What may be the stable oxidation state of the transition element with the following d electron configurations

Exercises · 4.5

3d^3

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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 an oxidation state for a transition element depends on how close the resulting d-electron count is to an empty ( $d^0$ ), half-filled ( $d^5$ ), or fully-filled ( $d^{10}$ ) subshell — or, within an octahedral ligand field, a half-filled $t_{2g}^3$ set. For $3d^3$ (vanadium), the stable state is +5, which empties the d-subshell completely ( $d^0$ ). For $3d^5$ (manganese), the stable state is +2, which preserves the half-filled $d^5$ configuration. For $3d^8$ (nickel), the stable state is +2, favoured chiefly by crystal field stabilisation energy rather than by reaching $d^0$ / $d^5$ / $d^{10}$ . For $3d^4$ (the configuration chromium would have if it followed the simple Aufbau pattern), the stable state is +3, which leaves a half-filled $t_{2g}^3$ set in an octahedral field.

Transition metals lose electrons from the $4s$ orbital first, then from the $3d$ orbital if more are removed. The stability of the resulting ion depends heavily on whether the $d$ -subshell ends up empty ( $d^0$ ), half-filled ( $d^5$ ), or fully-filled ( $d^{10}$ ) — these arrangements carry extra stabilisation from exchange energy and spherical symmetry. In an octahedral field, a half-filled $t_{2g}^3$ set (one electron in each of the three lower-energy orbitals) is similarly favoured. For each given ground-state $d$ -count, we ask: which oxidation state brings the ion closest to one of these especially stable arrangements, and is that oxidation state actually observed as the element's characteristic stable state?

$3d^3$ configuration (vanadium, atomic number 23)

Ground state: $[Ar]\,4s^2\,3d^3$ — five valence electrons in total.
- Losing the two $4s$ electrons and one $3d$ electron gives $M^{3+}$ with $d^2$ — not a specially stable configuration.
- Losing the two $4s$ electrons and two $3d$ electrons gives $M^{4+}$ with $d^1$ — also not special.
- Losing all five valence electrons gives $M^{5+}$ with $d^0$ — an empty, noble-gas-like subshell.
This $d^0$ state is realised for vanadium as $V^{5+}$ , e.g. in $\ce{V2O5}$ and the orthovanadate ion $\ce{VO4^{3-}}$ — the highest and most characteristic oxidation state of vanadium.

Watch out

A common mistake is to think that $d^5$ is the only stable configuration. $d^0$ (empty) and $d^{10}$ (full) are also very stable, because there is no electron-electron repulsion within the subshell and the ion has spherical symmetry. For early transition metals such as vanadium and chromium, high oxidation states that reach $d^0$ are common and genuinely stable, not just transient.
$3d^5$ configuration (manganese, atomic number 25)

Ground state: $[Ar]\,4s^2\,3d^5$ . Losing the two $4s$ electrons gives $Mn^{2+}$ with $d^5$ — exactly half-filled, with all five electrons unpaired and parallel-spin (maximum exchange energy). Manganese can also reach $d^0$ at +7 (as in permanganate, $\ce{MnO4^-}$ ), but that state is a strong oxidising agent, not the stable, resting state of manganese. The half-filled $d^5$ configuration of $Mn^{2+}$ resists both further oxidation and reduction, which is why +2 is manganese's characteristic stable oxidation state.
$3d^8$ configuration (nickel, atomic number 28)

Ground state: $[Ar]\,4s^2\,3d^8$ . Losing the two $4s$ electrons gives $Ni^{2+}$ with $d^8$ — neither half-filled nor fully-filled, so the $d^0$ / $d^5$ / $d^{10}$ rule does not apply here. Instead, $Ni^{2+}$ is favoured because $d^8$ gains substantial crystal field stabilisation energy in octahedral (and square planar) complexes. Higher states such as $Ni^{3+}$ ( $d^7$ ) are rare and strongly oxidising. So +2 is nickel's stable oxidation state — an example of CFSE, rather than the $d^0$ / $d^5$ / $d^{10}$ rule, deciding stability.
$3d^4$ configuration (chromium, treated by the idealised Aufbau pattern) …

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Method: Stability Based on Empty, Half-Filled, and Fully-Filled d-Orbitals (with CFSE as a Second Check)

This method uses two complementary ideas:

Exchange energy and symmetry — an empty ( $d^0$ ), half-filled ( $d^5$ ), or fully-filled ( $d^{10}$ ) subshell (or a half-filled $t_{2g}^3$ set in an octahedral field) is exceptionally stable.
Crystal field stabilisation energy (CFSE) — for configurations that don't reach one of those special counts, the oxidation state with the greatest CFSE in typical complexes tends to be the stable one.

Steps

Write the ground-state electron configuration of the neutral atom ( $3d^n\,4s^2$ for first-row transition metals, remembering the Cr and Cu exceptions).
Work out the d-electron count for successive oxidation states, always removing the $4s$ electrons before any $3d$ electrons.
Check whether any accessible oxidation state reaches $d^0$ , $d^5$ , $d^{10}$ , or (in an octahedral field) $t_{2g}^3$ .
If none does, use CFSE / known chemistry to identify the actual stable state.
Cross-check against the element's real, commonly observed chemistry.

