[Solution by Claude - not from source] $[NiCl_4]^{2-}$: $Ni^{2+}$ $d^8$, tetrahedral, 2 unpaired e$^-$ (para). $Ni(CO)_4$: $Ni(0)$ $d^{10}$ (dia). $[Ni(CN)_4]^{2-}$: square planar (dia). $[Ni(H_2O)_6]^{2+}$: octahedral $d^8$, 2 unpaired (para). $Ni(PPh_3)_4$: $d^{10}$ (dia). So A and D. Answer (3).