The electronic configuration of Cu(II) is $3d^9$ whereas that of Cu(I) is $3d^{10}$. Which of the following is correct? (i) Cu(II) is more stable (ii) Cu(II) is less stable (iii) Cu(I) and Cu(II) are equally stable (iv) Stability of Cu(I) and Cu(II) depends on nature of copper salts

> [!TLDR] > The key idea is that while a fully filled $3d^{10}$ subshell (Cu(I)) is stable, the higher charge and smaller size of Cu(II) give it a much larger hydration enthalpy in aqueous solution, which more than compensates for the energy cost of removing an extra electron. In aqueous medium, Cu(II) is more stable than Cu(I). 1. **Start with the electronic configurations.** Cu(I) has the configuration $3d^{10}$ — a completely filled d-subshell. Cu(II) has $3d^9$ — one electron short of a full d-subshell. A filled subshell is inherently more stable due to exchange energy and symmetry. So, in the gas phase, Cu(I) is more stable than Cu(II). If the question were about gaseous ions, Cu(I) would win. 2. **But the real world is not the gas phase — it's aqueous solution.** Most common chemistry of copper happens in water. Here, the stability of an ion depends not just on its electronic configuration, but also on its hydration enthalpy — the energy released when water molecules surround the ion. The hydration enthalpy depends on two factors: charge and ionic radius. Higher charge and smaller size both increase hydration enthalpy. 3. **Compare Cu(I) and Cu(II) in terms of charge and size.** Cu(II) has a +2 charge, while Cu(I) has only +1. Also, Cu(II) has a smaller ionic radius (about 73 pm) compared to Cu(I) (about 96 pm). The combination of higher charge and smaller size means Cu(II) has a much larger hydration enthalpy than Cu(I). 4. **The energy balance.** To go from Cu(I) to Cu(II), you need to remove one more electron — that costs ionization energy. But the huge hydration enthalpy of Cu(II) more than makes up for this cost. The net result is that in water, Cu(II) is thermodynamically more stable than Cu(I). 5. **Evidence from disproportionation.** Cu(I) in aqueous solution spontaneously disproportionates: $$2\text{Cu}^+(aq) \rightarrow \text{Cu}(s) + \text{Cu}^{2+}(aq)$$ This reaction is thermodynamically favorable (positive $

The electronic configuration of Cu(II) is 3d⁹ whereas that of Cu(I) is 3d¹⁰. Which of the following is correct

NCERT Exemplar · Q2

3d^9

(i) Cu(II) is more stable

(ii) Cu(II) is less stable

(iii) Cu(I) and Cu(II) are equally stable

(iv) Stability of Cu(I) and Cu(II) depends on nature of copper salts

Assam AhsecMCQ· 1mImportance★★★★★

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✓ Free question

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 key idea is that while a fully filled $3d^{10}$ subshell (Cu(I)) is stable, the higher charge and smaller size of Cu(II) give it a much larger hydration enthalpy in aqueous solution, which more than compensates for the energy cost of removing an extra electron. In aqueous medium, Cu(II) is more stable than Cu(I).

Start with the electronic configurations.

Cu(I) has the configuration $3d^{10}$ — a completely filled d-subshell. Cu(II) has $3d^9$ — one electron short of a full d-subshell. A filled subshell is inherently more stable due to exchange energy and symmetry. So, in the gas phase, Cu(I) is more stable than Cu(II). If the question were about gaseous ions, Cu(I) would win.
But the real world is not the gas phase — it's aqueous solution.

Most common chemistry of copper happens in water. Here, the stability of an ion depends not just on its electronic configuration, but also on its hydration enthalpy — the energy released when water molecules surround the ion. The hydration enthalpy depends on two factors: charge and ionic radius. Higher charge and smaller size both increase hydration enthalpy.
Compare Cu(I) and Cu(II) in terms of charge and size.

Cu(II) has a +2 charge, while Cu(I) has only +1. Also, Cu(II) has a smaller ionic radius (about 73 pm) compared to Cu(I) (about 96 pm). The combination of higher charge and smaller size means Cu(II) has a much larger hydration enthalpy than Cu(I).
The energy balance.

To go from Cu(I) to Cu(II), you need to remove one more electron — that costs ionization energy. But the huge hydration enthalpy of Cu(II) more than makes up for this cost. The net result is that in water, Cu(II) is thermodynamically more stable than Cu(I).
Evidence from disproportionation.

