ChemistryCBSE Class 12 · The d- and f-Block Elements

Oxidation States and Electrode Potentials of Transition Metals

Q: Why do transition elements show variable oxidation states?

Because the energies of the (n-1)d and ns orbitals are close, so different numbers of d electrons can be lost or shared along with the s electrons.

Q: Why is E°(Cu²⁺/Cu) positive?

Copper's high enthalpy of atomisation and ionisation enthalpies are not compensated by the hydration enthalpy of Cu²⁺, so forming Cu²⁺(aq) from Cu(s) is unfavourable.

Q: Why is Cr²⁺ reducing and Mn³⁺ oxidising?

Cr²⁺ becomes Cr³⁺, which has a stable d^3 (half-filled t_{2g}) configuration; Mn³⁺ becomes Mn²⁺, which has a stable d^5 configuration.

Q: Why can oxygen stabilise higher oxidation states than fluorine?

Oxygen can form multiple bonds with the metal atom, so fewer oxygen atoms are needed, as in Mn₂O₇ and MnO₄⁻.

Q: Which oxidation state is most common for 3d metals?

+2 is common for most (from loss of the 4s electrons); +3 is common at the start of the series.

Updated 15 October 2026 · Fact-checked

Transition metals show variable oxidation states because their $(n-1)d$ and $ns$ electrons have similar energies. Manganese reaches +7, the most in the 3d series. Their $E^\circ$ values depend on enthalpy of atomisation, ionisation enthalpy and hydration enthalpy, which explains why only copper has a positive $E^\circ(M^{2+}/M)$ and why Mn²⁺, Fe³⁺ and Cr³⁺ are especially stable.

Understand Oxidation States and Electrode Potentials of Transition Metals

Why oxidation states vary. In a transition element the $(n - 1) d$ and $n s$ orbitals are close in energy, so the atom can lose its $n s$ electrons and then a variable number of $d$ electrons. The oxidation states usually differ by one unit ( $F e^{2 +}$ and $F e^{3 +}$ , $C u^{+}$ and $C u^{2 +}$ ), unlike the p-block where they often differ by two. The greatest number of oxidation states appears in the middle of the 3d series: manganese shows every state from +2 to +7. At the start, scandium has too few electrons to lose (only +3); at the end, there are so many d electrons and so few unpaired orbitals that the range shrinks: zinc shows only +2. In the second and third series the heavier elements favour higher states ( $M o (V I)$ and $W (V I)$ are more stable than $C r (V I)$ ), the reverse of the p-block trend.

Ionisation enthalpy and stability. Ionisation enthalpies rise along the series, but irregularly. Configurations $d^{0}$ , $d^{5}$ and $d^{10}$ are specially stable, so removing an electron from them costs extra. That is why the third ionisation enthalpy of Mn ( $M n^{2 +}$ is $d^{5}$ ) and of Zn ( $Z n^{2 +}$ is $d^{10}$ ) is unusually high, and why $F e^{2 +}$ ( $d^{6}$ ) is easily oxidised to $F e^{3 +}$ ( $d^{5}$ ).

Highest oxidation states with oxygen and fluorine. The highest states are reached in oxides, fluorides and oxoanions: $M n_{2} O_{7}$ , $C r O_{3}$ , $V_{2} O_{5}$ , $M n O_{4}^{-}$ , $C r_{2} O_{7}^{2 -}$ . Oxygen does better than fluorine because it can form multiple bonds to the metal: the highest manganese fluoride is $M n F_{4}$ , but the highest oxide is $M n_{2} O_{7}$ . Fluorine stabilises high states through high lattice energy or bond enthalpy ( $V F_{5}$ , $C r F_{6}$ ). Low states such as 0 appear in carbonyls like $N i (C O)_{4}$ , where π-bonding helps. Oxides change from basic to amphoteric to acidic as the oxidation state rises: MnO is basic, $M n_{2} O_{7}$ acidic.

**Electrode potentials $E^{\circ} (M^{2 +} / M)$ .** Turning a solid metal into an aqueous ion involves three energies: enthalpy of atomisation (endothermic), the first two ionisation enthalpies (endothermic) and the hydration enthalpy of $M^{2 +}$ (exothermic). Their balance sets $E^{\circ}$ . The values generally become less negative across the series because ionisation enthalpies rise. Copper alone has a positive value (+0.34 V): its high atomisation enthalpy and low hydration enthalpy do not compensate for the energy needed to form $C u^{2 +}$ , so copper does not release hydrogen from dilute acids. Manganese is more negative than the trend suggests because $M n^{2 +}$ is the stable $d^{5}$ ion; zinc because $Z n^{2 +}$ is $d^{10}$ ; nickel because $N i^{2 +}$ has the most negative hydration enthalpy.

