# Why is PtCl4^2- square planar?

Dec 25, 2015

A good general rule is that if you have either square planar or tetrahedral, a low-spin complex generally forms square planar, and a high-spin complex generally forms tetrahedral. Platinum is not an exception to that statement.

To see why, we should consider nickel, which is in the same group, whose complexes are tetrahedral sometimes and square planar other times.

NICKEL (II) CONFIGURATION

$\text{Ni(II)}$'s cationic electron configuration is:

$\left[A r\right] \cancel{\textcolor{red}{4 {s}^{0}}} 3 {d}^{8}$

making it a ${d}^{8}$ metal.

With four ligands, the complex can either be tetrahedral or square planar, though not seesaw (with four identical weak-field ligands, why have one axial angle be ${180}^{\circ}$ but all the others be only ~90^@ or ~120^@? It makes more symmetrical sense to distribute the energy more evenly than that).

LIGAND FIELD STRENGTH

Now, how its $d$ orbitals fill up (${d}_{x y}$, ${d}_{x z}$, ${d}_{y z}$, ${d}_{{x}^{2} - {y}^{2}}$, and ${d}_{{z}^{2}}$) depends on the crystal field-strength of the ligand that binds to it. Thus, we should consider the fact that chloride is a weak-field ligand (weaker than water, which is around the middle of a typical spectrochemical series). Part of it is:

"I"^− < "Br"^− < "S"^(2−) < "SCN"^− < color(blue)("Cl"^−) < "NO"_3^− < "N"^(3−) <
. . . < color(blue)("H"_2"O") < . . . < "CN"^− ~~ "C"-="O"

That means its electrons generate little repulsion while chloride interacts with nickel, and the ligand-field splitting energy is small. That allows for square planar complexes to form, even though those are higher in energy.

MAJOR DIFFERENCE BETWEEN PLATINUM (II) AND NICKEL (II)

$\setminus m a t h b f \left(\text{Pt(II)}\right)$ has a similar electron configuration as $\setminus m a t h b f \left(\text{Ni(II)}\right)$, also acting as a $\setminus m a t h b f \left({d}^{8}\right)$ transition metal.

With platinum complexes, there is just something that promotes square planar more often than nickel complexes form square planar.

Platinum has bigger $\boldsymbol{d}$ orbitals than nickel does, which can hold more electron density more capably, because the electron density can be more spread out in a larger $d$ orbital.

Therefore, platinum can support a higher-energy structure such as the square planar structure.

AT THE END OF THE DAY...

A good general rule is that if you have either square planar or tetrahedral, a low-spin complex generally forms square planar, and a high-spin complex generally forms tetrahedral. Platinum is not an exception to that statement.

D ORBITAL SPLITTING FOR SQUARE PLANAR COMPLEXES

The $d$ orbitals look like this:

We should recognize that since the ligands lie on the axes:

• The $\setminus m a t h b f \left({d}_{{x}^{2} - {y}^{2}}\right)$ orbitals experience the most repulsions. It is highest in energy.
• The $\setminus m a t h b f \left({d}_{x y}\right)$ orbitals experience the second most. It is second highest in energy.
• The $\setminus m a t h b f \left({d}_{{z}^{2}}\right)$ orbitals experience the third most if you consider that there is a ring of electron density on the xy-plane. It is a little higher in energy than the ${d}_{y z}$ and ${d}_{x z}$ orbitals.
• The degenerate $\setminus m a t h b f \left({d}_{x z}\right)$ and $\setminus m a t h b f \left({d}_{y z}\right)$ are off the plane of the ligands, so they have the lowest-energy interactions with the ligands.

Thus, the square planar crystal field-splitting diagram is like so: