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Chapter 23: Problem 82

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}(\) chromate ion \(),\) and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of \(\mathrm{MnO}_{4}^{-}\) is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of \(565 \mathrm{nm}\). The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-}\) ? Explain. (e) The \(\mathrm{VO}_{4}{ }^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

Short Answer

Expert verified
The anions \(\mathrm{VO}_{4}^{3-}\), \(\mathrm{CrO}_{4}^{2-}\), and \(\mathrm{MnO}_{4}^{-}\) are isoelectronic, meaning they have the same number of electrons and similar electronic configurations. They do not exhibit \(d-d\) transitions due to empty \(3d\) orbitals in their central metal ions. Ligand-to-metal charge transfer (LMCT) transitions involve electron transfer from a ligand to a metal center, affecting the color of the complex. The wavelength of the LMCT transition for \(\mathrm{CrO}_{4}^{2-}\) (yellow) is larger than that for \(\mathrm{MnO}_{4}^{-}\) (violet). The colorless \(\mathrm{VO}_{4}{ }^{3-}\) ion likely absorbs light in the UV region of the electromagnetic spectrum due to the nature of LMCT transitions.

Step by step solution

01

(a) Meaning of Isoelectronic

Isoelectronic means that the species have the same number of electrons and the same electronic structure. In the case of the mentioned anions: \(\mathrm{VO}_{4}^{3-}\), \(\mathrm{CrO}_{4}^{2-}\), and \(\mathrm{MnO}_{4}^{-}\), they all have 34 electrons and similar electronic configurations. This is because they are formed by metal ions from the same period of the periodic table and have the same number of oxygen atoms in their structure.
02

(b) Expectation for d-d Transitions

We would not expect these anions to exhibit \(d-d\) transitions because their central metal ions (\(\mathrm{V}^{5+}\), \(\mathrm{Cr}^{6+}\), and \(\mathrm{Mn}^{7+}\)) have a completely empty \(3d\) orbital due to their high oxidation states. \(d-d\) transitions require an electron to be promoted from one \(d\) orbital to another within the same energy shell, which is not possible when there are no electrons in the \(d\) orbitals.
03

(c) Definition of LMCT Transition

A Ligand-to-Metal Charge Transfer (LMCT) transition refers to an electronic transition where an electron is promoted from a ligand's occupied orbital to an empty metal orbital. It is a type of charge transfer process where electron density is transferred from the ligand to the metal center, which can result in changes in the color or other spectroscopic properties of the complex.
04

(d) Wavelength Comparison of LMCT Transitions for \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{CrO}_{4}^{2-}\)

The wavelength of the LMCT transition for \(\mathrm{MnO}_{4}^{-}\) is given as 565 nm, which corresponds to violet color. The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. According to the visible light spectrum, yellow light has a longer wavelength than violet light. Therefore, the wavelength of the LMCT transition for chromate would be larger than that for permanganate.
05

(e) UV or IR Region for \(\mathrm{VO}_{4}{ }^{3-}\)'s Absorbed Light

The \(\mathrm{VO}_{4}{ }^{3-}\) ion is colorless, which means that the light absorbed by the LMCT transition in this ion must not be in the visible region of the electromagnetic spectrum. As LMCT transitions typically involve promoting an electron from the ligand's occupied orbital to an empty metal orbital, it is reasonable to expect that this absorption would occur in the ultraviolet (UV) region, since UV light has higher energy than infrared (IR) light. Therefore, the absorbed light would likely fall in the UV region of the electromagnetic spectrum.

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Most popular questions from this chapter

The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

The molecule methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at \(298 \mathrm{~K}\) for reactions of methylamine and en with \(\mathrm{Cd}^{2+}(a q):\) $$ \begin{array}{c} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \\ \Delta H^{\circ}=-57.3 \mathrm{~kJ} ; \quad \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-37.2 \mathrm{~kJ} \\\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=-56.5 \mathrm{~kJ} ; \quad \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-60.7 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons\) $$ \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) $$ Based on the value of \(K\) in part (a), what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic \(\left(\Delta H^{\circ}\right)\) and the entropic \(\left(-T \Delta S^{\circ}\right)\) contributions to \(\Delta G^{\circ}\) for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the "A Closer Look" box on the chelate effect, predict the sign of \(\Delta H^{\circ}\) for the following hypothetical reaction: $$ \begin{aligned} \left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) &+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \end{aligned} $$

Consider an octahedral complex \(\mathrm{MA}_{3} \mathrm{~B}_{3} .\) How many geometric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones?

The red color of ruby is due to the presence of Cr(III) ions at octahedral sites in the close-packed oxide lattice of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Draw the crystal-field splitting diagram for Cr(III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$
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