When acidified $K_2Cr_2O_7$ solution is added to $Sn^{2+}$ salts then $Sn^{2+}$ changes to (i) $Sn$ (ii) $Sn^{3+}$ (iii) $Sn^{4+}$ (iv) $Sn^{+}$

When acidified K₂Cr₂O₇ solution is added to Sn²⁺ salts then Sn²⁺ changes to (i) Sn (ii) Sn³⁺ (iii) Sn⁴⁺ (iv) S

NCERT Exemplar · Q18

K_2Cr_2O_7

(i)

Sn

(ii)

Sn^{3+}

(iii)

Sn^{4+}

(iv)

Sn^{+}

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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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Acidified $K_2Cr_2O_7$ is a strong oxidising agent. It oxidises $Sn^{2+}$ to its highest stable oxidation state, which is $Sn^{4+}$ . The correct option is (iii).

Concept and Intuition: Stability of Oxidation States

The key to this question lies in understanding two things: the oxidising power of dichromate and the stable oxidation states of tin.

Tin (Sn) is a group 14 element. Its common oxidation states are +2 and +4. The +4 state is more stable for tin, especially in aqueous solution, because of the inert pair effect — the tendency of the 5s² electrons to participate less readily in bonding as you go down the group. However, in the presence of a strong oxidising agent, the +2 state is easily converted to +4.

Now, acidified potassium dichromate ( $K_2Cr_2O_7$ ) is one of the most common oxidising agents in inorganic chemistry. In acidic medium, the dichromate ion ( $Cr_2O_7^{2-}$ ) gets reduced to $Cr^{3+}$ , and in the process, it oxidises whatever it reacts with. The half-reaction is:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

This reaction has a high reduction potential ( $E^\circ \approx +1.33\ \text{V}$ ), meaning it is a powerful oxidiser. So when you add it to a solution containing $Sn^{2+}$ , the tin will be oxidised — but to what?

Tin does not have a +3 oxidation state that is stable in water. The +1 state is also highly unstable. The only reasonable higher oxidation state for tin is +4. So $Sn^{2+}$ loses two electrons to become $Sn^{4+}$ .

Watch out

A common mistake is to think that $Sn^{2+}$ might get reduced to metallic $Sn$ (option (i)) because dichromate is an oxidising agent, not a reducing agent. It will never reduce $Sn^{2+}$ — it will oxidise it. Also, $Sn^{3+}$ is not a stable oxidation state for tin in aqueous solution.

Step-by-Step Reasoning

Identify the nature of the reagent. Acidified $K_2Cr_2O_7$ is a strong oxidising agent in acidic medium. The chromium in $Cr_2O_7^{2-}$ is in the +6 oxidation state, and it gets reduced to $Cr^{3+}$ (+3). This means it will accept electrons from whatever it reacts with.
Identify the possible change for tin. $Sn^{2+}$ is already in a +2 oxidation state. It can either:
- Gain electrons (be reduced) to $Sn$ (0) — but this would require a reducing agent, not an oxidising agent.
- Lose electrons (be oxidised) to a higher state. The only stable higher state for tin is $Sn^{4+}$ . …

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Let’s break this down.

🧪 The Core Concept

Acidified $K_2Cr_2O_7$ is a strong oxidizing agent. In acidic medium, the half-reaction is:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

$Sn^{2+}$ is a reducing agent — it tends to lose electrons and get oxidized. The stable higher oxidation state for tin is $Sn^{4+}$ .

The balanced reaction:

3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O

So the correct answer is (iii) $Sn^{4+}$ .

✗ Common Mistakes & How to Avoid Them

1. Choosing $Sn$ (option (i)) — confusing reduction with oxidation

Why it happens: Students see a strong oxidizing agent and think “it will oxidize the other substance,” but then mistakenly write the product as the reduced form of tin.
How to avoid: Always ask: Is $Sn^{2+}$ gaining or losing electrons?
- $Sn^{2+} \rightarrow Sn$ would be gain of 2 electrons (reduction).
- But $Sn^{2+}$ is the reactant here — it’s being oxidized, so it must lose electrons → go to a higher oxidation state.

2. Choosing $Sn^{3+}$ (option (ii)) — assuming an unstable oxidation state

Why it happens: Students think “+2 to +3 is a small step” or confuse with elements like iron ( $Fe^{2+} \rightarrow Fe^{3+}$ ).
How to avoid: Know the common stable oxidation states of tin:
- Tin shows +2 and +4 only (in its ionic compounds).
- $Sn^{3+}$ is not stable in aqueous solution.
- If an oxidizing agent acts on $Sn^{2+}$ , the only stable product is $Sn^{4+}$ .

3. Choosing $Sn^{+}$ (option (iv)) — ignoring charge balance and logic

Why it happens: Rushing or misreading the charge.
How to avoid:
- $Sn^{+}$ would mean oxidation state +1, which is even less stable than +3 for tin.
- Always check: oxidation means increase in oxidation number. From $+2$ to $+1$ is reduction, not oxidation. …

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Q.When acidified $K_2Cr_2O_7$ solution is added to $Sn^{2+}$ salts then $Sn^{2+}$ changes to

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

Concept and Intuition: Stability of Oxidation States

Step-by-Step Reasoning

Concept: Oxidation–Reduction (Redox) Reactions of Dichromate with Tin(II)

Steps

🧪 The Core Concept

✗ Common Mistakes & How to Avoid Them

1. Choosing $Sn$ (option (i)) — confusing reduction with oxidation

2. Choosing $Sn^{3+}$ (option (ii)) — assuming an unstable oxidation state

3. Choosing $Sn^{+}$ (option (iv)) — ignoring charge balance and logic

Q.When acidified K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ solution is added to Sn2+Sn^{2+}Sn2+ salts then Sn2+Sn^{2+}Sn2+ changes to

Concept and Intuition: Stability of Oxidation States

Step-by-Step Reasoning

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Q.When acidified $K_2Cr_2O_7$ solution is added to $Sn^{2+}$ salts then $Sn^{2+}$ changes to

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