Which of the following are paramagnetic?

A.  [NiCl₄]^2–

B.  Ni(CO)₄

C.[Ni(CN)₄]^2–

D. [Ni(H₂O)₆]²⁺

E. Ni(PPh₃)₄

Choose the correct answer from the options given below :

To determine which nickel complexes are paramagnetic we examine the oxidation state of nickel, its d‑electron count, and the ligand field produced by the surrounding ligands.

Nickel in its elemental form has the configuration [Ar] 3d⁸ 4s². In complexes the 4s electrons are lost first, so the d‑electron count depends only on the oxidation state:

$Ni^\{0\}:\; d^\{10\}$
$Ni^\{2+\}:\; d^\{8\}$

Paramagnetism arises when there are unpaired electrons in the d‑orbitals after the ligand field splitting.

Complex A: [NiCl₄]^2‑

The complex is tetrahedral. Chloride (Cl⁻) is a weak‑field ligand, producing a small crystal‑field splitting (Δ_t). For a d⁸ ion in a tetrahedral field the electron configuration is e⁴ t₂⁴ → two unpaired electrons remain. Hence the complex is paramagnetic.

Complex B: Ni(CO)₄

Carbon monoxide is a very strong‑field π‑acceptor ligand. The nickel is in the zero oxidation state (Ni⁰), giving a d¹⁰ configuration. All ten d‑electrons are paired regardless of the geometry, so the complex is diamagnetic.

Complex C: [Ni(CN)₄]^2‑

Cyanide (CN⁻) is a strong‑field ligand. The complex adopts a square‑planar geometry, which gives a large splitting that forces the d⁸ electrons to pair completely (configuration: d_xy² d_xz² d_yz² d_z²²). No unpaired electrons are left, so the complex is diamagnetic.

Complex D: [Ni(H₂O)₆]²⁺

Water is a weak��field ligand and the complex is octahedral. For a d⁸ ion in an octahedral weak‑field environment the electron filling is t_2g⁶ e_g², leaving two unpaired electrons in the e_g set. Therefore the complex is paramagnetic.

Complex E: Ni(PPh₃)₄

Triphenylphosphine (PPh₃) is a strong‑field ligand but the nickel remains in the zero oxidation state (Ni⁰, d¹⁰). All d‑electrons are paired, giving a diamagnetic complex.

Summarizing A and D only

To determine which nickel complexes are paramagnetic we examine the oxidation state of nickel, its d‑electron count, and the ligand field produced by the surrounding ligands.

Nickel in its elemental form has the configuration [Ar] 3d⁸ 4s². In complexes the 4s electrons are lost first, so the d‑electron count depends only on the oxidation state:

$Ni^\{0\}:\; d^\{10\}$
$Ni^\{2+\}:\; d^\{8\}$

Paramagnetism arises when there are unpaired electrons in the d‑orbitals after the ligand field splitting.

Complex A: [NiCl₄]^2‑

The complex is tetrahedral. Chloride (Cl⁻) is a weak‑field ligand, producing a small crystal‑field splitting (Δ_t). For a d⁸ ion in a tetrahedral field the electron configuration is e⁴ t₂⁴ → two unpaired electrons remain. Hence the complex is paramagnetic.

Complex B: Ni(CO)₄

Carbon monoxide is a very strong‑field π‑acceptor ligand. The nickel is in the zero oxidation state (Ni⁰), giving a d¹⁰ configuration. All ten d‑electrons are paired regardless of the geometry, so the complex is diamagnetic.

Complex C: [Ni(CN)₄]^2‑

Cyanide (CN⁻) is a strong‑field ligand. The complex adopts a square‑planar geometry, which gives a large splitting that forces the d⁸ electrons to pair completely (configuration: d_xy² d_xz² d_yz² d_z²²). No unpaired electrons are left, so the complex is diamagnetic.

Complex D: [Ni(H₂O)₆]²⁺

Water is a weak‑field ligand and the complex is octahedral. For a d⁸ ion in an octahedral weak‑field environment the electron filling is t_2g⁶ e_g², leaving two unpaired electrons in the e_g set. Therefore the complex is paramagnetic.

Complex E: Ni(PPh₃)₄

Triphenylphosphine (PPh₃) is a strong‑field ligand but the nickel remains in the zero oxidation state (Ni⁰, d¹⁰). All d‑electrons are paired, giving a diamagnetic complex.

Summarizing the analysis, only complexes A and D possess unpaired electrons and are thus paramagnetic.

The correct answer is A and D only.