Step 1: Start from the known physical fact: ferrocene is diamagnetic.
Ferrocene has no unpaired electrons, confirmed by experiment. This rules out any option that leaves electrons unpaired across two different-energy sets, and means all six $d$-electrons of $\mathrm{Fe^{2+}}$ ($d^6$) must sit fully paired.
Step 2: Use the geometry of each orbital relative to the two Cp rings.
Picture the rings as flat pentagons above and below the metal, with the sandwich axis as $z$. $d_{xy}$ and $d_{x^2-y^2}$ lie entirely in the equatorial $xy$ plane, roughly between the ring carbon positions, so they barely interact with the ring $\pi$ clouds: a spectator pair, lowest in energy. $d_{z^2}$ points directly up and down the $z$-axis toward the ring centroids, giving a small $\sigma$-type interaction, a middle-energy orbital. $d_{xz}$ and $d_{yz}$ tilt toward the ring carbons and interact most strongly with the filled ring $\pi$ orbitals, pushing this pair highest.
Step 3: Fill six electrons from the bottom up, respecting degeneracy.
The lowest doubly-degenerate pair ($d_{xy},d_{x^2-y^2}$) takes 4 electrons, the next single orbital ($d_{z^2}$) takes the remaining 2, and the highest doubly-degenerate pair ($d_{xz},d_{yz}$) is left empty. Every electron is paired, consistent with diamagnetism.
Step 4: Write the final configuration and confirm against options.
\[ d_{xy}^2 = d_{x^2-y^2}^2 < d_{z^2}^2 < d_{xz}^0 = d_{yz}^0 \]
Option (B) puts the near-nonbonding pair as highest energy, contradicting the weak-overlap argument. Option (C) removes the distinct level for $d_{z^2}$. Option (D) lists four levels instead of the required 2:1:2 degeneracy pattern. Only (A) survives.
Final Answer:
Option (A) is correct.
\[\boxed{d_{xy}^2 = d_{x^2-y^2}^2 < d_{z^2}^2 < d_{xz}^0 = d_{yz}^0}\]