Q.Explain why $[Fe(H_2O)_6]^{3+}$ has magnetic moment value of 5.92 BM whereas $[Fe(CN)_6]^{3-}$ has a value of only 1.74 BM.

The key idea is that the magnetic moment depends on the number of unpaired electrons, which is determined by the ligand field strength and the resulting high-spin vs. low-spin configuration.

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

In $[Fe(H_2O)_6]^{3+}$, Fe is in +3 state ($3d^5$). In $[Fe(CN)_6]^{3-}$, Fe is also +3 ($3d^5$). Both have the same $d^5$ configuration.

Step 2: Identify the ligand field and spin state.

$H_2O$ is a weak field ligand, causing a small crystal field splitting ($\Delta_o$). For $d^5$, electrons fill all five $d$-orbitals singly before pairing (Hund's rule), giving a high-spin configuration: $t_{2g}^3 e_g^2$ — 5 unpaired electrons.

$CN^-$ is a strong field ligand, causing a large $\Delta_o$. For $d^5$, pairing occurs, giving a low-spin configuration: $t_{2g}^5 e_g^0$ — only 1 unpaired electron.

Step 3: Calculate the magnetic moment. …

The magnetic moment of a transition metal complex tells us how many unpaired electrons are present. The formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) connects the experimental value directly to the electronic configuration. So when you see 5.92 BM and 1.74 BM, you're really being asked: why does the same metal ion, $Fe^{3+}$, have a different number of unpaired electrons in these two complexes?

The answer lies in the crystal field theory and the nature of the ligands.

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

   Iron in both complexes is in the +3 oxidation state. Atomic number of Fe is 26. $Fe^{3+}$ means we remove 3 electrons: $[Ar]3d^5$. So in both cases, we have a $d^5$ system.

2. Recall the crystal field splitting in an octahedral field.

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

3. The key: how do electrons fill these orbitals?

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

   • High-spin: Electrons fill all five orbitals singly first (Hund's rule), giving 5 unpaired electrons. This happens when $\Delta_o$ is small — it costs less energy to put an electron in a higher $e_g$ orbital than to pair up in a $t_{2g}$ orbital.
   • Low-spin: Electrons pair up in the $t_{2g}$ set first, giving only 1 unpaired electron. This happens when $\Delta_o$ is large — pairing energy is less than the energy needed to promote an electron to $e_g$.

4. Now, the ligands decide $\Delta_o$.

   Ligands are arranged in the spectrochemical series in order of increasing field strength:

   $I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < NO_2^- < CN^- < CO$

   • $H_2O$ is a weak field ligand. It produces a small $\Delta_o$.
   • $CN^-$ is a strong field ligand. It produces a large $\Delta_o$.

5. Apply to $[Fe(H_2O)_6]^{3+}$:

   Weak field → small $\Delta_o$ → high-spin configuration.

   The five electrons occupy all five orbitals singly: $t_{2g}^3 e_g^2$.

   Number of unpaired electrons, $n = 5$.

   Magnetic moment: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.

   This matches the given value exactly. …

Method: Crystal Field Theory (CFT) – Spin-Only Magnetic Moment Calculation

This method uses the spin-only formula to predict magnetic moment ($\mu$) based on the number of unpaired electrons ($n$):

Step-by-step reasoning for both complexes

Step 1: Determine the oxidation state and electron configuration of Fe

• Fe atomic number = 26
• Ground state: $[Ar]\, 3d^6 4s^2$

For $[Fe(H_2O)_6]^{3+}$:

• Ligand charge: $H_2O$ is neutral → overall +3 charge comes from Fe
• Fe is in +3 oxidation state → $Fe^{3+}$
• Electron configuration of $Fe^{3+}$: $[Ar]\, 3d^5$

For $[Fe(CN)_6]^{3-}$:

• Ligand charge: $CN^-$ is −1 each → total −6 from ligands
• Overall charge −3 → Fe must be +3 to balance: $Fe^{3+}$
• Same $3d^5$ configuration

Step 2: Identify ligand field strength and predict pairing

Step 3: Write the $d$-orbital filling (octahedral splitting)

High spin $Fe^{3+}$ (weak field):

• $t_{2g}$: ↑ ↑ ↑ (3 electrons)
• $e_g$: ↑ ↑ (2 electrons)
• Unpaired electrons = 5

Low spin $Fe^{3+}$ (strong field):

• $t_{2g}$: ↑↓ ↑ ↑ (5 electrons, all paired except 1 unpaired)
• $e_g$: empty
• Unpaired electrons = 1

Step 4: Apply spin-only formula

For $[Fe(H_2O)_6]^{3+}$ ($n = 5$): …

Here is a breakdown of the common mistakes students make on this exact question, and how to avoid them.

