Q.How can you explain higher stability of BCl3 as compared to TlCl3?

Step 1 - Boron: no inert pair effect

Boron has configuration $2s^22p^1$ with no intervening d or f electrons; it forms $\text{BCl}_3$ by using all three valence electrons in bonding ($sp^2$ hybridisation), and this $+3$ state is fully stable - there is essentially no inert pair effect at boron's small size/low period.

Step 2 - Thallium: strong inert pair effect

Thallium, a much heavier group 13 element ($6s^26p^1$), shows a pronounced inert pair effect: poor shielding by the filled $4f$ and $5d$ subshells means the $6s^2$ electron pair is held close to the nucleus and is reluctant to participate in bond formation.

Step 3 - Consequence for TlCl3

Consequently $\text{Tl}^{+1}$ (using only the single $6p$ electron) is the thermodynamically more stable oxidation state for thallium, while $\text{Tl}^{3+}$ compounds such as $\text{TlCl}_3$ are strong oxidising agents that are comparatively unstable and decompose, releasing chlorine and going to the more stable $\text{TlCl}$:

Step 4 - Comparison

Hence $\text{BCl}_3$ (boron fully using all 3 valence electrons, no inert pair effect) is markedly more stable than $\text{TlCl}_3$ (thallium's $6s^2$ pair resisting participation, +3 state disfavoured).

Contrast boron (period 2, no inert pair effect, +3 fully stable) with thallium (period 6, strong inert pair effect from poor 4f/5d shielding, +1 favoured); write the decomposition of TlCl3 to TlCl + Cl2 to demonstrate its instability.

• Attributing TlCl3's instability to simple "size" effects rather than specifically the inert pair effect.
• Thinking boron also shows an inert pair effect (it does not - the effect is significant only for heavier p-block elements).
• Failing to give the decomposition reaction that demonstrates TlCl3's instability.