= Solution
<platinum> is in group 10, so subtracting its <oxidation state> gives \b[$d^8$ for Pt(II) and $d^6$ for Pt(IV)]. Adding two <electrons> from each coordinated donor gives the local valence <electron counting> results
$$
\boxed{\begin{aligned}
\mathrm{cisplatin}:&\quad8+2(2)_{\rm NH_3}+2(2)_{\rm Cl}=16,\\
\mathrm{satraplatin}:&\quad6+2(2)_{\rm amine}+2(2)_{\rm Cl}+2(2)_{\rm OAc}=18.
\end{aligned}}
$$
Each <acetate> in the depicted structure is monodentate, binding through one <oxygen>. The 16-electron count of the square-planar complex is normal for a low-spin late-metal $d^8$ centre; it is not a missing ligand inferred from the <18-electron rule>.
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