** $E^{\circ} (M^{3 +} / M^{2 +})$ .** The very high positive value for Mn (+1.57 V) shows that $M n^{2 +}$ ( $d^{5}$ ) is very stable and $M n^{3 +}$ is a strong oxidising agent. Fe has a modest value (+0.77 V) because $F e^{3 +}$ is the stable $d^{5}$ ion. Cr has a negative value (−0.41 V): $C r^{3 +}$ with its half-filled $t_{2 g}$ level is stable, so $C r^{2 +}$ is a strong reducing agent. Both $C r^{2 +}$ and $M n^{3 +}$ are $d^{4}$ , yet one is reducing and the other oxidising: Cr²⁺ changes to $d^{3}$ (Cr³⁺), Mn³⁺ changes to $d^{5}$ (Mn²⁺).

Disproportionation. An ion may oxidise and reduce itself. $C u^{+}$ is unstable in water: $2 C u^{+} \to C u^{2 +} + C u$ , because the much more negative hydration enthalpy of $C u^{2 +}$ more than pays for its second ionisation enthalpy. $M n^{3 +}$ also disproportionates: $2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+}$ .

Oxidation States and Electrode Potentials of Transition Metals: formulas and key facts

Energy steps behind E°(M²⁺/M): $M (s) Δ_{a} H M (g) Δ_{i} H_{1} + Δ_{i} H_{2} M^{2 +} (g) Δ_{h y d} H M^{2 +} (a q)$; Large atomisation and ionisation, small hydration enthalpy → less negative (or positive) $E^{\circ}$ .
E°(M²⁺/M), 3d series (V): $T i - 1.63, V - 1.18, C r - 0.90, M n - 1.18, F e - 0.44, C o - 0.28, N i - 0.25, C u + 0.34, Z n - 0.76$; Irregular at Mn, Ni and Zn; Cu alone positive.
E°(M³⁺/M²⁺) (V): $C r - 0.41, M n + 1.57, F e + 0.77, C o + 1.97$; Negative: M²⁺ is reducing (Cr²⁺). Large positive: M³⁺ is oxidising (Mn³⁺, Co³⁺).
Disproportionation of Cu⁺: $2 C u^{+} (a q) \to C u^{2 +} (a q) + C u (s)$; $E^{\circ} = 0.52 - 0.16 = + 0.36$ V, so spontaneous.
Disproportionation of Mn³⁺: $2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+}$; Mn goes from +3 to +2 and +4.
Common oxidation states (3d): $S c + 3; T i + 2, + 3, + 4; V + 2 - + 5; C r + 2 - + 6; M n + 2 - + 7; F e + 2, + 3; C o + 2, + 3; N i + 2; C u + 1, + 2; Z n + 2$; The most stable or common states; Mn has the widest range.

How to solve Oxidation States and Electrode Potentials of Transition Metals questions

This topic is examined almost entirely through 'Account for' and 'Give reasons' questions, usually 1 mark each in a 3- or 5-mark set. A good answer names the configuration and the energy term responsible.

1Write the outer configuration of the atom and of each ion involved, e.g. $M n^{2 +}$ : $3 d^{5}$ , $C r^{3 +}$ : $3 d^{3}$ .
2Look for stable configurations ( $d^{0}$ , $d^{5}$ , $d^{10}$ , or half-filled $t_{2 g}^{3}$ ) on either side of the change.
3For an $E^{\circ} (M^{2 +} / M)$ question, name the three energies (atomisation, ionisation, hydration) and say which one is unusual.
4For an $E^{\circ} (M^{3 +} / M^{2 +})$ question, decide which ion is the stable one; a large positive value means $M^{3 +}$ is unstable (oxidising), a negative value means $M^{2 +}$ is unstable (reducing).
5For highest oxidation states, mention oxygen's ability to form multiple bonds (oxides, oxoanions) and fluorine's high lattice or bond enthalpy.
6For disproportionation, write the balanced equation and, if potentials are given, show $E_{ce l l}^{\circ} > 0$ .
7Keep each reason to one or two precise lines; the marking scheme looks for the key term.

