Q.Magnetic moment of $[MnCl_4]^{2-}$ is 5.92 BM. Explain giving reason.

The key idea is that the magnetic moment depends on the number of unpaired electrons, and the geometry of the complex determines the orbital contribution.

Step 1: Determine the oxidation state and electron configuration.

Mn in $[MnCl_4]^{2-}$: let oxidation state be $x$.

$x + 4(-1) = -2 \Rightarrow x = +2$.

Mn$^{2+}$ has $3d^5$ configuration.

Step 2: Identify geometry and ligand field.

$Cl^-$ is a weak field ligand, and with four ligands, the complex is tetrahedral (common for $d^5$ high-spin). In a tetrahedral field, the $d$-orbitals split into lower-energy $e$ and higher-energy $t_2$ sets.

Step 3: Determine unpaired electrons.

For a weak field tetrahedral $d^5$ ion, all five electrons remain unpaired (high-spin).

Number of unpaired electrons, $n = 5$. …

Why magnetic moment tells us the story

Magnetic moment is a direct window into the number of unpaired electrons in a transition metal complex. For first-row transition metals, orbital contributions are often "quenched" by the crystal field, so the observed moment closely follows the spin-only formula:

where $n$ is the number of unpaired electrons and BM stands for Bohr magneton. If you measure $\mu = 5.92$ BM, you can work backwards: $\sqrt{n(n+2)} = 5.92$ gives $n(n+2) \approx 35$, so $n = 5$. That is the maximum possible for a $d$ subshell — five unpaired electrons.

The question then becomes: how does a manganese complex end up with five unpaired electrons? That depends on the oxidation state of Mn, the geometry of the complex, and the ligand field strength.

Step-by-step reasoning

1. Find the oxidation state of manganese. The complex ion is $[MnCl_4]^{2-}$. Each chloride ligand carries a $-1$ charge. Let the oxidation state of Mn be $x$. Then:

So manganese is in the +2 oxidation state. The electronic configuration of neutral Mn ($Z = 25$) is $[Ar]\,3d^5 4s^2$. Removing two electrons (to form Mn²⁺) removes the 4s electrons first, leaving $[Ar]\,3d^5$.

2. Determine the geometry.

   The formula $[MnCl_4]^{2-}$ has four ligands. Four-coordinate complexes are either tetrahedral or square planar. For Mn²⁺, which is a $d^5$ system, tetrahedral geometry is far more common — especially with chloride, a weak-field ligand. Square planar geometry is typical for $d^8$ ions (like Ni²⁺, Pt²⁺), not $d^5$. So the geometry is tetrahedral.

3. Apply crystal field theory for a tetrahedral field.

   In a tetrahedral field, the $d$ orbitals split into two sets:

   • Lower energy: $e$ set ($d_{x^2-y^2}$, $d_{z^2}$)
   • Higher energy: $t_2$ set ($d_{xy}$, $d_{yz}$, $d_{zx}$) The splitting energy $\Delta_t$ is smaller than the octahedral splitting $\Delta_o$ — roughly $\Delta_t \approx \frac{4}{9} \Delta_o$. Chloride is a weak-field ligand (low in the spectrochemical series), so $\Delta_t$ is very small.

4. Fill the $d$ orbitals according to Hund's rule.

   With a small $\Delta_t$, the pairing energy is larger than the splitting energy. So electrons will not pair up — they occupy all five $d$ orbitals singly before any pairing occurs. The filling order:

   • $e$ set: two orbitals, each gets one electron (parallel spins)
   • $t_2$ set: three orbitals, each gets one electron (parallel spins) This gives five unpaired electrons — a high-spin $d^5$ configuration. …

Method: Spin-Only Magnetic Moment Formula

This problem uses the spin-only magnetic moment formula to determine the number of unpaired electrons and hence the geometry and electronic configuration of the complex.

Steps

Step 1: Recall the spin-only formula

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

where $n$ = number of unpaired electrons.

Step 2: Plug in the given value

Given $\mu = 5.92 \ \text{BM}$:

Squaring both sides:

Step 3: Solve for $n$

We look for an integer $n$ such that $n(n+2) \approx 35$.

• $n = 5$: $5 \times 7 = 35$ ✓

Thus, $n = 5$ unpaired electrons.

Step 4: Identify the metal ion and configuration

The complex is $[MnCl_4]^{2-}$.

• Mn atomic number = 25
• Mn in +2 oxidation state: $Mn^{2+}$ has electronic configuration $[Ar]\,3d^5$

Step 5: Relate $n$ to geometry

• For $3d^5$ with 5 unpaired electrons, the configuration is high-spin $d^5$ (all orbitals singly occupied). …

1. ✗ Mistake: Wrong oxidation state of Mn

Many students assume Mn is in +2 state without checking.

Why it happens: Mn is a transition metal; students often default to common states like +2.

How to avoid:

• Let the charge of the complex guide you.
• $[MnCl_4]^{2-}$: each Cl is $-1$, so total from 4 Cl = $-4$.
• Complex charge = $-2$.
• Let Mn oxidation state be $x$:

• Correct: Mn is in +2 state.

2. ✗ Mistake: Wrong electronic configuration of Mn²⁺

Students write Mn²⁺ as $[Ar]\,3d^5\,4s^0$ but then forget the d-orbital splitting in a tetrahedral field.

How to avoid:

• Mn (atomic number 25): $[Ar]\,3d^5\,4s^2$
• Mn²⁺: remove two 4s electrons → $[Ar]\,3d^5$
• In a tetrahedral field, the 5 electrons occupy orbitals according to Hund’s rule — all 5 are unpaired.

3. ✗ Mistake: Assuming octahedral geometry

Students see coordination number 4 and assume square planar or octahedral.

Why it happens: Many common complexes are octahedral; tetrahedral is less familiar.

How to avoid:

• $[MnCl_4]^{2-}$ has 4 ligands and no special ligand field stabilization (Mn²⁺ is d⁵, high spin).
• Cl⁻ is a weak field ligand → tetrahedral geometry is favoured.
• Key clue: Magnetic moment of 5.92 BM matches exactly with 5 unpaired electrons.

4. ✗ Mistake: Using wrong formula for magnetic moment

Students use $\mu = \sqrt{n(n+2)}$ but forget that $n$ = number of unpaired electrons.

How to avoid:

• For Mn²⁺ (d⁵, high spin tetrahedral): all 5 electrons unpaired.
• Formula:

• Always confirm: 5.92 BM is the spin-only value for 5 unpaired electrons.

5. ✗ Mistake: Ignoring orbital contribution

Some students add orbital contribution for d⁵ systems.

How to avoid: …