Question

The alkene ligand ($\pi$-C$_2$R$_4$) is both a '$\sigma$' donor and a '$\pi$' acceptor, similar to the CO ligand in metal carbonyls, and exhibits synergic bonding with metals. Correct order of C-C bond length in K[PtCl$_3$($\pi$-C$_2$R$_4$)] complexes in which R = H, F or CN is

• A. H > F > CN
• B. H > CN > F
• C. CN > F > H
• D. F > H > CN

Answer 1

Answer: (C)

CN > F > H

Answer 2

The bonding in metal-alkene complexes like K[PtCl$_3$($\pi$-C$_2$R$_4$)] involves synergic bonding, similar to metal carbonyls. This involves two main components:

1. $\sigma$-donation: The alkene donates electron density from its filled $\pi$ orbital to an empty d orbital of the metal.
2. $\pi$-back-donation: The metal donates electron density from its filled d orbital to the empty $\pi^*$ (antibonding) orbital of the alkene.

The $\pi$-back-donation is crucial for the C-C bond length in the alkene. When electron density is donated into the $\pi^*$ antibonding orbital of the C=C bond, it weakens the C-C $\pi$ bond and consequently increases the C-C bond length.

The extent of $\pi$-back-donation depends on the electron-accepting ability of the alkene. An alkene with more electron-withdrawing substituents (R) will have a lower energy (more stable) $\pi^*$ orbital, making it a better acceptor for electron density from the metal.

Let's compare the electron-withdrawing nature of the substituents R:

• R = H (Hydrogen): Hydrogen is neither strongly electron-donating nor electron-withdrawing.
• R = F (Fluorine): Fluorine is a highly electronegative atom and exerts a strong inductive electron-withdrawing effect (-I effect).
• R = CN (Cyano group): The cyano group is a strong electron-withdrawing group due to both its inductive effect (-I) and its resonance/mesomeric effect (-M), as it has an empty $\pi^*$ orbital that can accept electron density. In general, the -CN group is considered a stronger electron-withdrawing group than -F. For instance, comparing Hammett $\sigma_p$ values: $\sigma_p(\text{H}) = 0$, $\sigma_p(\text{F}) = +0.06$, $\sigma_p(\text{CN}) = +0.66$.

Based on the electron-withdrawing strength, the order is:

CN > F > H

A stronger electron-withdrawing R group makes the alkene a better $\pi$-acceptor.

Better $\pi$-acceptor alkene $\Rightarrow$ Greater extent of $\pi$-back-donation from metal to alkene $\pi^*$.

Greater $\pi$-back-donation $\Rightarrow$ More electron density in the C-C antibonding orbital.

More electron density in antibonding orbital $\Rightarrow$ Weaker C-C bond $\Rightarrow$ Longer C-C bond length.

Therefore, the order of C-C bond length will be directly proportional to the electron-withdrawing strength of R:

C-C bond length order: C$_2$(CN)$_4$ > C$_2$F$_4$ > C$_2$H$_4$

This corresponds to the order: CN > F > H.