Q.$[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral. Why?

The key idea is that the magnetic property depends on the oxidation state and ligand field strength, which determine the $d$-electron configuration and whether electrons remain unpaired.

1. In $[NiCl_4]^{2-}$, nickel is in the $+2$ oxidation state ($Ni^{2+}$), giving a $d^8$ configuration. $Cl^-$ is a weak field ligand, so the tetrahedral splitting is small. Electrons fill according to Hund's rule, leaving two unpaired electrons — hence paramagnetic.

2. In $[Ni(CO)_4]$, nickel is in the $0$ oxidation state ($Ni^0$), giving a $d^{10}$ configuration. $CO$ is a strong field ligand, causing all electrons to pair up. With no unpaired electrons, the complex is diamagnetic.

Why the geometry is the same but magnetism differs

Both complexes are tetrahedral — that much is true. But magnetism depends on unpaired electrons, not just shape. The key lies in how the ligands interact with the nickel ion’s d-orbitals.

Nickel in $[NiCl_4]^{2-}$ is in the +2 oxidation state: Ni²⁺ has the electron configuration $[Ar]\,3d^8$. In $[Ni(CO)_4]$, nickel is in the 0 oxidation state: Ni⁰ has $[Ar]\,3d^8 4s^2$, but in the complex, the 4s electrons are also involved in bonding — effectively, the d-electron count is 10 after accounting for ligand donation. Let’s walk through each.

1. Oxidation state and d-electron count

• $[NiCl_4]^{2-}$: Each Cl⁻ has a –1 charge. Four Cl⁻ give –4. The overall charge is –2, so Ni must be +2.

  Ni²⁺: $[Ar]\,3d^8$ — eight d-electrons.

• $[Ni(CO)_4]$: CO is a neutral ligand. Four CO molecules contribute 0 charge. The complex is neutral, so Ni is in 0 oxidation state.

  Ni⁰: $[Ar]\,3d^8 4s^2$ — CO is such a strong field ligand that its large crystal-field splitting favours pairing all electrons into the 3d subshell rather than leaving any unpaired. The two 4s electrons are promoted/paired into the 3d subshell, giving an effective 3d10 4s0 arrangement; the now-empty 4s and 4p orbitals are free to take part in $sp^3$ hybridisation.

2. Crystal field splitting in tetrahedral geometry

In a tetrahedral field, the d-orbitals split into two sets — inverted relative to octahedral:

• Lower energy: $d_{z^2}, d_{x^2-y^2}$ (the $e$ set)
• Higher energy: $d_{xy}, d_{xz}, d_{yz}$ (the $t_2$ set)

The splitting energy $\Delta_t$ is smaller than in octahedral complexes (about $\frac{4}{9}$ of $\Delta_o$). This means:

• Weak field ligands (like Cl⁻) produce a small $\Delta_t$, so electrons do not pair — they occupy orbitals singly (Hund’s rule).
• Strong field ligands (like CO) produce a larger $\Delta_t$, enough to force pairing.

3. Electron configuration in $[NiCl_4]^{2-}$ (weak field)

Ni²⁺ has 8 d-electrons. In a tetrahedral weak field:

• The $e$ set (2 orbitals) is lower in energy, so it fills first: $e^4$ (both orbitals doubly occupied, no unpaired electrons there).
• The remaining 4 electrons go into the higher-energy $t_2$ set (3 orbitals). Since the field is weak, Hund’s rule applies: 3 electrons occupy the 3 orbitals singly first, and the 4th pairs up with one of them: $t_2^4$ (2 orbitals singly occupied, 1 doubly occupied).

So there are two unpaired electrons (in the $t_2$ set) → paramagnetic.

4. Electron configuration in $[Ni(CO)_4]$ (strong field)

Here, Ni is in the 0 oxidation state, but the complex is formed by $sp^3$ hybridisation. The 4s and 4p orbitals hybridise to form four equivalent $sp^3$ orbitals, each accepting a lone pair from CO. The d-electrons remain in the 3d orbitals.

CO is a very strong field ligand — it causes a large $\Delta_t$. With an effective $3d^{10}$ configuration (10 d-electrons), in tetrahedral geometry:

• All 10 d-electrons fill the $e$ and $t_2$ sets completely: $e^4\, t_2^6$.
• Every electron is paired → no unpaired electrons → diamagnetic.

5. Summary of the difference

The geometry is tetrahedral in both, but the ligand strength and oxidation state change the electron distribution.

Concept: Magnetic Properties of Tetrahedral Complexes

The magnetic behaviour here follows from the electronic configuration the metal adopts in each ligand field — worked out below with the Crystal Field Theory (CFT) method.

