Coordination Chemistry and CFT: Unit Review

Introduction

Coordination chemistry becomes easier when three questions are kept distinct. How do ligands split metal d orbitals and stabilise a filling? How many electrons remain unpaired and what magnetic moment follows? How many spatial arrangements of the attached ligands exist? This review connects the answers in one workflow while preserving the limits of each model.

Core explanation

Start with formal charge. Assign ligand charges and solve for the metal oxidation state, then derive dⁿ by removing ns electrons before (n−1)d electrons. For a typical d-block ion, group number minus oxidation state is a useful formal shortcut. The d count belongs to the metal-ion bookkeeping convention; ligand donation does not simply add electrons to that count. Confirm coordination number and geometry before drawing split orbitals.

In an ideal octahedron, t₂g has three orbitals at −0.4Δₒ each and e g two at +0.6Δₒ. Signed orbital CFSE is (−0.4n t+0.6n e)Δₒ. d¹–d³ fill separate t₂g orbitals. d⁴–d⁷ can be high or low spin, depending on the competition between Δₒ and the additional pairing cost P. d⁸–d¹⁰ have no analogous elementary octahedral high/low-spin alternative. For a tetrahedron, lower e lies at −0.6Δ t and upper t₂ at +0.4Δ t, with Δ t often estimated near 4Δₒ/9 for corresponding metal–ligand conditions. Most familiar tetrahedral complexes are high spin.

An orbital CFSE calculation is not total complex formation energy. Pairing, metal–ligand bonding, solvation and lattice packing also matter. CFSE helps explain site preferences in spinels and irregular hydration or lattice trends, but a site with greater CFSE need not win when other terms differ. Jahn–Teller distortion reveals another limit of ideal symmetry: d⁹ Cu²⁺ often elongates an octahedron because uneven e g occupancy can be stabilised by splitting those levels. Pure point-charge CFT omits explicit covalency and the nephelauxetic reduction of electron-repulsion parameters; ligand-field theory supplies a richer account.

For magnetism, count unpaired electrons in the actual filling. The spin-only estimate μ so=√[n(n+2)] BM yields 1.73, 2.83, 3.87, 4.90 and 5.92 BM for n=1 through 5. A measured moment supports, but does not uniquely identify, a d count: high-spin d⁴ and d⁶ both have four unpaired electrons. Orbital contributions, especially for some Co²⁺ complexes, and coupling between metal centres can shift measured values. Spin crossover can make the moment temperature-dependent when two spin states are close in free energy.

For isomers, fix coordination connectivity first. Ionisation and linkage isomers change what binds or how it binds. Within each fixed branch, count ligand arrangements under rotation. Octahedral MA₄B₂ gives cis/trans; MA₃B₃ gives fac/mer. MA₂B₂C₂ gives five geometrical classes but six individual stereoisomers because the all-cis class has an enantiomeric pair. Square-planar MABCD gives three trans-pair arrangements and no optical pair in the ideal simple-ligand model. Chelates constrain donor adjacency: M(AA)₃ has Δ/Λ partners; M(AA)₂B₂ has trans plus a cis enantiomeric pair, giving three total stereoisomers.

Never infer isomer count from CFSE or magnetism alone. A cis and trans pair can have the same formal d count and similar simple orbital occupancy. Conversely, two complexes with the same ligand pattern can differ in spin state if their metal or ligand-field parameters differ. A complete answer labels which conclusions are formal, which are model predictions and which require experimental evidence.

Step-by-step reasoning

Use a fixed order: charge equation → oxidation state and d count → coordination geometry → orbital filling and pair comparison → signed CFSE → unpaired count and spin-only moment → connectivity branches → geometric classes → enantiomers. Check units and sign conventions at each calculation. Finally compare the prediction with supplied spectroscopy or magnetic data and state any idealisation.

Visual explanation

Draw one central formula box with two branches. The electronic branch contains the d-level diagram, a CFSE ledger and an unpaired-arrow count. The structural branch contains a labelled polyhedron, trans-pair signatures and mirror-image test. A final evidence box receives both branches plus measured colour or magnetism.

Real-world analogy

A building can be analysed by construction costs, number of occupants and room layouts. These are related but not interchangeable questions: knowing the occupant count does not fix the room arrangement. Likewise, energy, magnetism and isomerism describe different properties of one complex.

Real-world example

For [Co(en)₂Cl₂]⁺, Co³⁺ is formally d⁶ and the two en chelates occupy adjacent donor pairs. The two chlorides can be trans or cis; cis has a Δ/Λ pair. Magnetic evidence can then help test whether the cobalt centre is low spin, while the three-isomer structural count remains a separate conclusion.

Why?

Why does a single workflow start with oxidation state? Every later electronic step depends on d count. A misplaced ligand charge propagates into the filling, CFSE and magnetic moment, even if those later calculations are internally neat.

Common misconception

“One correct crystal-field diagram answers every coordination question.” It predicts an idealised orbital filling, not the complete bonding energy, measured moment with orbital effects, or spatial ligand-isomer count.

Worked example

An ideal octahedral [Cr(H₂O)₄Cl₂]⁺ problem gives Δₒ=16,000 cm⁻¹. Two chloride ligands total −2 and the ion is +1, so Cr is +3, d³. Fill t₂g³: CFSE=−1.2(16,000)=−19,200 cm⁻¹. Three unpaired electrons give μ so=√15≈3.87 BM. The ligand pattern MA₄B₂ has cis and trans forms, each achiral in the simple model: two stereoisomers. CFSE is an orbital contribution, not total formation enthalpy.

Quick check

1. Which octahedral orbital lies at +0.6Δₒ per electron? Answer: The higher e g set. 2. How many total simple stereoisomers does octahedral M(AA)₂B₂ have? Answer: Three: one trans form and two cis enantiomers.

Exam focus

State formal charges, draw filled orbital boxes, keep pairing cost separate from CFSE, append BM to moments and explicitly distinguish geometric from total stereoisomer counts. Explain any approximation for mixed ligands or distorted geometry.

Advanced insight

The most powerful use of the elementary model is cross-checking independent observations. If magnetism, spectral splitting and structural isomer evidence conflict, revisit oxidation state, symmetry and covalency before forcing an answer from a memorised table.

Summary

Charge and geometry establish the electronic problem; CFSE and pairing select candidate fillings; unpaired electrons predict a spin-only moment; ligand identity and symmetry determine isomer counts. Experimental data test the assumptions and reveal where richer ligand-field theory is needed.

Practice questions

1. For ideal octahedral low-spin d⁶, give orbital CFSE, unpaired count and moment. Answer: t₂g⁶e g⁰ gives CFSE −2.4Δₒ, zero unpaired electrons and a spin-only moment of zero BM. 2. A tetrahedral d³ complex has what occupancy and CFSE? Answer: It has e²t₂¹ with three unpaired electrons. CFSE=2(−0.6Δ t)+0.4Δ t=−0.8Δ t. 3. How many geometric and total stereoisomers exist for simple octahedral MA₂B₂C₂? Answer: Five geometric classes and six individual stereoisomers; the all-cis class contributes two enantiomers. 4. Why might measured Co²⁺ moment exceed the three-unpaired spin-only estimate? Answer: Orbital angular momentum may not be fully quenched and can add to the spin response. A larger moment does not require fractional or extra unpaired electrons.