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 following staements is true for the complex [Ni(PPh_(3))_(2)Br_(2)]?

Square planar complexes are generally formed by d^(8) ions with strong

1.	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 following staements is true for the complex [Ni(PPh_(3))_(2)Br_(2)]?
Answer» Hybridisation is the same as found in NICKEL `(+2)` complex with STRONG field ligands like CN. Hybridisation is the same as found in nickel (0) complex with strong field ligands like CO. Hybridisation is the same as found in nickel `(+2)` complex with weak field ligands like halide ions. (b) and (c ) both SOLUTION :`[Ni(PPh_(3))_(2)Br_(2)] implies sp^(3)`,tetrahedral `[Ni(CO)_(4)]""sp^(3)` `[NiCl_(4)]^(2-) "" implies sp^(3)`