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Chemistry 2e · Transition Metals and Coordination Chemistry

Spectroscopic and Magnetic Properties of Coordination Compounds

9 min read
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Numerical values (Planck's constant, speed of light, Avogadro's number, magnetic moments, spectrochemical series order) are standard reference values; verify against current sources before relying on them in assessments.
Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Why is [Cu(H2O)6]2+ blue while [Zn(H2O)6]2+ is colorless? Why does one iron complex get pulled into a magnetic field while another with the same metal barely notices it? The answers come from , which describes how ligands split the five d orbitals into different energy levels. The size of that split — the crystal field splitting energy Δ — decides which wavelengths of visible light a complex absorbs (its color) and how its d electrons are arranged (its magnetic behavior). This topic covers the octahedral splitting diagram, the that ranks ligands by field strength, high-spin vs low-spin configurations, the color-wheel logic of absorption, and the spin-only magnetic moment formula. Along the way: ruby and emerald are just chromium ions in different host crystals, and the color of blood depends on which ligand is bound to iron.

Why this matters

  • Everyday color: Gemstones (ruby = Cr³⁺ in Al₂O₃, emerald = Cr³⁺ in beryl, sapphire = Fe/Ti in corundum), paints, inks, and glass get their colors from d-electron transitions.
  • Biology and medicine: Hemoglobin turns bright red when oxygen binds iron and darker when it doesn't — a visible, ligand-controlled color change. MRI contrast agents are gadolinium complexes chosen for their large magnetic moments (7 unpaired electrons).
  • Analysis: UV-visible spectroscopy measures Δ directly, identifying metals, oxidation states, and ligand environments in solution.
  • Exams: Predicting color from absorbed wavelength, ranking ligands by the spectrochemical series, and computing spin-only moments are classic problems.

The college version

Core Concepts

Crystal field theory: the five d orbitals split

In a free metal ion the five d orbitals have identical energy (degenerate). When ligands approach along the axes of an octahedron, the two orbitals pointing directly at the ligands — dz2 and dx2-y2 (the eg set) — rise in energy because the ligand lone pairs repel them. The three pointing between the axes — (d{xy}), dxz, dyz (the t₂g set) — are lowered. The gap is the octahedral splitting energy Δoct; t₂g sits 2/5 Δoct below and eg (3/5\,\Delta{\text{oct}}) above the average energy.

The spectrochemical series: ranking ligands

Ligands differ in how strongly they split the d orbitals. The empirical order, weakest to strongest field:

I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO

Weak-field ligands (halides, water) give small Δoct; strong-field ligands (CN⁻, CO) give large Δoct. Splitting also grows with the metal's charge (Fe³⁺ splits more than Fe²⁺) and down a group (4d, 5d metals split more than 3d).

High spin vs low spin: filling the d orbitals

For d¹–d³ and d⁸–d¹⁰ the filling pattern is forced. For d⁴–d⁷ there is a choice, decided by the competition between Δoct and the electron pairing energy P:

  • If Δoct > P (strong field, e.g. CN⁻): electrons pair in the low t₂g level before occupying e_g → .
  • If Δoct < P (weak field, e.g. H₂O): electrons spread out per Hund's rule → .

The classic pair, both Fe(II) (d⁶): [Fe(CN)6]4- is low spin — all six electrons paired, zero unpaired, ; [Fe(H2O)6]2+ is high spin — four unpaired, . Tetrahedral complexes split much less (Δtet ≈ 49Δoct), so they are almost always high spin; square-planar geometry (common for d⁸: Pt²⁺, Ni²⁺) is low spin and diamagnetic.

Color: absorbed vs observed

A complex appears colored because it absorbs some wavelengths of visible light and transmits/reflects the rest. The absorbed photon promotes a d electron from t₂g to e_g; the transition energy equals the splitting:

Δoct = Ephoton = hcλ

where h is Planck's constant, c the speed of light, and λ the wavelength of maximum absorption. The color you see is the complementary of the absorbed one (opposite on the color wheel): absorb green (~500 nm) → look red-purple; absorb orange (~600 nm) → look blue. Consequences: (1) changing the ligand changes Δ, hence the observed color — why [Cu(H2O)6]2+ (pale blue) differs from [Cu(NH3)4]2+ (deep blue); (2) complexes with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺, Ag⁺) have no d–d transitions and are colorless.

