Question

Can ammonia complex with copper or chromium?

Answer 1

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_(xy)`, `d_(xz)`, `d_(yz)`, 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_(xy)`, `d_(xz)`, `d_(yz)`, 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_(2g)` (lower energy, i.e. `d_(xy)`, `d_(xz)`, and `d_(yz)`) 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.