Question

In the series Sc $\left( {Z = 21} \right)$ to Zn $\left( {Z = 30} \right)$ , the enthalpy of atomization of which element is the least?
A.Sc
B.Mn
C.Cu
D.Zn

Answer 1

To solve this question, we must find the electronic configurations of these elements and then then determine their metallic character. The element with the weakest metallic bonds has the lowest enthalpy for atomization.

Complete step by step answer:
Before we move forward with the solution of this question, let us first understand some important basic concepts.
Elements exist in their elemental state either in the form of a single, or as a molecule. If they are forming molecules, then we know that each bond has a certain amount of energy. To break these bonds, we need to supply an equivalent amount of energy to the system. Also, atomization means formation or breaking of molecules into its constituent atoms. Hence, we can say that the enthalpy of atomization can be understood as the amount of enthalpy change when a compound's bonds are broken and the component atoms are reduced to individual atoms.
Now, the enthalpy of atomization depends on certain factors, which directly relate to bond enthalpy. The enthalpy of atomization would be lower if the strength of the bonds formed is low. Bond strength is directly related to 2 main factors: the difference in the electronegativities of the constituent atoms and atomic size of the constituent atoms.
Metallic bonds can be described as the sharing of detached electrons. These bonds are different from covalent or ionic bonds. The valence electrons in metals can easily be delocalised due to their low ionization enthalpies. The strength of a metallic bond depends on the following factors:
1.Magnitude of positive charge held by the metal ion
2.Radius of the metal ion
3.Total number of delocalised electrons
The electronic configurations of the given elements can be given as:
Sc $\left( {Z = 21} \right)$ : $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^1}$
Mn $\left( {Z = 25} \right)$ : $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}$
Cu $\left( {Z = 29} \right)$ : $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^{10}}$
Zn $\left( {Z = 30} \right)$ : $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}$
Hence, we can see that there are no free electrons present in the valence shell of zinc. Hence, zinc has the weakest metallic bonds and as a result has the lowest atomization enthalpy.

Hence, Option D is the correct option.

Note:
Metallic bonds are not broken when the metal is heated into the melt state. Instead, these bonds are weakened, causing the ordered array of metal ions to lose their definite, rigid structure, and become liquid.