Oxidation States and Electrode Potentials of Transition Metals in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Fe²⁺ is oxidised to the stable d⁵ Fe³⁺ by air

Suppose a lab assistant prepares a pale green ferrous sulphate solution on Monday and finds it turning yellowish-brown by Friday because the bottle was left open. ( $E^{\circ}$ : $O_{2}, 4 H^{+} /2 H_{2} O = + 1.23$ V; $F e^{3 +} / F e^{2 +} = + 0.77$ V)

Show with $E^{\circ}$ values that dissolved oxygen can oxidise $F e^{2 +}$ , write the balanced ionic equation, and give the electronic reason for the stability of $F e^{3 +}$ .

Try it first, then show the solutionSolution

Cathode: $O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$ (1.23 V); anode: $F e^{2 +} \to F e^{3 +} + e^{-}$ (0.77 V).
$E_{ce l l}^{\circ} = 1.23 - 0.77 = + 0.46$ V, positive, so the reaction is spontaneous.
Balanced: $4 F e^{2 +} + O_{2} + 4 H^{+} \to 4 F e^{3 +} + 2 H_{2} O$ .
$F e^{2 +}$ is $3 d^{6}$ ; losing one electron gives $F e^{3 +}$ , $3 d^{5}$ , a stable half-filled set.

Answer: $E_{ce l l}^{\circ} = + 0.46$ V; $4 F e^{2 +} + O_{2} + 4 H^{+} \to 4 F e^{3 +} + 2 H_{2} O$ ; $F e^{3 +}$ is $d^{5}$ .

Memory hook: Green turns brown because iron prefers the half-filled d⁵ shell.

Tricks to remember Oxidation States and Electrode Potentials of Transition Metals

Short memory aids. Say each one aloud once, then use it on a question the same day.

Middle is mighty
Oxidation states rise to Mn (+7) in the middle of the 3d series and fall away on both sides: Sc +3 only, Zn +2 only.
Use it when: MCQs on maximum oxidation state.
d⁴ twins go opposite ways
Cr²⁺ runs to d³ (reducing); Mn³⁺ runs to d⁵ (oxidising). Each heads to its stable neighbour.
Use it when: Asked as a reason question in the 2023 and 2024 board papers (56/1/1).
Copper is the odd one up
Only Cu has positive $E^{\circ} (M^{2 +} / M)$ : high atomisation, low hydration. 'Copper stays out of acid.'
Use it when: Explaining why Cu does not liberate H₂.

Oxidation States and Electrode Potentials of Transition Metals: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Maximum oxidation state · 1 mark · MCQBoard level

Which element of the first transition series shows the largest number of oxidation states?
(a) Sc (b) Fe (c) Mn (d) Zn

Try it first, then show the solutionSolution

Mn ( $3 d^{5} 4 s^{2}$ ) can use all seven of its $3 d$ and $4 s$ electrons, showing +2 to +7.

Answer: (c) Mn.

Example 2: Zinc and scandium · 2 marks · Very short answerBoard level

Give reasons: (a) Zn, Cd and Hg are not regarded as transition elements. (b) Scandium shows only the +3 oxidation state.

Try it first, then show the solutionSolution

(a) A transition element has an incompletely filled d subshell in the atom or a common ion. Zn, Cd and Hg have $(n - 1) d^{10}$ in the ground state and in their common +2 ions, so they do not qualify (1 mark).
(b) Sc is $3 d^{1} 4 s^{2}$ . Losing all three outer electrons gives $S c^{3 +}$ with the noble gas configuration ( $d^{0}$ ); there are no further d electrons to use, and the +3 state is the stable one (1 mark).

Answer: (a) Full $d^{10}$ in atom and ion. (b) $S c^{3 +}$ is $d^{0}$ (noble gas core), so only +3.

Example 3: Three E° puzzles · 3 marks · Short answerHarder board level

Account for the following: (a) $E^{\circ} (M n^{3 +} / M n^{2 +})$ is much more positive than $E^{\circ} (C r^{3 +} / C r^{2 +})$ or $E^{\circ} (F e^{3 +} / F e^{2 +})$ . (b) $C r^{2 +}$ is strongly reducing while $M n^{3 +}$ is strongly oxidising, though both are $d^{4}$ . (c) $E^{\circ} (C u^{2 +} / C u)$ is positive (+0.34 V).

