JEE Mains Chemistry · The d- and f-Block Elements

Oxidation States and Electrode Potentials

Transition metals show many oxidation states because their (n−1)d and ns electrons are close in energy; the E° values of the M²⁺/M and M³⁺/M²⁺ couples then say which ions reduce water or acid, which are strong oxidants, and why Cu²⁺ rather than Cu⁺ survives in water.

All The d- and f-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-seven PYQs, twenty-two of them multiple choice, and one from 2026. Ten ask which oxidation states a metal shows and how their stability changes down a group; thirteen read E° values to decide which ion liberates hydrogen or is the strongest oxidant, several of them finishing with a magnetic moment; four ask why Cu²⁺, not Cu⁺, survives in water. The E° table below settles most of them.

Concept 1 of 3: Oxidation states of the 3d metals

A transition metal can lose its ns electrons and then some or all of its d electrons, one at a time, so its oxidation states differ in steps of one. The number of states is largest in the middle of the series, where there are the most electrons to lose and the most orbitals to hold them. Manganese reaches +7 by using all seven 3d and 4s electrons.

Definition

Most states: Mn (+2 to +7). Only one state: Sc (+3).
Early metals reach their group number (Ti +4, V +5, Cr +6, Mn +7). After Mn the highest states fall, and +2 becomes common.
Oxygen and fluorine stabilise high states. Mn reaches +7 only in $\mathrm{Mn_2O_7}$ ; its highest fluoride is $\mathrm{MnF_4}$ (+4). Oxygen can form multiple bonds to the metal, fluorine cannot.
Down a group the higher states become MORE stable: Mo(VI) and W(VI) are more stable than Cr(VI), so $\mathrm{CrO_3}$ is the strongest oxidant of the three. Ru and Os reach +8 ( $\mathrm{RuO_4}$ , $\mathrm{OsO_4}$ ).
This is the opposite of the p-block, where the lower state gets more stable down a group (Pb(II) over Pb(IV)).
The red of ruby is $\mathrm{Cr^{3+}}$ in $\mathrm{Al_2O_3}$ .

Metal	Oxidation states	Most stable in water	Highest fluoride and oxide
Sc	+3	+3	$\mathrm{ScF_3}$ , $\mathrm{Sc_2O_3}$ The only 3d metal with a single oxidation state besides 0.
Ti	+2, +3, +4	+4	$\mathrm{TiF_4}$ , $\mathrm{TiO_2}$
V	+2, +3, +4, +5	+4 (as $\mathrm{VO^{2+}}$ ) and +5	$\mathrm{VF_5}$ , $\mathrm{V_2O_5}$
Cr	+2, +3, +4, +5, +6	+3	$\mathrm{CrF_6}$ , $\mathrm{CrO_3}$
Mn	+2, +3, +4, +5, +6, +7	+2	$\mathrm{MnF_4}$ , $\mathrm{Mn_2O_7}$ Highest oxide (+7) and highest fluoride (+4) differ by 3.
Fe	+2, +3 (+4 and +6 rare)	+3 in air, +2 without it	$\mathrm{FeF_3}$ , $\mathrm{Fe_2O_3}$
Co	+2, +3, +4	+2	$\mathrm{CoF_3}$ , $\mathrm{Co_3O_4}$
Ni	+2, +3, +4	+2	$\mathrm{NiF_2}$ , NiO
Cu	+1, +2	+2	$\mathrm{CuF_2}$ , CuO
Zn	+2	+2	$\mathrm{ZnF_2}$ , ZnO

The number of states peaks at Mn; the ends of the series (Sc, Zn) show one.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q114Moderate

Example 1 · The d- and f-Block Elements · Oxidation States and Electrode Potentials

Given below are two statements: Statement I:

{CrO}_{3}

is a stronger oxidizing agent than

{MoO}_{3}

Statement II :

Cr(VI)

is more stable than

Mo(VI)

In the light of the above statements, choose the correct answer from the options given below

The d-block trend runs the other way from the p-block

In group 14 the lower state gets more stable down the group. In a d-block group the HIGHER state does. So 'Cr(VI) is more stable than Mo(VI)' is false, and that is exactly why

\mathrm{CrO_3}

is the stronger oxidant.

Scandium has no +4

Sc loses 3d¹4s² to reach the argon core and stops. A statement giving Sc a +4 state, oxidising or not, is false.

Drill 9 more on oxidation states of the 3d metals

Concept 2 of 3: E° values: which ions reduce acid and which oxidise

An E° value measures how much an ion wants electrons. A negative E°(M³⁺/M²⁺) means M²⁺ would rather lose an electron, so it reduces H⁺ to hydrogen. A large positive one means M³⁺ grabs electrons, so it is a strong oxidant. The stable shells explain the extremes: Cr²⁺ loses an electron to reach a half-filled t₂g³, and Mn³⁺ gains one to reach 3d⁵.

