Question

${PbCl}_4$ exists but ${PbBr}_4$ and ${PbI}_4$ do not because of:

• A. inability of bromine and iodine to oxidise ${Pb}^{2+}$ to ${Pb}^{4+}$
• B. bigger size of $B{r}^{-}$ and $I^{-}$
• C. more electropositive character of ${Br}_2$ and $I_2$
• D. none of the above

Answer 1

Answer: (A)

inability of bromine and iodine to oxidise ${Pb}^{2+}$ to ${Pb}^{4+}$

${PbCl}_4$ exists but ${PbBr}_4$ and ${PbI}_4$ do not because $Pb-I$ and $Pb-Br$ bonds initially formed during the reaction do not release enough energy to unpair $6s^2$ electrons and excite one of them to higher orbital to have four unpaired electrons around lead atom. So, bromine and iodine are unable to oxidise ${Pb}^{2+}$ to ${Pb}^{4+}$.