Which of the following d-electron distribution of the diamagnetic complex `[Ni(CN)]^(2-)` is possible according to the crystal field theory ?

Options

A. `(d_(yz))^(2) (d_(xy))^(2) (d_(z^(2)))^(2) (d_(xy))^(2)`
B. `(d_(xy))^(2) (d_(xz))^(2) (d_(yz))^(2) (d_(x^(2)-y^(2)))^(1) (d_(z^(2)))^(1)`
C. `(d_(xy))^(2) (d_(xy))^(2) (d_(yz))^(2) (d_(x^(2)-y^(2)))^(2)`
D. `(d_(xy))^(2) (d_(xz))^(2) (d_(yz))^(2) (d_(z^(2)))^(2)`

Answer

A

Explanation

For the period 4 transition metals, it is only Ni that tends to form square planer complexes, such as the tetracyanidonikelate (II) ion, `[Ni(CN)_(4)]^(2-)` . These complexesare diamagnetic. We can develop a crystal field diagram to see why this is so, even though both octahedral and tetrahedral geometries result in two unpaired electrons for the `d^(8)` configuration. If we start form the octahedral field withdrawn the ligands form the z-axis, the `d_(z)^(2)` orbital will no longer g=feel the electrostatic repulsion form axial ligands, hence, it will drop substantially in energy. The other two orbitals with a-z axis component (the `d_(xz)` and `d_(yz)`) will also undergo a decrease in energy. Conversely, with the withdrawl of the aial ligands, there will be a greater electrostatic attraction on the ligands in the plane and these will become closer to the metal ion. As a result, the `d_(x^(2)-y^(2))` and `d_(z^(2))` orbitals will increase substantially in energy.

As the `Ni^(2+)` complex with `CN^(-)` ligands is diamagnetic, the splitting of the `d_(x^(2)-y^(2))` and the `d_(xy)` orbitals must be greater than the pairing for this combination. In the square planar case, only low spin complexes have been found.