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In isolated condition $\text{ C}-\text{C }$ bond length of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ is x, than the bond length of $\text{ C}-\text{C }$ the bond of $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ in Zeise's salt is:
A) Greater than x
B) less than x
C) Equal to x
D) None of these

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Last updated date: 13th Jun 2024
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Answer
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Hint: The Zeise's salt is a coordination complex of platinum having a general anion as$\text{ }{{\left[ {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{PtC}{{\text{l}}_{\text{3}}} \right]}^{-1}}\text{ }$. The $\text{ Pt }$ metal is bonded to the three chloro and one ethylene ligand. The metal forms a sigma bond with the pi-electron density of the ethylene ligand. The back bonding may take place between the filled metal orbital and empty antibonding orbital of the carbon atom.

Complete step by step answer:
Zeise’s salt is a coordination complex of platinum. the salt which contains the $\text{ }{{\left[ {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{PtC}{{\text{l}}_{\text{3}}} \right]}^{-1}}\text{ }$ anions are known as the Zeise’s salt. Such salts may include potassium trichloro (ethylene) platinate (II) complex$\text{ K}\left[ \text{PtC}{{\text{l}}_{\text{3}}}\text{(}{{\text{C}}_{\text{2}}}{{\text{H}}_{4}}\text{)} \right]\text{.}{{\text{H}}_{\text{2}}}\text{O }$,$\text{ Na }\left[ \text{PtC}{{\text{l}}_{\text{3}}}\text{(}{{\text{C}}_{\text{2}}}{{\text{H}}_{4}}\text{)} \right]\text{.}{{\text{H}}_{\text{2}}}\text{O }$ etc. the Zeise's salt is a yellow, stable coordination complex that contains the ethylene ligand.

The Zeise’s salt has square planar geometry, where the three corners of the square are accommodated by the chloro ligand and one by the ethylene ligand. The alkene $\text{ C = C }$ is at the $\text{ 9}{{\text{0}}^{\text{0}}}\text{ }$ to the $\text{ PtC}{{\text{l}}_{\text{3}}}\text{ }$ plane.
The bonding between the platinum and the ligands can be explained based on the Molecular orbital theory.

The olefin $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ has the pi bond. This pi-electron density of the olefin acts as a ligand and overlaps with the $\text{ }\sigma \text{ }$ orbital of the platinum atom. This establishes a $\text{ }\sigma \text{ }$ bond between the metal and the ligand.
In Zeise’s salt, the metal has the filled $\text{ d }\!\!\pi\!\!\text{ }$ orbital and the carbon atom of the ethylene has the empty $\text{ }\!\!\pi\!\!\text{ * }$ orbital. The back bonding takes place between the metal orbital and empty pi orbital of the ligand. In this bonding, the electrons are transferred from the metal to ligand.

As a result, the distance between the carbon of the ethylene and metal decreases .The metal –olefin has the multiple bond character due to the reason of back bonding and thus this back bonding decreases the metal and carbon bond length. This ultimately increases the distance between the two carbon atoms of ethylene.

It is found that the $\text{ C}-\text{C }$ bond length in the Zeiss salt $\text{ (1}\text{.34 }\overset{0}{\mathop{\text{A}}}\,)\text{ }$increases than the $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$molecule$\text{ (1}\text{.4}-1.47\text{ }\overset{0}{\mathop{\text{A}}}\,)\text{ }$.
Thus, in isolated $\text{ C}-\text{C }$bond length in $\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{ }$ if considered as the ‘x’ then the bond length of $\text{ C}-\text{C }$ the bond in the Zeise’s salt is greater than the ‘x’.
So, the correct answer is “Option A”.

Note: The back bonding decreases the distance between the metal and the ethylene molecule. This bond is trans to the $\text{ Pt}-\text{Cl }$ bond and bond length increases .As a result, the bond between the $\text{ Pt}-\text{Cl }$ weakens. This is known as the trans effect. The back bonding seems similar to the metal carbonyl bonding .The back bonding is termed as the $\text{ }\!\!\mu\!\!\text{ }-$ bonding.