Question

$[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic even though both are tetrahedral. Why?

Hint:

\textbf{Weak Field Ligands (e.g., $Cl^-, F^-, H_2O$):} Usually high spin, no forced pairing. \textbf{Strong Field Ligands (e.g., $CN^-, CO, NH_3$):} Usually low spin, forced pairing occurs.

Answer 1

Step 1: Analyze $[NiCl_4]^{2-$}

• Oxidation State of Ni: $x + 4(-1) = -2 \Rightarrow x = +2$. So, $Ni^{2+}$.
• Configuration: Ni ($Z=28$) is $[Ar]3d^8 4s^2$. $Ni^{2+}$ is $[Ar]3d^8$.
• Ligand Nature: $Cl^-$ is a weak field ligand (Spectrochemical series). It does not cause pairing of electrons.
• Orbital Filling: The $3d^8$ electrons remain arranged as 3 pairs and 2 unpaired electrons according to Hund's rule.
• Hybridization: $sp^3$ (Tetrahedral).
• Result: Due to the presence of 2 unpaired electrons, it is Paramagnetic.

Step 2: Analyze $[Ni(CO)_4]$

• Oxidation State of Ni: $x + 4(0) = 0 \Rightarrow x = 0$. So, neutral Ni.
• Configuration: Ni ($Z=28$) is $[Ar]3d^8 4s^2$.
• Ligand Nature: $CO$ is a strong field ligand. It causes strong repulsion and forces electrons to pair up.
• Rearrangement: Under the influence of $CO$, the two $4s$ electrons are pushed into the $3d$ orbitals to pair up with the unpaired $3d$ electrons. The configuration becomes $3d^{10} 4s^0$.
• Hybridization: $sp^3$ (Tetrahedral).
• Result: All electrons are paired ($3d^{10}$). Due to the absence of unpaired electrons, it is Diamagnetic.