Question

In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest. i.e.126 kJ$mo{l^{ - 1}}$. Why?

Answer 1

Hint: The change in the enthalpy due to the breaking of bond between the atoms of the molecule to form atoms is known as enthalpy of atomization. The metallic bond is formed by the sharing of electrons by the metal cation.

Complete step by step answer:
The enthalpy of atomization is defined as the change in the enthalpy when the bond between the atoms of the compound is broken to give individual atoms in a gas phase.
The extent to which the metal can form metallic bonds determines the enthalpy of atomization. The more extensive is the metallic bonding, more will be the enthalpy of atomization.
The element from Scandium to zinc belongs to the third group element. Due to the presence of unpaired electrons in the valence shell of the element results in the sharing of electrons and forms stronger metallic bonding. The metallic bonding is observed in two metal atoms.
From scandium to zinc all the elements contain unpaired electrons in their valence shell except zinc whose atomic weight is 30. The electronic configuration of zinc is $[Ar]3{d^{10}}4{s^2}$. As in valence shell 10 electrons is present in 3d orbital and 2 electrons are present in 4s orbital, all the orbital contains paired electrons no unpaired electrons are present.
Therefore, the enthalpy of atomization of zinc is the lowest.

Note: In a diatomic molecule, the enthalpy of atomization is the same as the enthalpy of bond dissociation energy. The other factors which affect the strength of the metallic bond is number of delocalized electrons, the magnitude of the positive charge present on the metal cation and the ionic radius of the cation.