Rationalise the given statements and give chemical reactions.
a. Lead(II) chloride does not react with `Cl_(2)` to give `PbCl_(4)`.
b. Lead(IV) chloride is highly unstable towards heat.
c. Lead is known not to form an iodide, `Pbl_(4)`.

Explanation

(a) Lead (pb) in +2 oxidationstate i.e., pb (II) is more stable than in +4 oxidation state i.e., Pb (IV) this means that pb (II) chlride will not react with chlorine to form PB (IV) chloride
`PbCl_(2)(s)+Cl_(2)(g)rarrPbCl_(4)(g)`
(b) Lead in (II) oxidation state is more stable than in lead in (IV) oxidation state. Therefore lead (IV) chloride is high unstable to heat. it decomposes upon heating to form lead (II) chloride.
`PbCl_(4)overset("Heat")rarrPbCl_(2)+Cl_(2)` ltrbgt `Pb (IV) Pb(II)`
(c) Lead is not known to form `PBl_(4)` because `I^(-)` ion being a powerful reducting agent reduces `Pb^(4+)` ion to `Pb^(2+)` ion solution. These `PbI_(2)` is generally formed.
`Pb^(4+)+2I^(-)rarrPb^(2+)+I_(2)`
Pb (IV) Pb(II)