Question

Arrange the following compounds according to the increasing order of oxidation state of chlorine.

$HCl{O}_4,HCl{O}_3,HCl{O}_2,HClO$

• A. $HClO<HCl{O}_3<HCl{O}_2<HCl{O}_4$
• B. $HClO<HCl{O}_2<HCl{O}_3<HCl{O}_4$
• C. $HCl{O}_4<HCl{O}_2<HCl{O}_3<HClO$
• D. $HCl{O}_4<HCl{O}_3<HCl{O}_2<HClO$

Answer 1

The correct option is B $HClO<HCl{O}_2<HCl{O}_3<HCl{O}_4$

We know that the oxidation number of H and O is +1 and -2 respectively.

Since the sum of the oxidation number of each compound is zero, considering the oxidation number of Cl to be 'x',

For $HCl{O}_4\Rightarrow +1+x+4\times (-2)=0$

$\Rightarrow x-7=0$

$\Rightarrow x=7$

For $HCl{O}_3\Rightarrow +1+x+3\times (-2)=0$

$\Rightarrow x-5=0$

$\Rightarrow x=5$

For $HCl{O}_2\Rightarrow +1+x+2\times (-2)=0$

$\Rightarrow x-4=0$

$\Rightarrow x=4$

For $HClO\Rightarrow +1+x+(-2)=0$

$\Rightarrow x-1=0$

$\Rightarrow x=1$

Arranging compounds according to increasing oxidation number of chlorine gives

$\Rightarrow HClO<HCl{O}_2<HCl{O}_3<HCl{O}_4$