Question

Regarding magnetic properties of the complexes \(Ni(CO)_4\) [I], \(NiCl_4^{2-}\) [II], which of the following is correct?

• A. I = Diamagnetic, II = Paramagnetic
• B. I = Paramagnetic, II = Paramagnetic
• C. I = Diamagnetic, II = Diamagnetic
• D. I = Paramagnetic, II = Diamagnetic

Hint:

Strong field ligands (like CO) cause electron pairing → diamagnetism; weak field ligands (like Cl⁻) often give paramagnetism.

Answer 1

The Correct Option is A

Step 1: Oxidation states and d-electron counts.
In Ni(CO)₄, CO is neutral, giving Ni(0) with d¹⁰. In NiCl₄²⁻, four Cl⁻ ligands give Ni a +2 charge and a d⁸ configuration.

Step 2: Ligand field strength in Ni(CO)₄.
CO, a strong-field ligand, induces pairing, leading to a diamagnetic tetrahedral complex with all electrons paired.

Step 3: Ligand field strength in NiCl₄²⁻.
Cl⁻ is weak-field, so the tetrahedral d⁸ complex retains unpaired electrons, exhibiting paramagnetism.

Step 4: Magnetic summary.
Ni(CO)₄ is diamagnetic; NiCl₄²⁻ is paramagnetic.

Step 5: Conclusion.
I = Diamagnetic, II = Paramagnetic.