Question

The complex ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ is diamagnetic and the complex ${\left[NiC{l}_4\right]}^{2-}$ is paramagnetic. What can you conclude about their molecular geometries?

• A. Both complex have square planar geometries
• B. Both complexes have tetrahedral geometries
• C. ${\left[NiC{l}_4\right]}^{2-}$ has a square planar geometry while ${\left[Ni\right(CN)}_4{]}^{2-}$ has a tetrahedral geometry
• D. ${\left[NiC{l}_4\right]}^{2-}$ has a tetrahedral geometry while ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ has a square planar geometry

Answer 1

The correct option is D ${\left[NiC{l}_4\right]}^{2-}$ has a tetrahedral geometry while ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ has a square planar geometry

We have been asked about complexes in which Ni has a +2 oxidation state and coordination number 4. It is a $d^8$ configuration. $C{N}^{-}$ is a strong field ligand which means it forces pairing in the lower orbitals. $C{l}^{-}$ on the other hand, is a weak field ligand, which cannot force pairing of electrons.
$\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic due to which it forms an inner orbital complex. This tells us it will form a square planar complex.
$[NiC{l}_4{]}^{2-}$ is paramagnetic. Thus, it forms outer orbital complex which means it is tetrahedral in shape.