To what extent do the electronic configurations decide the stability of oxidation states in the first series o

Exercises · 4.4

Q. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

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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 …

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: …

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The stability of oxidation states in the first transition series is largely decided by the electronic configuration, specifically the ability to achieve half-filled or fully-filled $d$ -subshells, and the balance between ionization energy and hydration energy. The +2 state is generally stable for all elements, but the +3 state becomes less stable across the series, with exceptions like $\ce{Cr^{3+}}$ ( $d^3$ ) and $\ce{Mn^{2+}}$ ( $d^5$ ) being particularly stable due to exchange energy and symmetry.

The first transition series (Sc to Zn) shows a fascinating variety of oxidation states. The key to understanding their stability lies in the electronic configuration of the ions, but it is not the only factor. Let’s break this down.

1. The role of electronic configuration

The stability of an oxidation state is primarily determined by the energy required to remove electrons (ionization enthalpy) and the energy released when the ion is stabilized in a compound (e.g., hydration energy, lattice energy). However, the electronic configuration of the ion itself provides a powerful clue.

d

These configurations are not just arbitrary; they arise from the fact that electrons in half-filled or fully-filled subshells have lower energy because of exchange energy (Hund’s rule) and reduced electron-electron repulsion.

2. The +2 oxidation state: the baseline

For all elements in the first transition series, the +2 oxidation state is common. This is because the first two electrons removed are from the 4s orbital (which is filled before the 3d, but after removal, the 3d electrons remain). The +2 state corresponds to the configuration $3d^n$ (where $n$ is the number of $d$ -electrons left after removing two 4s electrons).

Example: $\ce{Fe^{2+}}$ has configuration $3d^6$ , $\ce{Co^{2+}}$ has $3d^7$ , etc.
The +2 state is stable for all, but its relative stability varies. For instance, $\ce{Mn^{2+}}$ ( $d^5$ ) is exceptionally stable because of the half-filled $d$ -subshell. Similarly, $\ce{Zn^{2+}}$ ( $d^{10}$ ) is very stable due to the fully-filled $d$ -subshell.

Watch out

A common mistake is to think that the +2 state is always the most stable. While it is common, it is not always the most stable for every element. For example, $\ce{Sc^{2+}}$ is not stable because $\ce{Sc^{3+}}$ ( $d^0$ ) is much more stable.

3. The +3 oxidation state: a tale of two halves

The +3 state becomes less stable as we move across the series from left to right. This is because removing a third electron (from the 3d subshell) requires more energy, and the stabilization from the resulting configuration may or may not compensate.

Early elements (Sc to Cr): The +3 state is stable. For example:
- $\ce{Sc^{3+}}$ ( $d^0$ ) is very stable.
- $\ce{Ti^{3+}}$ ( $d^1$ ) is stable.
- $\ce{V^{3+}}$ ( $d^2$ ) is stable.
- $\ce{Cr^{3+}}$ ( $d^3$ ) is particularly stable because the $d^3$ configuration has a half-filled $t_{2g}$ set in an octahedral field, giving extra crystal field stabilization energy (CFSE).
Middle elements (Mn to Fe): The +3 state is less stable. For example:
- $\ce{Mn^{3+}}$ ( $d^4$ ) is unstable and disproportionates into $\ce{Mn^{2+}}$ ( $d^5$ ) and $\ce{Mn^{4+}}$ ( $d^3$ ). The $d^4$ configuration is prone to Jahn-Teller distortion and is less stable than the $d^5$ of $\ce{Mn^{2+}}$ .
- $\ce{Fe^{3+}}$ ( $d^5$ ) is actually quite stable because of the half-filled $d$ -subshell. In fact, $\ce{Fe^{3+}}$ is more stable than $\ce{Fe^{2+}}$ in some environments (e.g., in acidic solution, $\ce{Fe^{3+}}$ is stable, but $\ce{Fe^{2+}}$ is easily oxidized).
Later elements (Co to Zn): The +3 state becomes increasingly unstable. For example:
- $\ce{Co^{3+}}$ ( $d^6$ ) is a strong oxidizing agent and is not stable in water (it oxidizes water to oxygen). However, in complexes like $\ce{[Co(NH3)6]^{3+}}$ , it is stabilized by strong field ligands.
- $\ce{Ni^{3+}}$ ( $d^7$ ) is very rare and unstable.
- $\ce{Cu^{3+}}$ ( $d^8$ ) is extremely rare.
- $\ce{Zn^{3+}}$ does not exist.

