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The complex \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) has five unpaired electrons, whereas \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}\) has only one. Using the ligand field model, depict the electron configuration for each ion. What can you conclude about the effects of the different ligands on the magnitude of \(\Delta_{0} ?\)

Short Answer

Expert verified
\( \mathrm{CN}^{-} \) causes a larger \( \Delta_{0} \) leading to electron pairing, unlike \( \mathrm{H}_2\mathrm{O} \).

Step by step solution

01

Identify the oxidation state of Mn

In both complexes, the manganese ion is the central metal atom. In \( [\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} \), the oxidation state of Mn is determined by the fact that each \( \mathrm{H}_2\mathrm{O} \) is neutral. Thus, Mn's charge is \( +2 \). For \( [\mathrm{Mn}(\mathrm{CN})_{6}]^{4-} \), each \( \mathrm{CN}^{-} \) ligand is \( -1 \), making the total charge \( -6 \). Therefore, Mn must be in the \( +2 \) oxidation state (since \( 6 \times (-1) + x = -4 \) gives \( x = +2 \)).
02

Determine the d-electron count of Mn

Manganese in the \( +2 \) state corresponds to the electron configuration \([\mathrm{Ar}] 3d^{5}\), since Mn is \([\mathrm{Ar}] 3d^{5} 4s^{2}\) in its neutral state and has lost two electrons.
03

Electron configuration for \( [\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} \)

\( \mathrm{H}_2\mathrm{O} \) is a weak field ligand, leading to a small \( \Delta_{0} \), the splitting between \( t_{2g} \) and \( e_{g} \) orbitals. Electrons prefer to occupy higher energy orbitals to minimize pairing. Thus, we have 5 unpaired electrons: \( t_{2g}^{3} e_{g}^{2} \).
04

Electron configuration for \( [\mathrm{Mn}(\mathrm{CN})_{6}]^{4-} \)

\( \mathrm{CN}^{-} \) is a strong field ligand, causing a larger \( \Delta_{0} \). Electrons pair in lower \( t_{2g} \) orbitals to minimize energy. The configuration is \( t_{2g}^{5} e_{g}^{0} \), with one unpaired electron.
05

Conclusion on the effect of ligands

\( [\mathrm{Mn}(\mathrm{CN})_{6}]^{4-} \) has a larger \( \Delta_{0} \) due to \( \mathrm{CN}^{-} \)'s strong field and ability to cause pairing. Meanwhile, in \( [\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} \), the weaker field of \( \mathrm{H}_2\mathrm{O} \) results in a small \( \Delta_{0} \), allowing more unpaired electrons. Thus, ligand strength affects electron pairing and configuration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Crystal Field Splitting
Crystal field splitting is a fundamental concept in Ligand Field Theory which helps us understand the electron configuration in transition metal complexes. In a coordinate complex, ligands, which are ions or molecules attached to a central metal atom, influence the energy levels of the d-orbitals in the metal.
When ligands approach the metal ion, their electric fields interact with the d-orbitals of the metal ion, causing these orbitals to split into two energy levels: the lower-energy set known as "t_{2g}" and the higher-energy set known as "e_{g}".
The energy difference between these two levels is termed as the "crystal field splitting energy," denoted by \(\Delta_0\).
  • Strong-field ligands, like \( \text{CN}^- \), cause a large \( \Delta_0 \) resulting in reduced unpaired electrons as the energy gap encourages electron pairing in the lower-energy orbitals.
  • Weak-field ligands, like \( \text{H}_2\text{O} \), result in a smaller \( \Delta_0 \), allowing electrons to spread out across both higher and lower energy levels, leading to more unpaired electrons.
Electron Configuration
Electron configuration describes the distribution of electrons in an atom or ion's orbitals. For transition metal complexes, the arrangement of electrons within the d-orbitals is particularly important as it determines the magnetic and spectroscopic properties of the complex.
In the ligand field model, electron configuration is influenced by the crystal field splitting. Electrons will first fill the lower-energy "t_{2g}" orbitals before moving to the higher-energy "e_{g}" orbitals. Strong-field ligands enforce a larger \( \Delta_0 \), forcing electrons to pair up in the "t_{2g}" orbitals, minimizing the overall number of unpaired electrons.
In the complex \([\text{Mn}({\text{H}_2\text{O}})_{6}]^{2+}\), \(\text{H}_2\text{O}\) being a weak-field ligand leads to an electron configuration of \(t_{2g}^{3} e_{g}^{2}\), resulting in five unpaired electrons. Conversely, in \([\text{Mn}(\text{CN})_{6}]^{4-}\), the strong field \( \text{CN}^- \) ligand leads to an electron configuration of \(t_{2g}^{5} e_{g}^{0}\), with only one unpaired electron.
Transition Metal Complexes
Transition metal complexes are formed when transition metal ions interact with ligands through coordinate covalent bonds. These complexes possess unique electronic structures owing to their partially filled d-orbitals, which are significantly influenced by the accompanying ligands.
The nature of the ligand significantly affects the properties of the complex. Strong-field ligands facilitate higher pairing within the d-orbitals by increasing \( \Delta_0 \) and usually result in low-spin configurations. Weak-field ligands, on the other hand, lead to high-spin configurations due to a smaller \( \Delta_0 \), resulting in more unpaired electrons.
  • In \([\text{Mn}({\text{H}_2\text{O}})_6]^{2+}\), the high-spin complex shows clearly unpaired electrons due to \( \text{H}_2\text{O} \)'s weak-field influence.
  • For \([\text{Mn}(\text{CN})_{6}]^{4-}\), the low-spin complex demonstrates mostly paired electrons, a result of the strong-field nature of \( \text{CN}^- \).
Transition metal complexes thus showcase diverse chemical behaviors, extensively utilized in catalysis, materials science, and bioinorganic chemistry.

