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Chapter 20: Problem 25

Compounds containing the \(\mathrm{Sc}^{3+}\) ion are colorless, whereas those containing the \(\mathrm{Ti}^{3+}\) ion are colored. Explain.

Short Answer

Expert verified
Compounds of \(Sc^{3+}\) are colorless because it has a full electron configuration with no available unpaired electrons. On the other hand, \(Ti^{3+}\) compounds are colored because it has one unpaired electron, which can be excited to a higher energy level by absorbing the energy of visible light.

Step by step solution

01

Coloration Theory

Color in compounds originates from the absorption of light of a specific wavelength. This absorption happens when the energy of the incoming light excites an electron to a higher energy level. The specific color observed depends on the wavelength of light absorbed, which is determined by the difference in energy between the electron's initial and final states. If there are no available energy transitions that match the energy of visible light, then the compound or solution will appear colorless to the human eye.
02

Electron Configurations

The electron configurations of each ion is crucial in understanding their color properties. \(Sc^{3+}\) has a configuration of [Ar] with no d or f electrons. On the other hand, \(Ti^{3+}\) has a configuration of [Ar] 3d1, meaning it has one unpaired d electron.
03

Explanation of Coloration

In compounds of \(Sc^{3+}\), there are no available unpaired electrons able to be excited to a higher energy level by visible light's energy, thus no color is observed. However, in compounds of \(Ti^{3+}\), the unpaired d electron can be excited to a higher energy level with the energy provided by visible light, thus causing color absorption and giving the compound its color.

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Most popular questions from this chapter

What are the systematic names for these ions and compounds: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+},\) (b) \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3},\) (c) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Br}_{2}\right]^{+},\) (d) \(\mathrm{Fe}(\mathrm{CO})_{5} ?\)

Give the oxidation numbers of the metals in these species: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) (c) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\)

The complex ion \(\left[\mathrm{Ni}(\mathrm{CN})_{2} \mathrm{Br}_{2}\right]^{2-}\) has a square planar geometry. Draw the structures of the geometric isomers of this complex.

Predict the number of unpaired electrons in these complex ions: (a) \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}(\mathrm{b})\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Draw structures of all the geometric and optical isomers of each of these cobalt complexes: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\) (e) \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\)
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