Question

Arrange $ICl, HI,$ $I _{2}$ and $HIO _{4}$ in decreasing order of oxidation state of iodine.

• A. $HIO _{4}> ICl > HI > I _{2}$
• B. $ICl > HIO _{4}> I _{2}> HI$
• C. $HIO _{4}> ICl > I _{2}> HI$
• D. $ICl > HIO _{4}> HI > I _{2}$

Answer 1

Answer: (C)

Oxidation number of iodine in

$\overset{x-1}{ICl} : x-1=0 \Rightarrow x=+1 $

$\overset{+1 x}{HI} :(+1)+x=0 \Rightarrow x=-1$

$\stackrel{x}{I}_{2} \,\,\,\,: x=0$

$\overset{+1 x-2}{H I O _{4}}:(+1)+x+4(-2)=0 \Rightarrow x=+7$

Thus decreasing order of oxidation number of iodine is

$HIO _{4}> ICl > I _{2}> HI$