It depends. If you are asking does something like "free" Cr
3+ (free Cr
3+ is really [Cr(H
20)
6]
3+ in water) float around in solution completely separate of anions like NO
3- like we depict when we have NaCl become "free" Na
+ (free Na
+ is really [Na(H
20)
6]
+ in water) and "free" Cl
- ions that float around independently of each other or if the two species involved in the formula remain "connected" in water, then it depends on whether or not the transition metal binds more strongly with the ligand or with H
20 (ignoring entropy effects). Here is an example written showing an equilibrium for a transition metal in water:
[Cu(H
20)
6]
2+ + 4 NH3 (aq)
[Cu(NH3)
4(H20)
2]
2+ + 4H
20 Keq = 1 x 10
13Originally before adding NH
3, Cu
2+ is "free" in solution because it binds to H
20 stronger than its original anion (which is a spectator ion that binds weakly to Cu
2+ and isn't shown in the equation). However, when we react this free Cu
2+ with NH
3, the result is that the transition metal now becomes part of a complex cation [Cu(NH3)
4(H20)
2]
2+ in aqueous solution as opposed to being by itself in solution.