Q.Why are low spin tetrahedral complexes not formed?

The key idea is that the crystal field splitting energy ($\Delta_t$) for a tetrahedral complex is inherently small — roughly $4/9$ of the octahedral splitting ($\Delta_o$). This small gap makes the high-spin configuration energetically far more favourable than pairing electrons.

Reasoning:

1. In a tetrahedral field, the $d$-orbitals split into a lower-energy $e$ set ($d_{x^2-y^2}, d_{z^2}$) and a higher-energy $t_2$ set ($d_{xy}, d_{yz}, d_{zx}$), with $\Delta_t \approx 0.44\Delta_o$.

2. Pairing two electrons in the same orbital costs the pairing energy ($P$). For a low-spin configuration to be stable, $\Delta_t$ must exceed $P$ — but $\Delta_t$ is too small to overcome $P$ for any common metal ion. …

The Core Idea: Why Spin State Depends on Geometry

The spin state of a metal complex (whether it is high spin or low spin) is decided by a simple energy competition: is it cheaper to pair two electrons in the same orbital, or to keep them unpaired in separate orbitals? The answer depends on two numbers:

• $\Delta$ — the crystal field splitting energy (the energy gap between the $t_{2g}$ and $e_g$ sets of d-orbitals).
• $P$ — the pairing energy (the energy cost of putting two electrons in the same orbital, which includes Coulomb repulsion and exchange energy loss).

If $\Delta > P$, the complex will be low spin — electrons prefer to pair up in the lower-energy orbitals rather than jump to the higher set. If $\Delta < P$, the complex will be high spin — electrons stay unpaired because pairing is too expensive.

Now here is the critical point: the magnitude of $\Delta$ depends heavily on geometry.

The Tetrahedral Field: A Weaker Split

In a tetrahedral complex, the metal ion is at the centre of a tetrahedron with four ligands at the corners. The d-orbitals split into two sets:

• The $e$ set ($d_{x^2-y^2}$, $d_{z^2}$) — lower in energy.
• The $t_2$ set ($d_{xy}$, $d_{xz}$, $d_{yz}$) — higher in energy.

The splitting energy is denoted $\Delta_t$. There is a well-known relationship between $\Delta_t$ and the octahedral splitting $\Delta_o$:

This is not an arbitrary number — it comes from the fact that in a tetrahedral field, the ligands approach along axes that are not directly aligned with the d-orbitals, so the electrostatic interaction is weaker. Also, there are only four ligands instead of six, which further reduces the field strength.

The consequence is immediate: $\Delta_t$ is always much smaller than $\Delta_o$ — typically less than half.

The Pairing Energy: A Fixed Cost

The pairing energy $P$ is an intrinsic property of the metal ion and its oxidation state. It does not change with geometry. For a given $d^n$ configuration, $P$ is a fixed number.

So the competition becomes: is $\Delta_t$ ever larger than $P$?

Step-by-Step Reasoning

1. Consider the maximum possible $\Delta_t$. Even with the strongest-field ligands (like CN$^-$ or CO), $\Delta_t$ is at most $\frac{4}{9}$ of the octahedral $\Delta_o$ for the same metal and ligands. For most metals, even the octahedral $\Delta_o$ is barely larger than $P$ for the $d^4$, $d^5$, $d^6$, and $d^7$ configurations where spin-state ambiguity exists.

2. Compare magnitudes. For a typical first-row transition metal like Fe$^{2+}$ ($d^6$), $\Delta_o$ for a strong-field ligand might be around 20,000–30,000 cm$^{-1}$, while $P$ is roughly 15,000–20,000 cm$^{-1}$. So $\Delta_o$ can exceed $P$ — low spin is possible in octahedral geometry. But $\Delta_t = \frac{4}{9} \Delta_o$ gives roughly 9,000–13,000 cm$^{-1}$, which is always less than $P$.

