Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

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Q. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

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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 regular decrease in reactivity from Sc to Cu is driven by increasing nuclear charge, which shrinks atomic size and raises ionization enthalpy, making electron loss progressively harder.

The question asks why the reactivity of transition elements falls almost regularly as you move from scandium (Sc) to copper (Cu) across the first transition series. Reactivity here means the tendency to lose electrons and form positive ions — a metal that reacts more readily gives up its electrons more easily.

The key lies in how the atomic structure changes across the series. As you go from Sc to Cu, each step adds one proton to the nucleus and one electron to the 3d subshell. The 3d electrons are poor at shielding the increasing nuclear charge from the outer 4s electrons. So the effective nuclear charge felt by the outermost electrons rises steadily.

This has two direct consequences:

Atomic size decreases — the electron cloud is pulled inward by the stronger nuclear attraction.
Ionization enthalpy increases — more energy is needed to remove an electron because it is held more tightly.

Since reactivity depends on how easily a metal can lose electrons, a higher ionization enthalpy means lower reactivity. The trend is not perfectly linear — there are small irregularities due to half-filled and fully filled d-subshell stability — but the overall direction is clear.

Let’s walk through the reasoning step by step.

Across the series, nuclear charge increases by one unit per element.

Sc has atomic number 21, Cu has atomic number 29. Each added proton pulls the electron cloud inward.
The added electrons go into the 3d subshell, which shields the 4s electrons poorly.

The 3d orbitals are more diffuse and penetrate less toward the nucleus than 4s orbitals. So the 4s electrons feel a progressively stronger pull from the nucleus.
Atomic radii decrease gradually from Sc to Cu.

For example, the metallic radius of Sc is about 164 pm, while that of Cu is about 128 pm. This contraction is called the lanthanoid contraction’s smaller cousin — it’s steady and predictable.
Ionization enthalpies (first, second, third) increase across the series.

The first ionization enthalpy of Sc is about 631 kJ/mol, while that of Cu is about 745 kJ/mol. Removing an electron becomes harder because the electron is held more tightly by the larger effective nuclear charge.
Reactivity is inversely related to ionization enthalpy.

A metal that requires more energy to lose an electron is less reactive. So as ionization enthalpy rises, reactivity falls. …

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Common Mistakes: Stability of Oxidation States in Transition Elements

Students often struggle with the statement: "Reactivity of transition elements decreases almost regularly from Sc to Cu." Here are the most frequent errors and how to avoid them.

✗ Mistake 1: Confusing "Reactivity" with "Stability of Oxidation States"

The error: Students think the statement is about how stable the +2 or +3 oxidation state is across the series.

Why it's wrong: Reactivity here refers to tendency to lose electrons (i.e., ease of oxidation), not the stability of a particular oxidation state once formed.

How to avoid: Remember:

Reactivity → how readily the metal atom loses electrons to form ions.
Stability of oxidation state → how likely that ion is to remain in that state without further change.

These are related but not the same.

✗ Mistake 2: Ignoring the Role of Nuclear Charge

The error: Students attribute the trend only to d-orbital filling, without considering the increasing nuclear charge.

Why it's wrong: As we move from Sc ( $Z=21$ ) to Cu ( $Z=29$ ), the nuclear charge increases by 8 units, while electrons are added to the inner 3d subshell. This causes:

Stronger attraction between nucleus and outer electrons
Higher ionization enthalpy
Less tendency to lose electrons

How to avoid: Always connect the trend to effective nuclear charge ( $Z_{\text{eff}}$ ). Write it out:

As $Z$ increases across the period, $Z_{\text{eff}}$ increases → electrons are held more tightly → ionization enthalpy increases → reactivity decreases.

✗ Mistake 3: Forgetting the "Almost Regularly" Part

The error: Students treat the trend as perfectly smooth (Sc > Ti > V > ... > Cu).

Why it's wrong: The trend is almost regular, not perfectly regular. There are small deviations due to:

Exchange energy stabilization (especially at half-filled and fully-filled d-subshells)
Anomalous electronic configurations (e.g., Cr: $3d^5 4s^1$ , Cu: $3d^{10} 4s^1$ )

How to avoid: Use the word "generally" or "almost" in your explanation. Mention that:

Cr and Cu have slightly higher stability due to half-filled/fully-filled d-orbitals
This makes them slightly less reactive than the simple trend would predict

✗ Mistake 4: Not Mentioning the 4s and 3d Energy Crossover

The error: Students treat 4s and 3d orbitals as having fixed energy ordering.

Why it's wrong: The energy of 4s and 3d orbitals crosses as nuclear charge increases:

For Sc–Mn: 4s is lower in energy than 3d
For Fe–Cu: 3d becomes lower than 4s

This affects how easily electrons are lost.

How to avoid: State clearly:

For earlier elements (Sc–Mn), the 4s electrons are lost first. For later elements (Fe–Cu), the 3d electrons are more tightly bound, making electron loss harder.

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Q.Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

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 …

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

Method: Electronic Configuration & Ionization Energy Trend Analysis

Steps

Common Mistakes: Stability of Oxidation States in Transition Elements

✗ Mistake 1: Confusing "Reactivity" with "Stability of Oxidation States"

✗ Mistake 2: Ignoring the Role of Nuclear Charge

✗ Mistake 3: Forgetting the "Almost Regularly" Part

✗ Mistake 4: Not Mentioning the 4s and 3d Energy Crossover

Q.Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

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