Q.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

Concept & Intuition

Disproportionation (also called dismutation) occurs when a single chemical species undergoes both oxidation and reduction. The key is to spot an element that appears in the same reactant in an oxidation state that is neither its highest nor its lowest possible value. That element then “disproportionates” into two different products: one with a higher oxidation state (oxidised) and one with a lower oxidation state (reduced).

To identify such a reaction, we assign oxidation states to every atom in the reactants and products, looking for an element that appears in only one reactant but in two different products with different oxidation states.

Step-by-step reasoning

1. Option (A): $2 \mathrm{H_2O_2} \rightarrow 2 \mathrm{H_2O} + \mathrm{O_2}$

   • In $\mathrm{H_2O_2}$, hydrogen is $+1$ (usual), so each oxygen must be $-1$ (since $2(+1) + 2x = 0 \Rightarrow x = -1$).
   • In $\mathrm{H_2O}$, oxygen is $-2$.
   • In $\mathrm{O_2}$, oxygen is $0$.
   • Oxygen in $\mathrm{H_2O_2}$ (oxidation state $-1$) is both reduced to $-2$ in water and oxidised to $0$ in oxygen.
   • This is a classic disproportionation of hydrogen peroxide. ✓

2. Option (B): $2 \mathrm{KMnO_4} \rightarrow \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$

   • In $\mathrm{KMnO_4}$, K is $+1$, O is $-2$, so Mn is $+7$.
   • In $\mathrm{K_2MnO_4}$, Mn is $+6$ (reduction).
   • In $\mathrm{MnO_2}$, Mn is $+4$ (further reduction).
   • In $\mathrm{O_2}$, oxygen is $0$ (oxidation of oxygen from $-2$).
   • Here, manganese is only reduced (from $+7$ to $+6$ and $+4$), not oxidised. The oxidation occurs on oxygen, not on the same element. So this is a decomposition, not a disproportionation. ✗

3. Option (C): $2 \mathrm{MnO_4^{-}} + 10 \mathrm{I^{-}} + 16 \mathrm{H^{+}} \rightarrow 2 \mathrm{Mn^{2+}} + 5 \mathrm{I_2} + 8 \mathrm{H_2O}$

   • Mn in $\mathrm{MnO_4^{-}}$ is $+7$; in $\mathrm{Mn^{2+}}$ it is $+2$ (reduction).
   • I in $\mathrm{I^{-}}$ is $-1$; in $\mathrm{I_2}$ it is $0$ (oxidation). …

The key idea is that disproportionation is a redox reaction where the same species (here $\mathrm{Cu}^{+}$) acts as both oxidising and reducing agent. To find the overall cell potential, we must combine the two relevant half‑reactions correctly — not simply subtract the given potentials, but use the relationship between Gibbs free energy and potential.

Why this approach works

We are given:

• $\mathrm{E}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^{\circ} = 0.34\ \mathrm{V}$ (the potential for $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$)
• $\mathrm{E}_{\mathrm{Cu}^{2+}/\mathrm{Cu}^{+}}^{\circ} = 0.15\ \mathrm{V}$ (the potential for $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^{+}$)

The disproportionation reaction is:

We need to find $E^\circ$ for this overall reaction. Since the number of electrons transferred in the two half‑reactions is different (2 vs 1), we cannot simply add or subtract the given potentials. Instead, we must convert each to $\Delta G^\circ$ (using $\Delta G^\circ = -nFE^\circ$), combine the free energies, and then convert back to $E^\circ$.

Step‑by‑step solution

1. Write the two half‑reactions we actually need

   For the disproportionation, one $\mathrm{Cu}^{+}$ is oxidised to $\mathrm{Cu}^{2+}$, and the other is reduced to $\mathrm{Cu}$. So the half‑reactions are:

   • Oxidation: $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+} + e^-$
   • Reduction: $\mathrm{Cu}^{+} + e^- \rightarrow \mathrm{Cu}$

   But we are not given these directly; we are given potentials involving $\mathrm{Cu}^{2+}/\mathrm{Cu}$ and $\mathrm{Cu}^{2+}/\mathrm{Cu}^{+}$. We can obtain the needed half‑reaction potentials by combining them.

2. Find $E^\circ$ for $\mathrm{Cu}^{+} + e^- \rightarrow \mathrm{Cu}$

   This is the reduction of $\mathrm{Cu}^{+}$ to $\mathrm{Cu}$. We can get it from the two given potentials using a free‑energy cycle.