Applying to the given configurations

1. $3d^3$ (vanadium)

Neutral atom: $3d^3\,4s^2$ (5 valence electrons).
Removing 2 electrons (both $4s$ ) → $M^{2+}$ , $d^3$ — not special.
Removing 3 electrons (2 $4s$ + 1 $3d$ ) → $M^{3+}$ , $d^2$ — not special.
Removing all 5 valence electrons → $M^{5+}$ , $d^0$ — empty subshell, very stable.
Stable oxidation state: +5 (vanadium in $\ce{V2O5}$ , $\ce{VO4^{3-}}$ ).

2. $3d^5$ (manganese)

Neutral atom: $3d^5\,4s^2$ .
Removing the 2 $4s$ electrons → $M^{2+}$ , $d^5$ — half-filled, very stable.
Stable oxidation state: +2 ( $Mn^{2+}$ ). Manganese can also reach $d^0$ at +7 ( $\ce{MnO4^-}$ ), but that state is a strong oxidant, not the resting stable state.

3. $3d^8$ (nickel)

Neutral atom: $3d^8\,4s^2$ .
Removing the 2 $4s$ electrons → $M^{2+}$ , $d^8$ — neither half- nor fully-filled, so the $d^0/d^5/d^{10}$ rule gives no special answer here.
$Ni^{2+}$ ( $d^8$ ) is nonetheless the stable state, because it gains large crystal field stabilisation energy in octahedral/square-planar complexes. Higher states ( $Ni^{3+}$ , $d^7$ ) are rare, strongly oxidising, and not stable in ordinary conditions.
Stable oxidation state: +2, decided by CFSE rather than by reaching $d^0$ / $d^5$ / $d^{10}$ .

4. $3d^4$ (chromium, idealised configuration)

No real first-row atom has $3d^4\,4s^2$ as its true ground state — chromium is anomalous, with a real ground state of $3d^5\,4s^1$ (already half-filled). The $3d^4$ given here is chromium's idealised, pre-exception configuration. …

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🧠 The Core Concept First

Stability of oxidation states in transition elements depends on electronic configuration — specifically, how close the ion can get to a half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) $d$ -subshell. These configurations have extra stability due to exchange energy and symmetry.

Rule of thumb:

Lose $4s$ electrons first (they are higher in energy once occupied).

Then lose $d$ electrons if needed.

Aim for $d^0$ , $d^5$ , or $d^{10}$ if possible.

✗ Mistake #1: Forgetting to remove $4s$ electrons first

What students do:

They look at $3d^3$ and say “stable oxidation state is +3” — but they forget that the ground state configuration of the atom is $4s^2 3d^3$ (for V, for example). So the first electrons lost are from $4s$ , not $3d$ .

How to avoid:

Always write the full ground state configuration first:

$3d^3$ atom → actually $4s^2 3d^3$
$3d^5$ atom → actually $4s^2 3d^5$
$3d^8$ atom → actually $4s^2 3d^8$
$3d^4$ atom → actually $4s^2 3d^4$ (but Cr is an exception: $4s^1 3d^5$ )

Then remove $4s$ electrons first.

✗ Mistake #2: Not checking for half-filled / fully-filled stability

What students do:

They stop at the first oxidation state that gives a noble gas configuration, missing the more stable half-filled or fully-filled $d$ subshell.

Example: For $3d^5$ (Mn, $4s^2 3d^5$ ):

Lose 2 electrons → $3d^5$ (half-filled) → +2 is the stable state
Lose 7 electrons → $3d^0$ → +7 exists (MnO $_4^-$ ) but is a strong oxidant, not the resting stable state

How to avoid:

For each configuration, check:

Can I reach $d^5$ ? → that oxidation state is stable.
Can I reach $d^{10}$ ? → that is also stable.
Can I reach $d^0$ ? → also stable for the early metals (e.g., V $^{5+}$ ); for later metals such high states exist only as strongly oxidising oxoanions.

✗ Mistake #3: Ignoring the $3d^4$ exception (Cr)

What students do:

They treat $3d^4$ as $4s^2 3d^4$ and predict +2 or +4.

Reality:

Cr ground state is $4s^1 3d^5$ (half-filled stability). So:

Lose 3 electrons → $3d^3$ → +3 is the stable state (a half-filled $t_{2g}^3$ set in an octahedral field)
Lose 6 electrons → $3d^0$ → +6 occurs in oxoanions (CrO $_4^{2-}$ , Cr $_2$ O $_7^{2-}$ ) but is strongly oxidising, not a resting stable state
Note: losing just 1 electron leaves $3d^5$ , but Cr(+1) is not a stable oxidation state in practice — a bare configuration count is not enough

How to avoid:

Memorise the two exceptions:

Cr: $4s^1 3d^5$ (not $4s^2 3d^4$ )
Cu: $4s^1 3d^{10}$ (not $4s^2 3d^9$ )

✗ Mistake #4: Confusing “stable” with “most common”

What students do:

They list every possible oxidation state, not the most stable one.

Example: For $3d^8$ (Ni, $4s^2 3d^8$ ): …

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Q.What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d33d^33d3, 3d53d^53d5, 3d83d^83d8 and 3d43d^43d4?

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Q.What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: $3d^3$ , $3d^5$ , $3d^8$ and $3d^4$ ?

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