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

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.