11th Class Chemistry The p-Block Elements-I

  • 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.

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