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5.1 Samples of sulfates, MSO4.nH2O, with M = Cr, Mn, Fe, Co, Ni, and Cu, were found unlabelled in the laboratory. Would it be possible to identify them by their colours alone?
The stoichiometry indicates M2+ salts, and unless the chromium containing sample was packed and sealed very carefully in an air tight container it would have oxidised to Cr(III), which would be a deep green colour, rather than the pale blue expected for Cr(II). Mn(II) salts are very pale pink in colour. Fe(II) salts are usually pale yellow or blue-green in colour (depending on extent of hydration), but are susceptible to oxidation to Fe(III) which is also pale yellow. (If chloride salts are expected the colour change is from pale green to brown on oxidation.) Hydrated Co(II) compounds are a characteristic pink-red, hydrated nickel(II) salts are green and hydrated copper salts are deep blue. Therefore, the colours would be a good starting point, but would need confirming by other measurements.
5.2 The UV-vis spectrum of [Ti(H2O)6]3+ in Figure 5.10 has a broad asymmetric peak at 20000 cm−1 (500 nm). Use the spectrochemical series to predict the position of the bands in the UV-vis spectrum of the [Ti(NH3)6]3+ ion. The UV-vis spectra of [Ti(NCS)6]3− and [Ti(urea)6]3− shows absorptions at 544 nm and 570 nm, respectively. Place the [NCS]‑ and urea ligands in the spectrochemical series.
NH3 is a stronger field ligand in the spectrochemical series than H2O, therefore the d-d band in [Ti(NH3)6]3+ is expected to be at higher wavenumber than in [Ti(H2O)6]3+. However, it has recently been shown that colourless [Ti(NH3)8]3+ is the best characterised titanium ammine complex (P. Woidy, A. J. Karttunen, S. S. Rudel and F. Kraus, Chem. Commun. 51 11826 (2015)).
The absorptions in [Ti(NCS)6]3− and [Ti(urea)6]3− at 544 nm and 570 nm, respectively, indicate that both of these ligands lie on the weak field side of water in the spectrochemical series. However, in the spectrochemical series in Section 5.5.3, the ordering is given as H2O < NCS−, and this acts as a good reminder that the sequence was originally based on Co(III) complexes, and the ordering is only a guide to expected behaviour.
5.3 Heating bright blue CuSO4.5H2O to 700 °C produces an almost colourless solid. Estimate the position of the sulfate anion in the spectrochemical series based on this observation.
In CuSO4.5H2O the copper is coordinated by the water rather than the sulfate, but on heating the water is lost, and the copper is coordinated by the sulfate ions. In the former the blue colour is due to an absorption at 12500 cm−1 (Figure 5.10). As it becomes colourless on heating the absorption band must move to lower wavenumber so that the absorption band falls outside (below) the visible region. Therefore, SO42− is a weaker field ligand in the spectrochemical series than water. As anhydrous CuF2 is also white, this indicates that SO42− is in a similar position to F− in the spectrochemical series.
5.4 Use the UV-vis spectral data for the selection of Cr3+ complexes in Table 5.4P to determine Δoct in each case and put the ligands in order of Δoct. Explain the order you obtain.
|
|
|
|
[CrF6]3− |
14900 |
22700 |
34400 |
[CrCl6]3− |
13200 |
18700 |
|
[Cr(CN)6]3− |
26700 |
32600 |
|
[Cr(H2O)6]3+ |
17400 |
24600 |
37800 |
[Cr(NH3)6]3+ |
21550 |
28500 |
|
5.4P UV-vis data for some octahedral chromium complexes.
As these are all d3 complexes, Δoct can be obtained directly from the first spin-allowed d-d transition and the order is: Cl− (13200 cm−1) < F− (14900 cm−1) < H2O (17400 cm−1) < NH3 (21550 cm−1) < CN− (26700 cm−1). As expected, this is the order of the spectrochemical series in Section 5.5.3.
