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From experiment, we know that \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is paramagnetic and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is diamagnetic. Using the ligand field model, depict the electron configuration for each ion, and use this model to explain the magnetic property. What can you conclude about the effect of these ligands on the magnitude of \(\Delta_{0} ?\)

Short Answer

Expert verified
F\(^{-}\) causes low \\(\Delta_0\\), leading to high spin, while \(\text{NH}_3\) causes high \\(\Delta_0\\), resulting in low spin.

Step by step solution

01

Determine Electron Configuration for Co

Start by identifying the electronic configuration of the ground state cobalt atom. Cobalt (Co) in its ground state has the electron configuration \[ ext{Co: [Ar] } 3d^7 4s^2 \]. For both complexes, cobalt is in the +3 oxidation state, meaning it loses 3 electrons. Electrons are removed from the 4s orbital first, then the 3d orbital. Thus, Co\(^{3+}\) has the configuration \[ ext{Co}^{3+}: [Ar] 3d^6 \].
02

Analyze the Ligand Field of [CoF6]3-

In the \([\text{CoF}_6]^{3-}\) complex, the F\(^{-}\) ligands are weak field ligands. They cause a small splitting of the d-orbitals (i.e., low \(\Delta_0\)). Hence, the electrons will occupy the orbitals in such a way as to minimize pairing energy. Thus, the configuration will be \[ t_{2g}^4 e_g^2 \,\] which is consistent with a high spin configuration, resulting in unpaired electrons and therefore a paramagnetic complex.
03

Analyze the Ligand Field of [Co(NH3)6]3+

In the \( [\text{Co(NH}_3)_6]^{3+} \) complex, the \(\text{NH}_3\) ligands are strong field ligands that create a large splitting of the d-orbitals (i.e., high \(\Delta_0\)). In such a situation, the energy cost of pairing electrons is less than the stabilization energy of occupying the lower energy d-orbitals. Hence, the electron configuration is \[ t_{2g}^6 e_g^0 \,\] leading to all electrons being paired and resulting in a diamagnetic complex.
04

Conclusion on Ligand Impact on Δ₀

The paramagnetic nature of \([\text{CoF}_6]^{3-}\) and the diamagnetic nature of \( [\text{Co(NH}_3)_6]^{3+} \) reflects the influence of ligand strength on \(\Delta_0\). Weak field ligands like \(\text{F}^-\) result in low \(\Delta_0\) and a high spin configuration, while strong field ligands like \(\text{NH}_3\) produce high \(\Delta_0\) leading to a low spin configuration.

