For M2+/M and M3+/M2+ systems the E∘ values for some metals are as follows:
| System | E∘ | System | E∘ |
|---|---|---|---|
| Cr2+/Cr | −0.9 V | Cr3+/Cr2+ | −0.4 V |
| Mn2+/Mn | −1.2 V | Mn3+/Mn2+ | +1.5 V |
| Fe2+/Fe | −0.4 V | Fe3+/Fe2+ | +0.8 V |
Use this data to comment upon:
- the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and
- the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Stability of Oxidation States
Stability of Oxidation States – From Intuition to Precision
Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).
The Core Intuition
An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.
Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.
The Precise Statement
Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.
This depends on three key factors:
- Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., Fe3+ with d5 is more stable than Fe2+ with d6 in some contexts).
- Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
- Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., Cu+ in aqueous solution gives Cu2+ and Cu).
Stability is relative — it depends on the environment. Mn2+ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.
Examples That Make It Concrete
Transition metals – Cr3+ (d3) and Mn2+ (d5) are exceptionally stable because half-filled/half-filled-like configurations have low energy. Cr2+ (d4) is easily oxidised to Cr3+ — it's unstable.
p-block elements – Pb2+ is stable, Pb4+ is a strong oxidising agent (unstable). Sn2+ is a reducing agent (easily oxidised to Sn4+), so Sn4+ is more stable for tin.
Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., Fe3+ is common but Fe2+ is more stable in acidic solution.
Do not confuse "stability" with "occurrence". Mn7+ (as MnO4−) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.
How to Think About It in Exams …
Why this formula?
Stability of Oxidation States: Why It Works
This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.
The Core Idea: Energy Minimisation
An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:
- Ionisation energy (energy needed to remove electrons)
- Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
- Electronic configuration (half-filled / fully-filled subshells)
There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.
Key Principle 1: Inert Pair Effect (for p-block elements)
Why it holds:
For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.
- Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
- Example: TlX3+ is a strong oxidising agent — it readily gains two electrons to become TlX+.
Derivation logic:
The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.
Key Principle 2: Half-Filled / Fully-Filled Subshell Stability
Why it holds:
A half-filled (d5, f7) or fully-filled (d10, f14) subshell has extra exchange energy and symmetry — making it unusually stable.
- Example: MnX2+ (d5) is more stable than MnX3+ (d4). FeX3+ (d5) is more stable than FeX2+ (d6).
Derivation logic:
The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.
Key Principle 3: Lattice Energy / Hydration Energy Compensation
For transition metals, stability of a particular oxidation state in aqueous solution depends on:
ΔG∘=ΔHhydration∘−ΔHionisation∘
Why it holds:
- Higher oxidation states have higher ionisation energy (harder to remove electrons).
- But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
- The balance determines which state is stable.
Example:
- CuX+ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
- CuX2+ is stable in water.
Key Principle 4: Disproportionation
Some oxidation states are unstable and spontaneously convert to two other states:
2CuX+Cu+CuX2+
Why it holds: …
Concept: Stability of Oxidation States — a more positive E∘ for M3+/M2+ means M3+ is a stronger oxidising agent (less stable in solution), while a more negative E∘ for M2+/M means the metal is more easily oxidised.
Reasoning:
- Stability of M3+: For Cr3+/Cr2+, E∘=−0.4 V (negative, so Cr3+ is not easily reduced — stable). For Fe3+/Fe2+, E∘=+0.8 V (positive, so Fe3+ readily accepts electrons — less stable). For Mn3+/Mn2+, E∘=+1.5 V (highly positive, so Mn3+ is a very strong oxidiser — least stable). Thus, Fe3+ is less stable than Cr3+ but more stable than Mn3+ in acid. …
The key idea is to use the standard electrode potentials to compare the tendency of M3+ to get reduced (stability in acid) and the ease of oxidation of M to M2+. For (i), Fe3+ is more stable than Mn3+ but less stable than Cr3+ in acid. For (ii), iron is harder to oxidise than chromium but easier than manganese.
