Free-Ion Terms: Russell–Saunders Coupling

Introduction

A diagram labelled t₂g³e g⁰ is a one-electron occupancy description. Electronic spectra involve transitions between many-electron states, so the combined orbital and spin angular momenta of all d electrons must be described. Russell–Saunders, or LS, coupling first combines individual orbital momenta into L and spins into S, then considers spin–orbit coupling to obtain J. The resulting term symbols organise free-ion states before ligand fields split them further.

Core explanation

Each d electron has orbital angular momentum quantum number l=2 and spin s=1/2. For several electrons, vector addition of their orbital momenta yields allowed total L values, while addition of spin momenta yields allowed S values. The Pauli exclusion principle restricts which combinations can occur for equivalent electrons in one d subshell. A term is written ^(2S+1)L, where L=0,1,2,3,4 corresponds to letters S, P, D, F, G, respectively. The superscript 2S+1 is the spin multiplicity. For example, ^3F has S=1 and L=3; ^1D has S=0 and L=2.

Spin–orbit coupling combines L and S into total J values ranging from L−S to L+S in integer steps. A ^3F term has L=3 and S=1, so possible J values are 2, 3 and 4. The corresponding levels can be labelled ^3F₂, ^3F₃ and ^3F₄. Which J lies lowest depends on shell occupancy and spin–orbit sign; the term label without J is often sufficient for first-pass ligand-field spectroscopy of 3d ions where interelectronic repulsion and crystal field dominate spin–orbit splitting.

One electron in d¹ has L=2 and S=1/2, so its free-ion term is ^2D, with possible J=3/2 and 5/2. For d², allowed terms include ^3F, ^3P, ^1G, ^1D and ^1S. These are not five different electron configurations; they are different ways two equivalent d electrons can combine their angular momenta while obeying Pauli antisymmetry. Their energies differ because of electron–electron repulsion, even before any ligand approaches.

A free-ion term has a number of microstates equal to (2S+1)(2L+1) before spin–orbit splitting. For ^3F this is 3×7=21; for ^1D it is 1×5=5. Summing the microstate counts over all terms of a configuration must equal the number of ways to place the electrons among ten d spin-orbitals, choose(10,n). For d² that count is choose(10,2)=45. This dimension check is a powerful way to catch missing or duplicate terms.

In a complex, spherical rotational symmetry is lost. A ligand field splits a free-ion L term into states labelled by molecular point-group irreducible representations. For example, a free-ion ^2D term in an octahedral field becomes ^2T₂g and ^2E g states. The superscript multiplicity can remain a useful label if spin–orbit mixing is modest. An absorption spectrum therefore connects many-electron states, not bare orbital boxes.

LS coupling works best when interelectronic repulsion dominates spin–orbit coupling, often a reasonable starting approximation for lighter transition-metal ions. For heavier metals and lanthanides, spin–orbit coupling can be comparable or dominant and a different coupling emphasis may be needed. The term-symbol notation still helps identify what is being mixed.

Step-by-step reasoning

Determine dⁿ and enumerate allowed electron microstates under Pauli. Combine individual spins to possible S and orbital angular momenta to possible L. Convert L to S/P/D/F/G letters and write ^(2S+1)L. Check the sum of term microstate dimensions against choose(10,n). If needed, add allowed J values and then consider how a ligand point group splits the term.

Visual explanation

Draw two d-electron arrows each carrying labels l=2 and s=1/2. Combine the two orbital arrows into L and the spin arrows into S, then join L and S into J. Beside this flow write the term symbol ^(2S+1)L J and a small table L=0→S, 1→P, 2→D, 3→F, 4→G.

Real-world analogy

A sports team has both a combined running direction and a combined strategy role; knowing only which individual positions are occupied does not describe the whole team state. Term symbols similarly summarise combined electron angular momenta rather than listing orbital occupancy alone.

Real-world example

The d¹ Ti³⁺ free ion has a ^2D term. In an octahedral aqua complex the ligand field separates that term into lower ^2T₂g and upper ^2E g components, providing a relatively simple spectral transition compared with a multi-term d² ion.

Why?

Why can d² give several spectral terms despite a fixed electron count? Two electrons can combine their spin and orbital momenta in several Pauli-allowed ways. Electron–electron repulsion gives those many-electron arrangements different energies.

Common misconception

“The letter D in ^2D means an electron occupies a d orbital.” Here D denotes total L=2. A d² configuration can produce F, P, G, D and S terms; the term letter is not the orbital subshell name.

Worked example

For a ^3F free-ion term, superscript 3 means 2S+1=3 and S=1. F means L=3. Allowed J values are 3−1 , 3−1 +1, …, 3+1, namely 2, 3 and 4. Before spin–orbit splitting, the term contains (2S+1)(2L+1)=3×7=21 microstates. A crystal field will further classify them by molecular symmetry.

Quick check

1. What are S and L for ^1G? Answer: Multiplicity one gives S=0; G corresponds to L=4.

Exam focus

Separate configuration dⁿ, free-ion term ^(2S+1)L and point-group state label. Check allowed J and microstate dimensions rather than interpreting term letters as orbital names.

Advanced insight

LS coupling is a hierarchy of energy scales. When spin–orbit coupling is no longer a small perturbation, L and S may not remain nearly good quantum numbers; J or more general mixed-state labels can become more useful for magnetic and spectral analysis.

Summary

Russell–Saunders coupling builds total L and S from individual d electrons, giving ^(2S+1)L terms and possible J levels. Pauli restrictions create multiple terms for one dⁿ configuration, which ligand fields then split.

Practice questions

1. What term symbol describes one free d electron before specifying J? Answer: ^2D, because l=2 gives L=2 (D) and one spin-half electron gives S=1/2, hence multiplicity 2. 2. How many microstates does ^1D contain? Answer: (2S+1)(2L+1)=1×5=5, since S=0 and L=2. 3. Why does choose(10,2) appear in a d² term check? Answer: There are ten available d spin-orbitals and two indistinguishable electrons occupy two distinct ones, giving 45 allowed microstates before grouping them into terms. 4. How many d spin-orbitals are available to a free ion? Answer: Ten: five spatial d orbitals times two spin possibilities.