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Rationalise the givenstatements and give chemical reaction : (i) lead (II) reacts with Cl_(2) givePbCl_(4) (ii) lead (IV) chloride is highly unstable towards heat (iii) lead is knownnot to form an iodide, PbI_(4) |
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Answer» Solution :(i) Due to inert pair EFFECT, Pb shows oxidation states of `+2`and `+4`. Since `Cl_(2)` is a STRONG oxidising agent, it oxidises `Pb^(2+)` to `Pb^(4+)` and hence `PbCl_(2)` reacts with `Cl_(2)` to FORM `PbCl_(4)`. `PbCl_(2) (s) + Cl_(2) (g) rarr PbCl_(4)(l)` (ii) Due TOGREATER stability of `+2` over `+4`oxidation state because of inert pair effect , lead (IV) chloride on heating decomposes to give lead (II) chloride and `Cl_(2)`. `PbCl_(4) (l) overset(Delta)rarrPbCl_(2) (s) + Cl_(2) (g)` (iii) Due to oxidising power of `Pb^(4+)` ions and reducingpower of `l^(-)` ion, `PbI_(4)`,does not exist. Alternatively, `Pb-I` bond initially formed during the reactiondoes not releaseenough energy to unpair`6r^(2)` electrons and exciteone of them to the higher`6P`-orbital to have fourunpairedelectronsaround lead atom needed for formationof `PbI_(4)`. |
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