Cu(I) in aqueous solution spontaneously disproportionates:

2\text{Cu}^+(aq) \rightarrow \text{Cu}(s) + \text{Cu}^{2+}(aq)

This reaction is thermodynamically favorable (positive $E^\circ$ ), which directly shows that Cu(II) is more stable than Cu(I) in water.

Watch out

A common mistake is to stop at the electronic configuration and conclude that Cu(I) with $3d^{10}$ must be more stable. That is true only for gaseous ions. In solution, the hydration effect reverses the stability order.

Tip

For transition metals, always check the medium. In aqueous solution, higher oxidation states are often stabilized by hydration or complexation, even if the gas-phase configuration suggests otherwise.

✓Final answer

The correct option is (i) Cu(II) is more stable .

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Assuming Full d-subshell Always Means Higher Stability

The Mistake: Students see that Cu(I) has a $3d^{10}$ configuration (completely filled d-subshell) and immediately assume it must be more stable than Cu(II) with $3d^9$ . They pick option (ii) "Cu(II) is less stable" without further thought.
Why it's Wrong: While a fully filled d-subshell does provide extra stability, it is not the only factor. In transition metals, the hydration enthalpy (energy released when ions dissolve in water) and lattice energy (for solid compounds) often outweigh the stability gained from a filled d-subshell.
How to Avoid: Remember the "exceptions" in transition metal chemistry. For copper, Cu(II) is actually more stable in aqueous solution and in most of its salts. Always check the real-world behavior of the element, not just the theoretical electron configuration.

Mistake 2: Ignoring the Role of Hydration Enthalpy

The Mistake: Students forget to consider what happens when the ion is in solution. They treat the stability as an isolated property of the ion in a vacuum.
Why it's Wrong: The question is about stability in a general chemical context. Cu(II) has a higher charge (+2) and a smaller ionic radius than Cu(I). This means it has a much higher hydration enthalpy (the energy released when water molecules surround the ion). This large energy release compensates for the loss of the filled d-subshell, making Cu(II) more stable in water.
How to Avoid: When comparing stability of different oxidation states of transition metals, always ask: "What is the environment?" For aqueous solutions, higher charge density often leads to greater stability due to hydration energy. Write down the formula: Stability = (Electronic Configuration Stability) + (Hydration/Lattice Energy).

Mistake 3: Confusing "Stability" with "Ease of Formation"

The Mistake: Students think that because Cu(I) is formed first (e.g., from Cu metal), it must be more stable.
Why it's Wrong: Cu(I) is often an intermediate that quickly disproportionates (self-oxidizes and reduces) into Cu(II) and Cu metal. The fact that Cu(I) is hard to isolate in water shows it is less stable under those conditions.
How to Avoid: Understand that stability refers to the tendency of a species to remain as it is, not how easily it is formed. A species can be formed easily but be highly unstable (like Cu(I) in water). Focus on the final, predominant species.

Mistake 4: Misinterpreting the "d^10" Exception

The Mistake: Students overgeneralize the stability of $d^{10}$ configurations. They apply it blindly to all elements (e.g., Zn, Cd, Hg) and assume it works the same for copper.
Why it's Wrong: For Zn, $d^{10}$ is indeed the most stable state (+2). But for Cu, the +2 state is more stable despite having $d^9$ . The difference is that Cu has a smaller nuclear charge and different ionic radii, making the +2 state energetically favorable in water.
How to Avoid: Treat each element as a unique case. Memorize the key exceptions: Cu(II) > Cu(I) in water, Fe(III) > Fe(II) in some complexes, etc. Don't rely on a single rule for all transition metals.

The Correct Answer and Why

The correct answer is (i) Cu(II) is more stable.

Reason: In aqueous solution and in most of its compounds, Cu(II) is more stable than Cu(I) due to the much higher hydration enthalpy of Cu(II) (which has a higher charge and smaller size). This energy gain outweighs the stability provided by the $3d^{10}$ configuration of Cu(I). Cu(I) compounds often disproportionate in water to give Cu(II) and Cu metal.

Key Takeaway: For copper, Cu(II) is the more stable oxidation state in water, despite having an incomplete d-subshell. Always consider the environment (aqueous vs. solid) and the energy contributions from hydration or lattice formation.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2025Set ANNUAL2 marks
$Q. Which one is more reducing Cr2+ or Fe2+ and why?$
›Reveal solutionSolution
Cr2+ readily loses an electron to become the more stable d3 Cr3+ (negative E°, strong reducing agent), whereas Fe2+ oxidising to Fe3+ is thermodynamically unfavourable (positive E°), so Fe2+ is only a weak reducing agent.