Magnetic properties: counting unpaired electrons

A complex with unpaired electrons is paramagnetic (weakly attracted into a magnetic field); all-paired is diamagnetic (weakly repelled). For first-row transition complexes the orbital contribution is small, so the measured moment follows the spin-only formula:

μ= n(n+2) BM

where n is the number of unpaired electrons and BM is the Bohr magneton. Measured μ → solve for n → assign high/low spin. Benchmarks: n = 1 → 1.73 BM; n = 3 → 3.87 BM; n = 4 → 4.90 BM; n = 5 → 5.92 BM.

Applications: gemstones and blood

Ruby is corundum (Al₂O₃) with ~1% Cr³⁺ substituting for Al³⁺; the d³ chromium absorbs yellow-green light, so ruby transmits red. Emerald is the same Cr³⁺ ion in beryl (Be₃Al₂Si₆O₁₈), where a slightly different crystal field shifts the absorption and the stone looks green. In hemoglobin, the iron porphyrin is bright red when O₂ (a strong field) binds and darker when it does not — ligand-field logic visible in your own veins.

How It Works / Step-by-Step Process

Predicting color and magnetism of a complex:

  1. Determine the metal's oxidation state and d-electron count.
  2. Identify the geometry (usually octahedral for 3d) and rank the ligands on the spectrochemical series.
  3. Decide high vs low spin: strong field → pair up.
  4. Fill the d orbitals; count unpaired electrons n.
  5. Predict magnetism: n = 0 diamagnetic; n > 0 paramagnetic with μ= n(n+2) BM.
  6. Predict color: d⁰/d¹⁰ colorless; otherwise absorb hc/Δ and report the .

Common Confusions

Do Not ConfuseWithDifference
Absorbed colorObserved colorAbsorb green (~500 nm) → looks red-purple; observed = complementary color
"All transition compounds are colored"Only those with d–d transitionsd⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Zn²⁺, Cu⁺) complexes are colorless
High spin = strong fieldHigh spin = weak fieldHigh spin happens when Δ< P (weak field); strong fields force pairing (low spin)
ParamagnetismFerromagnetismParamagnetic complexes are weakly attracted only inside a field; bulk iron's ferromagnetism is a collective solid-state effect
Number of d electronsNumber of unpaired electronsThe magnetic formula uses unpaired count n, not total d count (d⁶ Fe²⁺ can have 4 or 0 unpaired)
Δoct is a constantΔ depends on ligand, charge, geometrySame metal + different ligand → different color; tetrahedral splitting ≈ 4/9 of octahedral
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Inside a metal ion, the five electron rooms (d orbitals) start with equal energy, but when ligands move in, the rooms split: two become "expensive" (high energy) and three become "cheap" (low energy). If a light wave carries exactly the energy gap between them, an electron jumps up and the light gets eaten — that's why the complex has color, and the color you see is the opposite of the color eaten. Strong-gripping ligands like cyanide make a big gap; weak ones like water make a small gap. Electrons also decide whether to share rooms (pair up) or spread out — and every unpaired electron makes the complex act like a tiny magnet.

Worked example

Example 1: Measuring Δoct from an absorption spectrum

The d¹ complex [Ti(H2O)6]3+ absorbs visible light with a maximum at 500 nm. Calculate the octahedral splitting energy in kJ/mol.

Formula first — energy of one photon:

E = hcλ

with h = 6.626 × 10-34 J·s, c = 2.998 × 108 m/s, λ= 500 nm = 5.00 × 10-7 m:

E = (6.626 × 10-34 J·s)(2.998 × 108 m/s)5.00 × 10-7 m = 3.97 × 10-19 J

Per mole — multiply by Avogadro's number and convert to kJ:

E = 3.97 × 10-19 J × 6.022 × 1023 mol-1 × 1 kJ103 J = 239 kJ/mol

Answer: Δoct ≈ 239 kJ/mol. Because it absorbs green (~500 nm), the complex appears red-purple — the complementary color.

Example 2: High vs low spin and magnetic moment for two Fe(II) complexes

Predict the number of unpaired electrons and spin-only moment for [Fe(H2O)6]2+ and [Fe(CN)6]4-.

Step 1 — d count: Fe is +2 in both; Fe²⁺ is d⁶.

