Question

The correct statement among the following is:

• A. \(\mathrm{Ni(CO)_4}\) is diamagnetic and \([\mathrm{NiCl_4}]^{2-}\) and \([\mathrm{Ni(CN)_4}]^{2-}\) are paramagnetic.
• B. \(\mathrm{Ni(CO)_4}\) and \([\mathrm{NiCl_4}]^{2-}\) are diamagnetic and \([\mathrm{Ni(CN)_4}]^{2-}\) is paramagnetic.
• C. \([\mathrm{Ni(CN)_4}]^{2-}\) and \([\mathrm{NiCl_4}]^{2-}\) are diamagnetic and \(\mathrm{Ni(CO)_4}\) is paramagnetic.
• D. \(\mathrm{Ni(CO)_4}\) and \([\mathrm{Ni(CN)_4}]^{2-}\) are diamagnetic and \([\mathrm{NiCl_4}]^{2-}\) is paramagnetic.

Hint:

Strong-field ligands (CN\(^-\), CO) favour pairing and diamagnetism, while weak-field ligands (Cl\(^-\)) lead to paramagnetism.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The magnetic properties of coordination complexes depend on the oxidation state of the central metal and the field strength of the ligands.
Step 2: Detailed Explanation:
1. Ni(CO)\(_4\): Nickel is in the 0 oxidation state (\( d^{10} \)). Carbon monoxide (\( \text{CO} \)) is a very strong field ligand that causes pairing and \( sp^3 \) hybridisation. With all electrons paired in \( 3d \) and \( 4s \) orbitals involved in bonding, it is diamagnetic.
2. [Ni(CN)\(_4\)]\(^{2-}\): Nickel is in the +2 oxidation state (\( d^8 \)). Cyanide (\( \text{CN}^- \)) is a strong field ligand. It causes the 8 electrons to pair up in four \( d \)-orbitals, leaving one \( d \)-orbital empty for \( dsp^2 \) hybridisation (square planar). With no unpaired electrons, it is diamagnetic.
3. [NiCl\(_4\)]\(^{2-}\): Nickel is in the +2 oxidation state (\( d^8 \)). Chloride (\( \text{Cl}^- \)) is a weak field ligand. It does not cause pairing. The electrons follow Hund's rule, leaving 2 unpaired electrons. It undergoes \( sp^3 \) hybridisation (tetrahedral) and is paramagnetic.
Step 3: Final Answer:
Ni(CO)\(_4\) and [Ni(CN)\(_4\)]\(^{2-}\) are diamagnetic, while [NiCl\(_4\)]\(^{2-}\) is paramagnetic.