Question

The shape of \[{[PtC{l_3}({C_2}{H_4})]^ - }\] and hybridization of \[Pt\] respectively are:
A.Tetrahedral, \[s{p^3}\]
B.Trigonal bipyramidal, \[s{p^3}\]
C.Square planar, \[ds{p^2}\]
D.Square planar, \[{d^2}s{p^3}\]

Answer 1

Hint: The given compound is a transition metal complex and is known as zeise’s salt. The shape and hybridization of transition metal complexes are explained by VBT.

Complete step by step answer:
The given complex in question consists of \[Pt\] which is a transition metal and the complex is a transition metal complex. The name of this complex is Zeise’s salt. To find its hybridization we have to use VBT theorem.
VBT or Valence bond theory- it was given by Pauling. According to this theory
The central metal makes the available number of s, p or d orbital equal to its coordination number to form a covalent bond.
These orbitals that overlap with each other to give a net set of hybridized orbitals which are identical in energy.
The bonding takes place by overlapping a filled orbital containing a lone pair of electrons with a vacant orbital and forms a covalent coordinate bond. The magnetic moment and the coordination number of metal cations decide the hybridization and geometry of a complex.
The above given points are the some points of valence bond theory now we will apply these to our given complex.
The given complex is \[{[PtC{l_3}({C_2}{H_4})]^ - }\].
We can see that the coordination number is 4 so there are two possibilities .That the given complex may be square planar or tetrahedral.
Now we will calculate the oxidation number for \[Pt\].
\[
  Pt - 3 + 0 = - 1 \\
  Pt = + 2 \\
 \]
Electronic configuration for \[Pt\] is \[3{d^{10}}\].
Electronic configuration of \[P{t^{ + 2}}\] is \[3{d^8}\].
As we know Chlorine being a weak ligand could not pair the electrons but ethyl group leads to the pairing of electrons due to back bonding. It forms a pi-complex with platinum making it a square planar complex. There are 4 ligands available and $1d$, $1s$ and $2p$ orbitals are empty. After hybridization there will be a formation of $ds{p^2}$. As it is a diamagnetism complex, its shape will be square planar.

So, the correct answer is option C.

Note:
Platinum has bigger d-orbital than nickel. Its capacity to hold electron density is way better than nickel. Hence it always forms a square planar complex.
Weak field ligand does not lead to pairing of electrons making it paramagnetic whereas strong field ligand leads to pairing of electrons making it diamagnetic.