Assertion: [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂ are reducing in nature. Reason: Unpaired electrons are present in t

NCERT Exemplar · Q47

[Cr(H_2O)_6]Cl_2

(i) Assertion and reason both are true, reason is correct explanation of assertion.

(ii) Assertion and reason both are true but reason is not the correct explanation of assertion.

(iii) Assertion is true, reason is false.

(iv) Assertion is false, reason is true.

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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 assertion is true — both complexes are reducing — but the reason (unpaired electrons) is not the correct explanation. The real cause is the instability of the +2 oxidation state for Cr and Fe due to their electronic configurations, which drives them to lose electrons and reach a more stable +3 state.

Let’s break this down properly. The question tests your understanding of stability of oxidation states in transition metals — a concept that often gets confused with the mere presence of unpaired electrons.

1. What does “reducing in nature” mean here?

A reducing agent is a species that loses electrons easily — it gets oxidised itself while reducing something else. So when we say $[Cr(H_2O)_6]Cl_2$ and $[Fe(H_2O)_6]Cl_2$ are reducing, we mean the metal ion in the +2 state tends to get oxidised to +3 (or higher) by donating an electron.

The question is: why do these +2 ions want to lose an electron?

2. Check the electronic configurations

Cr in +2 state:

Atomic Cr: $[Ar]\,3d^5\,4s^1$

Cr $^{2+}$ : loses the 4s electron first → $[Ar]\,3d^4$

This is not a half-filled or stable configuration. Cr $^{3+}$ ( $3d^3$ ) is more stable because it has a half-filled $t_{2g}$ set in an octahedral field (though not half-filled overall, it’s still more stable than $d^4$ due to exchange energy and crystal field effects).
Fe in +2 state:

Atomic Fe: $[Ar]\,3d^6\,4s^2$

Fe $^{2+}$ : $[Ar]\,3d^6$

Fe $^{3+}$ : $[Ar]\,3d^5$ — half-filled d-subshell, which is exceptionally stable.

So in both cases, the +3 state is more stable than the +2 state. That’s why these +2 complexes readily lose an electron — they are reducing agents.

The driving force for reduction is the greater stability of the +3 oxidation state compared to +2, not merely the presence of unpaired electrons.

3. Now examine the reason given

The reason says: “Unpaired electrons are present in their d-orbitals.”

Cr $^{2+}$ ( $d^4$ ): in an octahedral field, high-spin → 4 unpaired electrons.
Fe $^{2+}$ ( $d^6$ ): high-spin → 4 unpaired electrons.

So the reason is factually true — unpaired electrons are present. …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Reducing Nature" with "Presence of Unpaired Electrons"

The Error:

Students often think that any compound with unpaired electrons must be reducing. They see unpaired $d$ -electrons in both $[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ and immediately mark option (i).

Why It's Wrong:

Reducing ability depends on how easily the metal can lose an electron (i.e., its tendency to get oxidized to a higher stable oxidation state). Unpaired electrons alone don't guarantee reducing power — the stability of the next higher oxidation state matters.

How to Avoid:

Always check: Can this ion easily go to a higher oxidation state?
For $[Cr(H_2O)_6]^{2+}$ : Cr(II) → Cr(III) is very easy because Cr(III) has a half-filled $t_{2g}^3$ configuration (stable).
For $[Fe(H_2O)_6]^{2+}$ : Fe(II) → Fe(III) is also easy because Fe(III) has a half-filled $d^5$ configuration (stable).
So both are reducing, but not because of unpaired electrons — it's because the next oxidation state is stable.

Mistake 2: Thinking "Unpaired Electrons = Reducing Agent" is Always True

The Error:

Students apply this logic blindly to all transition metal complexes.

Why It's Wrong:

Consider $[Mn(H_2O)_6]^{2+}$ — it has 5 unpaired electrons ( $d^5$ high spin), but it is not reducing because Mn(III) is less stable. Unpaired electrons don't guarantee reducing power.

How to Avoid:

Memorize the stability trends of common oxidation states:
- Cr(II) → Cr(III): reducing
- Fe(II) → Fe(III): reducing
- Mn(II) → Mn(III): not reducing (Mn(III) is unstable)
Reason: The stability of the product (higher oxidation state) determines reducing nature, not the number of unpaired electrons in the reactant.

Mistake 3: Misinterpreting the Assertion-Reason Relationship

The Error:

Students see both statements as true and pick (i) without checking if the reason correctly explains the assertion.

Why It's Wrong:

The reason says: "Unpaired electrons are present in their d-orbitals."

This is true — both Cr(II) and Fe(II) have unpaired electrons.

But this is not the correct explanation for why they are reducing. The correct explanation is the stability of the +3 oxidation state.

How to Avoid:

For Assertion-Reason questions, ask: "Does the reason directly cause the assertion?"
Here: Unpaired electrons ≠ reducing power. So reason is true but not the correct explanation.
Correct answer: (ii) — Assertion true, Reason true, but Reason is not the correct explanation.

Mistake 4: Forgetting the Electronic Configurations

The Error: …

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Q.Assertion: $[Cr(H_2O)_6]Cl_2$ and $[Fe(H_2O)_6]Cl_2$ are reducing in nature.
Reason: Unpaired electrons are present in their d-orbitals.

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

1. What does “reducing in nature” mean here?

2. Check the electronic configurations

3. Now examine the reason given

Method: Stability of Oxidation States – Electronic Configuration Analysis

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Reducing Nature" with "Presence of Unpaired Electrons"

Mistake 2: Thinking "Unpaired Electrons = Reducing Agent" is Always True

Mistake 3: Misinterpreting the Assertion-Reason Relationship

Mistake 4: Forgetting the Electronic Configurations

Q.Assertion: [Cr(H2O)6]Cl2[Cr(H_2O)_6]Cl_2[Cr(H2​O)6​]Cl2​ and [Fe(H2O)6]Cl2[Fe(H_2O)_6]Cl_2[Fe(H2​O)6​]Cl2​ are reducing in nature. Reason: Unpaired electrons are present in their d-orbitals.

1. What does “reducing in nature” mean here?

2. Check the electronic configurations

3. Now examine the reason given

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Q.Assertion: $[Cr(H_2O)_6]Cl_2$ and $[Fe(H_2O)_6]Cl_2$ are reducing in nature.
Reason: Unpaired electrons are present in their d-orbitals.

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