Why is the formation of lead (II) iodide precipitate favoured over that of the tetraiodo plumbate (II) complex ion in low concentrations of iodide?

I was wondering why

$$\ce{Pb2+ + 2I- <=> PbI2}$$

occurs in low $[\ce{I-}]$, however when $[\ce{I-}]$ is in excess, $\ce{PbI4^2-}$ is produced with

$$\ce{Pb2+ + 4I- <=> [PbI4]^2-}$$

Why is the formation of complex ion "favoured"? Furthermore, since the formation of insoluble precipitate will reduce entropy, wouldn't the formation of insoluble precipitate be favoured?

• @ Ong What do you mean by "Why is there no formation of complex ion "favoured"?" – Mockingbird Aug 4 '17 at 8:11
• @Mockingbird Hi there Mockingbird, my auto correct screwed up my sentence a bit. To correct myself, I wanted to know why formation of soluble tetraiodo plumbate(ii) ion complex is more observed than the formation of insoluble lead (ii) iodide (ie why the overall effect of added $\ce{I-}$ is the dissolving of ppt) – Ong Hai Xiang Aug 4 '17 at 9:21
• When $\ce{Cl^-}$ is in greater concentration, there are just enough ions to favor the 2nd reaction. – Mockingbird Aug 4 '17 at 10:07
• @Mockingbird is there any principle behind this? – Ong Hai Xiang Aug 4 '17 at 11:48
• Well, the rule of thumb is when you need to have grater number reactants for some reaction the greater concentration will work better. – Mockingbird Aug 4 '17 at 14:40