The type of hybrid orbitals used by chlorine atom in $C l O_{2}^{-}$ is:

s p^{3}

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s p^{2}

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s p

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d s p^{2}

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Solution

The correct option is A $s p^{3}$
Number of orbital involved in hybridisation by using the relation :
Stearic number : $\frac{1}{2}$ [V+M−C+A]
V= number of valence electrons
M= number of monovalent atoms surrounding the atom
C= Charge on cation.
A= Charge on anion.
Stearic number = $\frac{1}{2}$ [7+0+0+1] = 4
Stearic number = 4 means $s p^{3}$

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C l O_{2}^{-}

The type of hybrid orbitals used by the chlorine atom in $C l O_{2}^{-}$ is

C l O^{-}, C l O_{2}^{-}, C l O_{3}^{-}

C l O_{4}^{-}

The correct order of hybridization of the central atom in the following species $N H_{3}, [P t C l_{4}]^{2 -}$ , $P C l_{5}$ and $B C l_{3}$ is

s p

s p^{2}

s p^{3}

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