Question

hybridisation of AlCl3

Answer 1

sp2 will be its hybridization state

Answer 2

Answer:

The hybridization of $\mathrm{AlCl}_{3} \text { is } \mathrm{Sp}^{2}$

Explanation:  

Hybridization of any compound can be calculated as per the following formula:  

Hybridization $=\frac{\text { Group no of central atom }+\text { no.of monovalent atoms attached }+\text { charge }}{2}$

Here, the given compound is $\mathrm{AlCl}_{3}$

For $\mathrm{AlCl}_{3}$,

The group number of Al is = 3

Number of monovalent atoms attached to Al is = 3  

The Charge of the molecule of $\mathrm{AlCl}_{3}=0$

Thus, putting the values, we get,

Hybridization of $\mathrm{AlCl}_{3}=\frac{\text { Group no.of central atom }(\mathrm{Al})+\mathrm{no} . \text { of monovalent atoms attached to Al }+\text { charge }}{2}$

          $=\frac{3+3+0}{2}$

          =3  

which gives $\mathrm{Sp}^{2}$ configuration.