Question

\( [NiCl_4]^{2-} \) is paramagnetic while \( [Ni(CO)_4] \) is diamagnetic though both are tetrahedral. Why? [At. No. Ni=28]

Hint:

Weak field ligands usually produce high spin complexes; strong field ligands produce low spin complexes.

Answer 1

Step 1: Understanding the Concept:
Paramagnetism depends on the presence of unpaired electrons, which is determined by hybridization and ligand strength.
Step 2: Detailed Explanation:
1. \( [NiCl_4]^{2-} \): Ni is in \( +2 \) state (\( 3d^8 \)). \( Cl^- \) is a weak field ligand and cannot cause pairing. Hybridization is \( sp^3 \). It has 2 unpaired electrons in \( 3d \), making it paramagnetic.
2. \( [Ni(CO)_4] \): Ni is in \( 0 \) state (\( 3d^8 4s^2 \)). \( CO \) is a strong field ligand. It causes the \( 4s \) electrons to move into the \( 3d \) subshell to pair up with the \( 3d^8 \) electrons. This resulting \( 3d^{10} \) configuration has no unpaired electrons, making it diamagnetic.
Step 3: Final Answer:
The difference is due to strong-field CO causing electron pairing while weak-field \( Cl^- \) does not.