Which of the following is paramagnetic and squre planar ?

Options

A. `[Ni(CN)_(4)]^(2-)`
B. `[PtCI_(4)]^(2-)`
C. `[FeCI_(4)]^(-)`
D. `[NiCI_(4)]^(2-)`

Answer

B

Explanation

Coordination number 4 implies either tetrahedral geometry or square planar geometry. Remember that nearly all tetrahedral complexes are high-spin (small `Dalta`) while square planar complexes have a large energy gap (large `Delta`) between the `d_(x^(2)y-^(2))` orbital and the four lower-energyorbitals. `[NiC1_(4)]^(2-)` and `[FeC1_(4)]^(-)` are tetrahedral while `[PtC1_(4)]^(2-)` and `[Ni(CN)_(4)]^(2-)` are square planar complexes. `Pt^(2+) ([Xe]4d^(14)5d^(8))` has eight d electrons. Because `Delta` is larger in square planar complexes, all the electrons occupy the four lower energy d orbitals. There are no unpaired electrons, and the complex is diamagnetic.
`[PtC1_(4)]^(2+)`

The five d electrons of `Fe^(3+) ([Ar]3d^(5))` are distributed between the higher and lower energy orbitals as shown below:
`[FeC1_(4)]^(-)`

Because `Delta` is small in tetrahedral complexes, `[FeC1_(4)]^(-)` is high spin with five unpaired electrons.