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Describe the bonding in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\), using both crystal field theory and valence bond theory. Include the appropriate crystal field \(d\) -orbital energy-level diagram and the valence bond orbital diagram. Which model allows you to predict the number of unpaired electrons? How many do you expect?

Short Answer

Expert verified
No unpaired electrons, as predicted by crystal field theory.

Step by step solution

01

Understand the Coordination Complex

The coordination complex \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) consists of a manganese ion surrounded by six cyanide (CN鈦) ligands. The manganese ion has a charge of +3, which corresponds to a Mn(III) ion. This ion has an electronic configuration of \([\mathrm{Ar}] 3d^4\). Cyanide is known to be a strong field ligand.
02

Crystal Field Theory Analysis

According to Crystal Field Theory, the strong field CN鈦 ligands will cause a significant splitting of the \(d\)-orbitals in the manganese ion. In an octahedral field, the \(d\)-orbitals split into a lower-energy \(t_{2g}\) set and a higher-energy \(e_g\) set. Because CN鈦 is a strong field ligand, the splitting \(\Delta\) is large enough to allow the electrons to pair in the \(t_{2g}\) orbitals, resulting in the electron configuration \(t_{2g}^{4} e_{g}^{0}\).
03

Construct the Crystal Field Diagram

In the crystal field diagram for \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\), all four electrons occupy the lower-energy \(t_{2g}\) orbitals. The \(e_{g}\) orbitals remain empty. This configuration suggests that the complex is low-spin since the electrons pair up due to the strong field influence of CN鈦.
04

Valence Bond Theory Analysis

Valence Bond Theory considers hybridization to explain bonding. For \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\), manganese undergoes \(d^2sp^3\) hybridization to form the six equivalent hybrid orbitals, which overlap with CN鈦 ligand orbitals. The electron pairing in the \(d\)-orbitals aligns with the low-spin configuration suggested by the crystal field theory.
05

Construct the Valence Bond Orbital Diagram

In the valence bond diagram, the \(d^2sp^3\) hybrid orbitals form through the mixing of 3\(d\), 4\(s\), and 4\(p\) orbitals, and they accommodate the six ligand pairs provided by the cyanide ions. These orbitals are filled according to pairing, as determined by crystal field considerations.
06

Predict Number of Unpaired Electrons

Combining insights from both theories, especially focusing on the crystal field low-spin configuration, we deduce there are no unpaired electrons in \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\). The model that best predicts the number of unpaired electrons here is crystal field theory.

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Key Concepts

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

Crystal Field Theory
Crystal Field Theory (CFT) is a model that describes the electronic structure of coordination complexes. It focuses on the interaction between the central metal ion and the surrounding ligands. In \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\), CFT is used to explain how the presence of ligands affects the distribution of d-orbital electrons.
According to CFT, the ligands create an electrostatic field around the metal ion. This field causes a splitting of the metal's d-orbitals into two groups with different energies. In the case of \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\), which is an octahedral complex, these are the lower-energy \([t_{2g}]\) orbitals and the higher-energy \([e_g]\) orbitals.
The strength of the ligand, known as a strong field or weak field ligand, determines the extent of d-orbital splitting. Cyanide (CN鈦) is a strong field ligand, leading to a large gap between the \([t_{2g}]\) and \([e_g]\) orbital sets. This causes the electrons to pair within the lower \([t_{2g}]\) set, leading to a low-spin configuration.
Valence Bond Theory
Valence Bond Theory (VBT) is another model to explain the bonding in coordination complexes. VBT examines how hybridization of atomic orbitals occurs to allow bonding with ligands.
In \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\), VBT proposes that manganese undergoes \(d^2sp^3\) hybridization. This means that one \(s\), three \(p\), and two \(d\) orbitals from manganese mix to create six equivalent hybrid orbitals.
These hybrid orbitals overlap with the orbitals of the cyanide ligands. This leads to the formation of sigma (蟽) bonds between the metal ion and the ligands. The pairing of electrons in these hybrid orbitals aligns with the low-spin prediction from crystal field theory. Thus, VBT provides a comprehensive picture of orbital involvement in the formation of the coordination complex.
d-Orbital Splitting
The concept of d-orbital splitting is crucial in understanding the electronic structure of coordination complexes. It refers to how the degeneracy of d-orbitals is broken due to the presence of ligands.
For the \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\) complex, we have an octahedral arrangement where the \(t_{2g}\) orbitals (dxy, dxz, dyz) lie lower in energy, and the \(e_g\) orbitals (dz^2, dx^2-y^2) lie higher.
This splitting is represented by the crystal field splitting parameter \(\Delta\). Strong field ligands like CN鈦 cause a large \(\Delta\), which often results in low-spin complexes as electrons pair in the lower \(t_{2g}\) orbitals first.
Hybridization
Hybridization in coordination complexes refers to the combination of atomic orbitals to form new, equivalent orbitals for bonding with ligands.
In the \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\) complex, manganese performs \(d^2sp^3\) hybridization. This process allows manganese to form six equivalent hybrid orbitals that are capable of bonding with cyanide ligands.
Hybridization provides a clear view of the geometric arrangement of orbitals around the metal ion. It helps describe the sigma bond framework and can complement crystal field theory by explaining the hybrid orbitals' role in bonding.
Unpaired Electrons
Unpaired electrons are electrons that reside in an atomic or molecular orbital singly, without a matching paired electron. They carry a magnetic moment, making substances with unpaired electrons paramagnetic.
In the context of coordination complexes like \([\mathrm{Mn}(\mathrm{CN})_{6}]^{3-}\), the presence or absence of unpaired electrons is determined by the splitting of d-orbitals and the nature of the ligands.
Due to the strong field strength of the cyanide ligands, electron pairing occurs in the \(t_{2g}\) orbital, resulting in no unpaired electrons in this complex. The crystal field theory provides a reliable prediction of no unpaired electrons due to the low-spin configuration of the complex.

