Question

The diamagnetic species is :
At. No. Co = 27, Fe =26, Ni= 28]

• A. \( [Ni(CN)_4]^{2-} \)
• B. \( [NiCl_4]^{2-} \)
• C. \( [Fe(CN)_6]^{3-} \)
• D. \( [CoF_6]^{3-} \)

Hint:

Diamagnetic species have all their electrons paired and are repelled by a magnetic field, unlike paramagnetic species, which have unpaired electrons.

Answer 1

The Correct Option is A

The objective is to identify the diamagnetic coordination compound among the provided options.

1. Diamagnetism Defined:
A species is classified as diamagnetic if all its constituent electrons are paired, thereby exhibiting no unpaired electrons and a zero magnetic moment. The analysis requires evaluating each coordination compound to ascertain the number of unpaired electrons on the central metal ion, taking into account its oxidation state, electronic configuration, and the influence of ligand field effects.

2. Option A: [Ni(CN)\(_4\)]\(^{2-}\):
- Oxidation State: The cyanide ligand (CN\(^-\)) carries a charge of -1. With four such ligands, the total negative charge from ligands is -4. Given the overall charge of -2 for the complex, the oxidation state of Nickel (Ni) is calculated as +2 (since 4(-1) + x = -2, where x is the oxidation state of Ni, yielding x = +2).
- Electron Configuration: Nickel (atomic number 28) has an electronic configuration of [Ar] 4s\(^2\) 3d\(^8\). Upon forming the Ni\(^{2+}\) ion, two electrons are removed from the 4s orbital, resulting in the configuration [Ar] 3d\(^8\).
- Geometry and Ligand Field: [Ni(CN)\(_4\)]\(^{2-}\) is a four-coordinate complex. The presence of CN\(^-\), a strong field ligand, with Ni\(^{2+}\) (a d\(^8\) ion) favors a square planar geometry due to significant ligand field splitting. In this geometry, the d orbitals are split into five distinct energy levels: d\(_{xy}\), d\(_{xz}\), d\(_{yz}\), d\(_{x^2-y^2}\), and d\(_{z^2}\). The eight d-electrons are filled into the lower energy orbitals sequentially: (d\(_{xy}\))\(^2\) (d\(_{xz}\))\(^2\) (d\(_{yz}\))\(^2\) (d\(_{z^2}\))\(^2\). This leaves the d\(_{x^2-y^2}\) orbital vacant. Consequently, all electrons are paired.
- Conclusion: As there are no unpaired electrons, [Ni(CN)\(_4\)]\(^{2-}\) is identified as diamagnetic.

3. Option B: [NiCl\(_4\)]\(^{2-}\):
- Oxidation State: The chloride ligand (Cl\(^-\)) has a charge of -1. With four Cl\(^-\) ligands, the total negative charge is -4. With an overall complex charge of -2, the oxidation state of Nickel (Ni) is +2.
- Electron Configuration: Ni\(^{2+}\) has an electronic configuration of 3d\(^8\).
- Geometry and Ligand Field: [NiCl\(_4\)]\(^{2-}\) is a four-coordinate complex. Chloride (Cl\(^-\)) is a weak field ligand. For Ni\(^{2+}\) (d\(^8\)) in a four-coordinate environment, a tetrahedral geometry is typically observed, characterized by smaller ligand field splitting compared to square planar geometry. In a tetrahedral field, the d orbitals split into two sets: e (comprising d\(_{z^2}\) and d\(_{x^2-y^2}\)) and t\(_{2}\) (comprising d\(_{xy}\), d\(_{xz}\), and d\(_{yz}\)), with the e set being lower in energy. The eight d-electrons are filled as follows: (e)\(^4\) (t\(_{2}\))\(^4\). The t\(_{2}\) set contains three orbitals. Distributing four electrons into these three orbitals results in two paired electrons and two unpaired electrons (for instance, in the arrangement (d\(_{xy}\))\(^2\) (d\(_{xz}\))\(^1\) (d\(_{yz}\))\(^1\)).
- Conclusion: The presence of two unpaired electrons signifies that [NiCl\(_4\)]\(^{2-}\) is paramagnetic.

4. Option C: [Fe(CN)\(_6\)]\(^{3-}\):
- Oxidation State: The cyanide ligand (CN\(^-\)) contributes a total charge of -6. Given the overall charge of the complex is -3, the oxidation state of Iron (Fe) is calculated as +3 (from 6(-1) + x = -3, where x is the oxidation state of Fe, yielding x = +3).
- Electron Configuration: Iron (atomic number 26) has the ground state configuration [Ar] 4s\(^2\) 3d\(^6\). The Fe\(^{3+}\) ion has the configuration [Ar] 3d\(^5\).
- Geometry and Ligand Field: [Fe(CN)\(_6\)]\(^{3-}\) is a six-coordinate octahedral complex. Cyanide (CN\(^-\)) is a strong field ligand. In an octahedral field, the d orbitals split into two sets: t\(_{2g}\) (lower energy) and e\(_{g}\) (higher energy). For a d\(^5\) ion like Fe\(^{3+}\) in the presence of a strong field ligand, a large splitting occurs, resulting in a low-spin configuration. The electron distribution is (t\(_{2g}\))\(^5\) (e\(_{g}\))\(^0\). Within the three t\(_{2g}\) orbitals, five electrons are arranged such that there is one unpaired electron (e.g., two paired electrons in two orbitals and one electron in the third orbital).
- Conclusion: With one unpaired electron, [Fe(CN)\(_6\)]\(^{3-}\) is paramagnetic.

5. Option D: [CoF\(_6\)]\(^{3-}\):
- Oxidation State: The fluoride ligand (F\(^-\)) contributes a total charge of -6. With an overall complex charge of -3, the oxidation state of Cobalt (Co) is determined to be +3.
- Electron Configuration: Cobalt (atomic number 27) has the ground state configuration [Ar] 4s\(^2\) 3d\(^7\). The Co\(^{3+}\) ion has the configuration [Ar] 3d\(^6\).
- Geometry and Ligand Field: [CoF\(_6\)]\(^{3-}\) is an octahedral complex. Fluoride (F\(^-\)) is a weak field ligand. For a d\(^6\) ion like Co\(^{3+}\) coordinated to a weak field ligand, the ligand field splitting is small, leading to a high-spin configuration. The electron distribution is (t\(_{2g}\))\(^4\) (e\(_{g}\))\(^2\). The t\(_{2g}\) set contains four electrons, resulting in two paired and two unpaired electrons distributed among the three orbitals. The e\(_{g}\) set contains two electrons, each occupying one of the two orbitals, resulting in two unpaired electrons. In total, there are four unpaired electrons.
- Conclusion: The presence of four unpaired electrons indicates that [CoF\(_6\)]\(^{3-}\) is paramagnetic.

Final Determination:
The diamagnetic species among the given options is [Ni(CN)\(_4\)]\(^{2-}\), corresponding to option A.