Lead is not known to form an iodide. PbI4. Give reason
Pb being a metal would form ionic compounds. For that first the ions need to be formed, Pb would form the cation and Iodine, the anion.
Formation of a cation requires energy called Ionization energy which is endothermic process even for atoms like Na that lose their electron easily.
Iodine on the other hand releases energy (electron affinity) when it forms the anion.
The magnitude of ionization energy is always much larger in magnitude than electron affinity and electron affinity alone cannot provide sufficient energy to form ionic compounds. Lattice energy is the energy that provides the required energy to form the ionic compounds. This the energy released when the ionic lattice is formed.
Pb has 4 electrons in its outermost shell, two are in s orbital and 2 in p orbital. Due to d-block contraction the s orbital electrons are more strongly held than s electrons in lower periods in the same group. The s electrons are inert and therefore the 3rd and 4th ionization energy that requires the s electrons to be removed are exceptionally high. The lattice energy for formation of PbI4 cannot compensate for this high ionization energy and therefore PbI4 is not formed.
Formation of a cation requires energy called Ionization energy which is endothermic process even for atoms like Na that lose their electron easily.
Iodine on the other hand releases energy (electron affinity) when it forms the anion.
The magnitude of ionization energy is always much larger in magnitude than electron affinity and electron affinity alone cannot provide sufficient energy to form ionic compounds. Lattice energy is the energy that provides the required energy to form the ionic compounds. This the energy released when the ionic lattice is formed.
Pb has 4 electrons in its outermost shell, two are in s orbital and 2 in p orbital. Due to d-block contraction the s orbital electrons are more strongly held than s electrons in lower periods in the same group. The s electrons are inert and therefore the 3rd and 4th ionization energy that requires the s electrons to be removed are exceptionally high. The lattice energy for formation of PbI4 cannot compensate for this high ionization energy and therefore PbI4 is not formed.