Understanding the Concept: What the Potentials Tell Us
The two sets of potentials given — E∘ for M2+/M and E∘ for M3+/M2+ — are like two different thermometers for the same metal. The first tells us how easily the metal loses electrons to become M2+ (oxidation of the metal). The second tells us how easily M2+ loses another electron to become M3+ (oxidation of the +2 ion), or equivalently, how stable the M3+ ion is against being reduced back to M2+.
A more positive E∘ for M3+/M2+ means the half-reaction M3++e−→M2+ is more spontaneous — so M3+ is a stronger oxidising agent and is less stable in solution (it wants to grab an electron and become M2+). Conversely, a more negative E∘ for M3+/M2+ means M3+ is harder to reduce, hence more stable.
For the metal itself, a more negative E∘ for M2+/M means the oxidation M→M2++2e− is more spontaneous (since the reverse reduction is less favourable). So the metal is easier to oxidise.
Let's apply this logic step by step.
(i) Stability of Fe3+ in acid solution compared to Cr3+ and Mn3+
1. Compare the M3+/M2+ potentials.
The relevant values are:
- Cr3+/Cr2+: E∘=−0.4 V
- Fe3+/Fe2+: E∘=+0.8 V
- Mn3+/Mn2+: E∘=+1.5 V
2. Interpret the sign and magnitude.
A more positive E∘ means the reduction M3++e−→M2+ is more favourable. So Mn3+ (with +1.5 V) is the strongest oxidising agent among the three — it is most eager to accept an electron and become Mn2+. That makes Mn3+ the least stable in solution.
Fe3+ (+0.8 V) is less eager than Mn3+ but still quite willing to be reduced.
Cr3+ (−0.4 V) has a negative potential — the reduction is non-spontaneous under standard conditions. So Cr3+ is the most stable; it does not want to become Cr2+.
3. Order of stability.
Stability in acid solution: Cr3+>Fe3+>Mn3+.
A common mistake is to think a more positive E∘ for M3+/M2+ means M3+ is more stable. Actually, it means the opposite — it is more easily reduced, hence less stable.
4. Conclusion for (i).
Fe3+ is more stable than Mn3+ but less stable than Cr3+ in acid solution.
--- …
Method: Standard Electrode Potential Comparison for Redox Stability
This method uses the sign and magnitude of E∘ values to determine:
- Stability of a higher oxidation state (from M3+/M2+ data)
- Ease of oxidation of the metal (from M2+/M data)
Steps
Step 1: Understand what each E∘ tells you
| System | Meaning | More positive E∘ → |
|---|---|---|
| M2+/M | Tendency of M2+ to get reduced to M | M2+ is easier to reduce (less stable as ion) |
| M3+/M2+ | Tendency of M3+ to get reduced to M2+ | M3+ is easier to reduce (less stable as M3+) |
Step 2: Analyse stability of M3+ (part i)
-
Cr3+/Cr2+: E∘=−0.4 V
→ Negative E∘ means Cr3+ is not easily reduced to Cr2+
→ Cr3+ is stable in acid
-
Mn3+/Mn2+: E∘=+1.5 V
→ Large positive E∘ means Mn3+ is very easily reduced to Mn2+
→ Mn3+ is unstable (it will readily accept electrons)
-
Fe3+/Fe2+: E∘=+0.8 V
→ Positive E∘, but smaller than +1.5 V
→ Fe3+ is moderately stable — less stable than Cr3+, but more stable than Mn3+
Conclusion for (i):
Stability order in acid: Cr3+>Fe3+>Mn3+
Fe3+ is less stable than Cr3+ but much more stable than Mn3+.
Step 3: Analyse ease of oxidation of the metal (part ii)
Oxidation of metal: M→M2++2e−
This is the reverse of the given M2+/M reduction. …
Common Mistakes & How to Avoid Them
Mistake 1: Confusing "Stability" with "Reducing Power"
The error: Students often think a more negative E∘ for M3+/M2+ means the ion is more stable. They see Cr3+/Cr2+=−0.4 V and think Cr3+ is "unstable" because the value is negative.