Method: Crystal Field Theory (CFT) for Tetrahedral Complexes

Step 1: Identify the metal ion and its oxidation state

• In $[NiCl_4]^{2-}$:

  Ni is in +2 oxidation state → Ni²⁺

  Electronic configuration of Ni²⁺: $[Ar]\,3d^8$

• In $[Ni(CO)_4]$:

  Ni is in 0 oxidation state → Ni⁰

  Electronic configuration of Ni⁰: $[Ar]\,3d^8\,4s^2$

Step 2: Determine geometry and splitting pattern

Both complexes are tetrahedral.

In tetrahedral geometry, the $d$-orbitals split into:

• 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 small (about 4/9 of octahedral $\Delta_o$).

Step 3: Fill electrons — decide pairing or not

For $[NiCl_4]^{2-}$ (Ni²⁺, $3d^8$):

• $\Delta_t$ is small, and Cl⁻ is a weak field ligand (low spectrochemical series).
• Electrons do not pair — they occupy orbitals following Hund’s rule.
• Filling: $e$ set: ↑↓, ↑↓ (4 electrons) $t_2$ set: ↑, ↑, ↑↓ (4 electrons)
• Result: 2 unpaired electrons → paramagnetic

For $[Ni(CO)_4]$ (Ni⁰, $3d^8 4s^2$):

• CO is a very strong field ligand (high spectrochemical series).
• Large $\Delta_t$ forces pairing of electrons.
• Also, Ni⁰ uses $4s$ and $4p$ orbitals for bonding — the $3d$ orbitals are completely filled.
• Filling: $e$ set: ↑↓, ↑↓ $t_2$ set: ↑↓, ↑↓, ↑↓
• Result: No unpaired electrons → diamagnetic

Final Answer

Key takeaway: Even with the same geometry, ligand field strength determines whether electrons pair or remain unpaired, which decides the magnetic behavior.

Why This Confuses Students

Both complexes are tetrahedral, so students assume the same electronic configuration and magnetic behaviour.

But the ligand field strength is completely different — and that changes everything.

Common Mistake #1: Assuming tetrahedral always means $sp^3$

The error:

Thinking that because the shape is tetrahedral, the hybridisation must be $sp^3$ for both.

The truth:

• $[NiCl_4]^{2-}$ → $sp^3$ (weak field ligand $\text{Cl}^-$)
• $[Ni(CO)_4]$ → $sp^3$ but with strong field ligand CO, which forces pairing.

How to avoid:

Always check the ligand strength before deciding electron configuration.

• Weak field → high spin (more unpaired electrons)
• Strong field → low spin (pairing occurs)

Common Mistake #2: Forgetting the oxidation state of Ni

The error:

Using the wrong $d^n$ count.

How to fix:

• $[NiCl_4]^{2-}$: Ni is in +2 oxidation state → $Ni^{2+}$ is $d^8$
• $[Ni(CO)_4]$: CO is neutral, Ni is in 0 oxidation state → $Ni^0$ is $d^{10}$

Key result:

• $d^8$ with weak field → 2 unpaired electrons → paramagnetic
• $d^{10}$ → all electrons paired → diamagnetic

Common Mistake #3: Mixing up geometry and magnetic behaviour

The error:

Thinking tetrahedral complexes are always paramagnetic.

The truth:

Tetrahedral geometry does not guarantee paramagnetism — it depends on $d^n$ and ligand field.

How to avoid:

Memorise this shortlist for tetrahedral complexes:

For $d^{10}$, all orbitals are full — no unpaired electrons possible, regardless of ligand.

Common Mistake #4: Confusing pairing in tetrahedral vs square planar

The error:

Thinking CO can force pairing in tetrahedral $d^8$ like it does in square planar.

The truth:

In tetrahedral geometry, the $d$ orbitals split into two sets ($e$ and $t_2$) with a small energy gap.

Even with a strong field, you cannot pair electrons in $d^8$ tetrahedral — the $t_2$ set still has 3 orbitals, and you have 8 electrons.

Pairing only happens in square planar $d^8$ (like $[Ni(CN)_4]^{2-}$).

How to avoid:

• Tetrahedral $d^8$ → always paramagnetic (2 unpaired)
• Square planar $d^8$ → can be diamagnetic (if strong field)

Quick Summary Table

Final Exam Tip

When you see "tetrahedral" and "magnetism" together:

1. Find the oxidation state of the metal → get $d^n$
2. Check if $d^{10}$ → diamagnetic always
3. If $d^8$ → paramagnetic always (tetrahedral)
4. Never assume shape alone decides magnetism — electron count is king