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

To solve the problem, we need to identify which of the given coordination compounds is diamagnetic.

1. Understanding Diamagnetism:
A species is diamagnetic if all its electrons are paired, resulting in no unpaired electrons and a magnetic moment of zero. We need to examine each coordination compound to determine the number of unpaired electrons in the central metal ion, considering the oxidation state, electronic configuration, and ligand field effects.

2. Option A: [Ni(CN)\(_4\)]\(^{2-}\):
- Oxidation State: CN\(^-\) has a charge of -1, so 4 CN\(^-\) contribute -4. The overall charge is -2, so Ni’s oxidation state is +2 (4(-1) + x = -2, x = +2).
- Electron Configuration: Ni (atomic number 28) has a ground state configuration of [Ar] 4s\(^2\) 3d\(^8\). Ni\(^{2+}\) loses two electrons from the 4s orbital: [Ar] 3d\(^8\).
- Geometry and Ligand Field: [Ni(CN)\(_4\)]\(^{2-}\) is a 4-coordinate complex. Ni\(^{2+}\) (d\(^8\)) with CN\(^-\) (a strong field ligand) typically forms a square planar geometry due to the strong field splitting. In a square planar field, the d orbitals split as: d\(_{xy}\), d\(_{xz}\), d\(_{yz}\), d\(_{x^2-y^2}\), d\(_{z^2}\). The 8 d-electrons fill the lower orbitals: (d\(_{xy}\))\(^2\) (d\(_{xz}\))\(^2\) (d\(_{yz}\))\(^2\) (d\(_{z^2}\))\(^2\), leaving d\(_{x^2-y^2}\) empty. All electrons are paired.
- Result: No unpaired electrons, so [Ni(CN)\(_4\)]\(^{2-}\) is diamagnetic.

3. Option B: [NiCl\(_4\)]\(^{2-}\):
- Oxidation State: Cl\(^-\) has a charge of -1, so 4 Cl\(^-\) contribute -4. The overall charge is -2, so Ni’s oxidation state is +2.
- Electron Configuration: Ni\(^{2+}\) is 3d\(^8\).
- Geometry and Ligand Field: [NiCl\(_4\)]\(^{2-}\) is a 4-coordinate complex. Cl\(^-\) is a weak field ligand, and for Ni\(^{2+}\) (d\(^8\)), 4-coordinate complexes are typically tetrahedral (not square planar, as tetrahedral splitting is smaller). In a tetrahedral field, the d orbitals split as: e (d\(_{z^2}\), d\(_{x^2-y^2}\)) and t\(_{2}\) (d\(_{xy}\), d\(_{xz}\), d\(_{yz}\)), with e lower in energy. The 8 electrons fill as: (e)\(^4\) (t\(_{2}\))\(^4\). The t\(_{2}\) set has 3 orbitals, so 4 electrons distribute as 2 paired, 2 unpaired (e.g., (d\(_{xy}\))\(^2\) (d\(_{xz}\))\(^1\) (d\(_{yz}\))\(^1\)).
- Result: Two unpaired electrons, so [NiCl\(_4\)]\(^{2-}\) is paramagnetic.

4. Option C: [Fe(CN)\(_6\)]\(^{3-}\):
- Oxidation State: CN\(^-\) contributes -6, overall charge is -3, so Fe’s oxidation state is +3 (6(-1) + x = -3, x = +3).
- Electron Configuration: Fe (atomic number 26) is [Ar] 4s\(^2\) 3d\(^6\). Fe\(^{3+}\) is [Ar] 3d\(^5\).
- Geometry and Ligand Field: [Fe(CN)\(_6\)]\(^{3-}\) is a 6-coordinate octahedral complex. CN\(^-\) is a strong field ligand. In an octahedral field, d orbitals split into t\(_{2g}\) (lower) and e\(_{g}\) (higher). For d\(^5\) Fe\(^{3+}\) with a strong field ligand, the splitting is large, so it’s low spin: (t\(_{2g}\))\(^5\) (e\(_{g}\))\(^0\), with 5 electrons in t\(_{2g}\) (3 orbitals): (t\(_{2g}\))\(^5\) has 1 unpaired electron (e.g., 2 paired, 1 unpaired).
- Result: One unpaired electron, so [Fe(CN)\(_6\)]\(^{3-}\) is paramagnetic.

5. Option D: [CoF\(_6\)]\(^{3-}\):
- Oxidation State: F\(^-\) contributes -6, overall charge is -3, so Co’s oxidation state is +3.
- Electron Configuration: Co (atomic number 27) is [Ar] 4s\(^2\) 3d\(^7\). Co\(^{3+}\) is [Ar] 3d\(^6\).
- Geometry and Ligand Field: [CoF\(_6\)]\(^{3-}\) is octahedral. F\(^-\) is a weak field ligand. For d\(^6\) Co\(^{3+}\) with a weak field ligand, the splitting is small, so it’s high spin: (t\(_{2g}\))\(^4\) (e\(_{g}\))\(^2\). t\(_{2g}\) has 4 electrons (2 paired, 1 unpaired per orbital), and e\(_{g}\) has 2 electrons (1 unpaired per orbital), totaling 4 unpaired electrons.
- Result: Four unpaired electrons, so [CoF\(_6\)]\(^{3-}\) is paramagnetic.

Final Answer:
The diamagnetic species is [Ni(CN)\(_4\)]\(^{2-}\) (option A).