Answer:
In\[AlC{{l}_{3}}\], Al undergoes \[s{{p}^{3}}-\]hybridization
to give a trigonal planar structure. To form \[AlCl_{4}^{-},\] Al utilizes the
vacant \[{{p}_{z}}\] -orbital also and undergoes \[s{{p}^{3}}\]-hybridization
to give tetrahedral structure. \[C{{l}^{-}}\] ion provides a lone pair of
electrons to form a bond.
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