= 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>.
\b[$\mathrm{Ni(PPh_3)(CO)_3}$ is Ni(0), $d^{10}$, 18-electron and tetrahedral.] The count is $10+2+3(2)=18$. With only one distinct ligand in an $\mathrm{ML_3L'}$ <tetrahedral molecular geometry>, there is only one geometrical arrangement.
\b[$\mathrm{Rh(PPh_3)_2(CO)Cl}$ 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.
\b[$\mathrm{Pd[P(^{t}Bu)_2Ph]_2Cl_2}$ 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.
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