Concept: Magnetic Moment Calculation — The spin-only formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ = number of unpaired electrons) is used to infer the oxidation state, electron configuration, and whether the complex is high-spin or low-spin.

Step 1 — $K_4[Mn(CN)_6]$ ($\mu = 2.2$ BM)

$\mu \approx 2.2$ BM corresponds to $n \approx 1$ unpaired electron. Mn in this complex is in +2 state ($d^5$). CN$^-$ is a strong field ligand, so the complex is low-spin $t_{2g}^5$ — giving exactly 1 unpaired electron.

Step 2 — $[Fe(H_2O)_6]^{2+}$ ($\mu = 5.3$ BM)

$\mu \approx 5.3$ BM corresponds to $n \approx 4$ unpaired electrons. Fe is in +2 state ($d^6$). H$_2$O is a weak field ligand, so the complex is high-spin $t_{2g}^4 e_g^2$ — giving 4 unpaired electrons.

Step 3 — $K_2[MnCl_4]$ ($\mu = 5.9$ BM) …

The magnetic moment of a transition metal complex is a direct experimental window into its electronic structure. The spin-only formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) lets us work backwards: given $\mu$, we can deduce $n$, and from $n$, we can infer the metal’s oxidation state, the ligand field strength, and the geometry.

Let’s examine each complex one by one.

1. $K_4[Mn(CN)_6]$ — $\mu = 2.2$ BM

Step 1: Find the oxidation state of Mn.

Potassium is always +1, so $K_4$ contributes $4 \times (+1) = +4$. The complex ion is $[Mn(CN)_6]^{4-}$ (since the overall salt is neutral). CN⁻ is a −1 ligand, so six CN⁻ give −6. Let Mn have oxidation state $x$. Then:

$x + (-6) = -4 \implies x = +2$.

So Mn is in the +2 state.

Step 2: Determine the $d$-electron count.

Mn (atomic number 25) has electronic configuration $[Ar]\,3d^5 4s^2$. In the +2 state, it loses the two 4s electrons, leaving $3d^5$ — five $d$-electrons.

Step 3: Use the magnetic moment to find unpaired electrons.

The spin-only formula: $\mu = \sqrt{n(n+2)}$.

For $\mu = 2.2$ BM:

$2.2 \approx \sqrt{n(n+2)}$. Squaring: $4.84 \approx n(n+2)$.

Try $n=1$: $1 \times 3 = 3$ (too low). $n=2$: $2 \times 4 = 8$ (too high). So $n=1$ is the closest — the small deviation from 1.73 BM (theoretical for $n=1$) is due to orbital contribution.

Thus, there is 1 unpaired electron.

Step 4: Interpret the electronic configuration.

A $d^5$ system with only 1 unpaired electron means the electrons are paired as much as possible — this is a low-spin configuration. That happens only when the ligand field is strong. CN⁻ is a strong-field ligand (high in the spectrochemical series). The geometry is octahedral (six ligands), so the $d$-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy). For strong field, pairing energy is less than the splitting, so all five electrons go into $t_{2g}$: $(t_{2g})^5$, giving one unpaired electron (since $t_{2g}$ holds 6 electrons max, five means one unpaired).

2. $[Fe(H_2O)_6]^{2+}$ — $\mu = 5.3$ BM

Step 1: Oxidation state of Fe.

The complex ion is $[Fe(H_2O)_6]^{2+}$. Water is neutral, so the charge comes entirely from Fe. Thus Fe is in the +2 state.

Step 2: $d$-electron count.

Fe (atomic number 26) has $[Ar]\,3d^6 4s^2$. Fe²⁺ loses two 4s electrons, leaving $3d^6$.

Step 3: Find $n$ from $\mu$.

$\mu = 5.3$ BM. Try $n=4$: $\sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM. $n=5$: $\sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM. 5.3 is closer to 4.90, so $n=4$ unpaired electrons (the slight excess is again orbital contribution).

Step 4: Interpret.

A $d^6$ system with 4 unpaired electrons is high-spin. H₂O is a weak-field ligand (low in the spectrochemical series). In an octahedral field, weak field means the splitting is small, so electrons fill all five $d$-orbitals singly before pairing. The configuration is $(t_{2g})^4 (e_g)^2$: four unpaired electrons (two in $t_{2g}$ are paired, the other two in $t_{2g}$ are unpaired, plus two in $e_g$ are unpaired — total 4).

3. $K_2[MnCl_4]$ — $\mu = 5.9$ BM

Step 1: Oxidation state of Mn. …

Method: Spin-Only Magnetic Moment Formula & Electronic Configuration Analysis

This method uses the spin-only formula to determine the number of unpaired electrons, then infers the oxidation state, geometry, and whether the complex is high-spin or low-spin.