Step 2 — ligand field: H₂O is weak-field (small Δ); CN⁻ is strong-field (large Δ).

Step 3 — filling: [Fe(H2O)6]2+: Δ< P → high spin: t₂g⁴ e_g² → 4 unpaired. [Fe(CN)6]4-: Δ> P → low spin: t₂g⁶ e_g⁰ → 0 unpaired.

Step 4 — moments (μ= n(n+2) BM):

μ[Fe(H2O)6]2+ = 4(4+2) = 24 = 4.90 BM  (paramagnetic)

μ[Fe(CN)6]4- = 0(0+2) = 0 BM  (diamagnetic)

Answer: The aqua complex is paramagnetic with μ ≈ 4.90 BM; the cyano complex is diamagnetic — exactly how a measured moment distinguishes high- from low-spin complexes in the lab.

Example 3: Why is [Zn(NH3)4]2+ colorless?

Zn²⁺ is d¹⁰: all five d orbitals are fully occupied, so no d–d transition is possible (no empty d orbital to accept an electron), and no visible light is absorbed. Answer: d¹⁰ complexes — Zn²⁺, Cu⁺, Ag⁺ — are colorless regardless of ligand.

Key takeaways

  • Octahedral field splits d orbitals into t₂g (lower) and eg (higher); gap = (\Delta{\text{oct}}).
  • Spectrochemical series (weak → strong): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO.
  • High spin = weak field (Δ< P); low spin = strong field (Δ> P); only d⁴–d⁷ have the choice.
  • d⁶ pair: [Fe(H2O)6]2+ high spin (4 unpaired, paramagnetic); [Fe(CN)6]4- low spin (0 unpaired, diamagnetic).
  • Δtet ≈ 49Δoct → tetrahedral complexes almost always high spin; square planar (d⁸) is low spin.
  • Color = complementary of absorbed wavelength: Δoct = hc/λ. d⁰ and d¹⁰ complexes are colorless.
  • Spin-only moment: μ= n(n+2) BM; measured μ gives the unpaired-electron count.
  • Ruby/emerald/sapphire and oxy- vs deoxyhemoglobin are ligand-field color effects.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

  1. Why is [Cu(H2O)6]2+ blue? Which wavelength region does it absorb?

    Show answer

    It absorbs orange-red light around 600 nm (small Δ from weak-field water), so it transmits/reflects blue — the complementary color.

  2. Rank Cl⁻, CN⁻, and H₂O from weakest to strongest field.

    Show answer

    Cl⁻ < H₂O < CN⁻ (weakest to strongest per the spectrochemical series).

  3. How many unpaired electrons does a high-spin d⁵ complex have? A low-spin d⁶?

    Show answer

    High-spin d⁵: 5 unpaired (t₂g³ e_g²). Low-spin d⁶: 0 unpaired (t₂g⁶) — diamagnetic, like [Fe(CN)6]4-.

  4. Calculate the spin-only moment for a complex with 3 unpaired electrons.

    Show answer

    μ= 3(3+2) = 15 = 3.87 BM.

  5. Why are Zn²⁺ and Sc³⁺ complexes colorless?

    Show answer

    Zn²⁺ is d¹⁰ and Sc³⁺ is d⁰: no d–d transition is possible (no empty d orbital / no d electrons), so no visible absorption.

  6. A complex absorbs at 400 nm. What is Δoct in kJ/mol?

    Show answer

    E = hc/λ= (6.626 × 10-34)(2.998 × 108)/(4.00 × 10-7) = 4.97 × 10-19 J; per mole = 4.97 × 10-19 × 6.022 × 1023 = 2.99 × 105 J/mol = 299 kJ/mol.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Crystal field theory
Model in which ligands split the metal's d orbitals into different energy levels
Δoct
Energy gap between the t₂g and e_g sets in an octahedral complex
Spectrochemical series
Ranking of ligands by how strongly they split d orbitals
High spin
Configuration where Δ< P, so electrons spread out before pairing
Low spin
Configuration where Δ> P, so electrons pair in the lower set
Paramagnetic
Having unpaired electrons; weakly attracted into a magnetic field
Diamagnetic
All electrons paired; weakly repelled by a magnetic field
Complementary color
The color opposite the absorbed one on the color wheel

Sources & references

  1. openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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