Calculating Tetrahedral CFSE

Introduction

Tetrahedral CFSE uses the same bookkeeping idea as octahedral CFSE but a different diagram. The two e orbitals lie lower at −0.6Δ t each, while three t₂ orbitals lie higher at +0.4Δ t each. Confusing the labels or copying the octahedral coefficients reverses the predicted stabilisation. Small Δ t normally makes tetrahedral fillings high spin.

Core explanation

For a tetrahedral dⁿ ion, draw two lower e boxes and three upper t₂ boxes. Fill each available orbital singly with parallel spins before pairing, because Δ t is usually smaller than pairing energy. If n e electrons occupy e and n t occupy t₂, the orbital CFSE is (−0.6n e+0.4n t)Δ t. As in octahedral CFT, negative values indicate orbital energy lowering relative to the unsplit barycentre, while an extra pairing term belongs in any comparison of fillings with different pair counts.

The first electron gives e¹ and −0.6Δ t. The second goes into the other e orbital, giving e² and −1.2Δ t. The third occupies t₂, e²t₂¹, giving −1.2Δ t+0.4Δ t=−0.8Δ t. The fourth gives e²t₂² and −0.4Δ t. The fifth fills all five orbitals singly, e²t₂³, for zero orbital CFSE. This zero is an algebraic cancellation, not an absence of ligand–metal attraction.

After five electrons, pairing begins. The sixth electron pairs in lower e: e³t₂³ has −0.6Δ t. The seventh completes e⁴t₂³ and gives −1.2Δ t. The eighth pairs within one t₂ orbital: e⁴t₂⁴ gives −0.8Δ t. The ninth gives e⁴t₂⁵ and −0.4Δ t; the tenth gives e⁴t₂⁶ and zero. This symmetry of the sequence about d⁵ follows from filling and then pairing across a fixed two-plus-three level pattern.

The label “high spin” needs a precise meaning. It means maximizing unpaired electrons where the small gap makes pairing early unfavourable. For tetrahedral d⁷, e⁴t₂³ has three unpaired electrons, not seven: four e electrons form two pairs, while the three t₂ electrons remain separate. A spin-only moment estimate would therefore use n=3. Many tetrahedral Co(II) complexes are d⁷ and paramagnetic, consistent with this pattern.

To compare tetrahedral and octahedral stabilisation, express both in common units. If only octahedral Δₒ is supplied, Δ t≈4Δₒ/9 is an approximate same-metal, same-ligand conversion. For d³, tetrahedral e²t₂¹ has −0.8Δ t≈−0.356Δₒ, whereas octahedral t₂g³ has −1.2Δₒ. That orbital difference suggests octahedral preference, but it does not alone decide structure because bond, steric, lattice and solvation energies differ between geometries.

An experimentally observed tetrahedral colour cannot be calculated from a CFSE number alone. Absorption transition energies and intensities depend on allowed electronic states and selection rules. CFSE is a ground-state orbital-energy sum, not automatically the photon energy of a visible band.

Step-by-step reasoning

Confirm tetrahedral geometry and determine d count. Fill the two e orbitals and three t₂ orbitals singly before pairing. Multiply e occupancy by −0.6Δ t and t₂ occupancy by +0.4Δ t. Add the contributions, count unpaired electrons separately, and convert Δ t to Δₒ units only if the comparison warrants an approximation.

Visual explanation

Sketch two lower e boxes and three upper t₂ boxes across a short gap. Add arrows from d¹ through d⁵ one per box. In the d⁶ through d¹⁰ sketches, pair lower e first, then upper t₂. Write the energy contribution beside each row so a diagram directly becomes a sum.

Real-world analogy

Two cheaper desks and three slightly dearer desks can each be used privately before workers need to share. The first two workers occupy cheap desks, the next three take dearer ones, and later workers return to share the cheaper desks. The total price depends on occupancy at each level.

Real-world example

Tetrahedral [CoCl₄]²⁻ contains Co²⁺, hence d⁷. The standard high-spin filling e⁴t₂³ gives three unpaired electrons and orbital CFSE −1.2Δ t. This explains why its simple magnetic prediction differs from a d⁷ low-spin octahedral diagram.

Why?

Why is tetrahedral d⁵ CFSE zero? Two lower electrons contribute 2(−0.6Δ t)=−1.2Δ t, while three upper electrons contribute 3(+0.4Δ t)=+1.2Δ t. The two sums cancel about the barycentre.

Common misconception

“The tetrahedral lower level is t₂ because the octahedral lower level is t₂g.” Tetrahedral orientation reverses the order. Lower e contributes −0.6Δ t; upper t₂ contributes +0.4Δ t.

Worked example

An ideal tetrahedral d⁸ ion has e⁴t₂⁴. Its four lower electrons contribute −2.4Δ t and four upper electrons +1.6Δ t, so CFSE=−0.8Δ t. In t₂⁴, one orbital has a pair and two have one electron each, giving two unpaired electrons. If Δ t=9,000 cm⁻¹, the orbital CFSE is −7,200 cm⁻¹.

Quick check

1. What is the tetrahedral CFSE of e²t₂²? Answer: 2(−0.6Δ t)+2(+0.4Δ t)=−0.4Δ t. 2. What is the usual tetrahedral d⁷ unpaired count? Answer: Three, from e⁴t₂³.

Exam focus

Label tetrahedral orbitals e and t₂ without g, write the occupancy before substituting, and keep Δ t distinct from Δₒ. Do not confuse CFSE with total binding energy or optical transition energy.

Advanced insight

Tetrahedral complexes often show more intense d–d absorption than centrosymmetric octahedral complexes because a tetrahedron lacks an inversion centre, relaxing the Laporte parity restriction. This spectroscopic feature is separate from the magnitude of their ground-state CFSE.

Summary

Tetrahedral orbital CFSE is (−0.6n e+0.4n t)Δ t. The smaller splitting commonly favours high-spin filling; electron count and occupancy determine both signed stabilisation and unpaired-electron number.

Practice questions

1. Calculate tetrahedral CFSE and unpaired count for d³. Answer: e²t₂¹ gives 2(−0.6Δ t)+0.4Δ t=−0.8Δ t. Three electrons occupy separate orbitals and remain unpaired. 2. Calculate the same for tetrahedral d⁶. Answer: e³t₂³ gives 3(−0.6Δ t)+3(+0.4Δ t)=−0.6Δ t. One e orbital is paired; four electrons remain unpaired. 3. Why is e⁴t₂⁶ assigned zero CFSE? Answer: It is d¹⁰, filling all five orbitals twice. The weighted downward and upward energy shifts cancel exactly: −2.4Δ t+2.4Δ t=0.