$ 1.\,\,{{\left[ Ni{{(CN)}_{4}} \right]}^{2-}} $

$ 2.\,\,[Ni{{(CO)}_{4}}] $

$ 3.\,\,{{\left[ CO{{(N{{H}_{3}})}_{6}} \right]}^{3+}} $

$ 4.\,\,{{[NiC{{l}_{4}}]}^{2-}} $

Solution:  $ 4.\,\,{{[NiC{{l}_{4}}]}^{2-}} $

In (NiCl₄)^2-,

Here, Cl is a weak field ligand that does not aid in electron pairing, leaving nickel with two unpaired electrons, causing it to become paramagnetic with an sp3 structure.