# Green and red crystals of [Ni(Ph2EtP)2Br2]

From the reaction of $$\ce{NiBr2}$$ and $$\ce{Ph2EtP}$$, it is possible to isolate green and red crystals of $$\ce{[Ni(Ph2EtP)2Br2]}$$?

A) coordination number and geometries of the complexes?
B) possible isomeric structures for the green and red crystals.

I know that d orbitals effect the color but I don't know how to apply it to isomers.

The coordination number of $\ce{[Ni(Ph2EtP)2Br2]}$ is four;
thus, the complex may be tetrahedral

or square planar

(figures taken from here)

The complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains $\ce{Ni^2+}$, which has eight d electrons.

When filled with eight electrons, the tetrahedral structure is paramagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively small (4.4 Dq).

The corresponding square-planar structure is diamagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively large (10 Dq).

Therefore, the square planar complex absorbs light of higher energy (lower wavelength) than the tetrahedral complex.

The green crystals absorb red light, whereas the red crystals absorb green light. Thus, the red crystals absorb light of higher energy than the green crystals. Therefore, the red crystals correspond to the square-planar complex and the green crystals correspond to the tetrahedral complex.

Furthermore, since the complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains two different ligands, there are cis and trans isomers of the square-planar structure.

It's a well known experiment (J. Am. Chem. Soc. 1970, 92, 2691). Of course, it's possible to isolate green and red crystals $\ce{[Ni(Ph2EtP)2Br2]}$. The coordination number for both complexes is four.

The geometry of the complex with red crystals is square planar. It's diamagnetic. Hybridization of $\ce{Ni^2+}$ is $\ce{dsp^2}$ The geometry of the complex with green crystals is tetrahedral. It's paramagnetic. Hybridization of $\ce{Ni^2+}$ is $\ce{sp^3}$ Electron configuration of $\ce{Ni^2+}$: $\ce{[Ar] 4s^0 3d^8}$. We can explain the formation of the red crystals as follows:

On the other hand, we can explain the formation of the green crystals as follows:

As for possible isomers, there are two isomers (cis and trans) for the square planar complex since it has two kinds of ligands.