# Can ammonia complex with copper or chromium?

Apr 10, 2015

Yes. Ammonia acts as a nice, strong-field ligand for the metal-ligand complexation reaction with Cu(I), Cu(II), and Cr(III).

#### Explanation:

EX:
http://www.ncbi.nlm.nih.gov/pubmed/20030372 (Cu(I) and Cu(II) complexes)
http://alpha.chem.umb.edu/chemistry/ch371/documents/Labreportexample.pdf (hexamminechromium(III) nitrate)

Cu(I) is a transition metal that forms a ${d}^{10}$ tetrahedral complex (diamagnetic due to full occupation of the ${d}_{{z}^{2}}$, ${d}_{x y}$, ${d}_{x z}$, ${d}_{y z}$, and ${d}_{{x}^{2} - {y}^{2}}$ orbitals) with ammonia in Crystal Field Theory, while Cu(II) forms a ${d}^{9}$ octahedral complex (paramagnetic due to full occupation of the ${d}_{{z}^{2}}$, ${d}_{x y}$, ${d}_{x z}$, ${d}_{y z}$, and ${d}_{{x}^{2} - {y}^{2}}$ orbitals, minus one electron).

Because ammonia is a strong-field ligand, it more than likely causes a low spin state. All that means is that ammonia feels high metal-ligand repulsions with the transition metal, increasing the energy gap between the higher and lower energy d orbitals. The higher the gap, the more likely it is in a low-spin state, and vice versa.

The d-orbitals split into 2 ${e}_{g}$ (higher energy, i.e. ${d}_{{z}^{2}}$ and ${d}_{{x}^{2} - {y}^{2}}$) and 3 ${t}_{2 g}$ (lower energy, i.e. ${d}_{x y}$, ${d}_{x z}$, and ${d}_{y z}$) orbitals for octahedral complexes, whereas for tetrahedral complexes it depends on the metal-ligand spatial orientation with respect to the d orbitals.

Cr(III) forms a ${d}^{3}$ complex with ammonia, similarly, and forms a low spin octahedral complex that is paramagnetic.