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Key Concepts

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

Paramagnetic
Paramagnetism is a magnetic property observed in substances with unpaired electrons. When these unpaired electrons are subjected to an external magnetic field, they tend to align with it, causing the material to be attracted to the field. This property is seen in transition metal complexes like \( [\text{CoF}_6]^{3-} \). In this complex, cobalt is surrounded by fluoride ions, which are considered weak field ligands. As a result, the electrons in the d-orbitals do not completely pair up, leading to unpaired electrons.
This unpaired nature of electrons is what makes \( [\text{CoF}_6]^{3-} \) paramagnetic. These unpaired electrons create a net magnetic moment, allowing the complex to be attracted to an external magnetic field.
  • Unpaired electrons are present.
  • Attracted to magnetic fields.
  • High spin configurations.
Diamagnetic
In contrast to paramagnetism, diamagnetism is a property where all the electrons in a substance are paired. Substances exhibiting diamagnetism are repelled by external magnetic fields. This occurs because the paired electrons have opposite spins, effectively canceling out magnetic moments.
The complex \( [\text{Co(NH}_3)_6]^{3+} \) is an example of a diamagnetic substance. Here, cobalt is surrounded by ammonia ligands, which are strong field ligands. They cause a large splitting of the d-orbitals (\( \Delta_0 \)), encouraging the pairing of electrons.
As a result, all of the d-orbital electrons pair up, leading to no net magnetic moment and hence, a diamagnetic property.
  • All electrons are paired.
  • Repelled by magnetic fields.
  • Low spin configurations.
Electron Configuration
Electron configuration is a way of representing how electrons are distributed in an atom's orbitals. For transition metals, these configurations are crucial in determining their chemical and magnetic properties.
In the given problem, the electron configuration of cobalt in its +3 oxidation state is key. Cobalt initially has the electron configuration \([\text{Ar}] 3d^7 4s^2\). When it forms a +3 ion, it loses three electrons to become \([\text{Ar}] 3d^6\).
The electron configuration can further be split into the \(t_{2g}\) and \(e_g\) orbitals when considering the ligand field theory. For \([\text{CoF}_6]^{3-}\), it results in \(t_{2g}^4 e_g^2\) configuration, and for \([\text{Co(NH}_3)_6]^{3+}\), it results in \(t_{2g}^6 e_g^0\).
  • Important for understanding chemical behavior.
  • Affected by oxidation state.
  • Influences magnetic properties.
Magnetic Properties
The magnetic properties of a substance depend largely on its electron configuration and the type of ligands attached to the metal center. In transition metal complexes, the arrangement of electrons in the d-orbitals determines whether a complex is paramagnetic or diamagnetic.
By analyzing the electron distribution in the d-orbitals, one can predict the magnetic behavior. For example, a complex with unpaired electrons in its electron configuration will typically exhibit paramagnetic properties. Conversely, if all electrons are paired, as seen with strong field ligands like \( NH_3 \), the substance becomes diamagnetic.
These properties are part of what makes ligand field theory useful in predicting and explaining the characteristics of transition metal complexes.
  • Determined by d-orbital electron distribution.
  • Influenced by strength of ligands (weak vs. strong field).
  • Predictable via electron configuration.
Transition Metal Complexes
Transition metal complexes consist of a central metal ion bonded to a set of molecules or ions called ligands. These complexes are crucial in coordination chemistry, with their properties often determined by the metal's d-electron configuration and the nature of the ligands.
The ligands can be categorized based on their field strength: weak field ligands (such as \( F^-\)) result in smaller d-orbital splitting, whereas strong field ligands (like \( NH_3 \)) lead to greater splitting. This affects electron pairing and the resulting magnetic properties.
Ligand field theory helps explain how different ligands influence electronic structures and properties, such as paramagnetism or diamagnetism. Understanding these complexes is vital because they are involved in many biological processes and industrial applications.
  • Central metal ion surrounded by ligands.
  • Properties influenced by ligand type and field strength.
  • Applications in various fields.

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Most popular questions from this chapter

How many unpaired electrons are expected for high-spin and low-spin complexes of \(\mathrm{Fe}^{2+} ?\)

Experiments show that \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) is paramagnetic and has two unpaired electrons. The related complex \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{SCN})_{6}\right]\) is paramagnetic and has four unpaired electrons. Account for the magnetism of each compound using the ligand field model. Predict where the SCN - ion occurs in the spectrochemical series relative to CN \(^{-}\).

Give the formula and name of a square-planar complex of \(\mathrm{Pt}^{2+}\) with one nitrite ion $$\left(\mathrm{NO}_{2}^{-}\right.$$, which binds to \(\mathrm{Pt}^{2+}\) through \(\mathrm{N}\) ), one chloride ion, and two ammonia molecules as ligands. Are isomers possible? If so, draw the structure of each isomer, and tell what type of isomerism is observed.

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: \(\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) \quad K_{\mathrm{f}}=10^{8}\) \(\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) \(K_{f}=10^{18}\) The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\) \(\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{t} H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) and \(\Delta_{i} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} .\) Is the difference in \(\Delta_{r} H^{\circ}\) suffi- cient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

Give the oxidation number of the metal ion in each of the following compounds. (a) \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{SO}_{4}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (d) \(\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\)

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