 Solution

For an intact complex in the usual fast-conformational-averaging regime, counting distinct chemical shifts rather than individual split lines gives $C:9$, $D:11$, $E:10$ signals.

In $C$, the uncoordinated pyridinium arm permits the paired positions of a monosubstituted cyclopentadienyl ligand to average: its four ring hydrogens give two sets, $2+2$. The linker $\mathrm{C}{\mathrm{H}}_2$ gives one set, the four inequivalent pyridinium C–H positions give four, its N–H gives one, and the two symmetry-related bound acetonitrile methyl groups give one. Hence $2+1+4+1+1=9$. The exchangeable N–H resonance may be broadened or absent experimentally; nine is the structural count before exchange removes it.

In the rigid cis-dicarbonyl chelates $D$ and $E$, the different pyridyl-N and fourth-leg donors destroy the operation that related the two sides of the substituted ring. Each of the four Cp hydrogens is distinct, and the two linker hydrogens are diastereotopic. Together with four pyridine C–H sites, this gives $4+2+4=10$. The remaining bound acetonitrile adds one methyl signal to $D$, giving 11; $E$ has none, giving 10. The two enantiomers of a racemate have identical spectra in an achiral solvent and do not double these counts. This prediction assumes resolved resonances and no rapid chelate opening that averages inequivalent sites; variable-temperature proton nuclear magnetic resonance would test that assumption.