PbI 4 does not exist because

PbI4 does not exist because


5 Answers

askIITians Faculty 747 Points
8 years ago
The Iodide reduces the lead to Pb(II) and the Pb oxidizes the Iodide to Iodine (I2). Since iodide is not a strong enough reducing agent to reduce Pb(II) to Pb, the ionic compoundPbI2 is formed. Pb4+ , being a strong acid (strong oxidant) will take up electrons from soft base (have fairly good reducing power) I- to oxidise it to I2 and it self will reduce to Pb2+ to form PbI2
2061 Points
8 years ago
bigger size of iodine cause of the reducing nature of the iodine
thus it would reduce lead to form the compound PbI2
Abesh Bhar
11 Points
6 years ago
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.  
Aswin Sankar PD
19 Points
5 years ago
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.
13 Points
3 years ago
PbI4 does not exist because pb-I bond initially formed during the reaction does not release enough energy to unpair 6S electron and excite 1 electron around pb Central atom

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