A) \[Cl{{O}_{4}}^{-}<Cl{{O}_{3}}^{-}<Cl{{O}_{2}}^{-}<Cl{{O}^{-}}\]
B) \[Cl{{O}^{-}}<Cl{{O}_{2}}^{-}<Cl{{O}_{3}}^{-}<Cl{{O}_{4}}^{-}\]
C) \[Cl{{O}_{3}}^{-}<Cl{{O}_{4}}^{-}<Cl{{O}_{2}}^{-}<Cl{{O}^{-}}\]
D) \[Cl{{O}_{4}}^{-}<Cl{{O}_{3}}^{-}<Cl{{O}^{-}}<Cl{{O}_{2}}^{-}\]
Correct Answer: B
Solution :
As the smallest cation is most hydrated here \[Cl{{O}_{4}}^{-}\]the smallest central atom as it has +7 charge on it followed by \[Cl{{O}_{3}}^{-}(+5),Cl{{O}_{2}}^{-}(+3),Cl{{O}^{-}}(+1).\]So the order of hydration energy is \[Cl{{O}_{4}}^{-}>Cl{{O}_{3}}^{-}>Cl{{O}_{2}}^{-}>Cl{{O}^{-}}\] Hence, the correct option is [b].
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