Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?

Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?

Exercises · 4.2

Mn^{2+}

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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., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ 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., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ 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., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) 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

When asked "Explain the stability of oxidation states of [element]", follow this mental checklist:

Write the electronic configuration of the atom.
Write configurations for each possible oxidation state.
Look for half-filled, fully-filled, or inert pair effects.
Check if the state can disproportionate (common for +1 states of Cu, Au, and +3 states of Mn).
Mention the medium (acidic/alkaline) if relevant.

Tip

For d-block elements, remember: $d^0$ , $d^5$ , and $d^{10}$ are especially stable. For p-block, the inert pair effect makes lower oxidation states more stable as you go down the group.

The Bottom Line

Stability of an oxidation state is a measure of how strongly an atom holds onto that oxidation number — how hard it is to push it up or down. It's determined by electronic structure, the element's position in the periodic table, and the chemical environment. Master this, and you'll predict redox behaviour without memorising every reaction.

Stability of oxidation states among transition and inner-transition elements is discussed in the NCERT/CBSE Class 12 Chemistry chapter on d- and f-Block Elements, and ‘stability of oxidation states in transition elements’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Predicting which oxidation state is most stable is a reasoning skill regularly tested in competitive-exam inorganic chemistry MCQs.

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: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

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 ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

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:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

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:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds:

The free energy change $\Delta G^\circ$ for the reaction is negative. This happens when the intermediate oxidation state is less stable than the extremes.

Formula (for aqueous ions):

If $E^\circ_{\text{reduction}}$ for the higher state is more positive than for the lower state, disproportionation is spontaneous.

Summary Table: Why Each "Formula" Holds

Principle	Why it works	Key exam example
Inert pair effect	6s² electrons are too tightly bound	$\ce{Pb^{2+}}$ stable, $\ce{Pb^{4+}}$ oxidising
Half-filled stability	Extra exchange energy	$\ce{Mn^{2+}}$ > $\ce{Mn^{3+}}$
Hydration vs ionisation	Energy balance in solution	$\ce{Cu^{2+}}$ stable, $\ce{Cu^+}$ not
Disproportionation	$\Delta G < 0$ for intermediate state	$\ce{Cu^+}$ in water

Final Takeaway for Exams

Never memorise stability blindly. Always ask:

Is the electronic configuration special? (half-filled / inert pair)
Is the medium aqueous or solid? (hydration vs lattice)
Does the element belong to a heavier group? (inert pair effect)

The "formula" is really a balance of energies — and the reasoning is what gets you marks.

The stability of $Mn^{2+}$ over $Fe^{2+}$ toward oxidation is due to the extra stabilization from a half-filled d-subshell in $Mn^{2+}$ ( $3d^5$ ), which makes losing an electron to form $Mn^{3+}$ ( $3d^4$ ) energetically costly, whereas $Fe^{2+}$ ( $3d^6$ ) gains exchange energy upon oxidation to $Fe^{3+}$ ( $3d^5$ ), making it more favourable.

The key to this question lies in electronic configuration and exchange energy — a concept that explains why certain oxidation states are unusually stable or unstable.

The Concept: Stability of Oxidation States and the Half-Filled Shell

In transition metals, the stability of a particular oxidation state depends on how much energy is required to remove an electron. But there’s a subtlety: exchange energy — a quantum mechanical stabilization that arises when electrons have parallel spins in degenerate orbitals. The more unpaired electrons with parallel spins, the greater the exchange energy, and the more stable the configuration.

A half-filled d-subshell ( $d^5$ ) is especially stable because it maximizes the number of unpaired electrons (all five spins parallel), giving the highest possible exchange energy. This is the famous "half-filled shell stability" you’ve likely heard of.

Now, let’s apply this to $Mn$ and $Fe$ .

Step-by-Step Reasoning

Write the electronic configurations of the +2 ions
- $Mn$ (atomic number 25): $[Ar]\,3d^5\,4s^2$ $Mn^{2+}$ loses the two $4s$ electrons → $[Ar]\,3d^5$ This is a half-filled d-subshell — all five $3d$ orbitals are singly occupied with parallel spins.
- $Fe$ (atomic number 26): $[Ar]\,3d^6\,4s^2$ $Fe^{2+}$ loses the two $4s$ electrons → $[Ar]\,3d^6$ This has four unpaired electrons (Hund’s rule: five orbitals, six electrons → one orbital doubly occupied, four singly occupied).
What happens when each is oxidized to the +3 state?
- $Mn^{2+} \rightarrow Mn^{3+} + e^-$ $Mn^{3+}$ configuration: $[Ar]\,3d^4$ — four unpaired electrons. You are breaking a half-filled shell — losing the extra stabilization of $d^5$ . This requires a lot of energy.
- $Fe^{2+} \rightarrow Fe^{3+} + e^-$ $Fe^{3+}$ configuration: $[Ar]\,3d^5$ — five unpaired electrons. You are gaining a half-filled shell — the $Fe^{3+}$ state is stabilized by the maximum exchange energy.
Compare the exchange energy change