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Most popular questions from this chapter

As mentioned on page \(1047,\) transition metal organometallic compounds have found use as catalysts. One example is Wilkinson's catalyst, a rhodium compound \(\left[\mathrm{RhCl}\left(\mathrm{PR}_{3}\right)_{3}\right]\) used in the hydrogenation of alkenes. The steps involved in the catalytic process are outlined below. Indicate whether the rhodium compounds in each step have 18 - or 16 -valence electrons. (See Study Question \(34 .)\) Step \(1 .\) Addition of \(\mathrm{H}_{2}\) to the rhodium center of Wilkinson's catalyst. (For electron-counting purposes \(\mathrm{H}\) is considered a hydride ion, \(\left.\mathrm{H}^{-}, \text {a two-electron donor. }\right)\) Step \(2 .\) Loss of a PR \(_{3}\) ligand (a two-electron donor) to open a coordination site. (PR \(_{3}\) is a phosphine such as \(\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) triphenylphosphine.) Step \(3 .\) Addition of the alkene to the open site. Step 4. Rearrangement to add H to the double bond. (Here the \(-\mathrm{CH}_{2} \mathrm{CH}_{3}\) group is a two-electron donor and can be thought of as a \(\left[\mathrm{CH}_{2} \mathrm{CH}_{3}\right]^{-}\) anion for electron counting purposes.) Step \(5 .\) Loss of the alkane. Step \(6 .\) Regeneration of the catalyst. $$\text { Net reaction: } \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}$$

In water, the titanium(III) ion, \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\) has a broad absorption band centered at about \(500 \mathrm{nm}\). What color light is absorbed by the ion?

Determine whether the following complexes have a chiral metal center. (a) \(\left[\mathrm{Fe}(\mathrm{en})_{3}\right]^{2+}\) (b) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Br}_{2}\right]^{+}\) (c) fac-\(\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right]\) (d) square-planar \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{Cl})\left(\mathrm{NO}_{2}\right)\)

Name the following ions or compounds. (a) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right) \mathrm{Br}_{3}\right]^{-}\) (b) \(\operatorname{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{F}_{3}\) (d) \(\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{2+}\)

The following equations represent various ways of obtaining transition metals from their compounds. Balance each equation. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\) (b) \(\operatorname{Ti} \mathrm{Cl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s})\) (c) \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow\) \(\mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})\) (d) \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\)

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