 Solution

The methoxy substituent donates electron density into the coordinated arene by chemical resonance. This increases the electron density available at chromium for pi backbonding to CO, weakening C–O and lowering its stretching frequencies. The ester substituent withdraws electron density from the arene, reducing pi backbonding and raising the carbonyl stretching frequencies. For both observed bands, the methoxy-substituted compound has the lowest carbonyl frequency, the unsubstituted compound is intermediate, and the ester-substituted compound has the highest.

In the harmonic oscillator approximation,so the trend reflects an increasing effective C–O bond force constant rather than a change in the CO reduced mass. For example, comparison of the higher-frequency bands gives $k_{\mathrm{ester}}/k_{\mathrm{methoxy}}\approx (1992/1983{)}^2=1.0091$. This is only an approximate local-bond interpretation because the observed normal modes couple the three CO oscillators. In the idealized local $C_{3v}$ geometry they span $A_1+E$, giving two infrared-active frequencies. A substituted arene can lower the exact molecular symmetry and split the $E$ pair, even if that splitting is unresolved experimentally.