The type of hybrid orbitals used by chlorine atom in `ClO_(2)^(-)` is :

Options

A. `sp^3`
B. `sp^2`
C. sp
D. None of these

Answer

A

Explanation

First calculate the number of bonds in the molecule or ion as follows
No. of covalent bonds `=1/2` [Total no. of 2 electrons needed to complete the octet (or doublet in case of H)-Total no. of valence electrons]
`=1/2[18 xx 3 - (2 xx 6+7+1)]`
`=1/2 [24-20] =4/2=2`
Thus, the ion contains two covalent bonds. With this information draw the Lewis structure. Lewis structure of

According to VSEPR theory, shape of the ion is tetrahedral (two lone pairs, two `sigma`-bonds) and hybridisation state of Cl is `sp^3`.
Alternatively calculate the state of hybridisation of Cl as follows
Total no. of valence electrons = 20
Dividing 20 (9 lt 20 lt 56) by 8 and the remainder (if any) by 2.

Therefore state of hybridisation is `sp^3`.