Solveeit Logo

Question

Question: The shape of \({{[PtC{{l}_{3}}({{C}_{2}}{{H}_{4}})]}^{-}}\) and the hybridization of \(Pt\) respecti...

The shape of [PtCl3(C2H4)]{{[PtC{{l}_{3}}({{C}_{2}}{{H}_{4}})]}^{-}} and the hybridization of PtPt respectively are:
A. Tetrahedral, sp3s{{p}^{3}}
B. Trigonal pyramidal, sp3s{{p}^{3}}
C. Square planar, dsp2ds{{p}^{2}}
D. Square planar, d2sp3{{d}^{2}}s{{p}^{3}}

Explanation

Solution

When two atomic orbitals combine to form hybrid orbital in a molecule then redistribution of the energy of orbitals of individual atoms produces orbitals of equivalent energy and the new orbital formed are known as hybrid orbitals and the phenomenon is known as hybridization.

Complete step by step answer:
- According to valence bond theory the metal atom or ion in presence of ligands can use its outer orbitals for hybridization which yield a set of equivalent orbitals of definite geometry like octahedral, tetrahedral, square planar and so on. These hybrid orbitals are allowed to overlap with ligand orbitals which can easily donate electron pairs for bonding.
- The compound given is [PtCl3(C2H4)]{{[PtC{{l}_{3}}({{C}_{2}}{{H}_{4}})]}^{-}} known by the name zeise salt. It is a stable yellow color coordination complex which contains ethylene ligand and it has greater importance in organometallic chemistry as it is the first organometallic compound. It is generally a π\pi complex having a shape of square planar and according to valence bond theory hybridization in case of square planar will be dsp2ds{{p}^{2}}. Thus we can consider that shape of [PtCl3(C2H4)]{{[PtC{{l}_{3}}({{C}_{2}}{{H}_{4}})]}^{-}} and the hybridization of PtPt respectively are Square planar, dsp2ds{{p}^{2}}. The correct answer is option “C” .

Note: The shape and the hybridization of any compound can be concluded with the help of theory given by the German Physicists called valence bond theory. This theory explains the electronic structure of the molecule formed by the overlapping of atomic orbitals.