UV-Vis Spectroscopy of Coordination Compounds
Questions 1–4
Most transition metal complexes are brightly colored in solution. What is the fundamental electronic origin of color in these complexes?
Transition metal complexes are colored because d–d electronic transitions absorb photons in the visible region of the electromagnetic spectrum. The crystal field (or ligand field) splits the d orbitals into sets of different energy, and electrons can be promoted from the lower set to the upper set by absorbing light. The color we observe is the complementary color to the wavelengths absorbed.
![Absorption spectrum of [Ti(H2O)6]3+](../assets/figures/chapter20/fig_20_24.png)
The complex [Ti(H2O)6]3+ appears purple and shows a single broad absorption band near 500 nm. What electronic transition is responsible for this absorption?
Ti3+ is a d1 ion. In an octahedral crystal field, the five d orbitals split into a lower t2g set and an upper eg set. The single electron occupies a t2g orbital and absorbs a photon of energy Δoct to be promoted to the eg set. This is the simplest possible d–d transition, giving a single absorption band whose energy directly measures Δoct.
Two important selection rules govern the intensity of d–d transitions. Which statement correctly describes the Laporte and spin selection rules?
The Laporte (parity) selection rule states that transitions between states of the same parity are forbidden: g → g and u → u are forbidden, while g → u is allowed. Since all d orbitals are gerade (g), d–d transitions are Laporte-forbidden in centrosymmetric complexes. The spin selection rule forbids transitions where the total spin quantum number changes (ΔS ≠ 0). Both rules explain why d–d bands are typically weak (low molar absorptivity).
d–d transitions in octahedral complexes are Laporte-forbidden because they are g → g transitions. Yet octahedral complexes such as [Ti(H2O)6]3+ are still visibly colored. What mechanism allows these formally forbidden transitions to occur?
Vibronic coupling is the key mechanism. Asymmetric stretching vibrations of the metal–ligand bonds momentarily distort the octahedral geometry, temporarily removing the center of symmetry (inversion center). During these distorted configurations, the Laporte rule is partially relaxed, allowing d–d transitions to "steal" intensity. This is why octahedral d–d bands have low but nonzero molar absorptivities (typically ε ≈ 5–100 M−1cm−1).
Charge Transfer & Term Symbols
Questions 5–8
Charge-transfer (CT) transitions in metal complexes produce bands that are typically much more intense than d–d bands. What is a CT transition, and why is it more intense?
A charge-transfer transition involves the movement of electron density between metal-centered and ligand-centered orbitals. Because these orbitals have different parity (metal d orbitals are gerade, ligand π or σ orbitals are ungerade), CT transitions are Laporte-allowed. This makes them orders of magnitude more intense than d–d transitions, with molar absorptivities typically ε > 1000 M−1cm−1, producing vivid colors.
Two types of charge-transfer transitions are commonly observed in coordination compounds: LMCT and MLCT. Which of the following correctly distinguishes them with appropriate examples?
In LMCT (ligand-to-metal charge transfer), electron density moves from a filled ligand orbital to an empty or partially filled metal orbital. This is common with high-oxidation-state metals and easily oxidized ligands (e.g., MnO4−, CrO42−). In MLCT (metal-to-ligand charge transfer), electron density moves from a filled metal d orbital to an empty ligand π* orbital. MLCT is favored by low-oxidation-state metals and ligands with low-lying π* orbitals, such as bipyridine in [Ru(bpy)3]2+.
KMnO4 is intensely purple despite Mn being in the +7 oxidation state (d0 configuration). Since there are no d electrons, d–d transitions are impossible. What is the origin of its color?
In MnO4−, Mn(VII) has a d0 configuration, so no d–d transitions are possible. The intense purple color arises from LMCT transitions: electrons are transferred from the filled oxygen 2p orbitals (ligand) to the empty Mn 3d orbitals (metal). Because this is a parity-allowed transition, the molar absorptivity is very high (ε > 2000 M−1cm−1), producing the characteristic deep purple color even at low concentrations.
The figure above shows a Tanabe–Sugano diagram. What information does this diagram convey, and how is it used to interpret UV-Vis spectra of metal complexes?
A Tanabe–Sugano diagram plots the energies of all electronic states arising from a given dn configuration as a function of the crystal field strength ratio Δoct/B, where B is the Racah interelectronic repulsion parameter. The ground state is taken as the energy zero (x-axis). By measuring the ratios of observed absorption band energies, one can locate the value of Δoct/B on the diagram, assign each band to a specific electronic transition, and extract both Δoct and B for the complex.
Magnetism
Questions 9–12
The spin-only magnetic moment formula is μs.o. = √(n(n+2)) BM, where n is the number of unpaired electrons. What is the predicted spin-only magnetic moment for a d5 high-spin octahedral complex?
In a d5 high-spin octahedral complex (e.g., [Mn(H2O)6]2+), all five d electrons occupy separate orbitals with parallel spins according to Hund's rule: t2g3 eg2. This gives n = 5 unpaired electrons. Applying the spin-only formula: μs.o. = √(5 × 7) = √35 = 5.92 BM. This is the maximum possible spin-only moment for any dn configuration.
A chemist synthesizes a new Co3+ complex and wants to determine whether it is high-spin or low-spin. What experimental measurement would most directly distinguish between the two cases?
Magnetic susceptibility measurements (e.g., using a Gouy balance or SQUID magnetometer) directly report the number of unpaired electrons. For Co3+ (d6), the high-spin configuration is t2g4 eg2 with 4 unpaired electrons (μs.o. = √(4×6) = 4.90 BM), whereas the low-spin configuration is t2g6 with 0 unpaired electrons (μ = 0, diamagnetic). The difference is unambiguous and immediately diagnostic.
Three principal types of magnetic behavior are observed in chemical substances: paramagnetism, diamagnetism, and ferromagnetism. Which description correctly distinguishes all three?
Diamagnetic substances have all electrons paired and are weakly repelled by an external magnetic field. Paramagnetic substances contain unpaired electrons that align with the field, causing weak attraction; this alignment is lost when the field is removed. Ferromagnetic materials (Fe, Co, Ni) also have unpaired electrons, but neighboring spins align cooperatively in domains due to exchange interactions, producing strong, permanent magnetization that persists even after the external field is removed.
The complex [Fe(H2O)6]2+ has an experimentally measured magnetic moment of μ ≈ 5.1 BM. Fe2+ is a d6 ion. Is this complex high-spin or low-spin, and how many unpaired electrons does it have?
For Fe2+ (d6) in an octahedral field, the high-spin configuration is t2g4 eg2 with 4 unpaired electrons, giving μs.o. = √(4 × 6) = 4.90 BM. The low-spin configuration would be t2g6 with 0 unpaired electrons (μ = 0). The measured value of ~5.1 BM is close to the 4.90 BM prediction for 4 unpaired electrons, confirming the high-spin case. Water is a weak-field ligand, so Δoct is small and the high-spin configuration is favored.