Why it's wrong: Stability of a higher oxidation state (like +3) in acid solution depends on how easily it gets reduced to the +2 state. A more positive E∘ for M3+/M2+ means the M3+ ion is a stronger oxidising agent — it readily accepts electrons and gets reduced. This makes the +3 state less stable in solution.
How to avoid: Remember the rule:
- More positive E∘ for M3+/M2+ → M3+ is easily reduced → less stable
- More negative E∘ for M3+/M2+ → M3+ is hard to reduce → more stable
For the given data:
- Cr3+/Cr2+=−0.4 V → Cr3+ is most stable
- Fe3+/Fe2+=+0.8 V → Fe3+ is moderately stable
- Mn3+/Mn2+=+1.5 V → Mn3+ is least stable (strongest oxidiser)
Mistake 2: Misinterpreting "Ease of Oxidation" for Metals
The error: Students look at M2+/M values and think a more positive value means easier oxidation. They see Fe2+/Fe=−0.4 V and Cr2+/Cr=−0.9 V and conclude iron oxidises more easily.
Why it's wrong: Oxidation of a metal M to M2+ is the reverse of the reduction reaction M2++2e−→M. A more negative E∘ for M2+/M means the reduction is less favourable — so the reverse (oxidation) is more favourable.
How to avoid: Flip your thinking:
- More negative E∘ for M2+/M → metal is easier to oxidise (stronger reducing agent)
- Less negative E∘ for M2+/M → metal is harder to oxidise
For the given data:
- Mn2+/Mn=−1.2 V → easiest to oxidise
- Cr2+/Cr=−0.9 V → moderate
- Fe2+/Fe=−0.4 V → hardest to oxidise among the three
Mistake 3: Forgetting to Compare Both Systems Together
The error: Students answer part (i) using only M3+/M2+ data and part (ii) using only M2+/M data, without linking the two.
Why it's wrong: The question asks for comparative statements. For part (ii), "ease with which iron can be oxidised" refers to Fe→Fe2+, which uses Fe2+/Fe data. But the comparison with Cr and Mn must use their respective M2+/M values.
How to avoid: Write your answer in clear comparative sentences:
Part (i):
- Cr3+ is most stable (most negative E∘ for Cr3+/Cr2+)
- Fe3+ is moderately stable
- Mn3+ is least stable (most positive E∘)
Part (ii):
- Mn is easiest to oxidise (most negative E∘ for Mn2+/Mn)
- Cr is next …
Showing the 12 most recent of 19 on this concept.
- CBSE 2026Set 56/3/11 markMCQQ.In aqueous solution, Cr2O72− ion converts to which of the following in alkaline medium ? (A) Cr3+ (B) CrO42− (C) CrO (D) CrO3
›Reveal solutionSolution
In alkaline medium, dichromate (Cr2O72−) converts to chromate (CrO42−) without any change in oxidation state — it’s a simple acid-base equilibrium, not a redox reaction. The correct option is (B).
The key to this question lies in understanding that the conversion of dichromate to chromate is not a redox reaction — the oxidation state of chromium remains +6 throughout. Many students instinctively think of reduction to Cr3+ because they associate dichromate with strong oxidizing behaviour, but that only happens in acidic medium. In alkaline conditions, the chemistry is entirely different.
Let’s walk through the reasoning step by step.
-
Recall the oxidation state of chromium in dichromate.
In Cr2O72−, each oxygen is -2, so total from seven oxygens is -14. The ion has a -2 charge, so the sum of oxidation states of the two chromium atoms must be +12. Hence each Cr is in the +6 state.
-
Now consider the alkaline medium.
When you add a base (like NaOH) to a solution of K2Cr2O7, the dichromate ion reacts with hydroxide ions. The reaction is:
Cr2O72−+2OH−→2CrO42−+H2O
Notice that the oxidation state of Cr in CrO42− is also +6 (four oxygens at -2 give -8, charge -2, so Cr = +6). No electrons are transferred — this is an acid-base equilibrium, not a redox change.
