Take the coordination complex, #[Co(NH_3)_6]X_3#. This is a real complex consisting of a #Co^(3+)# ion that is bound to six ammine ligands that are distributed around the metal centre in an octahedral array. The ammine ligands certainly have lone electron pairs to donate to the metal centre (and the metal centre certainly has orbitals capable of accommodating them), but the overall charge of the complex is still #+3#; i.e. we represent the coordination complex as #[Co(NH_3)_6]^(3+)#.
And thus for this complex (a classic Werner complex), we would say that the #"SECONDARY valence"# of cobalt is #6#, and the #"PRIMARY valence"#, the charge on the cobalt ion, is #3+#. And this complex would require 3 equiv #Ag^+# for equivalence in a Mohr titration (to determine halide content). The chloride ligands are #"gegenions"#, counterions simply along the ride. Exchange (somehow - this is the province of experiment) the ligands (the coordinating species) to give say #[Co(NH_3)_5Cl]^(2+)#, the chloride ligand has now entered the primary sphere of coordination, and is directly bound to the metal centre. And thus only 2 equiv of silver halide would result in a Mohr titration if given #[Co(NH_3)_5Cl]Cl_2#. The primary valence is now #+2#; the secondary valence is still #+6#.
Note that this may seem trivial now, but cast yourself back to Werner's time. Could you simply account for KNOWN complexes with formulae such as #Co(NH_3)_5Cl_3# versus #Co(NH_3)_6Cl_3#. Was it simply an error? But hang on, #Co(NH_3)_5Cl_3#, #Co(NH_3)_6Cl_3# have consistently different chemical and physical properties. What's going on?
There should be good articles in your library, in #"J. Chem. Educ."# and elsewhere, that deal with Werner's approach. It is up to you to find them (your librarian will certainly help you). Also important in this context was Werner's correct interpretation of geometric isomerism observed for square-planar and octahedral coordination complexes. A square planar complex, say #Ni(PR_3)_2Cl_2#, UNLIKE a tetrahedral complex such as #CH_2Cl_2#, will generate 2 geometric isomers......which are...........?
Anyway I take it that this is a 2nd year inorganic chemistry question.
