Q.$[\text{Ni}(\text{CN})_4]^{2-}$ is square planar and diamagnetic, while $[\text{NiCl}_4]^{2-}$ is tetrahedral and paramagnetic, even though both contain $\text{Ni}^{2+}$. Explain this difference in terms of the field strength of $\text{CN}^-$ versus $\text{Cl}^-$ and the resulting hybridization at $\text{Ni}^{2+}$.

[!TLDR] Strong-field $\text{CN}^-$ forces $d^8$ pairing, freeing a $3d$ orbital for $dsp^2$ (square planar); weak-field $\text{Cl}^-$ does not, so $\text{Ni}^{2+}$ stays $sp^3$ (tetrahedral). [!ANSWER] $[\text{Ni}(\text{CN})_4]^{2-}$ is sq …

$\text{Ni}^{2+}$ has a $d^8$ configuration in both complexes. In $[\text{NiCl}_4]^{2-}$, $\text{Cl}^-$ is a weak-field ligand: it does not force any additional pairing beyond the $d^8$ ion's normal arrangement (which already has 2 unpaired electrons in its highest-energy orbitals), so none of the $3d$ orbitals is freed up for bonding. Nickel then uses its available empty $4s$ and $4p$ orbitals directly, hybridizing as $sp^3$ — giving a tetrahedral shape — and the 2 unpaired $d$-electrons remain unpaired, so the complex is paramagnetic. In $[\text{Ni}(\text{CN})_4]^{2-}$, $\text{CN}^-$ is a strong-field ligand: it forces the two normally-unpaired $3d$ electrons to pair up together within one orbital, which empties one $3d$ orbital completely. Nickel can now hybridize that one empty $3d$ orbital together with one $4s$ and two $4p$ orbitals as $dsp^2$ — giving a square planar shape — and with all $d$-electrons now paired, the complex is diamagnetic. The identical metal …

Compare the field strength of the two ligands; a strong-field ligand on a $d^8$ ion forces pairing and frees a $(n-1)d$ orbital for $dsp^2$/square-planar hybridization, while a weak-field ligand leaves the ion's normal …

Assuming coordination number 4 always predicts a single fixed geometry — for $d^8$ metal ions specifically, the ligand field strength (not just the co …