Q.$[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic whereas $[Fe(CN)_6]^{3-}$ is weakly paramagnetic. Explain.

Concept: Magnetic Moment Calculation — The number of unpaired electrons ($n$) determines the magnetic moment ($\mu = \sqrt{n(n+2)}$ BM), which depends on whether the ligand is weak-field (high-spin) or strong-field (low-spin).

Step 1: $Fe^{3+}$ has $3d^5$ configuration.

Step 2: $H_2O$ is a weak-field ligand — no pairing occurs. The five electrons occupy all five $d$ orbitals singly: $t_{2g}^3 e_g^2$ (high-spin). $n = 5$, so $\mu = \sqrt{5(7)} \approx 5.92$ BM — strongly paramagnetic. …

The key here is not just counting electrons — it’s understanding how the ligand environment reshapes the electron configuration of the central metal ion. Both complexes contain $\ce{Fe^{3+}}$, which has a $3d^5$ configuration. But the magnetic behaviour is dramatically different because the ligands control whether those five d-electrons remain unpaired or are forced into pairs.

Let’s walk through the reasoning step by step.

1. Determine the oxidation state and d-electron count.

   In both complexes, iron is in the +3 oxidation state.

   $\ce{Fe}$ (atomic number 26) has the ground-state configuration $[\ce{Ar}]\,3d^6 4s^2$.

   Losing three electrons (two from 4s and one from 3d) gives $\ce{Fe^{3+}}$: $3d^5$.

   So we have five electrons to place in the d-orbitals.

2. Recall the effect of ligand field splitting in octahedral complexes.

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

   • 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 $\Delta_o$ (or $10\,Dq$). The magnitude of $\Delta_o$ depends on the ligand: strong-field ligands (like $\ce{CN-}$) cause a large splitting; weak-field ligands (like $\ce{H2O}$) cause a small splitting.

3. Apply Hund’s rule vs. pairing energy.

   For a $d^5$ configuration, there are two possible arrangements:

   • High-spin: electrons occupy all five orbitals singly before any pairing occurs. This requires $\Delta_o$ to be smaller than the pairing energy ($P$).
   • Low-spin: electrons pair up in the $t_{2g}$ orbitals, leaving some orbitals empty. This happens when $\Delta_o > P$.

   $\text{If } \Delta_o < P \rightarrow \text{high-spin}; \quad \text{if } \Delta_o > P \rightarrow \text{low-spin}$

4. Analyse $[\ce{Fe(H2O)6}]^{3+}$.

   Water is a weak-field ligand. For $\ce{Fe^{3+}}$, $\Delta_o$ for $\ce{H2O}$ is about $13{,}700\ \text{cm}^{-1}$, while the pairing energy $P$ is roughly $17{,}600\ \text{cm}^{-1}$.

   Since $\Delta_o < P$, the electrons avoid pairing.

   The five d-electrons occupy all five orbitals singly:

This gives five unpaired electrons.

Magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\ \text{BM}$ — strongly paramagnetic.

5. Analyse $[\ce{Fe(CN)6}]^{3-}$. Cyanide is a strong-field ligand. For $\ce{Fe^{3+}}$, $\Delta_o$ for $\ce{CN-}$ is about $35{,}000\ \text{cm}^{-1}$, far above $P$. Now $\Delta_o > P$, so electrons pair up in the $t_{2g}$ orbitals before occupying $e_g$. The configuration becomes:

This means four electrons are paired in two orbitals, and one electron remains unpaired in the third $t_{2g}$ orbital. …

Method: Spin-Only Magnetic Moment Calculation Using Crystal Field Theory

Why This Method?

Magnetic moment depends on the number of unpaired electrons. Crystal Field Theory (CFT) explains how ligands affect electron pairing by splitting the d-orbitals into different energy levels.