- Why does this happen? Dichromate exists in equilibrium with chromate, and the position depends on pH. In acidic solution, the equilibrium shifts toward dichromate; in alkaline solution, it shifts toward chromate. The reaction is:
2CrO42−+2H+⇌Cr2O72−+H2O
Adding OH− removes H+, pulling the equilibrium to the left — producing chromate. …
-
- CBSE 2026Set 56/1/11 markMCQQ.Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — is not the correct explanation. The correct explanation lies in the small energy gap between 5f, 6d, and 7s orbitals, which allows many electrons to participate in bonding. So the answer is option (B).
The question tests your understanding of why actinoids (elements 90–103, from thorium to lawrencium) exhibit so many different oxidation states. Many students memorise that “actinoids show variable oxidation states” and also know they are radioactive, so they assume the second explains the first. That’s a trap.
Let’s break it down properly.
-
Is Assertion (A) true?
Yes. Actinoids display a remarkably wide range of oxidation states. For example, uranium shows +3, +4, +5, and +6; neptunium and plutonium go from +3 to +7. This is far more varied than most d-block elements. The reason is that the 5f, 6d, and 7s orbitals are very close in energy. Electrons from all three can be lost with relatively little energy cost, so many different oxidation numbers become accessible.
-
Is Reason (R) true?
Yes, actinoids are indeed radioactive. All actinoid nuclei are unstable and decay over time. So the reason statement is factually correct.
-
Does the radioactivity explain the wide range of oxidation states?
No. Radioactivity is a nuclear property — it depends on the instability of the nucleus (proton/neutron ratio, nuclear binding energy). Oxidation states are an electronic property — they depend on how easily electrons are lost from the outer orbitals. These two phenomena are completely independent.
Watch outA common mistake is to think that because both statements are true, the reason must be the explanation. But correlation is not causation. Radioactivity does not cause variable oxidation states; the orbital energy structure does.
-
What actually causes the wide range of oxidation states in actinoids? …
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- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following oxidation states is common for all lanthanoids?(a) +2(b) +3(c) +4(d) +5
›Reveal solutionSolution
All lanthanoids show a characteristic +3 oxidation state because it corresponds to a stable, similar electronic configuration achieved after losing the two 6s and one 4f (or 5d) electron.
Lanthanoids have the general electronic configuration [Xe] 4f^(1-14) 5d^(0-1) 6s2. Removal of the two 6s electrons and one more electron (from 4f or 5d) gives the Ln3+ ion, which is the most stable and commonly observed oxidation state across the entire series, from Ce to Lu.
…
- CBSE 2025Set 56/4/11 markMCQQ.The product of the oxidation of I− with MnO4− in alkaline medium is : (A) IO4− (B) I2 (C) IO− (D) IO3−
›Reveal solutionSolution
In alkaline medium, permanganate (MnO4−) oxidises iodide (I−) to iodate (IO3−), not to iodine or periodate. The balanced reaction shows I− loses 6 electrons to form IO3−, while MnO4− gains 3 electrons to form MnO2. The correct product is IO3−, option (D).
Why the medium matters
The oxidation state of iodine in its products depends heavily on the pH of the solution. Permanganate is a powerful oxidising agent, but its reduction product changes with medium:
- In acidic medium: MnO4−→Mn2+ (gains 5 electrons)
- In neutral/alkaline medium: MnO4−→MnO2 (gains 3 electrons)
This difference in electron gain per mole of permanganate directly affects how far it can oxidise iodide. In alkaline medium, permanganate is a milder oxidising agent (gains only 3 electrons) compared to acidic medium (gains 5 electrons). Yet it still oxidises I− all the way to IO3−, not stopping at I2.
Step-by-step reasoning
-
Identify the half-reactions
Iodide (I−) has oxidation state −1. The possible products given are:
- IO4−: iodine in +7 state
- I2: iodine in 0 state
- IO−: iodine in +1 state (hypoiodite)
- IO3−: iodine in +5 state (iodate)
In alkaline medium, permanganate reduces to MnO2 (manganese in +4 state, from +7 in MnO4−).
