A) \[ClO_{2}^{-}<C{{l}_{2}}O<Cl{{O}_{2}}\]
B) \[C{{l}_{2}}O<Cl{{O}_{2}}<ClO_{2}^{-}\]
C) \[Cl{{O}_{2}}<C{{l}_{2}}O<ClO_{2}^{-}\]
D) \[Cl{{O}_{2}}<ClO_{2}^{-}<ClO\]
Correct Answer: A
Solution :
In \[Cl{{O}_{2}},\]\[Cl\]is \[s{{p}^{2}}\]hybridised and the bond angle is\[118{}^\circ \]. In\[Cl{{O}_{2}},\]\[O\]is\[s{{p}^{3}}\]hybridised. However, due to repulsions between two big sized\[Cl\]atoms, the bond angle increases from \[109{}^\circ -{{28}^{1}}\] to\[111{}^\circ \]. In \[Cl{{O}_{2}},\]\[Cl\] atom is \[s{{p}^{3}}\]hybridised but due to \[lp-lp\] repulsions the angle decreases from\[109{}^\circ -{{28}^{1}}\]to\[103{}^\circ \]. Thus, the increasing order of bond angles is \[ClO_{2}^{-}<\]\[C{{l}_{2}}O<Cl{{O}_{2}}.\]
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