Animated Solution for Chemistry - Coordination Compounds: Which one of the following has a square planar geometry? (At. no. Co = 27, Ni = 28, Fe = 26, Pt = 78)

Why $[PtC{l}_4{]}^{2-}$ is Square Planar

When dealing with coordination compounds, predicting the geometry of a complex with a coordination number of 4 usually boils down to a simple check: is the ligand strong or weak?

If the ligand is weak, like the chloride ion ($C{l}^{-}$), it generally cannot overcome the pairing energy of the electrons in the central metal's d-orbitals. The electrons remain unpaired, and the metal utilizes outer s and p orbitals to form $s{p}^3$ hybridization, resulting in a tetrahedral geometry. This is exactly what happens with the 3d series metals in our options: Cobalt ($Co$), Iron ($Fe$), and Nickel ($Ni$). For instance, in $[NiC{l}_4{]}^{2-}$, the $N{i}^{2+}$ ion has a $3d^8$ configuration. The weak chloride ligands leave two electrons unpaired, leading to a paramagnetic, tetrahedral complex.

However, chemistry is full of fascinating exceptions, and Platinum ($Pt$) provides a perfect example. Platinum is not a 3d metal; it resides in the 5d transition series.

As we move down a group in the periodic table—from 3d to 4d to 5d—the spatial extent of the d-orbitals increases significantly. These larger orbitals extend further out from the nucleus, allowing for a much stronger and more effective overlap with the orbitals of the approaching ligands. Furthermore, the higher effective nuclear charge ($Z_{eff}$) of these heavier metals pulls the ligands closer.

Because of this enhanced interaction, the crystal field splitting energy ($\Delta$) increases dramatically. In fact, $\Delta$ increases by about 30% to 50% when moving from the 3d to the 4d series, and by another 30% to 50% from the 4d to the 5d series.

This splitting energy becomes so massive that it easily overcomes the pairing energy, regardless of the ligand's inherent strength. Therefore, for 4d and 5d metals like Palladium ($Pd$) and Platinum ($Pt$), almost all ligands—even weak ones like chloride—act as strong field ligands!

In the $[PtC{l}_4{]}^{2-}$ complex, the $P{t}^{2+}$ ion has a $5d^8$ configuration. Due to the immense crystal field splitting, these 8 electrons are forced to pair up completely in the lower energy d-orbitals.

This pairing leaves one inner d-orbital (specifically, the $d_{x^2-y^2}$ orbital) completely empty. Platinum eagerly utilizes this empty inner orbital, along with one s and two p orbitals, to undergo $ds{p}^2$ hybridization.

Whenever a complex exhibits $ds{p}^2$ hybridization, its geometry is strictly square planar. Thus, despite having weak chloride ligands, $[PtC{l}_4{]}^{2-}$ defies the basic rule and adopts a beautiful square planar structure.