 Solution

For the same pair of atoms, the harmonic approximation gives $\tilde{\nu }=(2\pi c{)}^{-1}\sqrt{k/\mu }$, where $k$ is the bond force constant and $\mu$ the reduced mass. The large shifts mainly reflect changes in $k$ from CO pi backbonding; the complexes are not distinguished simply by their total number of valence electrons.

$\mathrm{W}(\mathrm{CO}{)}_6$ has stronger net backbonding than $\mathrm{Cr}(\mathrm{CO}{)}_6$. Both are group-6 M(0), $d^6$, 18-electron complexes. The more spatially extended tungsten $5d$ orbitals provide effective overlap with CO ${\pi }^{\ast }$, consistent with the lower stretching frequency. As a simple same-mode force-constant comparison, $(1977/2000{)}^2=0.977$, about a 2.3% reduction.

$\mathrm{Ni}(\mathrm{CO}{)}_3(\mathrm{PM}{\mathrm{e}}_3)$ has appreciable but less C–O weakening than those hexacarbonyls. Ni(0) is $d^{10}$ and the complex is also 18-electron. The more contracted, lower-energy late-3d occupied orbitals give less effective donation into CO ${\pi }^{\ast }$ than the group-6 carbonyls in this comparison. Trimethylphosphine donates electron density and increases backbonding relative to an otherwise comparable less-donating nickel complex; that effect does not force its frequency below chromium's. Comparing different metals requires both orbital energies and overlap.

$[\mathrm{Cu}(\mathrm{CO}{)}_4{]}^{+}$ has very weak backbonding and a frequency above free CO. Cu(I) is $d^{10}$ and 18-electron here, but positive charge lowers its occupied orbital energies and suppresses backdonation. Sigma donation and the cation's electrostatic polarization can strengthen the internal C–O bond. Consequently the usual free-CO reference, about $2143\mathrm{c}{\mathrm{m}}^{-1}$, lies below this cation's band. A blue shift is compatible with carbon coordination; it does not imply absence of a metal–CO bond. Coupled normal modes and measurement conditions limit literal extraction of a local bond force constant from one band.