How can you explain higher stability of ${\text{BC}}{{\text{l}}_3}$ as compared to ${\text{TlC}}{{\text{l}}_3}$ ?

Hint: An effect is there in the elements of periodic table known as inert pair effect. In this effect the outermost two electrons of s-orbital remain unshared and we can see some changes in the behavior of elements during complex formations.

Complete step by step answer:
Inert pair effect: The effect seen in post-transition metals in which they have a tendency to make the outermost two electrons of the s-orbital unshared. As a result the two electrons get tightly held by the nucleus of an atom and this results in the no participation of those electrons in bond formation.
As we all know that boron and thallium are the elements of the same group so we expect the same properties but in reality they have some differences. They both exist with different oxidation states as their stable oxidation states.
${\text{B}}$ is stable in $ + 3$ oxidation state while ${\text{Tl}}$ is stable in $ + 1$ oxidation state.
Thallium shows both $ + 1{\text{ and + 3}}$ oxidation states but due to the inert pair effect it is stable in $ + 1$ oxidation state.
Therefore ${\text{BC}}{{\text{l}}_3}$ is more stable than ${\text{TlC}}{{\text{l}}_3}$.

Additional information:
The inert pair effect occurs in elements due to the decrease in bond energies down a group because as we move down a group we can see the increase in bond lengths. The post transition elements have two stable oxidation states but as we move down a group the lower oxidation state is seen to be more stable and this all happens due to the inert pair effect which resists the two electrons of outermost s-orbital to remain unshared.


Note:
The set of metallic elements which exist between metalloids and the transition metals in the periodic table are known as post – transition elements. These are basically soft metals having poor mechanical strength.