5.5 Two spin allowed transitions in [Fe(CN)6]4− are at 31000 and 37040 cm−1. Estimate Δoct for this complex.
An estimate of Δoct can be obtained by adding a quarter of the difference of the two transition energies to the lower one, giving 32510 cm−1. This is a large value of Δoct indicating that [Fe(CN)6]4− will be low spin.
5.6 Predict the differences in the LMCT spectra of the following groups of complexes.
(a) [VO4]3−, [CrO4]2−, [MnO4]−.
(b) [VO4]3−, [VS4]3−, [VSe4]3−.
(c) [CrO4]2−, [MoO4]2−.
The position of the LMCT bands in the spectra of these tetrahedral compounds is dependent on the energy gap between the full ligand based p orbitals (non-bonding in a σ only framework) and the empty metal based d orbitals as shown in Figure 5.6P
Figure 5.6P Schematic MO diagram for tetrahedral complexes showing ligand to metal charge transfer (LMCT) transitions.
(a) As the oxidation state increases from V(V) through Cr(VI) to Mn(VIII) the metal based d orbitals (1e and 2t2) will become lower in energy, and as the oxygen based p orbitals will not be so affected, the energy gap is smaller for the higher oxidation states. Therefore, the LMCT bands are at lower energy for the higher oxidation states. The alternative approach is that as the oxidation state increases, the metal becomes progressively easier to reduce, and therefore as the LMCT transition can be thought of as a transient reduction, the energy is lower for higher oxidation states.
The lowest energy LMCT bands are: [VO4]3− (36200 cm−1, 276 nm), [CrO4]2− (27000 cm−1, 370 nm), [MnO4]− (19000 cm−1, 526 nm)
(b) On going from oxide, through sulfide to selenide, the ligand based orbitals will increase in energy, and hence the energy gap reduces. Therefore, the LMCT transition will move to lower energy on going from oxide to sulfide to selenide. The lowest energy LMCT bands are: [VO4]3− (36200 cm−1, 276 nm), [VS4]3− (18600 cm−1, 538 nm), [VSe4]3- (15600 cm−1, 641 nm)
(c) On going from 3d Cr to 4d Mo the energy of the metal based d orbitals (e and t2) will increase, so the LMCT transition will move to higher energy, as a result whilst chromate is yellow, molybdate is colourless/white.
The lowest energy LMCT bands are: [CrO4]2− (27000 cm−1, 370 nm), [MoO4]2− (42700 cm−1, 234 nm).
5.7 The charge transfer transition in [IrCl6]3− is at 43100 cm−1 and for [IrBr6]3−at 36800 cm−1. Predict an optical electronegativity value for Ir(III) based on these data.
From Eqn 5.3 and Table 5.4 (χopt(Cl) = 3.0, χopt(Br) = 2.8)
43100 = 30000[3.0 - χopt(Ir)], ∴χopt(Ir) = 1.6
36800 = 30000[2.8 - χopt(Ir)], ∴χopt(Ir) = 1.6
5.8 Explain why many complexes of Ln2+ and Ln4+ are intensely coloured.
The pale colours of the Ln3+ complexes are due to the orbitally forbidden 4f ← 4f transitions as the 5d ← 4f transitions and charge transfer transitions tend to lie at high energy, beyond the visible region. The 4f ← 4f transitions in Ln2+ and Ln4+ will still be forbidden and therefore weak, so the intense colours must be due to the orbitally allowed 5d ← 4f and CT transitions moving into the visible part of the spectrum.
The intense colours of some Ce4+ compounds and complexes are due to LMCT transitions that occur in the visible part of the spectrum because the metal orbitals are held more tightly at lower energy in higher oxidation states, thus reducing the energy gap between the metal and ligand orbitals. The shift of the LMCT transitions for the Ln4+ ions into the visible part of the spectrum can also be explained using Eqn 5.3 as their optical electronegativities are significantly larger than those for Ln3+ ions.
The intense colour of some of the Ln2+ ions is because the 4fn−15d1 ← 4fn5d0 transitions are now in the visible part of the spectrum for the majority of the Ln2+ ions.