 Solution

For electron counting, use the ionic method: subtract the metal oxidation state from its group number, then add two electrons for each neutral phosphine ligand or carbonyl ligand and two for each chloride ligand.

$\mathrm{Ni}(\mathrm{PP}{\mathrm{h}}_3)(\mathrm{CO}{)}_3$ is Ni(0), $d^{10}$, 18-electron and tetrahedral. The count is $10+2+3(2)=18$. With only one distinct ligand in an $\mathrm{M}{\mathrm{L}}_3{\mathrm{L}}^{\prime }$ tetrahedral molecular geometry, there is only one geometrical arrangement.

$\mathrm{Rh}(\mathrm{PP}{\mathrm{h}}_3{)}_2(\mathrm{CO})\mathrm{Cl}$ is Rh(I), $d^8$, 16-electron and square planar, with the two phosphines trans. Its count is $8+2(2)+2+2=16$. A square planar molecular geometry is characteristic of this low-spin $4d^8$ configuration. Opposite triphenylphosphine groups avoid their largest mutual steric hindrance; CO and chloride occupy the other trans pair.

$\mathrm{Pd}[\mathrm{P}{(}^{\mathrm{t}}\mathrm{Bu}{)}_2\mathrm{Ph}{]}_2\mathrm{C}{\mathrm{l}}_2$ is Pd(II), $d^8$, 16-electron and square planar, with the bulky phosphines trans. Its count is likewise $8+4+4=16$. The two di-tert-butylphenylphosphine groups are particularly crowded if cis. These predictions concern the favored intact four-coordinate species; ligand dissociation can occur under other conditions.