• # question_answer 88)   Account for the following observations : (i) $AlC{{l}_{3}}$is a Lewis acid. (ii) Though fluorine: is mere electronegative than chlorine yet $B{{F}_{3}}$is a weaker Lewis acid than$BC{{l}_{3}}$. (iii) $Pb{{O}_{2}}$is a stronger oxidising agent than $Sn{{O}_{2}}$. (iv) The +1 oxidation state of thallium is more stable than its +3 state.

(i) $AlC{{l}_{3}}$is electron deficient molecule hence it acts as Lewis acid. (ii) In$B{{F}_{3}}$, boron has a vacant 2p-orbital and each fluorine has fully filled unutilized -orbitals. Fluorine transfer two electrons to vacant 2p-orbital of boron forming $p\pi -p\pi$bond. This is known as back bonding. This bond reduces the electron deficiency of boron atom hence its Lewis acid character decreases. This tendency of back bonding is maximum in $B{{F}_{3}}$ and of decreases from $B{{F}_{3}}$ to $BC{{l}_{3}}$ on account of increase in size of chlorine atom. (iii) In $Pb{{O}_{2}}$ and$Sn{{O}_{2}}$, the metals are present in (+4) state. In lead $P{{b}^{2+}}$ state is more stable than $P{{b}^{4+}}$state due to inert pair effect. Thus, $P{{b}^{4+}}$ tries to attain more stable state i.e., $P{{b}^{2+}}$, by acting as an oxidising agent. $P{{b}^{4+}}+2{{e}^{-}}\xrightarrow{{}}P{{b}^{2+}}$ In tin, $S{{n}^{2+}}$ state is less stable than$S{{n}^{4+}}$. Thus,$S{{n}^{2+}}$tries to attain more stable configuration i.e., $S{{n}^{4+}}$state by acting as a reducing agent. $S{{n}^{2+}}\xrightarrow{{}}S{{n}^{4+}}+2{{e}^{-}}$ (iv) The stability of +1 oxidation state is maximum in thallium amongst group 13 elements as the inert pair effect is maximum in thallium due to poor shielding effect of d-and f-electrons present in the inner energy shells on$n{{s}^{2}}$-electrons.