A) FOF < CIOCI < BrOBr
B) FOF < BrOBr < CIOCI
C) BrOBr < CIOCI < FOF
D) BrOBr < FOF < CIOCI
Correct Answer: A
Solution :
Order is FOF < CIOCI < BrOBr because electrons in the case of \[O{{F}_{2}}\] are nearer to fluorine due to high electronegativity of F compared to Cl to Br. The bonded pair of electrons, in \[C{{l}_{2}}O\]and \[B{{r}_{2}}O\] are closer to oxygen making the repulsion between them more and thereby reducing the Ip-lp. Repulsion on oxygen to same extent. Also, due to the bulkiness of Cl and Br, the angle CIOCI and BrOBr increase to such an extent that 109° 28', (the tetrahedral angle), is approache.
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