Square planar complexes are generally formed by `d^(8)` ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having `4d^(8)" and "5d^(8)` configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism.
Which of the ions do not form the square planar complexes amongst the following ions ?

Options

A. `Ni(+II)" and "Rh(+I)` with strong field ligands
B. `Rh(+I)" and "Au(+III)` with strong as well as weak field ligands.
C. `Pd(+II), Pt(+II)" and "Au(+III)` with strong as well as weak field ligands.
D. None of the above

Answer

D

Explanation

`Ni^(2+),Rh^(+),Au^(3+),Pd^(2+),Pt^(2+)` form square planar complexes.