Even though the molecule will have a structure that allows for it to exist in a planar conformation, there may be some/many that do not persist in a planar conformation due to steric effects, or complex three dimensional geometries.
Why is $\ce { (NH4)2 [CuCl4]}$ square planar complex but $\ce {Cs2 [CuCl4]}$ is tetrahedral even though both have same oxidation number of copper and same ligands?
inorganic chemistry - Why is (NH4)2 [CuCl4] square planar complex but ...
It is because of the fact that square planar complexes are formed by much strong ligands with d8-metal cation of 3d- series transition metals cation and 4d or 5d-series transition metal cation with either weak or strong ligands. The very strong ligands and 4d or 5d-series transition metal cations are responsible for higher crystal field splitting. Also, in square planar complexes of d8-metal ...
Why is the crystal field splitting energy larger for square planar than ...
However, the factors that affect the adoption of a square planar geometry vs tetrahedral geometry are the same, regardless of whether it is the metal centre or the ligand that is being compared.
Why is the geometry of tetracyanonickelate (II) square planar? Why is ...
I have come across these two splitting. Which of these is correct when there is crystal field splitting of ligand in square planar manner. Kindly provide reference of books and plausible explanatio...
The molecule $\ce { [PdCl4]^2-}$ is diamagnetic, which indicates a square planar geometry as all eight d electrons are paired in the lower-energy orbitals. However, $\ce { [NiCl4]^2-}$ is also $\mathr...
Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?