Which of the following statements is not correct ?

Options

A. `[CuCI_(4)]^(2-)` has a tetrahedral geometry and is paramagnetic
B. `[MnCI_(4)]^(2-)` has a tetrahedral geometry and is paramagnetic
C. `[NiCI_(4)]^(2-)` has a tetrahedral geometry and is paramagenitc
D. `[Cu(NH_(3))_(4)]^(2+)` has a square planar geometry and is paramagnetic

Answer

D

Explanation

(1) `Cu(3d^(10) 4s^(1)) rarr Cu^(2+) (3d^(9))` . Since `C1^(-)` is a weak field ligand, no rearrangement of electrons is possible. The hybridization is `sp^(3)` and the complex ion `[CuC1_(4)]^(2-)` is paramagnetic due to one unpaired electron.
(2) `Mn(3d^(5) 4s^(2)) rarr Mn^(2+) (3d^(5))` . Since `C1^(-)` is a weak field ligand, no rearrangement of electrons is possible. The hybridization is `sp^(3)` and the complex ion `[MnC1_(4)]^(2-)` is paramagnetic due to 5 unpaired electrons.
(3) `Ni(3d^(8) 4s^(2)) rarr Ni^(2+) (3d^(8))` . Since `C1^(-)` is a weak field ligand, no rerrangement of electrons is possible. Thus, the hybridization is `sp^(3)` and the complex ion `[NiC1_(4)]^(2-)` is paramagnetic due to two unpaired electrons.
(4) `Cu(3d^(10) 4s^(1)) rarr Cu^(2+) (3d^(9))` . Since `NH_(3)` is an intermediate field ligand and `Cu^(2+)` ion has relatively higher charge density possible. In this process, one electron is shifted form 3d to 4p subshell so as to provide one 3d orbital for `dsp^(2)` hybridization. Thus, the complex has square planer geometry and is paramagnetic due to one unpaired electron.