The Core Concept (Why the difference?)

The magnetic moment ($\mu$) depends on the number of unpaired electrons ($n$). The formula is:

• $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ has 5 electrons in the 3d orbital. $H_2O$ is a weak field ligand. It does not force pairing. The electrons remain unpaired (high spin). $n = 5$.

• $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ still has 5 electrons. $CN^-$ is a strong field ligand. It forces pairing. The electrons pair up, leaving only 1 unpaired electron (low spin). $n = 1$.

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to find the oxidation state of the metal first.

• The Error: Students directly count electrons from the neutral atom (e.g., $Fe$ has 8 electrons in 3d and 4s) without adjusting for the charge on the complex.
• Example of Error: Saying $Fe$ in $[Fe(CN)_6]^{3-}$ has 8 d-electrons.
• How to Avoid:
  1. Always calculate the oxidation state first.
  2. For $[Fe(CN)_6]^{3-}$: Let $x$ be the oxidation state of Fe. $CN^-$ has a -1 charge. So: $x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3$.
  3. $Fe^{3+}$ has lost 3 electrons. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$. For $Fe^{3+}$, remove the 4s electrons first, then one 3d electron: $[Ar] 3d^5$.
  4. Result: You always start with 5 d-electrons for $Fe^{3+}$.

Mistake 2: Confusing strong and weak field ligands.

• The Error: Thinking $H_2O$ is a strong field ligand or $CN^-$ is a weak field ligand.
• Example of Error: Predicting $[Fe(H_2O)_6]^{3+}$ will have paired electrons.
• How to Avoid:
  • Memorize the Spectrochemical Series (partial list is enough):
    • Strong field (low spin): $CN^- > CO > en > NH_3$
    • Weak field (high spin): $H_2O > F^- > Cl^- > Br^- > I^-$
  • Rule of thumb: $CN^-$ and $CO$ are almost always strong. $H_2O$ is borderline but usually weak for $3d$ metals. Halides are always weak.

Mistake 3: Forgetting the "Hund's Rule" logic for d5 configuration.

• The Error: Students correctly identify 5 electrons but then pair them incorrectly, leading to a wrong $n$ value.
• Example of Error: For $Fe^{3+}$ (d5), placing 2 electrons in one orbital and 3 in others, giving 3 unpaired electrons.
• How to Avoid:
  • Draw the d-orbital splitting diagram.
  • For weak field ($H_2O$): The energy gap ($\Delta_o$) is small. Electrons fill all 5 orbitals singly first (Hund's rule). Result: 5 unpaired electrons.
  • For strong field ($CN^-$): The energy gap ($\Delta_o$) is large. Electrons pair up in the lower energy $t_{2g}$ orbitals before going to the higher $e_g$ orbitals. For d5: 3 electrons fill $t_{2g}$ singly, then the next 2 pair up. Result: 1 unpaired electron.

Mistake 4: Using the wrong formula or miscalculating the square root.

• The Error: Using $\mu = \sqrt{n(n+1)}$ (which is for orbital angular momentum) or making arithmetic errors.
• Example of Error: For $n=5$, calculating $\sqrt{5 \times 7} = \sqrt{35} \approx 5.91$ (correct), but then writing 5.92 without checking. Or for $n=1$, calculating $\sqrt{1 \times 3} = \sqrt{3} \approx 1.73$ (correct), but writing 1.74.
• How to Avoid:
  • Use the spin-only formula: $\mu = \sqrt{n(n+2)}$.
  • Memorize the common values:
    • $n=1 \implies \mu = \sqrt{3} \approx 1.73 \text{ BM}$
    • $n=2 \implies \mu = \sqrt{8} \approx 2.83 \text{ BM}$ …