Try it first, then show the solutionSolution

(a) $M n^{2 +}$ is $3 d^{5}$ , a very stable half-filled configuration, so $M n^{3 +}$ ( $3 d^{4}$ ) readily gains an electron. For Fe, $F e^{3 +}$ is the $d^{5}$ ion, so the change is less favourable; for Cr, $C r^{3 +}$ ( $d^{3}$ , half-filled $t_{2 g}$ ) is the stable one (1 mark).
(b) $C r^{2 +}$ ( $d^{4}$ ) loses an electron to become $C r^{3 +}$ ( $d^{3}$ , half-filled $t_{2 g}$ ), so it is a reducing agent. $M n^{3 +}$ ( $d^{4}$ ) gains an electron to become $M n^{2 +}$ ( $d^{5}$ ), so it is an oxidising agent (1 mark).
(c) Copper's high enthalpy of atomisation and the low hydration enthalpy of $C u^{2 +}$ do not balance the energy needed to form $C u^{2 +} (g)$ ; the overall change $C u (s) \to C u^{2 +} (a q)$ is unfavourable, so $E^{\circ}$ is positive (1 mark).

Answer: (a) Mn²⁺ is d⁵. (b) Cr²⁺ → Cr³⁺ (d³) and Mn³⁺ → Mn²⁺ (d⁵). (c) High $Δ_{a} H$ and low $Δ_{h y d} H$ of Cu²⁺.

Example 4: Is Cu⁺ stable in water? · 3 marks · Short answerHarder board level

Given $E^{\circ} (C u^{+} / C u) = + 0.52$ V and $E^{\circ} (C u^{2 +} / C u^{+}) = + 0.16$ V: (a) Write the disproportionation reaction of $C u^{+}$ and show that it is spontaneous. (b) Calculate $Δ_{r} G^{\circ}$ for it. (c) Explain the result in terms of hydration enthalpy. ( $F = 96500$ C mol⁻¹)

Try it first, then show the solutionSolution

(a) $2 C u^{+} (a q) \to C u^{2 +} (a q) + C u (s)$ . Reduction $C u^{+} \to C u$ (cathode, 0.52 V); oxidation $C u^{+} \to C u^{2 +}$ (anode couple 0.16 V). $E_{ce l l}^{\circ} = 0.52 - 0.16 = + 0.36$ V > 0 (1 mark).
(b) $n = 1$ : $Δ_{r} G^{\circ} = - 1 \times 96500 \times 0.36 = - 34740$ J mol⁻¹ $\approx - 34.7$ kJ mol⁻¹ (1 mark).
(c) $C u^{2 +}$ is smaller and more highly charged, so its hydration enthalpy is much more negative than that of $C u^{+}$ ; this more than compensates for the second ionisation enthalpy of copper (1 mark).

Answer: (a) $E^{\circ} = + 0.36$ V, spontaneous. (b) About −34.7 kJ mol⁻¹. (c) Hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy.

Example 5: Oxides versus fluorides · 3 marks · Short answerHOTS / competency

(a) The highest fluoride of manganese is $M n F_{4}$ , but its highest oxide is $M n_{2} O_{7}$ . Explain. (b) Arrange MnO, $M n O_{2}$ and $M n_{2} O_{7}$ in order of increasing acidic character and justify. (c) Why is the highest oxidation state of a metal shown in its oxide or oxoanion rather than in its simple cation?

Try it first, then show the solutionSolution

(a) Oxygen can form multiple (π) bonds with the metal, so fewer oxygen atoms can stabilise a high oxidation state ( $M n_{2} O_{7}$ has Mn=O bonds). Fluorine forms only single bonds, and fitting seven F atoms round one Mn is not possible (1 mark).
(b) MnO (+2) < $M n O_{2}$ (+4) < $M n_{2} O_{7}$ (+7). As the oxidation state rises, the metal becomes more polarising and its oxide more covalent and acidic: MnO basic, $M n O_{2}$ amphoteric, $M n_{2} O_{7}$ acidic (1 mark).
(c) A simple $M n^{7 +}$ ion would need an impossibly large ionisation energy; in oxides and oxoanions such as $M n O_{4}^{-}$ the charge is shared through covalent bonds with oxygen (1 mark).

Answer: (a) Multiple bonding by oxygen. (b) MnO < MnO₂ < Mn₂O₇ in acidity. (c) Covalent bonding to O avoids forming a highly charged cation.

Where students lose marks in Oxidation States and Electrode Potentials of Transition Metals

The slips examiners see most often, and what to write instead.

