Which of the following will not act as oxidising agents? (i) CrO₃ (ii) MoO₃ (iii) WO₃ (iv) CrO₄²⁻

NCERT Exemplar · Q58

Q. Which of the following will not act as oxidising agents?

(i)

CrO_3

(ii)

MoO_3

(iii)

WO_3

(iv)

CrO_4^{2-}

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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 that the stability of the highest oxidation state decreases down a group in transition metals. For Group 6, $Cr(VI)$ is a strong oxidising agent, but $Mo(VI)$ and $W(VI)$ are very stable and do not act as oxidising agents. The correct answer is (ii) and (iii).

Let’s understand why this happens. In transition metals, the highest oxidation state (like +6 for Group 6 elements) becomes more stable as you go down the group. This is because the heavier elements (Mo, W) have larger atomic radii and their d-orbitals are more diffuse, making it harder to reduce them back to a lower state. So, $CrO_3$ and $CrO_4^{2-}$ (both $Cr(VI)$ ) are strong oxidising agents — they readily get reduced to $Cr(III)$ . But $MoO_3$ and $WO_3$ are quite inert; they don’t easily accept electrons.

Now, let’s check each option step by step.

Option (i): $CrO_3$

Chromium trioxide contains chromium in the +6 oxidation state. $Cr(VI)$ is highly unstable in the presence of reducing agents and even in water it tends to get reduced. For example, $CrO_3$ is a well-known oxidising agent used in organic chemistry (Jones reagent). It readily gets reduced to $Cr^{3+}$ (green). So, it will act as an oxidising agent.
Option (ii): $MoO_3$

Molybdenum trioxide has molybdenum in +6 state. But $Mo(VI)$ is much more stable than $Cr(VI)$ . In fact, $MoO_3$ is the stable form of molybdenum in air and does not readily oxidise other substances. It is not used as an oxidising agent. So, it will NOT act as an oxidising agent.
Option (iii): $WO_3$

Tungsten trioxide is even more stable. $W(VI)$ is the most stable oxidation state for tungsten. $WO_3$ is a yellow solid that does not get reduced easily. It is not an oxidising agent. So, it will NOT act as an oxidising agent.
Option (iv): $CrO_4^{2-}$ …

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Common Mistakes: Stability of Oxidation States (CrO₃, MoO₃, WO₃, CrO₄²⁻)

Mistake 1: Assuming all d-block oxides in high oxidation states are strong oxidising agents

Why it happens: Students memorise that Cr(VI) compounds (like CrO₃ and CrO₄²⁻) are strong oxidising agents. They then assume MoO₃ and WO₃ behave identically because Mo and W are in the same group.

The correct understanding:

Cr(VI) is unstable and readily gets reduced to Cr(III) — so CrO₃ and CrO₄²⁻ are oxidising agents.
Mo(VI) and W(VI) are much more stable due to poor shielding by 4f and 5f orbitals and lanthanoid contraction. Their higher oxidation states are stabilised, so MoO₃ and WO₃ do not act as oxidising agents under normal conditions.

How to avoid:

Remember: Stability of +6 oxidation state increases down Group 6: Cr < Mo < W
So Cr(VI) is oxidising, Mo(VI) and W(VI) are not.

Mistake 2: Confusing CrO₃ with CrO₄²⁻

Why it happens: Both contain Cr in +6 state, so students think they behave identically.

The correct understanding:

CrO₃ (chromium trioxide) is a strong oxidising agent — it oxidises alcohols, etc.
CrO₄²⁻ (chromate ion) is also an oxidising agent in acidic medium (converts to Cr₂O₇²⁻).
Both are oxidising agents — the question asks which will not act as oxidising agents, so both CrO₃ and CrO₄²⁻ are excluded from the answer.

How to avoid:

Treat all Cr(VI) species as oxidising agents unless stated otherwise.
Focus on the group trend, not just the oxidation number.

Mistake 3: Forgetting the role of the medium (acidic vs basic)

Why it happens: Students think CrO₄²⁻ is stable in all conditions.

The correct understanding:

In acidic medium, CrO₄²⁻ converts to Cr₂O₇²⁻, which is a strong oxidising agent.
In basic medium, CrO₄²⁻ is more stable but still can act as an oxidising agent under certain conditions.
For exam purposes, CrO₄²⁻ is considered an oxidising agent.

How to avoid: …

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Q.Which of the following will not act as oxidising agents?

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 of Higher Oxidation States Across the 3d, 4d, and 5d Series

Concept Behind the Method

Steps to Solve

Common Mistakes: Stability of Oxidation States (CrO₃, MoO₃, WO₃, CrO₄²⁻)

Mistake 1: Assuming all d-block oxides in high oxidation states are strong oxidising agents

Mistake 2: Confusing CrO₃ with CrO₄²⁻

Mistake 3: Forgetting the role of the medium (acidic vs basic)

Q.Which of the following will not act as oxidising agents?

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