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Chapter 21: Problem 15

Compounds of \(\mathrm{Sc}^{3+}\) are not colored, but those of \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) are. Why?

Short Answer

Expert verified
Compounds of Sc^3+ are colorless because they have no electrons in their 3d orbitals, preventing electron transitions between different energy levels and absorption of visible light. However, Ti^3+ and V^3+ compounds are colored because they have electrons in their 3d orbitals, allowing electron transitions between energy levels and absorption of visible light.

Step by step solution

01

Scandium (Sc)

Scandium has an atomic number of 21, and its electron configuration is \([Ar]3d^14s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving Sc a \([Ar]3d^0\) configuration.
02

Titanium (Ti)

Titanium has an atomic number of 22, so its electron configuration is \([Ar]3d^24s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving Ti a \([Ar]3d^1\) configuration.
03

Vanadium (V)

Vanadium has an atomic number of 23, and its electron configuration is \([Ar]3d^34s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving V a \([Ar]3d^2\) configuration.
04

Visualizing difference in energy levels between ions

In Sc^3+ ions, there are no electrons in the 3d orbital, which means there is no possibility of electron transitions between the lower energy levels and the higher energy levels of its d orbitals. Therefore, no visible light is absorbed, and Sc^3+ compounds appear colorless. However, Ti^3+ and V^3+ ions have electrons in their 3d orbitals, so electron transitions can occur between different energy levels of the 3d orbitals. These transitions absorb visible light, which results in colored compounds. In summary, compounds of Sc^3+ are colorless because there are no electrons in their 3d orbitals, while Ti^3+ and V^3+ compounds are colored due to electron transitions in their 3d orbitals.

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Most popular questions from this chapter

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

When concentrated hydrochloric acid is added to a red solution containing the \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex ion, the solution turns blue as the tetrahedral \(\mathrm{CoCl}_{4}{ }^{2-}\) complex ion forms. Explain this color change.

The complex ion Ru(phen) \({ }^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\mathrm{Ru}(\mathrm{phen})_{3}^{2+} ?\) b. Ru(phen) \(_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+}\) ). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

Acetylacetone (see Exercise 69, part a), abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac \(^{-}\), as shown below: Acetylacetone reacts with an ethanol solution containing a salt of europium to give a compound that is \(40.1 \% \mathrm{C}\) and \(4.71 \% \mathrm{H}\) by mass. Combustion of \(0.286 \mathrm{~g}\) of the compound gives \(0.112 \mathrm{~g}\) \(\mathrm{Eu}_{2} \mathrm{O}_{3}\). Assuming the compound contains only \(\mathrm{C}, \mathrm{H}, \mathrm{O}\), and \(\mathrm{Eu}\), determine the formula of the compound formed from the reaction of acetylacetone and the europium salt. (Assume that the compound contains one europium ion.)
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