= Solution
For the ideal <tetracyanonickelate(II)> geometries, the <square planar molecular geometry> has \b[$D_{4h}$], the <D4h molecular point group>, while the <tetrahedral molecular geometry> has \b[$T_d$], 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.
Back to article page
