The geometries of `[Ni(CO)_(4)]` and `[NiCI_(2)(PPh_(3))-(2)]` are

Options

A. tetrahedral
B. square planar
C. tetraheal and square planar repectively
D. square planer and tetrahedal respectively

Answer

A

Explanation

Oxidation state of Ni in the complex `[Ni(CO)_(4)]` is '0' . The ground state electron configuration of the Ni atom is `[Ar]3d^(8) 4s^(2)` , where three doubly occupied 3d orbitals and two dsingle occupide 3d . Since CO is the strongest possible ligands, it helps Ni atom to reshuffle its configuration. As a results, the 4s electrons pairup with the two unpaired 3d electrons giving an unoccupied 4s orbital. This vacant 4s orbital can be hybridized with three 4p orbital to give the tetrahedral `sp^(3)` hybrds. The orbital diagram for the complex molecule is:

`sp^(3)` hybrid orbitals accommodate four pairs of electrons form four CO molecules and the resulting tetrahedral complex is diamagnetic due to absence of absence of unpaired electrons. Several tetrahedral `Ni(+2)` complexes are known. These generally contain ligands, and often phosphine, phosphine oxide or arsine ligands as well, as in `[Ph_(4)As]_(2)^(+) [NiC1_(4)]^(2-)` , `[NiC1_(2)(PPh_(3))_(2)]` and `[NiBr_(2)(Ph_(3)AsO)_(2)]` . These complexes are typically intensely blue coloured and can be easily distinguished form square planar complexes both by the colour, and because they are paramagnetic. Note that `Pd(+2)` and `Pt(+2)` complexes are all square planar.