Definition

M²⁺/M: negative for every 3d metal except Cu (+0.34 V). So Cu does not liberate hydrogen from dilute acids.
Mn, Ni and Zn are more negative than the trend predicts: $\mathrm{Mn^{2+}}$ is 3d⁵, $\mathrm{Zn^{2+}}$ is 3d¹⁰, and $\mathrm{Ni^{2+}}$ has a very negative hydration enthalpy.
M³⁺/M²⁺: Ti, V and Cr are negative, so $\mathrm{Ti^{2+}}$ , $\mathrm{V^{2+}}$ and $\mathrm{Cr^{2+}}$ are reductants that liberate hydrogen from dilute acid.
Mn (+1.57) and Co (+1.97) are large and positive, so $\mathrm{Mn^{3+}}$ and $\mathrm{Co^{3+}}$ are strong oxidants. Fe (+0.77) is lower, because $\mathrm{Fe^{3+}}$ is already 3d⁵.
The hydration enthalpy of $\mathrm{Mn^{2+}}$ is the least negative of the 2+ ions: a 3d⁵ ion gains no crystal-field stabilisation.
In a combined question, find the ion from the E° clue, then count its d electrons as a free gaseous ion for the magnetic moment.

Metal	E° of M²⁺/M (V)	E° of M³⁺/M²⁺ (V)	What it means
Ti	−1.63	−0.37	Ti²⁺ is a reductant and liberates hydrogen
V	−1.18	−0.26	V²⁺ is a reductant and liberates hydrogen
Cr	−0.90	−0.41	Cr²⁺ is a strong reductant: it becomes Cr³⁺, d³
Mn	−1.18	+1.57	Mn³⁺ is a strong oxidant: it becomes Mn²⁺, d⁵
Fe	−0.44	+0.77	Fe³⁺ is a mild oxidant; lower than Mn because Fe³⁺ is d⁵
Co	−0.28	+1.97	Co³⁺ is the strongest oxidant of the series in water
Ni	−0.25	No simple Ni³⁺ in water	Ni²⁺ is the stable ion
Cu	+0.34	No Cu³⁺ in water	The only positive M²⁺/M value: Cu gives no hydrogen with dilute acid Cu has the highest M²⁺/M value of the 3d series.
Zn	−0.76	No Zn³⁺ in water	Zn²⁺ (d¹⁰) is the only ion

Negative M³⁺/M²⁺: the 2+ ion reduces acid. Large positive M³⁺/M²⁺: the 3+ ion is a strong oxidant.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q33Moderate

Example 2 · The d- and f-Block Elements · Oxidation States and Electrode Potentials

The standard electrode potential

\left( M^{3 +}/M^{2 +} \right)

for

V,Cr,Mn\&

Co are

- 0.26\text{ }V, - 0.41\text{ }V, + 1.57\text{ }V

and

+ 1.97\text{ }V

, respectively. The metal ions which can liberate

H_{2}

from a dilute acid are

Iron's M³⁺/M²⁺ value is not above manganese's

\mathrm{Fe^{3+}}

is already 3d⁵, so it gains little by taking an electron: +0.77 V.

\mathrm{Mn^{3+}}

reaches 3d⁵ by taking one: +1.57 V. A statement that iron's value is greater is false.

Count the free ion unless a complex is named

When a question pairs an E° clue with a magnetic moment 'in the gaseous state' or with no ligand named, count the free ion.

\mathrm{Co^{3+}}

is 3d⁶ with 4 unpaired electrons as a free ion, although its complex with water is low spin.

Drill 12 more on e° values: which ions reduce acid and which oxidise

Concept 3 of 3: Why Cu²⁺ is the stable copper ion in water

Cu⁺ has the tidy 3d¹⁰ configuration, so it looks like the more stable ion. In water it is not. The small, doubly charged Cu²⁺ attracts water so strongly that its hydration enthalpy is far more negative than that of Cu⁺, and this more than pays for the second ionisation enthalpy. So Cu⁺ disproportionates in water.

Definition

Disproportionation: $\mathrm{2Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)}$ .
The reason is the MORE negative hydration enthalpy of $\mathrm{Cu^{2+}}$ , not a smaller one.
Cu(I) compounds that are insoluble survive: CuCl, CuI, $\mathrm{Cu_2O}$ . They are white or colourless in the solid (except $\mathrm{Cu_2O}$ , red) because Cu⁺ is 3d¹⁰.
Cu²⁺ with iodide: $\mathrm{2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2(s) + I_2}$ . $\mathrm{Cu_2I_2}$ and CuI are the same white solid. $\mathrm{CuI_2}$ does not exist.
The iodine is titrated with thiosulphate: $\mathrm{I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6}$ (sodium tetrathionate).
So one $\mathrm{Cu^{2+}}$ uses one thiosulphate in the end: this is the iodometric estimation of copper.

