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As a higher wavenumber means a lower wavelength, which in turn means higher energy, the complex with the most effective backbonding would have the least wavenumber. The backbonding, as I know, is proportional to the number of electrons in the d-orbital, so all this says that as we go rightwards on the periodic table, wavenumbers should decrease, which clearly is not the case.
Where am I going wrong?

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The backbonding, as I know, is proportional to the number of electrons in the d-orbital

That's indeed part of it — and that explains why the early d-block metals, Sc and Ti, don't form (neutral) carbonyls at all. However, the backbonding also depends on how willing the metal is to share those d-electrons. $Z_\mathrm{eff}$ increases going across the 3d block, which leads to poorer overlap between metal 3d orbitals and the CO π* orbital, and a poorer match in terms of their energies.

Around the middle there is a sweet spot where there is enough electron density to satisfy the CO ligands (or, equivalently, to satisfy the 18 electron rule), and the metal $Z_\mathrm{eff}$ is not so high.

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