Exchange energy is proportional to the number of pairs of parallel-spin electrons: for $n$ electrons of the same spin, the number of such pairs is $\frac{n(n-1)}{2}$ .
- $Mn^{2+}$ ( $d^5$ : five parallel spins): exchange pairs = $\frac{5 \times 4}{2} = 10$
- $Mn^{3+}$ ( $d^4$ : four parallel spins): exchange pairs = $\frac{4 \times 3}{2} = 6$ Loss of 4 exchange pairs → oxidation is energetically unfavourable.
- $Fe^{2+}$ (high-spin $d^6$ : five spin-up electrons plus one spin-down): parallel-spin pairs = $\frac{5 \times 4}{2} = 10$ (the lone spin-down electron adds none)
- $Fe^{3+}$ ( $d^5$ : five parallel spins): exchange pairs = $10$ No exchange energy is lost at all — the electron removed is exactly the paired spin-down one, and its removal also relieves the electron–electron repulsion (pairing energy) of the doubly occupied orbital → oxidation is comparatively easy.

Watch out

A common mistake is to think that $Mn^{2+}$ is stable simply because it has a half-filled shell, without comparing the change in stability upon oxidation. The stability is relative — it’s the difference in exchange energy between the +2 and +3 states that matters.

Additional factor: Third ionization energy The third ionization energy (energy to remove an electron from the +2 ion) is higher for $Mn$ than for $Fe$ because removing an electron from a stable $d^5$ configuration disrupts the half-filled shell. This is consistent with the exchange energy argument.

Tip

You can remember this pattern: For $d^4$ , $d^5$ , $d^6$ , $d^7$ configurations, the $d^5$ state is always the most stable. So $Mn^{2+}$ ( $d^5$ ) resists oxidation, while $Fe^{2+}$ ( $d^6$ ) readily oxidizes to $Fe^{3+}$ ( $d^5$ ). Similarly, $Cr^{2+}$ ( $d^4$ ) is easily oxidized to $Cr^{3+}$ — there the driving force is the stability of the half-filled $t_{2g}^3$ set that $d^3$ attains in an octahedral field, a related but distinct argument.

The Final Picture

So, $Mn^{2+}$ compounds are more stable toward oxidation because:

$Mn^{2+}$ has a half-filled $d^5$ configuration with maximum exchange energy.
Oxidizing it to $Mn^{3+}$ ( $d^4$ ) loses that extra stabilization.
In contrast, $Fe^{2+}$ ( $d^6$ ) gains exchange energy when it becomes $Fe^{3+}$ ( $d^5$ ), making oxidation favourable.

✓Final answer

Mn^{2+}

Method: Electronic Configuration Analysis (Based on Exchange Energy & Half-Filled Stability)

This method uses the electronic configurations of the ions to explain relative stability toward oxidation.

Step 1: Write the ground-state electronic configurations

Mn²⁺:

Atomic number of Mn = 25

Mn²⁺ = $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$

→ $3d^5$ (half-filled d-subshell)
Fe²⁺:

Atomic number of Fe = 26

Fe²⁺ = $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$

→ $3d^6$ (one electron beyond half-filled)

Step 2: Identify the stability factor for each

Mn²⁺ has a half-filled $3d^5$ configuration.

This gives:
- Extra exchange energy (Hund’s rule: maximum number of parallel spins)
- Symmetrical distribution of electrons → lower energy, higher stability
Fe²⁺ has a $3d^6$ configuration.
- Lacks the special stability of half-filled or fully-filled subshells
- Losing one electron to form Fe³⁺ ( $3d^5$ ) actually gains the half-filled stability

Step 3: Compare the oxidation tendency

Ion	Configuration	Stability toward oxidation
Mn²⁺	$3d^5$ (half-filled)	Very stable — losing an electron destroys the half-filled stability
Fe²⁺	$3d^6$	Less stable — losing an electron gives the stable $3d^5$ configuration

Step 4: Conclusion

Mn²⁺ is more stable than Fe²⁺ toward oxidation to +3 state because Mn²⁺ already possesses the highly stable half-filled $3d^5$ configuration. Oxidising it to Mn³⁺ ( $3d^4$ ) would lose this stability.

In contrast, Fe²⁺ ( $3d^6$ ) can gain the half-filled stability by oxidising to Fe³⁺ ( $3d^5$ ), making Fe²⁺ more prone to oxidation.

Key takeaway for exams:

Half-filled and fully-filled subshells confer extra stability.
Mn²⁺ ( $3d^5$ ) → stable as is.
Fe²⁺ ( $3d^6$ ) → prefers to become Fe³⁺ ( $3d^5$ ).

Here are the common mistakes students make on this question, along with the correct conceptual approach to avoid them.

Mistake 1: Confusing the Trend in the 3d Series

The Mistake: Students often assume that stability of the +2 state increases across the series (from Sc to Zn) or that it follows a simple linear pattern. They might say "Mn is in the middle, so it should be less stable."

