Square Planar Complexes
In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The geometry is prevalent for transition metal complexes with d8configuration. This includes Rh(I), Ir(I), Pd(II), Pt(II), and Au(III). Notable examples include the anticancer drugs cisplatin [PtCl2(NH3)2] and carboplatin.
In principle, square planar geometry can be achieved by flattening a tetrahedron. As such, the interconversion of tetrahedral and square planar geometries provides a pathway for the isomerization of tetrahedral compounds. For example, tetrahedral nickel(II) complexes such as NiBr2(PPh3)2undergo this change reversibly.
The removal of a pair of ligands from the z-axis of an octahedron leaves four ligands in the x-y plane. Therefore, the crystal field splitting diagram for square planar geometry can be derived from the octahedral diagram. The removal of the two ligands stabilizes the dz2level, leaving the dx2-y2level as the most destabilized. Consequently, the dx2-y2remains unoccupied in complexes of metals with the d8configuration. These compounds typically have sixteen valence electrons (eight from ligands, eight from the metal).
In tetrahedral molecular geometry, a central atom is located at the center of four substituent atoms, which form the corners of a tetrahedron. The bond angles are approximately 109.5° when all four substituents are the same. This geometry is widespread, particularly for complexes where the metal has d0or d10electron configuration.
For example, tetrakis(triphenylphosphine)palladium(0), a popular catalyst, and nickel carbonyl, an intermediate in nickel purification, are tetrahedral. Many complexes with incompletely filled d-subshells are tetrahedral as well—for example, the tetrahalides of iron(II), cobalt(II), and nickel(II).
Tetrahedral complexes have ligands in all of the places that an octahedral complex does not. Therefore, the crystal field splitting diagram for tetrahedral complexes is the opposite of an octahedral diagram. The dx2−dy2and dz2orbitals should be equally low in energy because they exist between the ligand axis, allowing them to experience little repulsion. In contrast, the dxy,dyz, and dxzaxes lie directly on top of where the ligands go. This maximizes repulsion and raises energy levels.
Difference Between Square Planar And Tetrahedral Complexes In Tabular Form
|The square planar geometry has one central atom that is surrounded by the four constituent atoms.||One atom is located at the center of the four atoms, forming a structure similar to a tetrahedron.|
|The bond-angle between the atoms in the square planar geometry is 90 degrees.||The bond-angle between the ligands in the tetrahedral geometry is 109.5 degrees.|
|The coordination number of the complexes forming such molecular geometry is 4.||The complexes forming a tetrahedral geometry also have the coordination number of 4.|
|This type of geometry is mostly prevalent in the transition metal complexes.||The compounds having no lone pairs of electrons form a tetrahedral structure.|
|The square planar complexes form a four-tiered diagram in CFT.||Tetrahedral complexes form a two-tiered crystal field diagram.|
|The complexes forming square planar geometry has the electron configuration ending in d8.||The configuration of the electrons in tetrahedral complexes can be from d0 or d10.|
|Transitions metals such as Rh(I), Ir (II), etc has square planar geometry.||Methane or CH4 has the tetrahedral geometry where the Carbon atom is in the central position of the complex.|
|Square planar complexes are low spin as electrons tend to get paired instead of remaining unpaired.||Tetrahedral complexes are high spin because electrons in the complex tend to go the higher energy levels instead of pairing with other electrons.|