Q.Which is a stronger reducing agent $Cr^{2+}$ or $Fe^{2+}$ and why?

Concept: Stability of Oxidation States — the tendency of an ion to get oxidised depends on how stable its higher oxidation state is in the given medium.

Reasoning:

1. $Cr^{2+}$ has the configuration $3d^4$. It readily loses one electron to become $Cr^{3+}$ ($3d^3$), which is a half-filled $t_{2g}$ level in an octahedral field — this is an exceptionally stable configuration.

2. $Fe^{2+}$ has the configuration $3d^6$. It loses one electron to become $Fe^{3+}$ ($3d^5$), which is a half-filled $d$ subshell — also stable, but the energy gain is smaller than for $Cr^{2+} \to Cr^{3+}$.

3. The standard reduction potentials confirm this: …

Why this approach works

The question asks you to compare the reducing power of two transition metal ions. A reducing agent is a species that donates electrons and gets oxidized itself. So, the stronger the tendency of $M^{2+}$ to lose an electron and become $M^{3+}$, the stronger the reducing agent.

This tendency depends on the stability of the +3 oxidation state relative to the +2 state. In transition metals, stability is heavily influenced by electronic configuration — particularly the achievement of half-filled ($d^5$) or fully-filled ($d^{10}$) subshells, which are extra stable due to exchange energy and symmetry.

Let’s examine each ion.

1. Electronic configurations of the ions

   • Chromium ($Cr$, atomic number 24): $Cr^{2+}$: $[Ar]\,3d^4$ $Cr^{3+}$: $[Ar]\,3d^3$
   • Iron ($Fe$, atomic number 26): $Fe^{2+}$: $[Ar]\,3d^6$ $Fe^{3+}$: $[Ar]\,3d^5$

   Notice that oxidation is a $d^4 \rightarrow d^3$ change for chromium but a $d^6 \rightarrow d^5$ change for iron. In an aqueous (octahedral) medium, $d^3$ means a half-filled $t_{2g}$ level ($t_{2g}^3$) — an especially stable arrangement (see CFSE).

2. Which ion wants to lose an electron more?

   • For $Cr^{2+}$ to become $Cr^{3+}$, it loses one electron. The $d^4$ configuration of $Cr^{2+}$ (high-spin $t_{2g}^3e_g^1$) is not particularly stable. The resulting $Cr^{3+}$ ($d^3$) has a half-filled $t_{2g}$ set ($t_{2g}^3$), which in an aqueous medium is an especially stable arrangement — so the $d^4 \rightarrow d^3$ oxidation is strongly favoured. In fact, $Cr^{2+}$ is so eager to oxidize that it can even reduce water to hydrogen gas in acidic conditions.

   • For $Fe^{2+}$ to become $Fe^{3+}$, it also loses one electron. But $Fe^{2+}$ ($d^6$) is reasonably stable — it has a paired electron in one $t_{2g}$ orbital, but the configuration is not as unstable as $d^4$. More importantly, $Fe^{3+}$ ($d^5$) is extremely stable due to the half-filled subshell. So while $Fe^{2+}$ can be oxidized, the $d^6 \rightarrow d^5$ stability gain in water is smaller than chromium's $d^4 \rightarrow d^3$ gain — NCERT's own answer notes that in a medium like water, $d^3$ is more stable than $d^5$ (see CFSE).

   The critical point: $Cr^{2+}$ is much more unstable than $Fe^{2+}$, and its oxidation to $Cr^{3+}$ is highly favourable. This makes $Cr^{2+}$ a stronger reducing agent.

3. Standard reduction potentials confirm this

   The standard reduction potentials for the $M^{3+}/M^{2+}$ couples are:

   • $Cr^{3+} + e^- \rightarrow Cr^{2+}$: $E^\circ = -0.41\ \text{V}$ …

Method: Standard Electrode Potential (E° Value) Comparison

This method uses the standard reduction potential ($E^\circ$) to determine reducing strength. A more negative $E^\circ$ means the species is more easily oxidised — hence a stronger reducing agent.

Steps

1. Write the relevant half-reactions for oxidation of each ion to its common stable state.

   • For $Cr^{2+}$:

• For $Fe^{2+}$:

2. Interpret the sign:

   • A more negative $E^\circ$ means the reverse reaction (oxidation) is more favourable.
   • $Cr^{2+}$ has $E^\circ = -0.41 \, \text{V}$ (negative), while $Fe^{2+}$ has $E^\circ = +0.77 \, \text{V}$ (positive).

3. Compare reducing power:

   • $Cr^{2+}$ is more easily oxidised to $Cr^{3+}$ than $Fe^{2+}$ is to $Fe^{3+}$.
   • Therefore, $Cr^{2+}$ is a stronger reducing agent.