3. Check all $d^n$ configurations. The only configurations that can potentially show low-spin behaviour are $d^4$, $d^5$, $d^6$, and $d^7$ (where there is a choice between pairing in the lower set or occupying the higher set). For each of these, the tetrahedral splitting is simply too small. For $d^8$, $d^9$, and $d^{10}$, there is no spin-state ambiguity anyway — the ground state is fixed regardless of $\Delta$. …

Concept: Crystal Field Theory (CFT) and Spin States in Tetrahedral Complexes

Method: Crystal Field Splitting Analysis

Step 1: Recall the crystal field splitting pattern for tetrahedral complexes

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

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

The splitting energy is denoted as $\Delta_t$ (or $10 Dq_t$).

Step 2: Compare $\Delta_t$ with pairing energy ($P$)

For a low spin configuration to occur, the crystal field splitting energy must be greater than the pairing energy:

However, for tetrahedral complexes:

where $\Delta_o$ is the octahedral splitting energy.

Step 3: Apply the numerical comparison

Since $\Delta_t$ is only about 44% of $\Delta_o$, it is always much smaller than the pairing energy $P$ for any $d^n$ configuration.

Step 4: Draw the conclusion …

Why Are Low Spin Tetrahedral Complexes Not Formed?

This question tests your understanding of crystal field theory (CFT) and how splitting energy ($\Delta_t$) compares to pairing energy (P) in tetrahedral geometry.

Common Mistakes & How to Avoid Them

1. Confusing Tetrahedral and Octahedral Splitting

Mistake: Students often assume tetrahedral splitting ($\Delta_t$) is large, like octahedral splitting ($\Delta_o$).

Why it’s wrong:

In tetrahedral complexes, the crystal field splitting is much smaller:

Since $\Delta_t$ is small, it is almost always less than the pairing energy (P) for any metal ion.

How to avoid:

• Memorise the ratio: $\Delta_t \approx 0.44 \Delta_o$
• Always compare $\Delta_t$ with $P$ — if $\Delta_t < P$, electrons will not pair (high spin is favoured).

2. Forgetting That Pairing Energy Is Always Positive

Mistake: Thinking that pairing can happen “for free” in tetrahedral complexes.

Why it’s wrong:

Pairing two electrons in the same orbital costs energy (the pairing energy P). Since $\Delta_t$ is small, the energy gained by pairing (which would require $\Delta_t > P$) is never enough.

How to avoid:

• Write the condition for low spin: $\Delta > P$
• For tetrahedral: $\Delta_t \ll P$ always → low spin impossible.

3. Ignoring the d-Orbital Splitting Pattern

Mistake: Treating tetrahedral splitting like octahedral (e.g., thinking $t_{2g}$ is lower in energy).

Why it’s wrong:

In tetrahedral geometry, the splitting is inverted:

• $e$ set ($d_{x^2-y^2}, d_{z^2}$) → lower energy
• $t_2$ set ($d_{xy}, d_{yz}, d_{zx}$) → higher energy

This inversion does not change the fact that $\Delta_t$ is small — but students sometimes misapply the orbital labels and get confused.

How to avoid:

• Draw the tetrahedral splitting diagram correctly:

  Energy
    ↑
    |   t₂ (higher)
    |   ↑ Δ_t (small)
    |   e  (lower)
  

• Remember: the magnitude matters more than the order for spin state.

4. Assuming Low Spin Exists for d⁴, d⁵, d⁶, d⁷ in Tetrahedral

Mistake: Applying octahedral rules directly to tetrahedral complexes.

Why it’s wrong:

In octahedral, low spin is possible for $d^4$ to $d^7$ (e.g., $[Co(NH_3)_6]^{3+}$ is low spin $d^6$).

In tetrahedral, no known example exists because $\Delta_t$ is too small.

How to avoid:

• Memorise: All tetrahedral complexes are high spin (except for very rare cases with extremely strong field ligands — but these are not in the JEE/NEET syllabus).
• For exam purposes: “Low spin tetrahedral complexes are not formed.”

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