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Arrange the following in increasing oxidation number of iodine - $$\text{I}_2$$, HI, $$\text{HIO}_4$$ and ICl
Correct Answer: 0
PLACEHOLDER ANSWER - the answer is an order of compounds. The oxidation number of iodine is $$-1$$ in HI, 0 in $$\text{I}_2$$, $$+1$$ in ICl (Cl is more electronegative) and $$+7$$ in $$\text{HIO}_4$$. So the order is $$\text{HI} < \text{I}_2 < \text{ICl} < \text{HIO}_4$$.
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