= 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:
$$
\begin{aligned}
\mathrm A:\quad&4(15)-12(4)=12\ \mathrm{skeletal\ electrons}=6\ \mathrm{pairs},\\
\mathrm B:\quad&3(16)-12(3)=12\ \mathrm{skeletal\ electrons}=6\ \mathrm{pairs}.
\end{aligned}
$$
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 \b[<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 \b[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 $\mathrm{Fe_4}$ 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.<Structural Characterization of the Cyclopentadienyliron Carbonyl Tetramer, Fe4(eta5-C5H5)4(CO)4. A Cubane-Type Structure Containing a Completely Bonding Tetrametal Cluster Unit by Melvin A. Neuman and others>[1972, pp.3383–3388, tetrahedral iron core and face-capping <carbonyls>]{ref}
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