4. Higher oxidation states: the role of oxygen

Higher oxidation states (like +4, +5, +6, +7) are possible only when the metal is bonded to highly electronegative elements like oxygen or fluorine. This is because these ligands can stabilize the high positive charge through strong covalent bonding (e.g., in oxoanions).

Examples:
- $\ce{Cr^{6+}}$ in $\ce{CrO4^{2-}}$ (chromate) and $\ce{Cr2O7^{2-}}$ (dichromate) is stable, but $\ce{Cr^{6+}}$ as a simple ion does not exist. The $d^0$ configuration of $\ce{Cr^{6+}}$ is stable.
- $\ce{Mn^{7+}}$ in $\ce{MnO4^-}$ (permanganate) is stable, again with $d^0$ configuration.
- $\ce{Fe^{6+}}$ in $\ce{FeO4^{2-}}$ (ferrate) is known but is a strong oxidizing agent and less stable.

Tip

A useful shortcut: For a given element, the most stable oxidation state is often the one that gives a $d^0$ , $d^5$ , or $d^{10}$ configuration. For example, $\ce{Cr}$ is most stable as $\ce{Cr^{3+}}$ ( $d^3$ ) and $\ce{Cr^{6+}}$ ( $d^0$ ), while $\ce{Mn}$ is most stable as $\ce{Mn^{2+}}$ ( $d^5$ ) and $\ce{Mn^{7+}}$ ( $d^0$ ).

5. The trend across the series

Let’s summarize the stability of common oxidation states for the first transition series:

Element	Common oxidation states	Most stable state(s)	Reason
Sc	+3	+3	$d^0$ configuration

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Method: Electronic Configuration & Stability of Oxidation States in 3d Series

Method Name: The Half-Filled and Fully-Filled d-Orbital Stability Rule (also called the Exchange Energy & Symmetry Principle)

Core Concept (Why this works)

Stability of an oxidation state depends on the electronic configuration of the ion. Two configurations confer exceptional stability:

$d^5$ (half-filled) — maximum exchange energy, symmetrical distribution
$d^{10}$ (fully-filled) — complete subshell, spherical symmetry
$d^0$ (empty) — also stable (noble gas core)

Any deviation from these requires more energy to achieve, making those states less stable or forcing the element to adopt alternative oxidation states.

Step-by-Step Method

Step 1: Write the ground state electronic configuration of the neutral atom (3d series)

General: $[Ar]\, 3d^{1-10}\, 4s^{2}$ (except Cr & Cu anomalies)

Step 2: Identify the common oxidation states

Remove 4s electrons first, then 3d electrons as needed.

Step 3: Check if the ion’s configuration is $d^0$ , $d^5$ , or $d^{10}$

If yes → that oxidation state is highly stable
If no → that state is less stable or tends to change

Step 4: Compare stability across the series

Early elements (Sc–Mn): higher oxidation states stable (reach $d^0$ or $d^5$ )
Later elements (Fe–Zn): lower oxidation states stable (reach $d^5$ or $d^{10}$ )

Illustrated Examples

Element	Neutral Config	Common States	Stable State	Reason
Sc	$3d^1 4s^2$	+3	Sc $^{3+}$ = $d^0$	Noble gas core
Cr	$3d^5 4s^1$	+3, +6	Cr $^{3+}$ = $d^3$ (half-filled $t_{2g}^3$ — extra CFSE)	Cr $^{6+}$ = $d^0$ also occurs in oxyanions, but is strongly oxidising
Mn	$3d^5 4s^2$	+2, +4, +7	Mn $^{2+}$ = $d^5$	Half-filled — very stable
Fe	$3d^6 4s^2$	+2, +3	Fe $^{3+}$ = $d^5$	Half-filled more stable than Fe $^{2+}$ ( $d^6$ )
Co	$3d^7 4s^2$	+2, +3	Co $^{2+}$ = $d^7$	Neither $d^5$ nor $d^{10}$ ; +3 is less stable

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Here is a breakdown of the common mistakes students make on this topic, structured for clarity and exam accuracy.