   Given:

   • (1) $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ $E_1^\circ = 0.34\ \mathrm{V}$, $n_1 = 2$
   • (2) $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^{+}$ $E_2^\circ = 0.15\ \mathrm{V}$, $n_2 = 1$

   We want:

   • (3) $\mathrm{Cu}^{+} + e^- \rightarrow \mathrm{Cu}$ $E_3^\circ = ?$, $n_3 = 1$

   Notice that (1) = (2) + (3). So:

Cancel $-F$:

$E_3^\circ = 0.53\ \mathrm{V}$ …

Concept & Intuition

Disproportionation is a redox reaction where the same element in one oxidation state simultaneously gets oxidized (to a higher state) and reduced (to a lower state). This is possible only if the element’s oxidation number in the species is intermediate — not the highest nor the lowest possible for that element. If the element is already in its maximum oxidation state, it can only be reduced (no oxidation possible), so disproportionation cannot occur. Similarly, if it’s in its minimum state, it can only be oxidized. So the trick is: check the oxidation state of the central (or key) element in each species. If every species in a group has that element at its highest (or lowest) possible oxidation state, then none can disproportionate.

Let’s examine each option.

1. Option (A): $\mathrm{F}_2, \mathrm{ClO}^{-}, \mathrm{Cl}_2, \mathrm{ClO}_2^{-}$

• $\mathrm{F}_2$: Fluorine has oxidation state 0. Fluorine’s possible states are 0 and –1 (it is the most electronegative element, never positive). Since 0 is not the lowest (–1 is), $\mathrm{F}_2$ can disproportionate (e.g., $\mathrm{F}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{OF}^- + \mathrm{F}^- + \mathrm{H}_2\mathrm{O}$, though this is rare; more commonly $\mathrm{F}_2$ can be reduced to $\mathrm{F}^-$ and oxidized to $\mathrm{OF}_2$? Actually, fluorine’s only negative state is –1, and it can form positive compounds with oxygen, so 0 is intermediate. So $\mathrm{F}_2$ can disproportionate.)
• $\mathrm{ClO}^-$: Chlorine is +1. Chlorine’s states range from –1 to +7. +1 is intermediate → can disproportionate (e.g., $3\mathrm{ClO}^- \rightarrow 2\mathrm{Cl}^- + \mathrm{ClO}_3^-$).
• $\mathrm{Cl}_2$: Chlorine is 0, intermediate → can disproportionate (e.g., $\mathrm{Cl}_2 + 2\mathrm{OH}^- \rightarrow \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O}$).
• $\mathrm{ClO}_2^-$: Chlorine is +3, intermediate → can disproportionate. So (A) contains species that can disproportionate → not the answer.

2. Option (B): $\mathrm{P}_4, \mathrm{H}_2\mathrm{O}_2, \mathrm{BrO}_2^-, \mathrm{NO}_2$

• $\mathrm{P}_4$: Phosphorus is 0. Phosphorus states: –3 to +5. 0 is intermediate → can disproportionate (e.g., $\mathrm{P}_4 + 3\mathrm{OH}^- + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{PH}_3 + 3\mathrm{H}_2\mathrm{PO}_2^-$).
• $\mathrm{H}_2\mathrm{O}_2$: Oxygen is –1. Oxygen states: –2 (in oxides, water), 0 (in $\mathrm{O}_2$), –1 (peroxides). –1 is intermediate → can disproportionate (e.g., $2\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2$).
• $\mathrm{BrO}_2^-$: Bromine is +3. Bromine states: –1 to +7. +3 is intermediate → can disproportionate.
• $\mathrm{NO}_2$: Nitrogen is +4. Nitrogen states: –3 to +5. +4 is intermediate → can disproportionate (e.g., $2\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HNO}_3 + \mathrm{HNO}_2$). So (B) also contains species that can disproportionate.

3. Option (C): $\mathrm{Cr}_2\mathrm{O}_7^{2-}, \mathrm{F}_2, \mathrm{MnO}_4^{-}, \mathrm{ClO}_4^{-}$

• $\mathrm{Cr}_2\mathrm{O}_7^{2-}$: Chromium is +6. Chromium’s highest common state is +6 (also +3, +2). +6 is the maximum → cannot be oxidized further, so no disproportionation. …