= Solution
\b[<zinc> reduction is required in A, C and F.] Reading the drawn ligands with consistent ionic bookkeeping gives
$$
\boxed{\begin{array}{c|c|c}
\text{step}&\text{nickel change}&\text{electrons accepted per Ni}\\\hline
\mathrm A&\mathrm{Ni^{II}\to Ni^0}&2\\
\mathrm C&\mathrm{Ni^{II}\to Ni^I}&1\\
\mathrm F&\mathrm{Ni^I\to Ni^0}&1
\end{array}}
$$
In A the two <chlorides> of the precursor are removed and neutral <solvent> donors occupy the reduced centre. In C the <bromide> is removed while the <aryl ligand> remains and the neutral <alkyne> coordinates; the drawn neutral intermediate has Ni(I). In F an anionic alkoxide-derived product is released and the neutral solvated Ni(0) complex is regenerated. <zinc> supplies <electrons> by oxidation to Zn(II); halide and <alkoxide> ions can be taken up in zinc-containing products.
A is <precatalyst> activation. The two one-electron reductions C and F together require two <electrons> per catalytic turnover, corresponding to one Zn atom if its oxidation ends at $+2$. The diagram does not establish whether each formal reduction is a single elementary electron-transfer event or a sequence of surface reactions and ligand transfers.
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