Why? — The Concept Behind

• Stability of oxidation states:
  • $Cr^{2+}$ has an unstable $d^4$ configuration (high-spin in aqueous solution) and readily loses one electron to give $Cr^{3+}$ ($d^3$), whose half-filled $t_{2g}^3$ level is especially stable in aqueous solution. …

Here is a breakdown of the common mistakes students make on this specific question, along with the correct reasoning to avoid them.

The Core Concept (The "Why")

To compare reducing power, you must compare the tendency to get oxidized (lose electrons). A stronger reducing agent is one that is more easily oxidized.

For $Cr^{2+}$ and $Fe^{2+}$, the key is the stability of the +3 oxidation state in aqueous solution.

• $Cr^{2+}$ is easily oxidized to $Cr^{3+}$ because $Cr^{3+}$ has a half-filled $t_{2g}$ configuration ($d^3$), which is extra stable.
• $Fe^{2+}$ is harder to oxidize to $Fe^{3+}$ because $Fe^{3+}$ has a half-filled $d$ subshell ($d^5$), but the difference in stability between $Fe^{2+}$ ($d^6$) and $Fe^{3+}$ ($d^5$) is not as dramatic as the jump from $Cr^{2+}$ ($d^4$) to $Cr^{3+}$ ($d^3$).

Result: $Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$.

Common Mistake #1: Confusing "Reducing Agent" with "Oxidizing Agent"

• The Mistake: Students think a stronger reducing agent is one that gains electrons easily (like $Fe^{3+}$). They might say "$Fe^{2+}$ is a stronger reducing agent because it becomes $Fe^{3+}$ which is stable."
• Why it's wrong: A reducing agent donates electrons (gets oxidized itself). You are comparing the reactant ($Cr^{2+}$ vs $Fe^{2+}$), not the product.
• How to Avoid:
  • Memorize the definition: "Reducing agent → gets oxidized (loses e⁻)."
  • Ask yourself: "Which of these two ions wants to lose an electron more?" The one that loses it more easily is the stronger reducing agent.

Common Mistake #2: Relying Only on the "Half-filled d⁵" Rule for $Fe^{3+}$

• The Mistake: Students see that $Fe^{3+}$ has a half-filled $d^5$ configuration and immediately assume $Fe^{2+}$ is a very strong reducing agent because it "wants" to become $d^5$.
• Why it's wrong: While $d^5$ is stable, the difference in stability between $Fe^{2+}$ ($d^6$) and $Fe^{3+}$ ($d^5$) is not as large as the difference between $Cr^{2+}$ ($d^4$) and $Cr^{3+}$ ($d^3$). The $Cr^{2+}$ to $Cr^{3+}$ jump involves moving from an unstable $d^4$ to a stable $d^3$ (half-filled $t_{2g}$). The $Fe^{2+}$ to $Fe^{3+}$ jump involves moving from a stable $d^6$ to a very stable $d^5$.
• How to Avoid:
  • Compare the starting configuration. $Cr^{2+}$ ($d^4$) is less stable than $Fe^{2+}$ ($d^6$). An unstable starting point makes it more eager to change.
  • Use the standard reduction potential ($E^\circ$) as a numeric check. The more negative the $E^\circ$ for $M^{3+}/M^{2+}$, the stronger the reducing agent.
    • $E^\circ(Cr^{3+}/Cr^{2+}) = -0.41\ \text{V}$ (negative → strong reducing agent)
    • $E^\circ(Fe^{3+}/Fe^{2+}) = +0.77\ \text{V}$ (positive → weak reducing agent)

Common Mistake #3: Ignoring the Aqueous Medium

• The Mistake: Students discuss stability in terms of solid-state or gas-phase configurations, forgetting the question is about aqueous solution.
• Why it's wrong: Hydration energy and ligand field effects in water significantly alter the stability of oxidation states. For example, $Cr^{2+}$ is rapidly oxidized by air in water, while $Fe^{2+}$ is more stable in acidic solution.
• How to Avoid:
  • Always assume aqueous medium unless stated otherwise for such comparisons.
  • Mention hydration energy in your answer: $Cr^{3+}$ has a high hydration energy due to its small size and high charge, which stabilizes it in water.

Common Mistake #4: Giving a Vague or Incomplete "Why"

• The Mistake: Writing only: "$Cr^{2+}$ is a stronger reducing agent because it has a $d^4$ configuration."
• Why it's wrong: This is incomplete. You must explain why $d^4$ makes it a strong reducing agent.
• How to Avoid: …