Questions: Color and Spectroscopy of Coordination Compounds

Question 1

A complex absorbs strongly in the orange region (~600 nm) of the visible spectrum. What color does the complex appear?

• A. Orange
• B. Blue
• C. Green
• D. Red

Answer: (B)

The observed color is the complement of the absorbed color. Orange light (~600 nm) and blue light (~450 nm) are complementary colors. When the complex absorbs orange, it transmits and reflects all other wavelengths, but the dominant perceived color is blue. This is why the color wheel is so useful in coordination chemistry: find the absorbed color on the wheel, and the opposite side gives you the observed color.

Question 2

The Laporte selection rule states that d-d transitions in centrosymmetric (octahedral) complexes are formally forbidden because the transition does not involve a change in parity (g → g).

• T. True
• F. False

Answer: True

The Laporte rule requires that allowed electronic transitions must involve a change in the orbital angular momentum quantum number (Δl = ±1), which translates to a change in parity: gerade (g) to ungerade (u) or vice versa. In an octahedral complex, d-orbitals are all gerade (even under inversion), so d-d transitions are g → g and formally Laporte-forbidden. This is why d-d transitions are relatively weak (molar absorptivity ε typically 1-100 M⁻¹cm⁻¹) compared to fully allowed transitions (ε > 1000). The transitions are not completely forbidden because vibrational coupling temporarily destroys the inversion center, making them weakly allowed.

Question 3

Charge-transfer transitions in coordination compounds are typically much more intense than d-d transitions because they are Laporte-allowed.

• T. True
• F. False

Answer: True

Charge-transfer (CT) transitions involve electron movement between ligand orbitals and metal orbitals — for example, from a ligand-centered orbital (ungerade) to a metal d-orbital (gerade), or vice versa. These transitions involve a change in parity (u → g or g → u) and are therefore Laporte-allowed, giving molar absorptivities of 1000-50,000 M⁻¹cm⁻¹. This is why compounds like permanganate (MnO₄⁻) are so intensely colored — their purple color arises from ligand-to-metal charge transfer, not weak d-d transitions.

Question 4

[Mn(H₂O)₆]²⁺ is nearly colorless despite having five d-electrons. Explain why, referencing both the spin and Laporte selection rules.

Model answer: Mn²⁺ is d⁵ high-spin with all five electrons having parallel spins (one in each d-orbital). Any d-d transition would require flipping the spin of the promoted electron to pair with an electron in the destination orbital, violating the spin selection rule (ΔS = 0 required). These spin-forbidden transitions are extremely weak. Combined with the Laporte selection rule (d-d transitions in octahedral complexes are parity-forbidden), the absorption is doubly forbidden, making [Mn(H₂O)₆]²⁺ nearly colorless with a very pale pink tint visible only in concentrated solutions.

This is a beautiful example of selection rules in action. The faint pink color that IS visible comes from spin-orbit coupling, which weakly relaxes the spin selection rule. But the double prohibition (spin-forbidden AND Laporte-forbidden) makes d⁵ high-spin octahedral complexes the weakest absorbers among all d-electron configurations.

Question 5

Why do tetrahedral complexes often have more intense d-d absorption bands than their octahedral counterparts, even though tetrahedral Δ is smaller?

Model answer: Tetrahedral complexes lack an inversion center, so the Laporte selection rule does not strictly apply. In an octahedral complex, d-d transitions are formally parity-forbidden (g → g) and rely on vibrational coupling to gain intensity. In a tetrahedron, there is no inversion symmetry, so the d-orbitals mix partially with p-orbitals (which have opposite parity), and the transition gains some formally allowed character. This relaxation of the Laporte rule means tetrahedral d-d bands are typically 10-100 times more intense than octahedral ones, even though the absorption energy may be lower due to the smaller Δ_tet.

This explains an otherwise puzzling observation: CoCl₄²⁻ (tetrahedral, intense blue) is deeply colored while [Co(H₂O)₆]²⁺ (octahedral, pale pink) is barely colored, despite cobalt being in the same oxidation state in both.