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The anion \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) is paramagnetic, but when \(\mathrm{CN}^{-}\) ions are added, the product, \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-},\) is diamagnetic. Explain this observation. $$\begin{array}{l} \left[\mathrm{NiCl}_{4}\right]^{2-}(\mathrm{aq})+4 \mathrm{CN}^{-}(\mathrm{aq}) \longrightarrow \\ \text { paramagnetic } \end{array}$$

Short Answer

Expert verified
\([\text{NiCl}_4]^{2-}\) is paramagnetic due to weak field ligands, while \([\text{Ni(CN)}_4]^{2-}\) is diamagnetic due to strong field ligands pairing all electrons.

Step by step solution

01

Understand Paramagnetism and Diamagnetism

A paramagnetic substance has unpaired electrons which cause it to be attracted by an external magnetic field, whereas a diamagnetic substance has all of its electrons paired and is not attracted by a magnetic field.
02

Electron Configuration of Nickel in \([\text{NiCl}_4]^{2-}\)

Nickel in \([\text{NiCl}_4]^{2-}\) is in the +2 oxidation state, giving it an electronic configuration of \([\text{Ar}]3d^8\). The four \(\text{Cl}^-\) ligands create a tetrahedral geometry, which results in weak field splitting.
03

Analyze Tetrahedral Ligand Field Splitting

In tetrahedral complexes, ligands cause weak field splitting of the \(3d\) orbitals, resulting in the configuration \(t_2^4e^4\) with two unpaired electrons remaining in \([\text{NiCl}_4]^{2-}\), making it paramagnetic.
04

Electron Configuration of Nickel in \([\text{Ni(CN)}_4]^{2-}\)

When \(\text{CN}^-\) ions replace \(\text{Cl}^-\) ions, they form a square planar complex. \(\text{CN}^-\) is a strong field ligand causing strong field splitting in \(\text{Ni}^{2+}\), leading to a reconfiguration into \(t_{2g}^6e_g^2\) (or \(d_{xy},d_{yz},d_{zx}\)) with all electrons paired.
05

Resulting Magnetic Properties

Since \([\text{Ni(CN)}_4]^{2-}\) has all paired electrons due to strong field splitting by \(\text{CN}^-\) ligands in a square planar arrangement, it is diamagnetic, unlike \([\text{NiCl}_4]^{2-}\), which retains unpaired electrons and is paramagnetic.

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Key Concepts

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

Paramagnetism
Paramagnetism is the property of being attracted to a magnetic field. This occurs in substances that have unpaired electrons in their electron configuration. These unpaired electrons produce small magnetic moments, which align with external magnetic fields, causing the substance to be attracted to it.
For example, the complex \( [\text{NiCl}_4]^{2-} \) is paramagnetic. This is because in its electron configuration, electrons are distributed in a way where some remain unpaired. A paramagnetic behavior implies that there are still free spins contributing to the magnetic properties. The more unpaired electrons there are, the stronger the paramagnetic response will be.
Knowing whether a complex is paramagnetic is crucial for understanding its magnetic properties and the nature of its bonding.
Diamagnetism
Diamagnetism refers to the property of being repelled by a magnetic field. This occurs when all electrons in a substance are paired. The paired electrons have spins that cancel each other out, leading to no net magnetic moment.
Unlike paramagnetic materials, diamagnetic substances do not have unpaired electrons and are therefore not attracted to a magnetic field. The complex \( [\text{Ni(CN)}_4]^{2-} \) is an example of a diamagnetic substance. Its electron configuration does not have any unpaired electrons, thanks to the strong field effect of the cyanide ligands which cause all electrons to be paired.
Understanding diamagnetism helps in analyzing the electron pairing and the types of interactions a compound will have in a magnetic field.
Electron Configuration
Electron configuration is crucial in determining the magnetic properties of a substance. It defines the arrangement of electrons in an atom or a molecule, particularly in its outer shells and orbitals.
The nickel ion in \( [\text{NiCl}_4]^{2-} \) has the electronic configuration of \([\text{Ar}]3d^8\). In this configuration, unpaired electrons remain because of the weak field splitting by chloride ligands.
However, in \( [\text{Ni(CN)}_4]^{2-} \), the electrons get rearranged due to the strong field splitting by the cyanide ions. This leads to a different configuration, \( t_{2g}^6e_g^2 \) (in a simplified approach, this means electrons occupy the lower energy \(d\) orbitals completely), resulting in all electrons being paired. This rearrangement is significant in transiting from paramagnetic to diamagnetic properties when ligands change within a complex.
Tetrahedral vs Square Planar Complexes
Tetrahedral and square planar are types of geometries that metal complexes can adopt, determined largely by their ligands and the metal ion's electron configuration.
The \( [\text{NiCl}_4]^{2-} \) complex has a tetrahedral shape, usually associated with weak field ligands like chloride that do not cause a significant difference in the energy levels of \(d\) orbitals, resulting in unpaired electrons (and paramagnetism).
Conversely, \( [\text{Ni(CN)}_4]^{2-} \) forms a square planar configuration due to the presence of strong field \( \text{CN}^- \) ligands. This geometry typically causes significant splitting of the \(d\) orbitals, allowing all electrons to pair up (diamagnetism).
Recognizing the type of geometry helps predict the magnetic and chemical properties of a compound. Tetrahedral complexes are likely to be paramagnetic, while square planar ones with strong ligands are more apt to be diamagnetic.

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Most popular questions from this chapter

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (c) \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) (b) \(\mathrm{VOCl}_{3}\) (d) \(\left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{2+}\)

Give a systematic name or the formula for the following: (a) \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{CuCl}_{4}\right]\) (b) \(\mathrm{Mo}(\mathrm{CO})_{6}\) (c) tetraaquadichlorochromium(III) chloride (d) aquabis(ethylenediamine)thiocyanatocobalt(III) nitrate

Draw all possible geometric isomers of the following. (a) \(\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) (b) \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})(\mathrm{Br})\left(\mathrm{SCN}^{-} \text {is bonded to } \mathrm{Pt}^{2+}\right.\) through S) (c) \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\left(\mathrm{NO}_{2}^{-} \text {is bonded to } \mathrm{Co}^{3+}\right.\) through N) (d) \(\left[\mathrm{Co}(\mathrm{en}) \mathrm{Cl}_{4}\right]^{-}\)

How many geometric isomers are possible for the square-planar complex ion \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{CN}) \mathrm{Cl}_{2}\right]^{-} ?\)

A platinum-containing compound, known as Magnus's green salt, has the formula \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]\) (in which both platinum ions are \(\mathrm{Pt}^{2+}\) ). Name the cation and the anion.

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