Question

Account for the following statement:
The enthalpy of atomisation of zinc \[(Zn)\] is lowest in \[3d\] series of the transition elements.

Answer 1

Hint: Enthalpy of atomisation is the energy released or absorbed during the metal – metal bond formation. And this enthalpy of atomisation depends on how strong or weak the bond is between two metals.

Complete answer: To solve this question we have to keep in mind or have to look for the electronic configuration of the atom.
Since, we know that zinc is the element of the \[3d\] series, with an atomic number of \[30\] .
Now, this electronic configuration will determine the strength and the possibility of metal –metal bonding.
So writing the valence electronic configuration of zinc \[(Zn)\] - \[3{d^{10}}4{s^2}\]
Now, we can see that the valence shell of the zinc is completely filled and is symmetrical.
So being symmetrical, there is no or very less chances of metal-metal bonding.
So, if the metal-metal bond is weak then , the enthalpy of atomisation will also be very low , hence , this is the reason why zinc has lowest enthalpy of atomisation in the \[3d\] series i.e. due to complete symmetrical valence shells.

Note:
Valence shells are completely filled and have no lone pairs of electrons. Also we know that lone pairs of electrons help in the bond formation between the elements. Since, there are no free electrons or no lone pairs of electrons are present in zinc, so no bonds or very weak bonds will be formed. Otherwise in general, the transition elements have high enthalpy of atomisation because they have more number of unpaired electrons, which help in the easy bond formation and hence stronger metallic bonds with the transition elements and hence, high enthalpy of atomisation.