Q.How would you account for the increasing oxidising power in the series $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$?

The key idea is the stability of higher oxidation states across a period. As we move from V to Mn, the +6 and +7 states become progressively less stable and more strongly oxidising.

1. All three ions contain the metal in a high oxidation state: V(+5), Cr(+6), and Mn(+7).
2. Across the first transition series, the effective nuclear charge increases, making it harder for the metal to retain its high oxidation state — it tends to gain electrons more readily.
3. The standard reduction potentials confirm this trend: $E^\circ (VO_2^+/VO^{2+}) \approx +1.00\ \text{V}$ $E^\circ (Cr_2O_7^{2-}/Cr^{3+}) \approx +1.33\ \text{V}$ …

The question asks you to explain why the oxidising power — the ability to gain electrons and get reduced — increases in the series $VO_2^+$, $Cr_2O_7^{2-}$, $MnO_4^-$. This is a classic trend in d-block chemistry, and the answer lies in the oxidation states of the central metal atoms and the stability of their reduced forms.

Concept first: Oxidising power is measured by the standard reduction potential ($E^\circ$). A higher (more positive) $E^\circ$ means the species is more readily reduced, hence a stronger oxidising agent. For oxoanions of transition metals in their highest oxidation states, the trend depends on:

• The charge density on the central metal ion (higher oxidation state → stronger pull for electrons).
• The stability of the reduced product in the given medium (here, acidic medium).

Let’s break it down step by step.

1. Identify the oxidation states of the central metal atoms.

   In $VO_2^+$, vanadium is in the +5 oxidation state.

   In $Cr_2O_7^{2-}$, chromium is in the +6 state.

   In $MnO_4^-$, manganese is in the +7 state.

   So the series is: +5 < +6 < +7.

   As the oxidation state increases, the metal ion becomes more electron-deficient and more eager to accept electrons — this directly increases oxidising power.

2. Consider the reduction half-reactions in acidic medium.

   The standard reduction potentials (at 298 K, 1 M H⁺) are:

   • $VO_2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H_2O$; $E^\circ = +1.00\ \text{V}$
   • $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$; $E^\circ = +1.33\ \text{V}$
   • $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$; $E^\circ = +1.51\ \text{V}$

   The $E^\circ$ values increase from +1.00 V to +1.33 V to +1.51 V. This is the quantitative proof: a more positive $E^\circ$ means a stronger oxidising agent.

3. Why does $E^\circ$ increase with oxidation state?

   The key is the stability of the reduced species.

   • $VO_2^+$ reduces to $VO^{2+}$ (V⁴⁺), which is relatively stable but not as stable as the others in acidic medium.
   • $Cr_2O_7^{2-}$ reduces to $Cr^{3+}$ (Cr³⁺), which has a half-filled $t_{2g}^3$ configuration — this is extra stable due to exchange energy and crystal field stabilisation.
   • $MnO_4^-$ reduces to $Mn^{2+}$ (Mn²⁺), which has a half-filled $d^5$ configuration — this is exceptionally stable (high exchange energy, symmetrical electron distribution).

   The greater the stability of the reduced form, the more the equilibrium shifts toward reduction, giving a higher $E^\circ$.

4. Also consider the role of pH. …

Method: Standard Electrode Potential (E° Value) Comparison

This method uses the standard reduction potential ($E^\circ$) to explain the trend in oxidising power. A higher (more positive) $E^\circ$ means the species is more easily reduced — hence a stronger oxidising agent.

Steps:

1. Identify the half-reactions for each species in acidic medium:

   • $VO_2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H_2O$
   • $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
   • $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

2. Recall the standard reduction potentials ($E^\circ$ values):

   • $VO_2^+$: $E^\circ \approx +1.00\ \text{V}$
   • $Cr_2O_7^{2-}$: $E^\circ \approx +1.33\ \text{V}$
   • $MnO_4^-$: $E^\circ \approx +1.51\ \text{V}$

3. Compare the $E^\circ$ values:

   • $E^\circ$ increases from $VO_2^+$ to $MnO_4^-$.
   • A higher $E^\circ$ means a greater tendency to gain electrons (be reduced).

4. Conclude the trend in oxidising power:

   • Since $E^\circ(MnO_4^-) > E^\circ(Cr_2O_7^{2-}) > E^\circ(VO_2^+)$, the oxidising power increases as: …

Here are the most common mistakes students make on this question, along with the conceptual fixes to avoid them.

Mistake 1: Confusing "Oxidising Power" with "Stability of the Ion"

• The Error: Students think $MnO_4^-$ is the strongest oxidising agent because it is the most stable ion. This is backwards.
• The Fix: Oxidising power is the ease of reduction. A strong oxidising agent is unstable in its current high oxidation state and wants to get reduced.
  • $MnO_4^-$ has Mn in the +7 state. This is a very high, unstable oxidation state for a 3d metal.
  • $VO_2^+$ has V in the +5 state. This is relatively more stable for Vanadium.
  • Key Concept: Higher oxidation states become less stable as you move right across a period (Sc → Zn). Therefore, the tendency to get reduced (oxidising power) increases.

Mistake 2: Ignoring the Role of Electronic Configuration

• The Error: Students memorise the trend ($VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$) without linking it to the $d$-orbital configuration after reduction.
• The Fix: Look at what the metal gets reduced to.
  • Mn(VII) in $MnO_4^-$ gets reduced to Mn(II) ($d^5$). This is a half-filled $d$-subshell — an exceptionally stable configuration. The huge stability of the product drives the reaction forward strongly.
  • Cr(VI) in $Cr_2O_7^{2-}$ gets reduced to Cr(III) ($d^3$). This is a half-filled $t_{2g}$ level in an octahedral field — also stable, but less so than $d^5$.
  • V(V) in $VO_2^+$ gets reduced to V(IV) ($d^1$) or V(III) ($d^2$). These are not particularly stable configurations.
  • Conclusion: The greater the stability of the reduced form, the greater the oxidising power of the original ion.

Mistake 3: Forgetting the Role of Oxygen and Charge

• The Error: Students compare only the central metal atom's oxidation state (e.g., +7 > +6 > +5) and stop there.
• The Fix: The oxidising power is also influenced by the number of oxygens and the overall charge on the ion.
  • $MnO_4^-$ has a single negative charge and 4 oxygens. The Mn=O bonds are highly polarised, making the Mn centre very electron-deficient (electrophilic).
  • $Cr_2O_7^{2-}$ has a double negative charge and 7 oxygens for 2 Cr atoms. The electron density is slightly better distributed.
  • $VO_2^+$ has a positive charge and only 2 oxygens. It is the least electron-deficient of the three. …