Integrated Problems: CFSE, Magnetism and Isomers

Introduction

An advanced coordination question may ask for oxidation state, d count, crystal-field stabilisation, magnetic moment and isomers of the same formula. Each answer depends on earlier decisions, but the tasks should not be conflated. Charge determines formal d count; geometry and splitting determine filling; filling determines moment and CFSE; ligand identities and symmetry determine isomer count.

Core explanation

Consider a six-coordinate mixed-ligand ion [Cr(H₂O)₄Cl₂]⁺. Water is neutral and two chlorides contribute −2, so x−2=+1 gives Cr³⁺. Chromium is in group 6, giving formal d³. An approximately octahedral d³ metal has t₂g³e g⁰ in the elementary model. All three t₂g orbitals are singly occupied, so there are three unpaired electrons. The spin-only moment is √[3(3+2)]=√15≈3.87 BM. The ideal octahedral orbital CFSE is −1.2Δₒ.

The formula also has ligand pattern MA₄B₂ with A=H₂O and B=Cl. The two chlorides can be cis or trans. Rotation identifies all positions within each class, so there are two geometrical isomers. With these simple monodentate ligands, neither class has a distinct optical partner, giving two individual stereoisomers. This count does not follow from d³; a d⁶ metal with the same MA₄B₂ ligand pattern would also have the same simple cis/trans count.

If a question supplies Δₒ=15,000 cm⁻¹ as an idealised splitting parameter, CFSE is −1.2(15,000)=−18,000 cm⁻¹ in energy-equivalent units, about −215 kJ mol⁻¹ using 0.01196 kJ mol⁻¹ per cm⁻¹. This is the orbital-splitting contribution relative to the barycentre, not the full formation enthalpy of the complex. Because four waters and two chlorides are not all identical, actual orbital degeneracies can split further below perfect octahedral symmetry. An exam’s single Δₒ usually signals that the ideal octahedral approximation is intended.

A different integrated example, [Co(NH₃)₄Cl₂]⁺, contains Co³⁺ d⁶. Its two geometric isomers are again cis and trans. But now d⁶ can be high or low spin, so a supplied Δₒ and P or measured magnetic moment is needed to choose an electron filling. If Δₒ>P in the one-P model, low-spin t₂g⁶ gives signed orbital CFSE −2.4Δₒ and zero spin-only moment. If Δₒ<P, high-spin t₂g⁴e g² gives −0.4Δₒ and four unpaired electrons. The geometrical isomer count remains two in either simplified spin assignment.

This demonstrates two useful independence checks. First, changing spin state does not automatically change a fixed ligand pattern’s cis/trans count. Second, cis/trans isomers can differ subtly in actual spectral splitting because their symmetry is different, even though a simplified problem gives one Δₒ for both. State the assumptions that permit a common ideal calculation.

The safest solution order is charge → d count → geometry → orbital filling → CFSE and moment → structural/optical isomers. If the problem gives measured moment, use it to test the proposed filling; if it gives spectral splitting, ensure its units match pairing energy. A contradictory set of clues should prompt rechecking oxidation state and ligand charges before inventing new chemistry.

Step-by-step reasoning

Write an oxidation-state equation from ligand charges and complex charge. Derive dⁿ from group number. Draw the stated or inferred geometry and decide whether high/low-spin alternatives exist. Calculate signed CFSE and n-based moment. Independently classify ligand pattern and enumerate geometric plus optical isomers. Finish with a consistency check against all given data and explicit approximation limits.

Visual explanation

Draw a flow diagram with two branches after formula and geometry. One branch leads through d count and orbital filling to CFSE and magnetism. The other leads through ligand labels and symmetry to isomer count. The branches reunite in a final box labelled “one chemically consistent answer.”

Real-world analogy

Solving a travel puzzle requires the traveller’s identity, route cost and seating arrangement, but those are different questions. Knowing seat arrangement does not determine ticket price. Likewise, ligand positions, electron spin and orbital energy describe different aspects of one complex.

Real-world example

Cr(III) ammine and aqua complexes can be prepared as distinct cis/trans forms while retaining a d³ magnetic pattern. Their ligand substitution may be slow enough to study separate forms; this kinetic fact does not alter the formal two-isomer count.

Why?

Why should isomer counting come after, but separately from, CFSE calculation? Both require an established coordination geometry, yet they use different inputs: orbital occupancy for CFSE and ligand identity plus spatial symmetry for isomers. Mixing them can produce false dependencies.

Common misconception

“A low-spin complex has fewer isomers because its electrons are paired.” Pairing changes electronic state, not the number of distinct placements of a fixed set of ligands in the stated geometry.

Worked example

For [Cr(H₂O)₄Cl₂]⁺ with ideal Δₒ=15,000 cm⁻¹: x−2=+1 gives Cr³⁺ d³; t₂g³ has CFSE −1.2Δₒ=−18,000 cm⁻¹ and n=3, so μ so≈3.87 BM. Four waters and two chlorides give cis and trans octahedral geometries, both achiral in this simple model. Final answers are two stereoisomers, a paramagnetic moment estimate, and the stated ideal CFSE.

Quick check

1. Does cis/trans placement of the two chlorides change formal Cr³⁺ d³ count? Answer: No. Formula and oxidation state remain the same in both isomers. 2. Why might real mixed-ligand orbital energies differ from one ideal Δₒ diagram? Answer: Different ligands lower exact octahedral symmetry and can split levels that are degenerate in the idealised model.

Exam focus

Show a coherent chain of charge, d count, filling, energy, moment and isomer classification. State the approximation behind a single Δₒ for mixed ligands and do not mistake CFSE for total complex stability.

Advanced insight

Integrated evidence can test a structural assignment: a moment inconsistent with a proposed spin state or a spectrum showing lower symmetry may reveal an incorrect geometry, unexpected oxidation state or more detailed ligand-field behaviour.

Summary

Multi-step coordination problems become manageable when electronic and stereochemical branches are solved separately from the same formula. Cross-check the resulting d count, CFSE, moment and isomer tally for consistency.

Practice questions

1. For [Co(NH₃)₄Cl₂]⁺, what is the formal d count and simple geometric-isomer count? Answer: Co is +3 because two chlorides contribute −2 to a +1 ion. Group 9 minus 3 gives d⁶. The MA₄B₂ pattern gives cis and trans, two geometrical forms. 2. If this Co³⁺ d⁶ example is low spin, give CFSE and spin-only moment. Answer: Ideal low-spin t₂g⁶ has orbital CFSE −2.4Δₒ and n=0, so the spin-only moment is zero BM. 3. If the same d⁶ example is high spin, how many unpaired electrons and what orbital CFSE result? Answer: High-spin t₂g⁴e g² has four unpaired electrons and orbital CFSE −0.4Δₒ; its cis/trans count remains two.