A more negative (or less positive) standard reduction potential for the M3+/M2+ couple means the M2+ ion more readily gives up an electron, i.e. it is a stronger reducing agent.

$E^\circ(Cr^{3+}/Cr^{2+}) = -0.41\ V$ (negative) — this means Cr2+ is easily oxidised to Cr3+. This is because Cr3+ has the extra-stable half-filled t2g3 (d3) configuration, so the Cr2+ → Cr3+ oxidation is thermodynamically very favourable.

$E^\circ(Fe^{3+}/Fe^{2+}) = +0.77\ V$ (positive) — this means Fe3+ is comparatively stable and does NOT readily accept an electron; conversely, Fe2+ is only a weak reducing agent, oxidising to Fe3+ with difficulty (Fe3+ does have a stable half-filled d5 configuration too, but not enough to make the potential negative).

Since Cr2+ has a much more negative reduction potential for its oxidised/reduced couple, it loses its electron far more readily than Fe2+ does.

✓Final answer
$Cr2+ is the stronger (more) reducing agent, because oxidation of Cr2+ to the very stable d3 Cr3+ is thermodynamically much more favourable (E° = -0.41 V) than oxidation of Fe2+ to Fe3+ (E° = +0.77 V).$
AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL2 marks
$Q. Why is Cr2+ reducing and Mn3+ oxidizing when both have d4 configuration? OR Out of Cu+ and Cu2+, which ion is more stable in aqueous solution and why?$
›Reveal solutionSolution

Both options are explained by which resulting ion (after losing/gaining an electron, or by hydration) is thermodynamically more stable.

Option 1 — Cr²⁺ reducing, Mn³⁺ oxidising (both d⁴):

$Cr^{2+}$ (3d⁴) tends to lose one electron and get oxidised to $Cr^{3+}$ (3d³), because the $d^3$ configuration ( $t_{2g}^3$ , all three lower-energy orbitals singly occupied) is a particularly stable, symmetric half-filled- $t_{2g}$ arrangement in an octahedral field. This driving force to reach a more stable $d^3$ state makes $Cr^{2+}$ a strong reducing agent.

$Mn^{3+}$ (3d⁴) tends to gain one electron and get reduced to $Mn^{2+}$ (3d⁵), because the exactly half-filled $d^5$ configuration has extra stability (maximum exchange energy, all five d orbitals singly occupied). This strong tendency to gain an electron makes $Mn^{3+}$ a strong oxidising agent.

So although both starting ions have the same $d^4$ configuration, they move in opposite directions (Cr²⁺ loses an electron, Mn³⁺ gains one) because each is driven toward a more stable configuration — $d^3(t_{2g}^3)$ for chromium and $d^5$ (half-filled) for manganese.

Option 2 — Cu⁺ vs Cu²⁺ stability in water:

Isolated $Cu^+$ (3d¹⁰, fully-filled d subshell) looks like it should be the more stable ion electronically. However, in aqueous solution, stability is governed by overall thermodynamics, not just electronic configuration. $Cu^{2+}$ , being smaller and more highly charged, has a much larger (more negative) hydration enthalpy than $Cu^+$ . This extra hydration energy released more than compensates for the additional (second) ionisation energy required to remove an electron from $Cu^+$ to form $Cu^{2+}$ .

As a result, $Cu^{2+}(aq)$ is thermodynamically more stable than $Cu^+(aq)$ , and $Cu^+$ actually disproportionates in aqueous solution: $2Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s)$ — direct evidence that $Cu^{2+}$ is the more stable aqueous species.

✓Final answer
$Option 1: Cr²⁺ is reducing (\to stable d³ Cr³⁺); Mn³⁺ is oxidising (\to stable half-filled d⁵ Mn²⁺). Option 2: Cu²⁺ is more stable than Cu⁺ in water, because its far greater hydration enthalpy outweighs the extra ionisation energy needed.$
AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL2 marks
$Q. Answer the following questions (any two): When HCl reacts with finely powdered iron, it forms ferrous chloride, and not ferric chloride. Explain, why?$
›Reveal solutionSolution

HCl dissolves iron by a simple acid-metal displacement reaction (Fe loses 2 electrons to H+), which can only reach the Fe2+ state; reaching Fe3+ would require a stronger oxidizing acid than HCl.

When finely powdered iron reacts with dilute hydrochloric acid, the reaction is a straightforward metal-acid displacement (redox) reaction:

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)↑

Here, Fe is oxidised by H⁺ ions: Fe → Fe²⁺ + 2e⁻, while 2H⁺ + 2e⁻ → H2. HCl (specifically, the H⁺ ion) is only a MODEST/weak oxidizing agent — it is capable of oxidising Fe only up to the Fe²⁺ (ferrous) state, which is the more easily accessible oxidation state for iron in this kind of simple acid dissolution.

