Why is PtCl4^2- square planar?
1 Answer
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
#[Ar]cancel(color(red)(4s^0)) 3d^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
LIGAND FIELD STRENGTH
Now, how its
#"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)
With platinum complexes, there is just something that promotes square planar more often than nickel complexes form square planar.
Platinum has bigger
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
We should recognize that since the ligands lie on the axes:
- The
#\mathbf(d_(x^2 - y^2))# orbitals experience the most repulsions. It is highest in energy. - The
#\mathbf(d_(xy))# orbitals experience the second most. It is second highest in energy. - The
#\mathbf(d_(z^2))# 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_(yz)# and#d_(xz)# orbitals. - The degenerate
#\mathbf(d_(xz))# and#\mathbf(d_(yz))# 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: