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A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahedral complex. Name some possible metal ions for which this would be true.

Short Answer

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Possible metal ions for which high-spin octahedral complexes have two more unpaired electrons than low-spin octahedral complexes are Cr(III) with a d4 electron configuration and Fe(II) with a d6 electron configuration.

Step by step solution

01

Understanding high-spin and low-spin complexes

High-spin and low-spin complexes refer to the electronic configuration of metal ions in coordination complexes. The difference between them lies in the pairing of electrons in the d orbitals of the metal ion. For high-spin complexes, the electrons preferentially occupy different d orbitals, maximizing unpaired electrons. On the other hand, in low-spin complexes, the electrons are more likely to pair up within the same d orbital, minimizing unpaired electrons. This distinction occurs due to the difference in ligands' crystal field splitting energy (CFSE).
02

Determining electron configuration

In an octahedral complex, the crystal field splitting causes the energy levels to split into two sets of orbitals: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). In low-spin complexes, the CFSE is large, meaning the energy difference between the two sets of orbitals is significant, and it is more favorable for electrons to pair up in the \(t_{2g}\) orbitals before occupying the higher energy \(e_g\) orbitals. In high-spin complexes, the CFSE is small, and electrons will occupy all the \(t_{2g}\) orbitals with parallel spins before pairing up in the same orbital or occupying the \(e_g\) orbitals.
03

Identifying possible metal ions

To find possible metal ions for which high-spin octahedral complexes have two more unpaired electrons than low-spin octahedral complexes, we will examine the d-electron configurations of various metal ions and determine if their electron configurations meet the criteria. 1. d4: In a high-spin octahedral complex, all four electrons would occupy the \(t_{2g}\) orbitals (3 unpaired electrons). In a low-spin octahedral complex, two electrons would pair up in the \(t_{2g}\) orbitals, leaving two unpaired electrons. 2. d6: In a high-spin octahedral complex, the 6 electrons would occupy all the \(t_{2g}\) orbitals and one of the \(e_g\) orbitals (4 unpaired electrons). In a low-spin octahedral complex, the \(t_{2g}\) orbitals would be fully occupied by paired electrons, leaving no unpaired electrons. From the d4 and d6 configurations, we can see that they meet the criteria outlined in the exercise, having two more unpaired electrons in their high-spin octahedral complexes than in their low-spin counterparts.
04

Possible metal ions

Considering the electron configurations that meet the given criteria, the following metal ions can be used: 1. Cr(III) (d4 electron configuration) 2. Fe(II) (d6 electron configuration) These metal ions would exhibit the desired difference in unpaired electron count when forming high-spin and low-spin octahedral complexes.

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Most popular questions from this chapter

a. In the absorption spectrum of the complex ion [Cr(NCS) \(\left._{6}\right]^{3-}\), there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{~cm}^{-1}\). Given \(1 \mathrm{~cm}^{-1}=\) \(1.986 \times 10^{-23} \mathrm{~J}\), what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) is predicted to be \(180^{\circ}\). What is the hybridization of the \(\mathrm{N}\) atom in the \(\mathrm{NCS}^{-}\) ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ} \mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}\) undergoes substitution by ethylenediammine (en) according to the equation \(\left[\mathrm{Cr}(\mathrm{NCS})_{6}\right]^{3-}+2 \mathrm{en} \longrightarrow\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}+4 \mathrm{NCS}^{-}\) Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit geometric isomerism? Does \(\left[\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}\right]^{+}\) exhibit optical isomerism?

There are three salts that contain complex ions of chromium and have the molecular formula \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\). Treating \(0.27 \mathrm{~g}\) of the first salt with a strong dehydrating agent resulted in a mass loss of \(0.036 \mathrm{~g}\). Treating \(270 \mathrm{mg}\) of the second salt with the same dehydrating agent resulted in a mass loss of \(18 \mathrm{mg}\). The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to \(100.0-\mathrm{mL}\) portions of \(0.100 M\) solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 \(\mathrm{mg} \mathrm{AgCl} ;\) another, \(2870 \mathrm{mg} \mathrm{AgCl}\); the third, \(4300 \mathrm{mg} \mathrm{AgCl}\). Two of the salts are green and one is violet. Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

Name the following coordination compounds. a. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\) c. \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\) b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

How many bonds could each of the following chelating ligands form with a metal ion? a. acetylacetone (acacH), a common ligand in organometallic catalysts: b. diethylenetriamine, used in a variety of industrial processes: c. salen, a common ligand for chiral organometallic catalysts: d. porphine, often used in supermolecular chemistry as well as catalysis; biologically, porphine is the basis for many different types of porphyrin- containing proteins, including heme proteins:

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 $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

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