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Periodicity - Colored complexes (HL only)

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Transition metals form colored compounds due to the partially filled dd-orbitals. In an isolated gaseous ion, the five dd-orbitals are degenerate (equal in energy).

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When ligands approach the central metal ion to form a complex, the repulsion between the lone pairs on the ligands and the electrons in the dd-orbitals causes the orbitals to split into two sets of different energy levels. In octahedral complexes, these are the t2gt_{2g} (lower energy: dxyd_{xy}, dyzd_{yz}, dxzd_{xz}) and ege_g (higher energy: dz2d_{z^2}, dx2βˆ’y2d_{x^2-y^2}) levels.

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The energy difference between these levels is denoted as Ξ”E\Delta E. When light shines on the complex, an electron can be promoted from a lower energy dd-orbital to a higher energy dd-orbital (a dβˆ’dd-d transition) by absorbing a photon of energy equal to Ξ”E\Delta E.

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The color observed is the complementary color to the wavelength of light absorbed. For example, if a complex absorbs blue light, it will appear orange.

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The magnitude of Ξ”E\Delta E, and thus the color, depends on four factors: 1. The identity of the metal ion (nuclear charge), 2. The oxidation state of the metal (higher oxidation state usually leads to larger Ξ”E\Delta E), 3. The nature of the ligand (Spectrochemical series), and 4. The geometry of the complex (e.g., octahedral vs. tetrahedral).

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The Spectrochemical Series ranks ligands by their ability to split dd-orbitals: Iβˆ’<Brβˆ’<S2βˆ’<Clβˆ’<Fβˆ’<OHβˆ’<H2O<SCNβˆ’<NH3<en<NO2βˆ’<CNβˆ’<COI^- < Br^- < S^{2-} < Cl^- < F^- < OH^- < H_2O < SCN^- < NH_3 < en < NO_2^- < CN^- < CO.

πŸ“Formulae

Ξ”E=hΞ½\Delta E = h\nu

Ξ”E=hcΞ»\Delta E = \frac{hc}{\lambda}

c=Ξ½Ξ»c = \nu\lambda

πŸ’‘Examples

Problem 1:

Explain why [Cu(H2O)6]2+[Cu(H_2O)_6]^{2+} is blue, while [Zn(H2O)6]2+[Zn(H_2O)_6]^{2+} is colorless.

Solution:

Cu2+Cu^{2+} has a 3d93d^9 configuration, meaning it has a partially filled dd-subshell. Electrons can undergo dβˆ’dd-d transitions by absorbing specific wavelengths of visible light, reflecting the complementary blue color. Zn2+Zn^{2+} has a 3d103d^{10} configuration; its dd-subshell is completely full, so no dβˆ’dd-d transitions are possible.

Explanation:

Colored complexes require partially filled dd-orbitals (between d1d^1 and d9d^9). Ions with d0d^0 (like Sc3+Sc^{3+}) or d10d^{10} (like Zn2+Zn^{2+} or Cu+Cu^+) are colorless.

Problem 2:

Compare the splitting energy Ξ”E\Delta E and the wavelength of light absorbed by [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+} and [Fe(CN)6]3βˆ’[Fe(CN)_6]^{3-}.

Solution:

CNβˆ’CN^- is a stronger field ligand than H2OH_2O according to the spectrochemical series. Therefore, [Fe(CN)6]3βˆ’[Fe(CN)_6]^{3-} will have a larger Ξ”E\Delta E than [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+}. Since Ξ”E=hcΞ»\Delta E = \frac{hc}{\lambda}, a larger Ξ”E\Delta E corresponds to the absorption of a shorter wavelength (higher energy) of light.

Explanation:

Stronger ligands cause greater repulsion and larger splitting of the dd-orbitals, shifting the absorption toward the violet/blue end of the spectrum.

Colored complexes (HL only) - Revision Notes & Key Formulas | IB Grade 12 Chemistry