Question

Describe the change in hybridisation (if any) of the Al atom in the following reaction: $AlCl_{3}+Cl^{-} \longrightarrow AlCl_{4}^{-}$

Answer 1

The given chemical reaction is as follows.

${ AlCl }_{ 3 }\quad +\quad { Cl }^{ - }\quad \rightarrow \quad { AlCl }_{ 4 }^{ - }$

Electronic configuration of Al

$Z\quad =\quad 13\quad =\quad { 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }_{ x }^{ 1 }$

Ground state electronic configuration =$\quad { 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }_{ x }^{ 1 }$

Excited electronic configuration =$\quad { 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 1 }{ 3 }p_{ x }^{ 1 }{ 3 }p_{ y }^{ 1 }$

Hence in $Al{ Cl }_{ 3 }$, Al undergoes ${ sp }^{ 2 }$ hybridization to give  planar triangular structure. In $Al{ Cl }_{ 4 }$, the empty 3p orbital is also involved so that there is hybridization in $s{ p }^{ 3 }$ and its shape is tetrahedral.