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Most popular questions from this chapter

Cobalt(III) trifluoroacetylacetonate, \(\mathrm{Co}(\mathrm{tfac})_{3}\), is a six-coordinate, octahedral metal chelate in which three planar, bidentate tfac ligands are attached to a central Co atom: (a) Draw all possible diastereoisomers and enantiomers of \(\mathrm{Co}(\mathrm{tfac})_{3-}\) (b) Diastereoisomers \(\mathrm{A}\) and \(\mathrm{B}\) have dipole moments of \(6.5 \mathrm{D}\) and \(3.8 \mathrm{D}\), respectively. Which of your diastereoisomers is \(\mathrm{A}\) and which is \(\underline{B}\) ? (c) The isomerization reaction \(A \longrightarrow B\) in chloroform solution has first-order rate constants of \(0.0889 \mathrm{~h}^{-1}\) at \(66.1^{\circ} \mathrm{C}\) and \(0.0870 \mathrm{~min}^{-1}\) at \(99.2^{\circ} \mathrm{C}\). What is the activation energy for the reaction? (d) Draw a crystal field energy-level diagram for \(\mathrm{Co}(\mathrm{tfac})_{3}\), and predict its magnetic properties. (In this complex, tfac is a strong-field ligand.)

Give a valence bond description of the bonding in each of the following complexes. Include orbital diagrams for the free metal ion and the metal ion in the complex. Indicate which hybrid orbitals the metal ion uses for bonding, and specify the number of unpaired electrons. (a) \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{NiBr}_{4}\right]^{2-}\) (tetrahedral) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) (low-spin) (d) \(\left[\mathrm{MnCl}_{6}\right]^{3-}\) (high-spin)

What is the formula, including the charge, for each of the following complexes? (a) An iron(III) complex with six water ligands (b) A nickel(II) complex with two ethylenediamine and two bromide ligands (c) A platinum(II) complex with two chloride and two ammonia ligands

The \(\mathrm{Ni}^{2+}(a q)\) cation is green, but \(\mathrm{Zn}^{2+}(a q)\) is colorless. Explain.

On the periodic table below, locate the transition metal atom or ion with the following electron configurations. Identify each atom or ion. (a) An atom: \([\mathrm{Ar}] 3 d^{5} 4 s^{2}\) (b) \(\mathrm{A} 2+\) ion: \([\mathrm{Ar}] 3 d^{8}\) (c) An atom: \([\mathrm{Kr}] 4 d^{10} 5 s^{1}\) (d) \(\mathrm{A} 3+\) ion: \([\mathrm{Kr}] 4 d^{3}\)

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