Chemistry - Coordination Chemistry Question with Solution | TestHub

Understand $\pi$-backbonding: CO is a strong $\pi$-acceptor ligand, while NH₃ is purely a $\sigma$-donor with no $\pi$-accepting capability.

Effect of substituting CO with NH₃: As the number of NH₃ ligands increases from (I) to (VI), electron density on the central Cr atom increases.

Cr–C bond shortening: This excess electron density increases $\pi$-backdonation from Cr into the ${\pi }^{\ast }$ orbitals of the remaining CO ligands.

Conclusion: Stronger backbonding from (I) to (VI) makes the Cr–C bond stronger and shorter, resulting in the correct bond length order: I > II > III > IV > V > VI.