Step 1: Determine the oxidation state and d-electron count

• $[Fe(H_2O)_6]^{3+}$

  Fe atomic number = 26

  Fe³⁺ = 26 − 3 = 23 electrons → $d^5$ configuration

• $[Fe(CN)_6]^{3-}$

  Same Fe³⁺ → also $d^5$ configuration

Step 2: Identify ligand field strength and orbital splitting

• $H_2O$ is a weak field ligand → small $\Delta_o$ (crystal field splitting energy)
• $CN^-$ is a strong field ligand → large $\Delta_o$

For an octahedral complex:

• Weak field → high-spin (electrons fill all orbitals singly before pairing)
• Strong field → low-spin (electrons pair up in lower $t_{2g}$ orbitals first)

Step 3: Write the electron distribution in $t_{2g}$ and $e_g$ orbitals

Step 4: Apply the spin-only formula

where $n$ = number of unpaired electrons, BM = Bohr magneton.

• For $[Fe(H_2O)_6]^{3+}$ ($n = 5$): …

Here is a breakdown of the common mistakes students make when explaining the magnetic behavior of $[Fe(H_2O)_6]^{3+}$ and $[Fe(CN)_6]^{3-}$, along with how to avoid them.

Mistake 1: Forgetting to Check the Oxidation State First

The Mistake: Students jump straight to the ligand (H₂O vs. CN⁻) without first calculating the oxidation state of Iron. This leads to using the wrong electronic configuration (e.g., assuming Fe is in +2 state or using the wrong d-count).

How to Avoid:

Always start with the charge balance.

• For $[Fe(H_2O)_6]^{3+}$: The complex has a +3 charge. H₂O is neutral. Therefore, $x + 0 = +3 \implies \text{Fe is in } \mathbf{+3} \text{ state}$.
• For $[Fe(CN)_6]^{3-}$: The complex has a -3 charge. CN⁻ has a -1 charge. Therefore, $x + 6(-1) = -3 \implies x - 6 = -3 \implies \text{Fe is in } \mathbf{+3} \text{ state}$.

Key Result: Both complexes contain Fe(III), which has a $d^5$ configuration. The difference in magnetism is not due to a different number of electrons, but due to how those 5 electrons are arranged.

Mistake 2: Confusing Strong Field vs. Weak Field Ligands

The Mistake: Students memorize that CN⁻ is a strong field ligand and H₂O is a weak field ligand, but they apply this rule blindly without considering the geometry or the specific metal ion. They might also forget that H₂O can act as a strong field ligand for some metals (like Co³⁺), but not for Fe³⁺.

How to Avoid:

• Memorize the Spectrochemical Series: I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < N₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < CH₃CN < py < NH₃ < en < bipy < phen < NO₂⁻ < PPh₃ < CN⁻ ≈ CO.
• Apply it to the specific metal: For Fe³⁺ (a $d^5$ system), H₂O is a weak field ligand (causes high spin), while CN⁻ is a strong field ligand (causes low spin).

Mistake 3: Drawing the Wrong Crystal Field Splitting Diagram

The Mistake: Students draw the $t_{2g}$ and $e_g$ levels but place the electrons incorrectly. Common errors include:

• Forgetting Hund's rule (electrons must occupy all orbitals singly before pairing).
• Pairing electrons in the $t_{2g}$ set for the weak field case.
• Not pairing electrons in the $t_{2g}$ set for the strong field case.

How to Avoid:

Draw the diagram step-by-step.

For $[Fe(H_2O)_6]^{3+}$ (Weak Field, High Spin):

• $\Delta_o$ is small.
• Fill all 5 orbitals singly first.
• Configuration: $t_{2g}^3 e_g^2$.
• Number of unpaired electrons: 5.

For $[Fe(CN)_6]^{3-}$ (Strong Field, Low Spin):

• $\Delta_o$ is large.
• Electrons pair up in the lower energy $t_{2g}$ orbitals before occupying $e_g$.
• Configuration: $t_{2g}^5 e_g^0$.
• Number of unpaired electrons: 1.

Mistake 4: Confusing "Paramagnetic" with "Diamagnetic"

The Mistake: Students might say a complex with 1 unpaired electron is diamagnetic, or they might forget that "weakly paramagnetic" still means it has unpaired electrons (just fewer of them).

How to Avoid:

• Paramagnetic: Has at least one unpaired electron. Attracted to a magnetic field.
• Diamagnetic: All electrons are paired. Repelled by a magnetic field.
• Strongly paramagnetic: Many unpaired electrons (here, 5). …