-
Balance the oxidation half-reaction
Iodide going to iodate:
I−→IO3−
Balance oxygen with water (alkaline medium):
I−+3H2O→IO3−+6H+
Balance charge: left side has −1, right side has −1+6=+5. Add 6 electrons to right:
I−+3H2O→IO3−+6H++6e−
In alkaline medium, add OH− to neutralise H+:
I−+6OH−→IO3−+3H2O+6e−
So each I− loses 6 electrons to become IO3−.
-
Balance the reduction half-reaction
Permanganate to manganese dioxide in alkaline medium:
MnO4−→MnO2
Balance oxygen with water:
MnO4−+2H2O→MnO2+4OH−
Balance charge: left −1, right −4. Add 3 electrons to left:
MnO4−+2H2O+3e−→MnO2+4OH−
So each MnO4− gains 3 electrons.
-
Combine the half-reactions
To equalise electrons: multiply reduction half by 2 (gives 6 electrons gained) and oxidation half by 1 (gives 6 electrons lost):
2MnO4−+4H2O+6e−→2MnO2+8OH−
I−+6OH−→IO3−+3H2O+6e−
Adding:
2MnO4−+I−+4H2O+6OH−→2MnO2+IO3−+3H2O+8OH−
Cancel 3H2O from both sides and 6OH− from both sides: …
- CBSE 2025Set 56/6/11 markMCQQ.For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature.
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — does not explain this property. The correct answer is (B).
The question tests your understanding of why actinoids exhibit variable oxidation states. The key is to separate two distinct facts: actinoids are radioactive, and they do show many oxidation states — but the radioactivity is not the cause of the oxidation state variability.
Let’s break this down.
-
Why do actinoids show a wide range of oxidation states?
The 5f, 6d, and 7s orbitals in actinoids are very close in energy. This means electrons can be removed from any of these orbitals with relatively little energy cost. As you move across the actinoid series, the 5f orbitals gradually become more stable, but early actinoids (like Th, Pa, U, Np, Pu) can lose anywhere from 3 to 7 electrons. For example, uranium shows +3, +4, +5, and +6; plutonium shows +3, +4, +5, +6, and +7. This is the real reason for the wide range — it’s an electronic structure effect, not a nuclear one.
-
What about radioactivity?
Yes, all actinoids are radioactive — their nuclei are unstable and decay over time. But radioactivity is a nuclear property, while oxidation states depend on electron configuration. A nucleus decaying does not directly change how many electrons an atom can lose or gain in a chemical reaction. So while both statements are factually true, the reason does not explain the assertion. …
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- CBSE 2025Set ANNUAL1 markQ.What is the common oxidation state of Lanthanoids?
›Reveal solutionSolution
All lanthanoids overwhelmingly favour the +3 oxidation state, since their poorly-bonding 4f electrons are not readily involved, leaving the same outer 5d/6s electrons available across the series.
Across the entire lanthanide series, the +3 oxidation state is by far the most common and stable one, shown by essentially every lanthanoid. This is because the 4f electrons are deeply buried and well-shielded, taking little part in bonding, while the outer 5d0−16s2 electrons are readily lost to give the stable Ln3+ ion. Occasional +2 or +4 states occur only for a few elements whe …
- CBSE 2024Set A11 markMCQQ.Which of the following pair of metal oxides are amphoteric?(a) V2O5, Cr2O3(b) Mn2O7, CrO3(c) V2O5, V2O4(d) CrO, V2O5
›Reveal solutionSolution
V2O5 and Cr2O3 are the amphoteric pair — option (a).
For transition-metal oxides, the character changes from basic (low oxidation state) through amphoteric to acidic (high oxidation state). Cr2O3 (Cr in +3) is amphoteric — it dissolves in acids to give Cr3+ salts and in alkali to give chromite. V2O5 (V in +5) is chiefly acidic but is genuinely amphoteric, dissolving in both acids and alka …
- CBSE 2024Set ANNUAL1 markMCQQ.Which of the following oxidation state is common for all lanthanoids ?(a) +2(b) +3(c) +4(d) +5
›Reveal solutionSolution
Every lanthanoid shows the +3 oxidation state as its characteristic and most stable state, even though a few also show +2 or +4 in special cases.