To oxidise iron further, all the way to Fe³⁺ (ferric), a STRONGER oxidizing agent is required — such as concentrated (oxidizing) HNO3, or chlorine gas (Cl2) reacting directly with iron, both of which have the oxidizing power to remove the additional electron needed to reach Fe³⁺. Ordinary dilute HCl simply lacks this oxidizing strength, so the product remains FeCl2, not FeCl3.

✓Final answer
$Fe + 2HCl \to FeCl2 + H2 — dilute HCl is only a weak (non-oxidizing beyond H⁺/H2) acid, capable of oxidising Fe only to Fe2+ (FeCl2); reaching Fe3+ (FeCl3) needs a stronger oxidizing agent, such as conc. HNO3 or Cl2 gas.$
AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. Why is Cr2+ reducing and Mn3+ oxidizing when both have d4 configuration?$
›Reveal solutionSolution

Although both Cr2+ and Mn3+ have a d4 configuration, they behave oppositely because each is being "pulled" towards a different, more stable neighbouring configuration: Cr2+ loses an electron to reach the stable t2g³ arrangement of Cr3+, while Mn3+ gains an electron to reach the extra-stable half-filled d5 configuration of Mn2+.

Cr2+ is a reducing agent

Cr2+ has the electronic configuration [Ar]3d4. When it loses one electron, it becomes Cr3+, which has the configuration [Ar]3d3, corresponding (in an octahedral field) to the arrangement t2g³eg⁰. A t2g³ configuration, with three electrons singly occupying the three lower-energy t2g orbitals, is particularly stable. Because Cr3+ (d3) is markedly more stable than Cr2+ (d4), Cr2+ readily loses an electron to become Cr3+ — that is, Cr2+ is easily oxidised, making it a good reducing agent.

Mn3+ is an oxidising agent

Mn3+ also has the configuration [Ar]3d4. When it gains one electron, it becomes Mn2+, with configuration [Ar]3d5 — a half-filled d-subshell, in which all five d orbitals are singly occupied. A half-filled d-subshell is exceptionally stable due to maximum exchange energy and symmetric charge distribution. Because Mn2+ (d5) is much more stable than Mn3+ (d4), Mn3+ readily accepts an electron to become Mn2+ — that is, Mn3+ is easily reduced, making it a good oxidising agent.

Summary

Both ions start at d4, an inherently less stable configuration, but each moves towards stability in the opposite direction: Cr2+ moves "down" by oxidation (losing an electron) to reach the stable d3 (t2g³) state, while Mn3+ moves "up" by reduction (gaining an electron) to reach the stable, half-filled d5 state.

✓Final answer
$Cr2+ (d4 \to d3 on oxidation) is reducing because Cr3+'s t2g³ configuration is extra stable. Mn3+ (d4 \to d5 on reduction) is oxidising because Mn2+'s half-filled d5 configuration is extra stable.$

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Q.The electronic configuration of Cu(II) is $3d^9$ whereas that of Cu(I) is $3d^{10}$ . Which of the following is correct?

Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

The Core Intuition

The Precise Statement

Examples That Make It Concrete

How to Think About It in Exams

The Bottom Line

Why this formula?

Stability of Oxidation States: Why It Works

The Core Idea: Energy Minimisation

Key Principle 1: Inert Pair Effect (for p-block elements)

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Key Principle 3: Lattice Energy / Hydration Energy Compensation

Key Principle 4: Disproportionation

Summary Table: Why Each "Formula" Holds

Final Takeaway for Exams

Method: Stability Analysis Using Electronic Configuration & Hydration Energy

Step 1: Compare electronic configurations

Step 2: Consider the real-world stability in aqueous medium

Step 3: Check disproportionation tendency

Step 4: Conclusion

Mistake 1: Assuming Full d-subshell Always Means Higher Stability

Mistake 2: Ignoring the Role of Hydration Enthalpy

Mistake 3: Confusing "Stability" with "Ease of Formation"

Mistake 4: Misinterpreting the "d^10" Exception

The Correct Answer and Why

Q.The electronic configuration of Cu(II) is 3d93d^93d9 whereas that of Cu(I) is 3d103d^{10}3d10. Which of the following is correct?

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Q.The electronic configuration of Cu(II) is $3d^9$ whereas that of Cu(I) is $3d^{10}$ . Which of the following is correct?

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