Iodometry of copper(II)

\mathrm{2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2} \qquad \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}}

Worked example

A solution containing 0.02 mol of

\mathrm{CuSO_4}

is treated with excess KI. How many moles of iodine are released, and how many moles of

\mathrm{Na_2S_2O_3}

are needed to titrate it?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q127Moderate

Example 3 · The d- and f-Block Elements · Oxidation States and Electrode Potentials

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A):

Cu^{2 +}

in water is more stable than

Cu^{+}

. Reason (R): Enthalpy of hydration for

Cu^{2 +}

is much less than that of

Cu^{+}

. In the light of the above statements, choose the correct answer from the options given below:

The hydration enthalpy of Cu²⁺ is larger, not smaller

Cu²⁺ is stable in water BECAUSE its hydration enthalpy is much more negative than that of Cu⁺. A reason that says it is 'much less' than that of Cu⁺ is false, even when the assertion beside it is true.

Cu₂I₂ and CuI are one compound

\mathrm{Cu_2I_2}

is only CuI written for two copper atoms. When both appear as options, they name the same white precipitate.

Drill 3 more on why cu²⁺ is the stable copper ion in water

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Why Cu²⁺ is the stable copper ion in water
Iodometry of copper(II)
$\mathrm{2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2} \qquad \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}}$

Reference tables (2)

Oxidation states of the 3d metals10 rows

Metal	Oxidation states	Most stable in water	Highest fluoride and oxide
Sc	+3	+3	$\mathrm{ScF_3}$ , $\mathrm{Sc_2O_3}$ The only 3d metal with a single oxidation state besides 0.
Ti	+2, +3, +4	+4	$\mathrm{TiF_4}$ , $\mathrm{TiO_2}$
V	+2, +3, +4, +5	+4 (as $\mathrm{VO^{2+}}$ ) and +5	$\mathrm{VF_5}$ , $\mathrm{V_2O_5}$
Cr	+2, +3, +4, +5, +6	+3	$\mathrm{CrF_6}$ , $\mathrm{CrO_3}$
Mn	+2, +3, +4, +5, +6, +7	+2	$\mathrm{MnF_4}$ , $\mathrm{Mn_2O_7}$ Highest oxide (+7) and highest fluoride (+4) differ by 3.
Fe	+2, +3 (+4 and +6 rare)	+3 in air, +2 without it	$\mathrm{FeF_3}$ , $\mathrm{Fe_2O_3}$
Co	+2, +3, +4	+2	$\mathrm{CoF_3}$ , $\mathrm{Co_3O_4}$
Ni	+2, +3, +4	+2	$\mathrm{NiF_2}$ , NiO
Cu	+1, +2	+2	$\mathrm{CuF_2}$ , CuO
Zn	+2	+2	$\mathrm{ZnF_2}$ , ZnO

The number of states peaks at Mn; the ends of the series (Sc, Zn) show one.

E° values: which ions reduce acid and which oxidise9 rows

Metal	E° of M²⁺/M (V)	E° of M³⁺/M²⁺ (V)	What it means
Ti	−1.63	−0.37	Ti²⁺ is a reductant and liberates hydrogen
V	−1.18	−0.26	V²⁺ is a reductant and liberates hydrogen
Cr	−0.90	−0.41	Cr²⁺ is a strong reductant: it becomes Cr³⁺, d³
Mn	−1.18	+1.57	Mn³⁺ is a strong oxidant: it becomes Mn²⁺, d⁵
Fe	−0.44	+0.77	Fe³⁺ is a mild oxidant; lower than Mn because Fe³⁺ is d⁵
Co	−0.28	+1.97	Co³⁺ is the strongest oxidant of the series in water
Ni	−0.25	No simple Ni³⁺ in water	Ni²⁺ is the stable ion
Cu	+0.34	No Cu³⁺ in water	The only positive M²⁺/M value: Cu gives no hydrogen with dilute acid Cu has the highest M²⁺/M value of the 3d series.
Zn	−0.76	No Zn³⁺ in water	Zn²⁺ (d¹⁰) is the only ion

Negative M³⁺/M²⁺: the 2+ ion reduces acid. Large positive M³⁺/M²⁺: the 3+ ion is a strong oxidant.

Watch out for (6)

The d-block trend runs the other way from the p-block
→ Oxidation states of the 3d metals
Scandium has no +4
→ Oxidation states of the 3d metals
Iron's M³⁺/M²⁺ value is not above manganese's
→ E° values: which ions reduce acid and which oxidise
Count the free ion unless a complex is named
→ E° values: which ions reduce acid and which oxidise
The hydration enthalpy of Cu²⁺ is larger, not smaller
→ Why Cu²⁺ is the stable copper ion in water
Cu₂I₂ and CuI are one compound
→ Why Cu²⁺ is the stable copper ion in water

Test yourself on The d- and f-Block Elements

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Related notes

Oxides of Transition Metals — oxidation state and acidity →