Step 1 — Recall the spin-only formula

The magnetic moment $\mu$ (in BM) is given by:

where $n$ = number of unpaired electrons.

Step 2 — Calculate $n$ for each complex

Step 3 — Determine oxidation state and d-electron count

• $K_4[Mn(CN)_6]$: K⁺ (×4) → total +4; CN⁻ (×6) → total –6. Overall neutral, so $4(+1) + x + 6(-1) = 0 \Rightarrow x = +2$.

  → Mn²⁺: $d^5$ configuration.

• $[Fe(H_2O)_6]^{2+}$: H₂O is neutral; charge +2 → Fe²⁺: $d^6$ configuration.

• $K_2[MnCl_4]$: K⁺ (×2) → +2; Cl⁻ (×4) → –4. So Mn: $+2 + x - 4 = 0 \Rightarrow x = +2$.

  → Mn²⁺: $d^5$ configuration.

Step 4 — Compare $n$ with possible spin states

Here are the common mistakes students make when analyzing magnetic moment data for coordination compounds, along with how to avoid each.

1. Forgetting to Account for Strong vs. Weak Field Ligands

Mistake:

Students assume all complexes are high-spin or low-spin based only on the metal ion, ignoring the ligand field strength.

Example from the table:

• $K_4[Mn(CN)_6]$ has $\mu = 2.2$ BM — this is low-spin $d^5$ (1 unpaired electron).
• $K_2[MnCl_4]$ has $\mu = 5.9$ BM — this is high-spin $d^5$ (5 unpaired electrons). Both are Mn(II), but CN⁻ is strong field and Cl⁻ is weak field.

How to avoid:

• Memorize the spectrochemical series: CN⁻ > en > NH₃ > H₂O > F⁻ > Cl⁻ > Br⁻ > I⁻
• For each complex, first identify the ligand strength, then decide high-spin or low-spin.

2. Using the Wrong Formula for $\mu$

Mistake:

Using $\mu = \sqrt{n(n+2)}$ when spin-orbit coupling is significant, or using it for $n=0$ incorrectly.

Correct approach:

• For most first-row transition metals, use:

where $n$ = number of unpaired electrons.

• For $n=0$, $\mu = 0$ BM (diamagnetic).
• For $n=5$, $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM — matches $K_2[MnCl_4]$ (5.9 BM).

How to avoid:

• Always calculate the theoretical $\mu$ from $n$ and compare with the experimental value.
• If the experimental value is slightly lower (e.g., 2.2 BM instead of 1.73 BM for 1 e⁻), it may indicate orbital contribution — but for exams, stick to the spin-only formula unless told otherwise.

3. Misidentifying the Oxidation State of the Metal

Mistake:

Assuming the charge on the complex is the metal's oxidation state.

Example:

• $K_4[Mn(CN)_6]$: K⁺ is +1 each, so total +4 from K. Complex ion charge = $[Mn(CN)_6]^{4-}$. CN⁻ is -1 each, so:

So Mn is in +2 oxidation state, not +4.

How to avoid:

• Write the charge balance equation carefully.
• Remember: ligands like CN⁻, Cl⁻, H₂O have known charges.
• For $[Fe(H_2O)_6]^{2+}$, Fe is clearly +2.

4. Confusing Geometry with Magnetic Behavior

Mistake:

Assuming tetrahedral complexes are always high-spin and octahedral are always low-spin.

Truth:

• Tetrahedral complexes are almost always high-spin because $\Delta_t$ is small.
• Octahedral complexes can be high-spin or low-spin depending on ligand strength.

Example from table:

• $K_2[MnCl_4]$ is tetrahedral → high-spin $d^5$ → 5 unpaired e⁻ → $\mu \approx 5.9$ BM.
• $K_4[Mn(CN)_6]$ is octahedral with strong field CN⁻ → low-spin $d^5$ → 1 unpaired e⁻ → $\mu \approx 1.73$ BM (experimental 2.2 BM due to some orbital contribution).

How to avoid:

• For tetrahedral: always assume high-spin.
• For octahedral: check ligand strength before deciding.

5. Ignoring Orbital Contribution to $\mu$

Mistake:

Expecting exact match with $\sqrt{n(n+2)}$ and panicking when it's off.

Example:

• $[Fe(H_2O)_6]^{2+}$: $\mu = 5.3$ BM. Theoretical for high-spin $d^6$ (4 unpaired e⁻):

The experimental 5.3 BM is higher — this indicates orbital contribution (common for $d^6$ high-spin Fe²⁺).

How to avoid:

• Know that $\mu_{\text{spin-only}}$ is a lower bound. …