The species that contain a vacant d x2 - y2 orbital is :
[Ni(CN)4]2-
All four species are nickel complexes, so the question turns on the oxidation state, the d-electron count and above all the geometry, because geometry decides where dx2-y2 sits in energy.
[Ni(CN)4]2- is the answer. Nickel is Ni(II), a d8 ion, and cyanide is a strong-field ligand, so the complex is square planar and diamagnetic. In a square-planar field the ligands lie along the x and y axes, pointing straight at the lobes of dx2-y2. That orbital is therefore pushed far above the other four, and the eight d electrons fill dxy, dxz, dyz and dz2 completely, leaving dx2-y2 empty.
[Ni(en)3]2+ is octahedral Ni(II), d8. In an octahedral field the configuration is t2g6eg2, and the two eg electrons occupy dz2 and dx2-y2 singly. The orbital is half filled, not vacant, which is why the complex is paramagnetic with two unpaired electrons.
[NiCl4]2- is tetrahedral Ni(II), d8, because chloride is a weak-field ligand. The tetrahedral splitting places e (dz2, dx2-y2) below t2, so the e set is filled and dx2-y2 carries electrons.
[Ni(CO)4] is tetrahedral Ni(0), a d10 system. Every d orbital is doubly occupied, so none is vacant.
Hence the species with a vacant dx2-y2 orbital is [Ni(CN)4]2-.
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