PbCl4 exist, but PbBr4 and PbI4 do not because of

PbCl4 exists but PbBr4 and PbI4 do not because Pb−I and Pb−Br bonds initially formed during the reaction do not release enough energy to unpair 6s2 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 Pb2+ to Pb4+.

PbCl4 exist, but PbBr4 and PbI4 do not because of

PbCl4 exists but PbBr4 and PbI4 do not because Pb−I and Pb−Br bonds initially formed during the reaction do not release enough energy to unpair 6s2 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 Pb2+ to Pb4+.

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PbCl₄ exist, but PbBr₄ and PbI₄ do not because of

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Br^--- and I^--- are bigger in size

Inability of bromine and iodine to oxidise Pb²⁺ to Pb⁴⁺

Chlorine is a gas

More electropisitive character of Br₂ and I₂

answer is A.

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Detailed Solution

PbCl₄ exists but PbBr₄ and PbI₄ do not because Pb−I and Pb−Br bonds initially formed during the reaction do not release enough energy to unpair 6s² 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 Pb²⁺ to Pb⁴⁺.

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