In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in MoO₃ and W(VI) in

NCERT Exemplar · Q52

MoO_3

(i) Cr(VI) is more stable than Mo(VI) and W(VI).

(ii) Mo(VI) and W(VI) are more stable than Cr(VI).

(iii) Higher oxidation states of heavier members of group-6 of transition series are more stable.

(iv) Lower oxidation states of heavier members of group-6 of transition series are more stable.

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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 key idea is the stability of higher oxidation states increases down a group in transition series. Cr(VI) is a strong oxidising agent because it is less stable and readily reduces, while Mo(VI) and W(VI) are more stable and thus not strongly oxidising. The correct option is (iii).

The question asks why Cr(VI) in dichromate is a powerful oxidising agent in acidic medium, but Mo(VI) in $MoO_3$ and W(VI) in $WO_3$ are not. This is a classic example of a trend in the transition series — the stability of higher oxidation states.

Understand the trend in stability of oxidation states.

In the d-block, as you move down a group (from 3d → 4d → 5d series), the higher oxidation states become more stable. This happens because the heavier transition metals have larger atomic radii and their d-orbitals are more diffuse, which allows them to share electrons more effectively with oxygen and form strong bonds. For example, Cr(VI) is less stable than Mo(VI) and W(VI).
Relate stability to oxidising power.

A species that is less stable in a high oxidation state tends to gain electrons easily (i.e., it is a strong oxidising agent). Conversely, a species that is more stable in a high oxidation state resists reduction and is a weak oxidising agent.
- Cr(VI) (as $Cr_2O_7^{2-}$ ) is relatively unstable in acidic medium and readily reduces to Cr(III), making it a strong oxidiser.
- Mo(VI) and W(VI) are much more stable in their +6 states, so they do not readily accept electrons — hence they are not strong oxidising agents.
Evaluate the given options.
- (i) Cr(VI) is more stable than Mo(VI) and W(VI) — this is false; it's the opposite.
- (ii) Mo(VI) and W(VI) are more stable than Cr(VI) — this is true in terms of stability, but the question asks why Cr(VI) is a strong oxidiser while the others are not. The reason is the trend in stability, not just the fact itself. …

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Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing “Stability” with “Oxidising Power”

The Error: Students see that Cr(VI) is a strong oxidising agent and assume this means Cr(VI) is more stable than Mo(VI) or W(VI). They then pick option (i).
Why it’s wrong: An oxidising agent is strong because it is unstable and wants to get reduced. If Cr(VI) were more stable, it would not readily accept electrons.
How to Avoid: Remember the inverse relationship:
- High stability → Weak oxidising agent (hard to reduce).
- Low stability → Strong oxidising agent (easy to reduce).
- Since Cr(VI) is a strong oxidiser, it is less stable than Mo(VI) and W(VI).

Mistake 2: Forgetting the Trend Down the Group

The Error: Students memorise that “higher oxidation states are stable for heavier elements” but then apply it incorrectly. They might pick option (iii) but fail to connect it to the oxidising power logic.
Why it’s wrong: The correct trend is that stability of the +6 oxidation state increases down Group 6 (Cr < Mo < W). This means Mo(VI) and W(VI) are more stable, not less.
How to Avoid: Draw a mental arrow:
- Down Group 6: Stability of +6 state increases.
- Therefore: Oxidising power of +6 state decreases.
- So: Cr(VI) is the strongest oxidiser; W(VI) is the weakest.

Mistake 3: Misinterpreting the “Why” Behind the Trend

The Error: Students think the trend is random or just based on atomic size. They don’t connect it to the inert pair effect or shielding.
Why it’s wrong: The actual reason involves the poor shielding by d- and f-electrons in heavier elements. For Mo and W, the 4d and 5d orbitals are more diffuse, making the +6 state more stable relative to Cr’s compact 3d orbitals.
How to Avoid: Link the concept to periodicity:
- For 3d series (Cr): Smaller size, poor shielding → +6 state is highly oxidising.
- For 4d/5d series (Mo, W): Larger size, better shielding → +6 state is more stable and less oxidising.

Mistake 4: Picking the Wrong Option Due to Wording …

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Q.In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in $MoO_3$ and W(VI) in $WO_3$ are not because ___________.

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: Stability Trend in Transition Metal Oxidation States

Steps:

Mistake 1: Confusing “Stability” with “Oxidising Power”

Mistake 2: Forgetting the Trend Down the Group

Mistake 3: Misinterpreting the “Why” Behind the Trend

Mistake 4: Picking the Wrong Option Due to Wording …

Q.In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in MoO3MoO_3MoO3​ and W(VI) in WO3WO_3WO3​ are not because ___________.

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Q.In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in $MoO_3$ and W(VI) in $WO_3$ are not because ___________.

Unlock everything free for 14 days