Q.The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

The key idea is that the crystal field splitting caused by different ligands determines whether the $d^5$ configuration of Mn(II) is high-spin or low-spin.

1. Identify the metal ion and its $d$-electron count. Mn(II) has the electronic configuration $[Ar]\,3d^5$. In an octahedral field, the five $d$-electrons must be placed in the $t_{2g}$ and $e_g$ orbitals.

2. Consider the ligand field strength. $\ce{H2O}$ is a weak-field ligand, causing a small crystal field splitting ($\Delta_o$). $\ce{CN-}$ is a strong-field ligand, causing a large $\Delta_o$. …

The Core Idea: Crystal Field Splitting and Electron Pairing

In an octahedral complex, the five $d$ orbitals split into two sets: the lower-energy $t_{2g}$ set ($d_{xy}, d_{xz}, d_{yz}$) and the higher-energy $e_g$ set ($d_{z^2}, d_{x^2-y^2}$). The energy gap between them is called $\Delta_o$ (or $10 Dq$).

The key question for a $d^5$ ion like $\ce{Mn^{2+}}$ is: when you place the fifth electron, does it:

• Pair up in the $t_{2g}$ set (overcoming the pairing energy $P$), or
• Go singly into the higher $e_g$ orbital?

The answer depends on whether $\Delta_o > P$ (strong field → low-spin) or $\Delta_o < P$ (weak field → high-spin).

Step-by-Step Reasoning

1. Identify the metal ion and its $d$ count.

   $\ce{Mn^{2+}}$ has the electronic configuration $[\ce{Ar}]\,3d^5$. In both complexes, the metal is in the +2 oxidation state, so we are dealing with a $d^5$ system.

2. Recognize the ligand field strength.

   • $\ce{H2O}$ is a weak field ligand — it lies low in the spectrochemical series. It produces a small $\Delta_o$.
   • $\ce{CN-}$ is a strong field ligand — it lies very high in the spectrochemical series. It produces a large $\Delta_o$.

3. Apply Hund's rule vs. the pairing energy.

   For a $d^5$ ion in a weak field ($\Delta_o$ small):

   • The first three electrons go into the three $t_{2g}$ orbitals, all unpaired (Hund's rule).
   • The fourth and fifth electrons go into the two $e_g$ orbitals, also unpaired, because it costs less energy to place them in higher orbitals than to pair them up in the $t_{2g}$ set.
   • Result: 5 unpaired electrons — the high-spin configuration $(t_{2g})^3(e_g)^2$.

   For a $d^5$ ion in a strong field ($\Delta_o$ large):

   • The first three electrons go into the $t_{2g}$ orbitals, unpaired.
   • The fourth and fifth electrons now find it energetically cheaper to pair up in the $t_{2g}$ orbitals (since $\Delta_o > P$) than to jump to the $e_g$ level.
   • Result: 1 unpaired electron — the low-spin configuration $(t_{2g})^5(e_g)^0$. …

Method: Crystal Field Theory (CFT) Analysis of d-orbital Splitting

Why this method works

Crystal Field Theory explains how the arrangement of ligands around a central metal ion affects the energies of its d-orbitals. The number of unpaired electrons depends on whether the ligand field is weak (high-spin) or strong (low-spin).

Step 1: Identify the metal ion and its d-electron count

• Manganese (Mn) atomic number = 25
• Mn²⁺: remove 2 electrons → electron configuration: $[\text{Ar}]\,3d^5$
• Number of d-electrons = 5

Step 2: Determine the geometry and splitting pattern

Both complexes are octahedral (6 ligands).

In an octahedral field, the five d-orbitals split into:

• Lower energy: $t_{2g}$ (three orbitals: $d_{xy}, d_{xz}, d_{yz}$)
• Higher energy: $e_g$ (two orbitals: $d_{x^2-y^2}, d_{z^2}$)

The energy gap between them is called $\Delta_{\text{oct}}$ (or $10\,Dq$).