Mistake: Saying 'Cr²⁺ is reducing because it is d⁴'
Mn³⁺ is also d⁴ but oxidising, so the reason is incomplete.
Do this: Always name the product configuration: Cr²⁺ → Cr³⁺ ( $d^{3}$ , half-filled $t_{2 g}$ ); Mn³⁺ → Mn²⁺ ( $d^{5}$ ).
Usually costs: 1 mark.
Mistake: Explaining Cu's positive E° by ionisation enthalpy alone
Do this: Mention all three: high enthalpy of atomisation, high ionisation enthalpies, and hydration enthalpy that does not compensate.
Usually costs: ½ to 1 mark.
Mistake: Writing configurations with 4s electrons left in the ion
Students remove d electrons first.
Wrong working
$F e^{2 +}$ : $3 d^{4} 4 s^{2}$
Right working
$F e^{2 +}$ : $3 d^{6}$ (4s electrons are lost first)
Do this: Remove 4s electrons before 3d when forming cations.
Usually costs: The reasoning mark that depends on it.
Mistake: Unbalanced disproportionation equations
Wrong working
$M n^{3 +} \to M n^{2 +} + M n O_{2}$
Right working
$2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+}$
Do this: Balance atoms, then charge with $H^{+}$ and $H_{2} O$ .
Usually costs: 1 mark.

Exam tips

Learn the E° table for the 3d series; questions quote values and ask for the reason behind an irregularity.
In 'Account for' sets, number your answers to match the parts and keep each to one or two lines.
Use the words 'half-filled', 'completely filled', 'enthalpy of atomisation' and 'hydration enthalpy' explicitly.
When asked for 'stable oxidation states', give both the state and the configuration of the ion.
Write balanced ionic equations for disproportionation; charge balance is checked.

