 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, 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 ${\mathrm{C}}_5{\mathrm{H}}_5^{-}$ with six pi electrons in its three occupied ring combinations, hence a six-electron donor. The same coordination ligand is assigned five electrons in neutral counting.