Lanthanoids (Ce to Lu) have the general electronic configuration [Xe]4f1−145d0−16s2. Losing the two 6s electrons and one 4f/5d electron gives the stable, half-filled/fully-filled-favouring Ln3+ ion, which is why +3 is the predominant and universally shown oxidation state across the whole series. A handful of lanthanoids additi …
- CBSE 2023Set 56/1/11 markMCQQ.The most common and stable oxidation state of a Lanthanoid is : (A) + 2 (B) + 3 (C) + 4 (D) + 6
›Reveal solutionSolution
Lanthanoids overwhelmingly prefer the +3 oxidation state due to the stability gained from losing the two 6s and one 5d/4f electron, achieving a configuration analogous to noble gases or half-filled/filled f-subshells. The answer is (B) +3.
Why Lanthanoids Love +3: Electronic Configuration and Stability
The lanthanoid series (elements 57–71: La through Lu) sits in the f-block, where the 4f orbitals are being progressively filled. To understand their oxidation state preference, we need to look at what electrons are available and what configurations become stable upon ionization.
A typical lanthanoid has the general electronic configuration:
[Xe]4f0−145d0−16s2
The 6s electrons are outermost and easiest to remove. The 5d and 4f orbitals are close in energy, so sometimes one electron occupies 5d instead of 4f. When a lanthanoid forms a cation, it loses electrons in a specific order: 6s electrons go first, then 5d, then 4f (because 4f is more tightly held, being an inner orbital).
Step-by-Step Reasoning
-
First ionization removes 6s electrons
All lanthanoids have two 6s electrons. Removing both gives a +2 state, but this is rarely the stopping point because the resulting ion still has relatively accessible 5d or 4f electrons.
-
Third electron removal: the key to +3 stability
After losing the two 6s electrons, removing one more electron (from 5d if occupied, otherwise from 4f) produces the +3 oxidation state. This configuration turns out to be remarkably stable across the entire series.
Why? The resulting Ln3+ ion achieves one of several favorable electronic arrangements:
- For La (4f0): [Xe] — a noble gas configuration.
- For Gd (4f7): half-filled f-subshell with all spins parallel (exchange energy stabilization).
- For Lu (4f14): completely filled f-subshell.
- For others: partially filled 4f with reasonable stability.
-
Why not +2?
The +2 state does exist for a few lanthanoids (Eu, Yb) where it leads to half-filled or filled f-subshells (4f7 for Eu²⁺, 4f14 for Yb²⁺), but these are exceptions, not the rule. Most lanthanoids find +2 too reducing and unstable in aqueous solution.
-
Why not +4 or higher?
Removing a fourth electron means breaking into the tightly held 4f subshell (which is shielded and contracted). The ionization energy jumps dramatically. Only Ce commonly shows +4 (because Ce⁴⁺ achieves 4f0=[Xe]), and even that is a strong oxidizing agent. Higher states like +6 are virtually unknown in lanthanoids—the 4f electrons are too stable to remove. …
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- CBSE 2023Set 56/2/11 markMCQQ.The oxidation state of Fe in [Fe(CO)5] is (A) +2 (B) 0 (C) +3 (D) +5
›Reveal solutionSolution
Carbonyl (CO) is a neutral ligand that does not contribute any charge. With five neutral CO ligands, the overall complex is neutral, so Fe must be in the 0 oxidation state. The correct option is (B).
Why this is a trick question — and how to see through it
Most students memorise that transition metals in coordination compounds usually show positive oxidation states like +2 or +3. Iron especially is famous for Fe(II) and Fe(III). So when you see
[Fe(CO)5], the instinct is to guess +2 or +3. That instinct is wrong here — and the reason is beautiful.The key is to ask: What charge does each ligand bring?
CO (carbonyl) is a neutral ligand. It donates a lone pair to the metal but carries no net charge. If every ligand is neutral, and the overall complex is neutral (no square brackets with a superscript charge), then the metal must be in the zero oxidation state.
This is not a rare exception — it is a whole class of compounds called metal carbonyls, where metals often exist in low or zero oxidation states. CO is a strong field ligand that stabilises these low states through back-bonding.
