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Question: Statement: Aluminium chloride \(\text{ AlC}{{\text{l}}_{\text{3}}}\text{ }\) is a Lewis acid because...

Statement: Aluminium chloride  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } is a Lewis acid because it can donate the electron.
If a given statement is true enter 1 if false enter 0.

Explanation

Solution

Lewis suggested the acid-base concept. According to which the Lewis base is a chemical species that can donate the electron pair and Lewis base is a chemical species that can easily accept the electron pair.  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } has an empty orbital and accommodates the electron pair in an empty orbital.

Complete answer:
Lewis acid is an ion or a molecule that accepts the nonbonding valence electrons in its empty orbital. Lewis base is an ion or molecule which can donate an electron pair.
Aluminium trichloride  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } is an electron-deficient species.it has three electrons in the valence shell. The electronic configuration of aluminium is  Al = [Ne]3s23p13d\text{ Al = }\left[ \text{Ne} \right]\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{1}}}\text{3}{{\text{d}}^{\text{0 }}}
In  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } aluminium loses its three valence shell electrons. Two electrons from the 3s orbitals and one electron from the 3p orbital. When bonded with chlorine  Al3+ \text{ A}{{\text{l}}^{\text{3+ }}} has empty 3s, 3p, and 3d orbitals. These empty orbitals are used to accommodate the nonbonding pair of the electron which are donated by the Lewis base.
Let’s consider a reaction of  Al3+ \text{ A}{{\text{l}}^{\text{3+ }}} the water molecule. The  Al3+ \text{ A}{{\text{l}}^{\text{3+ }}} ion is an electron-deficient species and bonds three chloride molecules. The bonding between the aluminium and chloride is as shown below,

One orbital of the  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } is empty. This orbit can accommodate the nonbonding electron pair of electrons. Thus  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } acts as a Lewis acid because it can accept an electron.
Thus the given statement is false.

Hence, ‘0’ is the correct answer.

Note: Lewis acid is used as a catalyst in Friedel craft reaction.  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } accepts the lone pair from chloride ion and form the  AlCl4 \text{ AlCl}_{4}^{-}\text{ } ion. This forms a carbonium ion which further acts as a strong Lewis acid. Thus Friedel craft acylation and alkylation reactions are facilitated by  AlCl3 \text{ AlC}{{\text{l}}_{\text{3}}}\text{ } the catalyst.