Account for the following observations:
$(A)$ $AlCl_3$ is a Lewis acid.
$(B)$ Though fluorine is more electronegative than chlorine yet $BF_3$ is a weaker Lewis acid than $BCl_3$.
$(C)$ $PbO_2$ is a stronger oxidizing agent than $SnO_2$.
$(D)$ The $+1$ oxidation state of thallium is more stable than its $+3$ state.

(N/A) In $AlCl_3$,the octet of $Al$ is incomplete as it has $6$ electrons and accepts a pair of electrons. Electron acceptors are Lewis acids.
$(B)$ In $BF_3$,boron has a vacant $2p$ orbital and fluorine has one of the $2p$ orbitals completely filled. Both have the same energy and can overlap effectively to give $p\pi-p\pi$ back bonding.
While such type of bonding is not possible in $BCl_3$ as there is no effective overlapping between the $2p$-orbital of boron and $3p$-orbital of chlorine.
Therefore,the electron deficiency of $B$ in $BF_3$ is reduced,making it a weaker Lewis acid than $BCl_3$.
$(C)$ In $PbO_2$ and $SnO_2$,both lead and tin are present in the $+4$ oxidation state. Due to the inert pair effect,$Pb^{2+}$ is more stable than $Pb^{4+}$,making $PbO_2$ a strong oxidizing agent that readily reduces to $Pb^{2+}$. $SnO_2$ is more stable in the $+4$ state.
$(D)$ $Tl^{+}$ is more stable than $Tl^{3+}$ due to the inert pair effect,which is the reluctance of the $ns^2$ electrons to participate in bonding as we move down the group.