 Solution

For the ideal tetracyanonickelate(II) geometries, the square planar molecular geometry has $D_{4h}$, the D4h molecular point group, while the tetrahedral molecular geometry has $T_d$, the Td molecular point group. Collinearity of each Ni–C–N unit means that including the terminal nitrogen atoms does not lower either point group.

For the square, take the Ni–C directions as $+x,+y,-x,-y$ and $z$ perpendicular to the plane. For the tetrahedron, use the conventional cubic axes with ligand directions proportional to $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. These choices fix the irreducible representation labels used below.