 Solution

Neutral counting assigns five electrons to cyclopentadienyl ligand and two to CO. Thus each $\mathrm{FeCpCO}$ fragment has $8+5+2=15$ valence electrons and contributes three skeletal electrons; each $\mathrm{CoCpCO}$ fragment has sixteen and contributes four:For A the six-pair four-vertex count is the nido count derived from a five-vertex trigonal bipyramid. Removing an axial vertex leaves a tetrahedron: the three equatorial vertices plus the remaining axial vertex. For B the corresponding arachno fragment is obtained by removing both axial vertices from that same parent, leaving a three-metal triangle. Small clusters require specifying the parent and removed vertices; mechanically calling every nido fragment a square pyramid is wrong.

The Fe tetramer is indeed known to have a tetrahedral $\mathrm{F}{\mathrm{e}}_4$ core, with four face-capping CO groups. The carbon atoms of those carbonyls and the iron atoms form interpenetrating tetrahedra, giving a cubane-like Fe–C connectivity. Bridging rather than terminal CO changes the distribution of local donation, not the total electron count above.^[ref]