Step 3: Apply the ligand strength to decide pairing

Step 4: Fill the d-orbitals using Hund’s rule

For $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ (weak field, small $\Delta$)

• Electrons fill all five orbitals singly before pairing (Hund’s rule)
• Configuration: $t_{2g}^3\, e_g^2$
• All 5 electrons are unpaired → high-spin

For $[\text{Mn}(\text{CN})_6]^{4-}$ (strong field, large $\Delta$)

• Large $\Delta$ makes pairing energetically favourable
• Electrons pair up in the lower $t_{2g}$ orbitals first …

Here are the common mistakes students make when explaining the difference in unpaired electrons (high-spin vs low-spin $d^5$) using Crystal Field Theory (CFT) for this specific question, along with how to avoid each.

Mistake 1: Confusing High-Spin and Low-Spin Configurations

The Mistake:

Students often incorrectly assign the electron configuration. They might say the hexacyanoion ($[\text{Mn(CN)}_6]^{4-}$) has five unpaired electrons because they forget that $\text{CN}^-$ is a strong field ligand, or they incorrectly apply Hund's rule to the low-spin case.

Why it happens:

Students memorize "$\text{H}_2\text{O}$ is weak field, $\text{CN}^-$ is strong field" but fail to apply the consequence: strong field ligands cause a large crystal field splitting ($\Delta$), forcing electrons to pair up in the lower $t_{2g}$ orbitals before occupying the higher $e_g$ orbitals.

How to Avoid:

• Always draw the $d$-orbital splitting diagram. For an octahedral complex, draw the $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.
• Apply the pairing rule: For $\text{Mn}^{2+}$ ($d^5$):
  • Weak field ($\text{H}_2\text{O}$): Small $\Delta$. Electrons fill all five orbitals singly first (Hund's rule). Result: 5 unpaired electrons (high-spin).
  • Strong field ($\text{CN}^-$): Large $\Delta$. Electrons pair up in the $t_{2g}$ set first. Result: 1 unpaired electron (low-spin).
• Memorize the ligand strength series but always verify by drawing the diagram.

Mistake 2: Forgetting the Oxidation State of Manganese

The Mistake:

Students calculate the $d$-electron count incorrectly. They might treat the metal as $\text{Mn}^0$ or $\text{Mn}^{3+}$ instead of $\text{Mn}^{2+}$.

Why it happens:

The question mentions "hexaquo manganese(II) ion" ($[\text{Mn(H}_2\text{O)}_6]^{2+}$) and "hexacyanoion" ($[\text{Mn(CN)}_6]^{4-}$). Students forget to determine the oxidation state of Mn in the cyano complex.

How to Avoid:

• Always find the oxidation state first.
  • For $[\text{Mn(CN)}_6]^{4-}$: $\text{CN}^-$ has a charge of $-1$. Total ligand charge = $6 \times (-1) = -6$. Complex charge = $-4$. So, $\text{Mn} + (-6) = -4 \implies \text{Mn} = +2$.
• Confirm the $d^n$ configuration: $\text{Mn}^{2+}$ is $d^5$ (Mn atomic number = 25; $\text{Mn}^{2+}$ loses 2 electrons from 4s, leaving 5 in 3d).

Mistake 3: Stating the Configurations Without the Energy Argument

The Mistake:

Students write "high-spin" and "low-spin" as memorised labels but cannot say why the strong-field ligand makes pairing worthwhile — the answer earns little credit without the energy reasoning.

Why it happens:

The link between the splitting diagram and the Crystal Field Stabilisation Energy (CFSE) is skipped.

How to Avoid:

• Explain using CFSE:
  • For $d^5$ in a weak field (high-spin $t_{2g}^3 e_g^2$): CFSE $= 3 \times (-0.4\Delta_o) + 2 \times (+0.6\Delta_o) = 0$ — no net stabilization relative to the spherical field.
  • For $d^5$ in a strong field (low-spin $t_{2g}^5 e_g^0$): CFSE $= 5 \times (-0.4\Delta_o) = -2\Delta_o$.
  • When $\Delta_o$ is large (as with $\text{CN}^-$), this $2\Delta_o$ stabilization outweighs the extra pairing energy, so the electrons pair up — the low-spin arrangement is the lower-energy one. For $\text{H}_2\text{O}$, $\Delta_o$ is too small to pay for pairing, so the high-spin arrangement wins.

Mistake 4: Ignoring the Role of Ligand Field Strength

The Mistake: …