The Core Concept (The "Why")

The stability of an oxidation state in transition elements is not decided solely by the electronic configuration of the free ion. Instead, it is a balance between:

Ionization Enthalpy: The energy required to remove electrons.
Hydration/ Lattice Enthalpy: The energy released when the ion is stabilized in a compound (by water molecules or anions).
Electronic Configuration: The stability of the resulting $d^n$ configuration (e.g., $d^0$ , $d^5$ , $d^{10}$ are extra stable).

The key mistake is thinking the configuration alone is the deciding factor.

Common Mistake #1: Ignoring the Role of the Environment

The Error: Students claim that $\ce{Fe^{3+}}$ is more stable than $\ce{Fe^{2+}}$ only because it has a half-filled $d^5$ configuration.

Why it's wrong: While $d^5$ is stable, the stability of $\ce{Fe^{3+}}$ in water is also due to its high hydration enthalpy (the energy released when water molecules surround the ion). In a solid like $\ce{FeCl3}$ , the stability is due to lattice enthalpy.

How to Avoid: Always mention the thermodynamic factors (hydration/lattice enthalpy) alongside the electronic configuration. The configuration provides a tendency, but the environment provides the actual stability.

Correct Example:

"The stability of $\ce{Fe^{3+}}$ in aqueous solution is due to its half-filled $d^5$ configuration ( $t_{2g}^3 e_g^2$ ) and its high hydration enthalpy, which compensates for the high third ionization enthalpy."

Common Mistake #2: Confusing "Stability" with "Existence"

The Error: Students say $\ce{Mn^{2+}}$ is "stable" because it has a $d^5$ configuration, but then fail to explain why $\ce{Mn^{3+}}$ is unstable.

Why it's wrong: $\ce{Mn^{2+}}$ is stable because it is resistant to further oxidation (it is hard to remove another electron from a half-filled $d^5$ subshell). However, $\ce{Mn^{3+}}$ is unstable because it can easily gain an electron to become $\ce{Mn^{2+}}$ (the $d^5$ state).

How to Avoid: Distinguish between:

Stability towards oxidation: How hard it is to remove an electron.
Stability towards reduction: How hard it is to gain an electron.

Correct Example:

" $\ce{Mn^{2+}}$ ( $d^5$ ) is stable because it is resistant to oxidation. $\ce{Mn^{3+}}$ ( $d^4$ ) is unstable because it is easily reduced to the more stable $d^5$ state."

Common Mistake #3: Overgeneralizing the "Half-filled" Rule

The Error: Students claim that all stability follows from the half-filled $d^5$ rule, ignoring the fact that $\ce{Cr^{2+}}$ ( $d^4$ ) is a strong reducing agent because it is easily oxidized to $\ce{Cr^{3+}}$ ( $d^3$ — a half-filled $t_{2g}$ level, not $d^5$ ), and that $\ce{Fe^{3+}}$ ( $d^5$ ) is not always stable.

Why it's wrong: The $d^5$ stability is a tendency, not a law. For example, $\ce{Co^{3+}}$ ( $d^6$ ) is actually more stable in some complexes (like $\ce{[Co(NH3)6]^{3+}}$ ) than $\ce{Co^{2+}}$ ( $d^7$ ), even though $d^6$ is not half-filled.

How to Avoid: Use the rule as a guide, not a rigid rule. Always check the actual oxidation states observed in compounds.

Correct Example:

"While $\ce{Mn^{2+}}$ ( $d^5$ ) is very stable, $\ce{Fe^{3+}}$ ( $d^5$ ) is stable in acidic medium but can be reduced to $\ce{Fe^{2+}}$ in the presence of strong reducing agents. The stability is context-dependent."

Common Mistake #4: Forgetting the +2 and +3 States are the Norm

The Error: Students focus only on the exotic states (like $\ce{Cr^{6+}}$ or $\ce{Mn^{7+}}$ ) and forget that the +2 and +3 states are the most common for the first transition series. …

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Q.To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

1. The role of electronic configuration

2. The +2 oxidation state: the baseline

3. The +3 oxidation state: a tale of two halves

4. Higher oxidation states: the role of oxygen

5. The trend across the series

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