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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} $
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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.
answered Apr 4, 2014 by meena.p
 

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