From the magnetic behaviour of [NiCl4]2- (paramagnetic) and [Ni(CO)4] (diamagnetic), choose the correct geometry and oxidation state.
- [NiCl4]2- : Ni(0), tetrahedral [Ni(CO)4] : Ni(0), square planar
- [NiCl4]2- : NiII, tetrahedral [Ni(CO)4] : Ni(0), tetrahedral
- [NiCl4]2- : NiII, tetrahedral [Ni(CO)4] : NiII, square planar
- [NiCl4]2- : NiII, square planar [Ni(CO)4] : Ni(0), square planar
Answer
(2) [NiCl4]2- : NiII, tetrahedral [Ni(CO)4] : Ni(0), tetrahedral
Solution
In [NiCl4]2- nickel is Ni2+ (3d8), and since Cl- is a weak-field ligand there is no pairing:
[figure]
Hence it is sp3, tetrahedral, with 2 unpaired electrons (paramagnetic).
In [Ni(CO)4] nickel is Ni(0), 3d10, because strong-field CO pairs the 4s electrons into 3d:
[figure]
Hence it is sp3, tetrahedral and diamagnetic.
Answer: (2)