Here it is
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Bond length∝  1/Bond order

In KO2  is O₂ ^- whereas in O₂AsF₄ it is O₂⁺

Now if we calculate bond order isO₂ and O₂⁺

O₂ has 2 bond order
O₂has =1
O₂⁺ has bond order of 1.5

Borazine is more polar in nature due highly EN atom N and it can also make H bonds .
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Now bond angles SOX2 where X  is F Cl Br I
Size order of halogen is
F<Cl<Br<I
as the size of the atoms increases, hybridization tendency decreases  due to more energy difference of hybridized orbitals and halogen orbitals overlapping with hybridized orbitals of S.If pure orbitals are overlapping then bond angle approaches closer to 90 degree.