Browse Questions

The type of hybrid orbitals used by the chlorine atom in $ClO^{-}_2$ is

$(A)\;SP^3 \\ (B)\;SP^2 \\(C)\;SP \\(D)\;\text{none of these}$

We can calculate the number of orbitals involved in hybridisation by using the reaction.
Number of orbital involved in hybridisation by using the relation:
Number of orbital involved in hybridisation $= \large\frac{1}{2} $$(V+M -C+A) Where V= number of valence electrons M= number of monovalent atoms surrounding the atom C= Charge on cation. A= Charge on anion. Number of orbitals involved in hybridisation =\large\frac{1}{2}$$(7+0+0+1)=4$
Since 4 orbitals are involved in hybridisation so it is $SP^3$ hybridised.
Hence A is the correct answer.