Q.Why is $Cr^{2+}$ reducing and $Mn^{3+}$ oxidising when both have $d^4$ configuration?

Concept: Stability of Oxidation States — The tendency depends on the electronic configuration of the ion and the energy gain from achieving a half-filled or fully filled $d$ subshell.

Reasoning:

1. $Cr^{2+}$ has $d^4$ configuration. It can lose one electron to become $Cr^{3+}$ ($d^3$), which is a half-filled $t_{2g}$ set — a stable arrangement. So $Cr^{2+}$ is easily oxidised, acting as a reducing agent.

2. $Mn^{3+}$ also has $d^4$ configuration. It can gain one electron to become $Mn^{2+}$ ($d^5$), which is a half-filled $d$ subshell — an extra stable configuration. So $Mn^{3+}$ is easily reduced, acting as an oxidising agent. …

The key to understanding this lies in the stability of half-filled and fully-filled orbitals. In transition metals, the $d$-subshell has a special stability when it is exactly half-filled ($d^5$) or fully-filled ($d^{10}$). This stability arises from the symmetrical distribution of electrons and the exchange energy gained when electrons have parallel spins.

Both $Cr^{2+}$ and $Mn^{3+}$ have the same $d^4$ configuration, but they achieve stability in opposite directions.

1. Identify the electronic configurations.

   Chromium ($Cr$, atomic number 24) in its +2 state loses two electrons. The ground state configuration of Cr is $[Ar]3d^5 4s^1$. Removing the 4s electron and one 3d electron gives $Cr^{2+}$ as $[Ar]3d^4$.

   Manganese ($Mn$, atomic number 25) in its +3 state loses three electrons. The ground state configuration of Mn is $[Ar]3d^5 4s^2$. Removing the two 4s electrons and one 3d electron gives $Mn^{3+}$ as $[Ar]3d^4$.

   So, both ions have the same $d^4$ configuration.

2. Analyse the tendency of $Cr^{2+}$.

   $Cr^{2+}$ can lose one electron to become $Cr^{3+}$, which has a $d^3$ configuration. In an octahedral field (such as water), $d^3$ corresponds to a half-filled $t_{2g}$ set ($t_{2g}^3$) — a stable arrangement because all three $t_{2g}$ orbitals are singly occupied with parallel spins.

   Losing an electron is oxidation, so $Cr^{2+}$ itself acts as a reducing agent (it gets oxidised to $Cr^{3+}$).

   The reaction is: $Cr^{2+} \rightarrow Cr^{3+} + e^-$.

   The driving force is the stability of the half-filled $t_{2g}$ set in $Cr^{3+}$.

3. Analyse the tendency of $Mn^{3+}$.

   $Mn^{3+}$ can gain an electron to become $Mn^{2+}$, which has a $d^5$ configuration. This is the half-filled $d^5$ configuration, which is exceptionally stable due to maximum exchange energy and spherical symmetry.

   Alternatively, $Mn^{3+}$ could lose an electron to become $Mn^{4+}$ ($d^3$). But the stability of $d^5$ is far greater than that of $d^3$.

   So $Mn^{3+}$ tends to gain an electron (i.e., it acts as an oxidising agent) to become $Mn^{2+}$.

   The reaction is: $Mn^{3+} + e^- \rightarrow Mn^{2+}$. …

Method: Electronic Configuration & Stability Analysis (Based on Exchange Energy & Half-Filled Stability)

This method uses the electronic configuration of the ions and the stability of half-filled/totally filled subshells to predict their tendency to gain or lose electrons.

Step 1: Write the ground state electronic configurations

• Cr (Z = 24):

  $[Ar]\, 3d^5\, 4s^1$

  (Exception: half-filled $d^5$ is extra stable)

• Cr$^{2+}$: Remove two electrons (first from 4s, then from 3d)

  $\rightarrow [Ar]\, 3d^4$

• Mn (Z = 25):

  $[Ar]\, 3d^5\, 4s^2$

• Mn$^{3+}$: Remove three electrons (two from 4s, one from 3d)

  $\rightarrow [Ar]\, 3d^4$

Both have $d^4$ configuration — but their stability differs.

Step 2: Identify the stable target configuration for each

• Cr$^{2+}$ can lose one more electron to become Cr$^{3+}$

  $\rightarrow$ Cr$^{3+}$ has $[Ar]\, 3d^3$ — half-filled $t_{2g}^3$ in octahedral field (stable).

• Mn$^{3+}$ can gain one electron to become Mn$^{2+}$

  $\rightarrow$ Mn$^{2+}$ has $[Ar]\, 3d^5$ — half-filled $d^5$ (extra stable).

Step 3: Predict the tendency (reducing vs oxidising)

| Ion | Configuration | Tends to | Reason |

|-----|---------------|----------|--------| …

Here are the common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Assuming Same Configuration Means Same Stability

The Error: Students see that both $\text{Cr}^{2+}$ and $\text{Mn}^{3+}$ have a $d^4$ configuration and assume they should behave similarly. They forget that the nuclear charge (and the ionic charge) is different — even though the electron count is the same.

How to Avoid:

• Always check the atomic number. Cr (Z=24) and Mn (Z=25). $\text{Cr}^{2+}$ has 22 electrons (24 − 2) and $\text{Mn}^{3+}$ also has 22 electrons (25 − 3) — the two ions are isoelectronic, which is exactly why both are $d^4$. What differs is the nuclear charge holding those 22 electrons: 25 protons in Mn versus 24 in Cr.
• Remember the trend: Higher nuclear charge (more protons) pulls the $d$-electrons in more tightly, making further electron loss harder. For $\text{Mn}^{3+}$ this works together with the pull toward the half-filled $d^5$ configuration: gaining one electron gives the exceptionally stable $\text{Mn}^{2+}$ ($d^5$), so $\text{Mn}^{3+}$ readily gains an electron and acts as an oxidising agent.

Mistake 2: Forgetting the Half-Filled and Fully-Filled Stability Rule

The Error: Students know the rule but fail to apply it to the products of the redox reaction. They only look at the $d^4$ configuration of the starting ion.

How to Avoid:

• Look at what the ion wants to become. The driving force for redox is often the stability of the product.
  • $\text{Cr}^{2+}$ (d⁴): It can lose one electron to become $\text{Cr}^{3+}$, which has a half-filled $t_{2g}$ set ($d^3$). This is a very stable configuration. So, $\text{Cr}^{2+}$ is a reducing agent (it gives away an electron).
  • $\text{Mn}^{3+}$ (d⁴): It can gain one electron to become $\text{Mn}^{2+}$, which has a half-filled $d^5$ configuration. This is exceptionally stable. So, $\text{Mn}^{3+}$ is an oxidising agent (it accepts an electron).

Mistake 3: Confusing the Direction of Electron Transfer

The Error: Students mix up which ion gives electrons (reducing agent) and which takes electrons (oxidising agent).

How to Avoid:

• Use the mnemonic "OIL RIG": Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). …