The same procedure can be applied to any transition metal complex. For example, consider the complex [Cu(NH3)4]2+. Because ammonia is a neutral ligand, Cu is in the 2+ oxidation state. Copper (II), in group 11 of the periodic table has 11 electrons in its valence shell, minus two, leaving it with 9 d-electrons (3d9). In the neutral complex [Rh(OH)3(H2O)3], Rh is in the +3 oxidation state and is in group 9, so the electron count is 4d6. Zinc(II) in group 12 would have 10 d-electrons in [Zn(NH3)4]2+, a full shell, and manganese (VII) has zero d-electrons in MnO4-. Nickel carbonyl, Ni(CO)4, contains the neutral CO ligand and Ni in the zero oxidation state. Since Ni is in group 10, we count the electrons on Ni as 3d10.
A frequent source of confusion about electron counting is the fate of the s-electrons on the metal. For example, our electron counting rules predict that Ti is 3d1 in the octahedral complex [Ti(H2O)6]3+. But the electronic configuration of a free Ti atom, according to the Aufbau principle, is 4s23d2. Why is the Ti3+ ion 3d1 and not 4s1? Similarly, why do we assign Mn2+ as 3d5 rather than 4s23d3? The short answer is that the metal s orbitals are higher in energy in a metal complex than they are in the free atom because they have antibonding character. We will justify this statement with a MO diagram in the next section.