Can ammonia complex with copper or chromium?

1 Answer

Answer:

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.