Step-by-step reasoning
1. Identify the charge on each ligand.
CO is carbon monoxide — a neutral molecule. In coordination chemistry, neutral ligands contribute 0 to the oxidation state calculation. Other examples: NH₃, H₂O, PPh₃.
2. Identify the overall charge on the complex.
The formula is written as
[Fe(CO)5]— no superscript charge. That means the complex is neutral: overall charge = 0.3. Set up the oxidation state equation.
Let the oxidation state of Fe be x.
Each CO contributes 0. There are 5 CO ligands.
So:
x+5(0)=0
4. Solve for x.
x=0
That is the entire calculation — it takes one line once you know the rule.
Watch outA common mistake is to treat CO as if it were a charged ligand like CN⁻ or Cl⁻. CO is not cyanide — it is neutral. Do not assign it a −1 charge. Also, do not confuse this with ferrocene or other organometallics where the ligand (like cyclopentadienyl) is anionic. …
- CBSE 2023Set 56/2/11 markMCQQ.Which of the following characteristics of transition metals is associated with their catalytic activity ? (A) Paramagnetic nature (B) Colour of hydrated ions (C) High enthalpy of atomisation (D) Variable oxidation states
›Reveal solutionSolution
The catalytic activity of transition metals arises primarily from their ability to adopt variable oxidation states, which allows them to form intermediate complexes and lower activation energy. The correct option is (D).
Why this question tests a core idea
Catalysis is about providing an alternative reaction pathway with a lower activation energy. For a substance to be a good catalyst, it must be able to temporarily bind to reactants, change its own electronic state, and then release the products. Transition metals excel at this because they can change their oxidation state easily — often by ±1 — without breaking down. This flexibility lets them shuttle electrons to and from reactants, stabilising transition states that would otherwise be too high in energy.
The other options — paramagnetism, colour, and high enthalpy of atomisation — are important properties of transition metals, but they don't directly explain catalytic activity. Let's see why.
Step-by-step reasoning
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Paramagnetic nature (A)
Paramagnetism arises from unpaired electrons. While many transition metal ions are paramagnetic, this property has no direct role in catalysis. A catalyst doesn't need unpaired electrons to speed up a reaction — it needs to form bonds with reactants and then break them. Paramagnetism is a consequence of electronic configuration, not a cause of catalytic behaviour.
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Colour of hydrated ions (B)
The colour of transition metal complexes comes from d–d transitions — electrons jumping between split d orbitals when they absorb visible light. This is fascinating, but it's a spectroscopic property. Colour tells us about the electronic structure of the ion, but it doesn't help the ion catalyse a reaction. A colourless catalyst can be just as effective.
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High enthalpy of atomisation (C)
This refers to the energy required to convert a solid metal into isolated gaseous atoms. Transition metals have high enthalpies of atomisation because of strong metallic bonding (due to unpaired d electrons contributing to bonding). This property is related to the strength of the metal lattice, not to its ability to change oxidation states during a catalytic cycle. In fact, a very high enthalpy of atomisation might make it harder for the metal to leave the lattice and participate in solution-phase catalysis.
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Variable oxidation states (D) …
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- CBSE 2023Set ANNUAL1 markMCQQ.Which of the following lanthanoid ions in solution is a good oxidizing agent ?(a) Eu2+(b) Yb2+(c) Sm2+(d) Tb4+
›Reveal solutionSolution
+3 is the overwhelmingly preferred oxidation state across the whole lanthanide series, so an unusual +4 ion like Tb⁴⁺ tends to gain an electron and revert to +3 — making it a good oxidising agent.
Across the lanthanide series, +3 is by far the most stable and common oxidation state (arising from the overall energetics of the whole series, not just an individual ion's own f-subshell configuration). Ions that deviate from +3 — whether to +2 or +4 — tend to revert back to +3, and in doing so they act as either reducing or oxidising agents:
- +2 lanthanide ions (Eu²⁺, Sm²⁺, Yb²⁺) tend to lose an electron to revert to +3 — they act as reducing agents. …
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