Due to the absorption of a part of the light in the visible region of the spectrum, transition metal — Inorganic Chemistry Chemistry Question
Colours of complexes
Due to the absorption of a part of the light in the visible region of the spectrum, transition metal complexes are often coloured. In this problem, we shall focus on simple cases in which this absorption is caused by d-d transitions.
Titanium complex
Titanometry is a reductometric method that uses blue-violet aqueous solution of titanium(III) chloride. The colour of the solution is caused by the presence of octahedral particles [Ti(H2O)6]3+. The spectrum features absorption band with a maximum at 20 300 cm−1:
[VISUAL]
Draw the electron configuration of the ground and the excited state of the [Ti(H2O)6]3+ ion into the schemes. [VISUAL]
Model Answer
Ground state: t2g^1 eg^0 (one electron in the lower t2g orbital group). Excited state: t2g^0 eg^1 (one electron promoted to the upper eg orbital group).
Predict the colour of the complex. Consider the absorption of the light at 20 300 cm−1.
Model Answer
The wavenumber of 20 300 cm−1 corresponds to the wavelength of 493 nm which means the absorption of the blue-green light. The colour of the complex is the complementary one, i.e. orange-red.
In fact, there is a second absorption band in the spectrum. This band shows itself as a shoulder at 17 400 cm−1. Explain the colour of the complex based on the actual spectrum.
Model Answer
The complex absorbs visible light in the range from 493 to 575 nm, i.e. blue-green to yellow-green. The complex is purple.
The presence of two bands in the spectrum is caused by the fact that the [Ti(H2O)6]3+ particle is not a regular octahedron; it is rather an elongated octahedron. This elongation causes further splitting of the d-orbitals.
Draw the electron configuration of the ground (a) and the excited states (b) and (c) into the schemes. [VISUAL]
Model Answer
Ground state (a) shows the single d-electron in the lowest split d-orbital level. Excited states (b) and (c) represent the configuration when the electron is promoted to higher energy levels of the elongated octahedron splitting diagram.
Write the equations of reactions (1) to (3).
Model Answer
(1) 2 CoCl2 + 3 F2 → 2 CoF3 + 2 Cl2
(2) CoF3 + 3 KF → K3[CoF6]
(3) 4 CoF3 + 2 H2O → 4 HF + 4 CoF2 + O2
Write the equation of reaction (4).
Model Answer
(4) 4 CoCl2 + 4 NH4Cl + 20 NH3 + O2 → 4 [Co(NH3)6]Cl3 + 2 H2O
The common name of the [Co(NH3)6]Cl3 complex is luteochloride. It has two absorption bands at the wavenumbers of 21 050 cm−1 and 29 400 cm−1 in the near UV and visible part of the spectrum. Predict the colour of the complex. Find the relation between the colour and its common name.
Model Answer
The wavenumbers correspond to the wavelengths of 475 nm (blue light) and 340 nm (UV region). The second band has no effect on the observed colour and the complex is orange. Luteus means yellow in Latin (it refers to the yellow-orange colour of the complex).
Explain why the K3[CoF6] complex is high-spin and paramagnetic, while the [Co(NH3)6]Cl3 complex is low-spin and diamagnetic.
Model Answer
Due to their position in the spectrochemical series, fluoride ions (F−) cause only small splitting, which leads to a high-spin configuration with four unpaired electrons. Ammonia molecules (NH3) cause greater splitting, which means that all the electrons in the t2g orbitals pair up and a low-spin configuration is formed.
Draw the electron configuration of the ground state and excited states provided that the net spin of the particle is not changed when excited. [VISUAL]
Model Answer
Electron configurations correspond to high-spin d6 configuration of Co(III) within an elongated octahedral split environment.
The wavenumbers of the bands corresponding to these excitations are 11 400 cm−1 and 14 500 cm−1. Predict the colour of the [CoF6]3− ion.
Model Answer
The wavenumbers correspond to the wavelengths of 877 nm (IR region) and 690 nm (red light). The first band has no effect on the observed colour and the complex is blue-green.