Practice questions on Oxidation States and Electrode Potentials of Transition Metals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In which of the following does manganese show its highest oxidation state?
(a) $M n O_{2}$ (b) $K_{2} M n O_{4}$ (c) $K M n O_{4}$ (d) $M n_{3} O_{4}$
Hint
K is +1 and O is −2.
Answer
(c) $K M n O_{4}$ , Mn +7.
Q2 · 2 marks · Very short answerBoard level
Which metal of the 3d series has a positive $E^{\circ} (M^{2 +} / M)$ value? Why can this metal not liberate hydrogen from dilute hydrochloric acid?
Hint
Compare with $E^{\circ} (H^{+} / H_{2}) = 0$ .
Answer
Copper (+0.34 V). Its reduction potential is higher than that of hydrogen (0 V), so $C u^{2 +} / C u$ cannot reduce $H^{+}$ ; Cu is below hydrogen in the electrochemical series.
Q3 · 3 marks · Short answerHarder board level
Explain why the $E^{\circ} (M^{2 +} / M)$ values of Mn, Ni and Zn are more negative than expected from the general trend.
Hint
Configuration for two, hydration for one.
Answer
Mn: $M n^{2 +}$ is $d^{5}$ , extra stable (half-filled). Zn: $Z n^{2 +}$ is $d^{10}$ , extra stable (completely filled). Ni: $N i^{2 +}$ has the most negative hydration enthalpy in the series, which releases more energy on forming the aqueous ion.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $C u^{+}$ ions are unstable in aqueous solution.
Reason (R): The hydration enthalpy of $C u^{2 +}$ is much more negative than that of $C u^{+}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Disproportionation.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Give reasons: (a) The third ionisation enthalpy of manganese is exceptionally high. (b) $F e^{3 +}$ is more stable than $F e^{2 +}$ . (c) Transition elements show variable oxidation states.
Hint
Write the configurations.
Answer
(a) The third electron must be removed from $M n^{2 +}$ ( $3 d^{5}$ ), a stable half-filled configuration. (b) $F e^{3 +}$ is $3 d^{5}$ (half-filled, stable); $F e^{2 +}$ is $3 d^{6}$ . (c) The $(n - 1) d$ and $n s$ electrons are close in energy, so different numbers of them can take part in bonding.
Q6 · 2 marks · Very short answerHarder board level
Write the balanced equation for the disproportionation of $M n^{3 +}$ in aqueous solution and state the oxidation states of Mn in each species.
Hint
Products: $M n^{2 +}$ and $M n O_{2}$ .
Answer
$2 M n^{3 +} + 2 H_{2} O \to M n^{2 +} + M n O_{2} + 4 H^{+}$ . Mn: +3 in $M n^{3 +}$ , +2 in $M n^{2 +}$ , +4 in $M n O_{2}$ .
Q7 · 3 marks · Short answerHOTS / competency
Using $E^{\circ} (C r^{3 +} / C r^{2 +}) = - 0.41$ V and $E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77$ V: (a) Which of $C r^{2 +}$ and $F e^{2 +}$ is the stronger reducing agent? (b) Will $F e^{3 +}$ oxidise $C r^{2 +}$ ? Write the reaction and calculate $E_{ce l l}^{\circ}$ and $Δ_{r} G^{\circ}$ . ( $F = 96500$ C mol⁻¹)
Hint
The more negative reduction potential belongs to the better reducing agent's couple.
Answer
(a) $C r^{2 +}$ , since its couple has the lower (negative) potential. (b) Yes: $F e^{3 +} + C r^{2 +} \to F e^{2 +} + C r^{3 +}$ ; $E_{ce l l}^{\circ} = 0.77 - (- 0.41) = 1.18$ V; $Δ_{r} G^{\circ} = - 96500 \times 1.18 = - 113.9$ kJ mol⁻¹, spontaneous.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student tabulates the oxides of manganese: MnO (green), $M n_{3} O_{4}$ , $M n_{2} O_{3}$ , $M n O_{2}$ (black) and $M n_{2} O_{7}$ (a dark oily liquid). She notes that MnO dissolves in acids, $M n O_{2}$ reacts with both strong acids and strong bases, and $M n_{2} O_{7}$ dissolves in water to give an acid.
(a) Give the oxidation state of Mn in MnO, $M n O_{2}$ and $M n_{2} O_{7}$ . (1 mark)
(b) Classify these three as acidic, basic or amphoteric and state the trend. (1 mark)
(c) Why is $M n_{2} O_{7}$ covalent while MnO is ionic? (1 mark)
(d) Why can manganese reach +7 with oxygen but only +4 with fluorine? (1 mark)
Hint
Higher oxidation state, smaller and more polarising metal centre.
Answer
(a) +2, +4, +7. (b) MnO basic, $M n O_{2}$ amphoteric, $M n_{2} O_{7}$ acidic; acidic character increases with oxidation state. (c) In the +7 state the metal is highly polarising, so its bonds with oxygen are covalent; in +2 the bonding is largely ionic. (d) Oxygen forms multiple bonds with Mn, stabilising +7 with few atoms; fluorine forms only single bonds, and $M n F_{4}$ is the highest fluoride.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Highest oxidation state shown by Mn?Show answer
+7 (as in $K M n O_{4}$ , $M n_{2} O_{7}$ ).
2.Why does Sc show only +3?Show answer
$S c^{3 +}$ has the noble gas configuration ( $d^{0}$ ).
3.Which 3d metal has positive $E^{\circ} (M^{2 +} / M)$ ?Show answer
Copper (+0.34 V).
4.Which is the stronger oxidant: Mn³⁺ or Fe³⁺?Show answer
Mn³⁺ ( $E^{\circ}$ +1.57 V vs +0.77 V).
5.Highest fluoride vs oxide of Mn?Show answer
$M n F_{4}$ vs $M n_{2} O_{7}$ .
6.Equation for Cu⁺ disproportionation?Show answer
$2 C u^{+} \to C u^{2 +} + C u$ .

Oxidation States and Electrode Potentials of Transition Metals: frequently asked questions

Why do transition elements show variable oxidation states?

Because the energies of the (n-1)d and ns orbitals are close, so different numbers of d electrons can be lost or shared along with the s electrons.

Why is E°(Cu²⁺/Cu) positive?

Copper's high enthalpy of atomisation and ionisation enthalpies are not compensated by the hydration enthalpy of Cu²⁺, so forming Cu²⁺(aq) from Cu(s) is unfavourable.

Why is Cr²⁺ reducing and Mn³⁺ oxidising?

Cr²⁺ becomes Cr³⁺, which has a stable d^3 (half-filled t_{2g}) configuration; Mn³⁺ becomes Mn²⁺, which has a stable d^5 configuration.

Why can oxygen stabilise higher oxidation states than fluorine?

Oxygen can form multiple bonds with the metal atom, so fewer oxygen atoms are needed, as in Mn₂O₇ and MnO₄⁻.

Which oxidation state is most common for 3d metals?

+2 is common for most (from loss of the 4s electrons); +3 is common at the start of the series.