Question

Order of boiling point and melting point of $Zn>Cd>Hg$ respectively is
A) $Zn>Cd>Hg$
B) $Hg>Cd>Zn$
C) $Hg=Cd=Zn$
D) None of these

Answer 1

Hint: The relative strength of the four intermolecular forces is: ionic > hydrogen bonding > dipole-dipole > van der waals dispersion forces. The influence of these attractive forces will depend on the functional groups present.

Complete answer:
Because $Hg,Zn$ and $Cd$ every one of the electrons in $d-$ subshell is paired. Thus the metallic bonds available in them are feeble. That is the reason they are soft metals and have low melting and boiling point.
The boiling point of $Zn$ is ${907}^{\circ }C$ , $Cd$ is ${766.8}^{\circ }C$ and $Hg$ is ${357}^{\circ }C$ and melting point of $Zn$ is ${120}^{\circ }C$ $Cd$ is ${321}^{\circ }C$ and $Hg$ is $-{39}^{\circ }C$ .Therefore, the order boiling point and melting point is $Zn>Cd>Hg$.
The correct option is $(A)$.

Additional Information:
- $Hg,Zn,Cd$ are altogether broadly utilized in electric and electronic applications, just as in different alloys. The first two individuals from the gathering share comparable properties as they are strong metals under standard conditions. Mercury is the lone metal that is a fluid at room temperature. While zinc is vital in the biochemistry of living organisms, cadmium and mercury are both exceptionally harmful
-metallic bonds are formed when an inflexible, definite lattice of metal cations share a sea of delocalized valence electrons.

Note:
 $Zn,Hg,Cd$ are not transition elements on the account of their electronic configuration. The overall representation of the electronic configuration of these three elements is $(n-1)d^{10}n{s}^2$ . The orbitals of these elements are totally filled when they are in their ground state just as in their overall oxidation state.