3+ (d4), is 3260 kJ/mol. For iron, which goes from Fe2+ (d6) to Fe3+ (d5), it is only 2962 kJ/mol.
The standard electrode potentials E°(M3+/M2+) tell the same story:
Mn: +1.57 V, so Mn3+ is eager to gain an electron and return to d5. It is a strong oxidising agent.
Fe: +0.77 V, much lower, because Fe3+ is already d5.
Co: +1.97 V, so Co3+ is also a strong oxidising agent in water.
Cr: −0.41 V, so Cr2+ readily loses an electron. NCERT explains this as d4 going to d3, which has a half-filled t2g level.
Worked question: which is the stronger oxidising agent in water, Mn3+ or Fe3+? Compare E°(M3+/M2+): +1.57 V against +0.77 V. Mn3+ has the more positive value, so it is reduced more readily and is the stronger oxidising agent.
Why does Cu+ disproportionate in water?
NCERT states that many copper(I) compounds are unstable in water and undergo 2Cu⁺ → Cu²⁺ + Cu, because the hydration enthalpy of Cu2+ is much more negative. You can check this with numbers from the NCERT Electrochemistry table: E°(Cu+/Cu) = +0.52 V and E°(Cu2+/Cu) = +0.34 V.
First you need E°(Cu2+/Cu+). Potentials cannot be added directly, but Gibbs energies can, since ΔG° = −nFE°:
Cu2+ → Cu (n = 2): −2F(0.34)
This equals Cu2+ → Cu+ (n = 1) plus Cu+ → Cu (n = 1): −F(E) − F(0.52)
So 0.68 = E + 0.52, giving E°(Cu2+/Cu+) = +0.16 V.
In the disproportionation, Cu+ is reduced to Cu (+0.52 V) and oxidised to Cu2+ (the reverse of +0.16 V). E°cell = 0.52 − 0.16 = +0.36 V. A positive value means the reaction is spontaneous. With log K = 0.36/0.059 ≈ 6.1, K is about 1.3 × 10⁶, so very little Cu+ survives in water.
Do oxygen and fluorine stabilise high states?
Yes, and oxygen does it better. NCERT gives MnF4 as the highest manganese fluoride but Mn2O7 as the highest oxide, because oxygen can form multiple bonds to the metal. Going down a group, the trend is the reverse of the p-block. Mo(VI) and W(VI) are more stable than Cr(VI), so dichromate is a strong oxidising agent while MoO3 and WO3 are not. Compare this with heavy p-block elements in our post on the inert pair effect.
Which mistakes cost the most marks?
Oxidation numbers trip up strong students too. Garnett and Treagust (1992) interviewed 32 final-year school chemistry students and found many could not identify redox equations, often because they applied definitions of oxidation and reduction in the wrong setting. Watch for these:
Assuming the highest state is the most stable. For iron, +3 is common but +6 (ferrate) decomposes.
Adding or subtracting electrode potentials with different n. Go through ΔG, as in the copper example.
Forgetting the ligand charge. CN⁻ is −1, CO and H2O are 0.
A practice routine for this chapter
Write the 3d and 4s configuration of each element from Sc to Zn, then each M2+ and M3+ ion.
For ten compounds, including two complexes and one mixed oxide, write the oxidation-state equation in one line.
Explain Mn3+, Co3+ and Cr2+ using E°(M3+/M2+) and the d count, in two sentences each.
If configurations and ionisation enthalpies feel shaky, go back one step. The Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise electronic configuration before returning to this chapter.
Tonight, write the oxidation state of the metal in KMnO4, K2MnO4, MnO2 and MnSO4, and explain which one is easiest to keep in water.