Why it’s wrong: The stability of the +2 state increases from left to right across the first half of the series (Sc < Ti < V < Cr < Mn) — the rising third ionisation enthalpy makes the d-electrons progressively harder to remove — and it peaks at Manganese ( $d^5$ ); beyond Mn the simple pattern breaks. The key is the electronic configuration of the ions, not just the position in the periodic table.

How to Avoid:

Focus on the half-filled d-orbital rule. The stability of an oxidation state is determined by the electronic configuration of the ion, not the neutral atom.
Write the configurations:
- $Mn^{2+}$ : $[Ar]\,3d^5$ (half-filled, extra stable)
- $Fe^{2+}$ : $[Ar]\,3d^6$ (not half-filled)
- $Mn^{3+}$ : $[Ar]\,3d^4$ (loses the half-filled stability)
- $Fe^{3+}$ : $[Ar]\,3d^5$ (gains half-filled stability)
Conclusion: $Mn^{2+}$ is stable because it already has a half-filled d-subshell. To oxidize it to $Mn^{3+}$ , you must destroy this stable configuration. For $Fe^{2+}$ , oxidation to $Fe^{3+}$ creates a stable half-filled configuration, making it easier.

Mistake 2: Ignoring the "Exchange Energy" or "Stabilization" Argument

The Mistake: Students simply state "half-filled is stable" without explaining why it is stable in terms of energy. They might also confuse it with the "fully-filled" (d¹⁰) case.

Why it’s wrong: The examiner expects a reason based on exchange energy (a quantum mechanical stabilization). A half-filled d⁵ configuration has the maximum number of parallel spins (Hund's rule), leading to the highest exchange energy and thus the lowest energy (most stable) state.

How to Avoid:

Use the correct terminology: Mention exchange energy or symmetrical distribution of electrons.
Explain the energy change:
- $Mn^{2+} (d^5) \rightarrow Mn^{3+} (d^4)$ : Loss of exchange energy (destabilization).
- $Fe^{2+} (d^6) \rightarrow Fe^{3+} (d^5)$ : Gain of exchange energy (stabilization).
Key phrase: "The $d^5$ configuration of $Mn^{2+}$ has extra stability due to high exchange energy, making it resistant to further oxidation."

Mistake 3: Forgetting to Compare Both Sides of the Equation

The Mistake: Students only talk about the stability of $Mn^{2+}$ and forget to explain why $Fe^{2+}$ is less stable. They might say "Mn²⁺ is stable" without contrasting it with Fe²⁺.

Why it’s wrong: The question explicitly asks "Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ ?" This is a comparative question.

How to Avoid:

Always frame the answer as a comparison:
- For Mn: $Mn^{2+}$ (d⁵) is stable. $Mn^{3+}$ (d⁴) is less stable.
- For Fe: $Fe^{2+}$ (d⁶) is less stable. $Fe^{3+}$ (d⁵) is more stable.
Conclusion: Therefore, $Mn^{2+}$ resists oxidation, while $Fe^{2+}$ readily oxidizes to $Fe^{3+}$ .

Mistake 4: Using the Wrong Ion or Configuration

The Mistake: Students write the configuration of the neutral atom (e.g., Mn: $[Ar]\,3d^5 4s^2$ ) instead of the ion ( $Mn^{2+}$ : $[Ar]\,3d^5$ ). Or they confuse $Mn^{2+}$ with $Mn^{3+}$ .

Why it’s wrong: The stability of an oxidation state depends on the ion's configuration, not the atom's.

How to Avoid:

Always remove the 4s electrons first. For transition metals, the 4s orbital is filled before the 3d, but when forming ions, the 4s electrons are lost first.
Write the ion configurations explicitly:
- $Mn^{2+}$ : $[Ar]\,3d^5$
- $Fe^{2+}$ : $[Ar]\,3d^6$
- $Fe^{3+}$ : $[Ar]\,3d^5$

Summary Table for Quick Revision

Aspect	Common Mistake	Correct Approach
Trend	Assume linear stability across series	Check electronic configuration of the ion
Reason	Just say "half-filled"	Explain exchange energy / symmetry
Comparison	Only discuss Mn²⁺	Compare both Mn²⁺ and Fe²⁺
Config	Use neutral atom config	Remove 4s electrons first; write $d^n$ for the ion

Final Answer to the Question (for reference):

$Mn^{2+}$ has a $3d^5$ configuration (half-filled), which is highly stable due to maximum exchange energy and symmetrical distribution. To oxidize it to $Mn^{3+}$ ( $3d^4$ ), this stable configuration is destroyed. In contrast, $Fe^{2+}$ ( $3d^6$ ) is less stable, and its oxidation to $Fe^{3+}$ ( $3d^5$ ) creates a stable half-filled configuration. Hence, $Mn^{2+}$ is more resistant to oxidation than $Fe^{2+}$ .

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Q.Why are Mn2+Mn^{2+}Mn2+ compounds more stable than Fe2+Fe^{2+}Fe2+ towards oxidation to their +3 state?

The Concept: Stability of Oxidation States and the Half-Filled Shell

Step-by-Step Reasoning

The Final Picture

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Q.Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?

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