Which of the following statements is not correct? (i) Copper liberates hydrogen from acids. (ii) In its higher

NCERT Exemplar · Q17

Q. Which of the following statements is not correct?

(i) Copper liberates hydrogen from acids.

(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.

(iii)

Mn^{3+}

and

Co^{3+}

are oxidising agents in aqueous solution.

(iv)

Ti^{2+}

and

Cr^{2+}

are reducing agents in aqueous solution.

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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 to check each statement against known trends in transition-metal chemistry — particularly the stability of oxidation states and their redox behaviour in aqueous solution. The incorrect statement is (i), because copper does not liberate hydrogen from acids (it lies below hydrogen in the electrochemical series).

Statement (i): Copper liberates hydrogen from acids.

This is a classic point. The electrochemical series places hydrogen above copper — meaning copper has a lower tendency to get oxidised than hydrogen does. For a metal to displace hydrogen from an acid, its standard reduction potential must be more negative than $0\ \text{V}$ (or at least less positive than hydrogen’s). Copper’s standard reduction potential for $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ is $+0.34\ \text{V}$ , which is positive. So copper cannot reduce $\text{H}^+$ to $\text{H}_2$ under normal conditions.

Watch out

A common mistake is to think that because copper reacts with concentrated nitric acid (producing NO₂, not H₂), it reacts with all acids. That’s wrong — the reaction with nitric acid is due to the oxidising nature of the nitrate ion, not the acidity. Copper does not liberate hydrogen from any acid.
Statement (ii): In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.

Manganese in its +7 oxidation state (as in $\text{MnO}_4^-$ ) is very stable with oxygen. Similarly, $\text{MnF}_4$ and $\text{MnF}_3$ exist, though $\text{Mn(VII)}$ fluoride is not stable (fluorine cannot stabilise the +7 state as well as oxygen does). The statement says “with oxygen and fluorine” — it does not claim all higher states are stable with both, but that stable compounds exist. This is correct: $\text{Mn}_2\text{O}_7$ (with oxygen) and $\text{MnF}_4$ (with fluorine) are known.
Statement (iii): $\text{Mn}^{3+}$ and $\text{Co}^{3+}$ are oxidising agents in aqueous solution.

Both ions have a strong tendency to gain an electron and get reduced to a lower oxidation state.
- $\text{Mn}^{3+}$ is unstable in water and readily disproportionates: $2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+$ . It acts as an oxidising agent. …

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Here are the common mistakes students make on this question, broken down by each option, along with the conceptual fix.

Mistake 1: Assuming Copper behaves like other transition metals

The Trap: Students see "Copper" and think it will react with acids like Zinc or Magnesium to give $H_2$ gas.

The Reality: Copper is below hydrogen in the electrochemical series. It cannot displace $H^+$ from acids.

Why? $E^\circ_{Cu^{2+}/Cu} = +0.34V$ (positive), meaning the reduction of $Cu^{2+}$ is easier than the reduction of $H^+$ .
The correct fact: Copper reacts only with oxidising acids (like conc. $HNO_3$ or hot conc. $H_2SO_4$ ), but these produce $NO_2$ or $SO_2$ , not $H_2$ .
How to avoid: Memorise the reactivity series and the position of Cu. If a metal has a positive standard reduction potential, it will not liberate $H_2$ from dilute acids.

Mistake 2: Forgetting the "Fluorine Exception" for Manganese

The Trap: Students recall that Mn in +7 state (like $MnO_4^-$ ) is a strong oxidising agent, so they assume it cannot form stable compounds with any non-metal.

The Reality: The statement says "with oxygen and fluorine." This is a specific exception.

Why? Oxygen and fluorine are the most electronegative elements. They can stabilise very high oxidation states.
- Oxygen: $Mn_2O_7$ (Mn in +7) is stable (though explosive).
- Fluorine: $MnF_4$ (Mn in +4) and $K_2MnF_6$ (Mn in +4) are stable. Mn in +7 with F is not stable, but the statement says "higher oxidation states" (not necessarily the highest), so +4 qualifies.
How to avoid: Remember the "Electronegativity Rule": High oxidation states are stabilised by highly electronegative atoms (O, F). Do not confuse "oxidising power" with "stability of the compound."

Mistake 3: Confusing Oxidising vs. Reducing Agents

The Trap: Students mix up which species gets oxidised/reduced.

The Logic:

Oxidising agent: Gets reduced itself (its oxidation number decreases). It takes electrons from others.
Reducing agent: Gets oxidised itself (its oxidation number increases). It gives electrons to others.

For Option (iii): $Mn^{3+}$ and $Co^{3+}$

Why they are oxidising agents: They want to gain an electron to become $Mn^{2+}$ and $Co^{2+}$ (more stable $d^5$ and $d^6$ configurations).
Common error: Thinking $Mn^{3+}$ is stable because it is a common ion. It is actually unstable in water and readily reduces.
How to avoid: Check the electronic configuration of the product. If the product has a half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) d-subshell, the reactant is likely an oxidising agent.

For Option (iv): $Ti^{2+}$ and $Cr^{2+}$ …

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Q.Which of the following statements is not 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 …

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 & Stability Analysis

Steps:

Mistake 1: Assuming Copper behaves like other transition metals

Mistake 2: Forgetting the "Fluorine Exception" for Manganese

Mistake 3: Confusing Oxidising vs. Reducing Agents

Q.Which of the following statements is not correct?

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