PbI₄ does not exist because

bigger size of iodine cause of the reducing nature of the iodine

thus it would reduce lead to form the compound PbI2

Here in PbI₄, Pb has +4 oxidation state. Now the electronic configuration of Pb is [Xe]4f¹⁴5d¹⁰6s²6p². So we can see that there is 14 f electrons which have very poor shielding effect. So the effective nuclear charge on the outer s & p electrons are increased. So they are stongly attracted towards nucleus and hence becomes inert. So the Pb⁴⁺ becomes unstable and shows stong oxidising property. So it reduced to Pb²⁺. On the other hand, I^- shows high reducing property. So the association of these two ions are quite impossible. Hence PbI₄ does not exist.  

PbI4 does not exist because, During the formatio of Pb-I enough amount of energy is not released so as to unpair one of the 6s^2 electron and excite one of them to higher orbital to have a lone electron around the central atom.

PbI4 does not exist because pb-I bond initially formed during the reaction does not release enough energy to unpair 6S²  electron and excite 1 electron around pb Central atom