= Solution
Let $z$ be normal to the Cp ring through its centre. The five <carbon> $p_z$ orbitals form <symmetry-adapted linear combinations> with angular patterns $m=0$, $|m|=1$, and $|m|=2$. The all-in-phase $a_1$ combination has sigma symmetry about this axis and can interact with $d_{z^2}$. The $e_1$ pair matches $(d_{xz},d_{yz})$ by pi symmetry, while the $e_2$ pair matches $(d_{xy},d_{x^2-y^2})$ by delta symmetry. Rotating within either degenerate pair changes the drawing but not the physical interaction.
For the usual eta-five geometry, \b[the <metal> $d$ interactions are ranked qualitatively pi > delta > sigma]. The $d_{xz},d_{yz}$ lobes overlap effectively with the occupied ring combinations. The delta interaction is weaker and involves the higher, initially unoccupied Cp combinations; the $d_{z^2}$ interaction with the ring is particularly weak because its axial and equatorial parts contribute little net overlap in this geometry. Thus the statement that a <sigma bond> is always stronger than a <pi bond> cannot be transferred indiscriminately to this delocalized ring–<metal> overlap problem. <metal> $s$ and $p_z$ orbitals also have the symmetry of the $a_1$ <coordination ligand> combination and contribute to overall bonding, so weak $d_{z^2}$ overlap does not mean the entire sigma-symmetry channel is absent.
In ionic <electron counting>, Cp is $\mathrm{C_5H_5^-}$ with six pi <electrons> in its three occupied ring combinations, hence \b[a six-electron donor]. The same <coordination ligand> is assigned five <electrons> in neutral counting.
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