Question

The oxidation number of chlorine in $ HClO_{4} $ is

• A. + 7
• B. + 5
• C. + 3
• D. - 1

Answer 1

Answer: (A)

$HClO _{4}$
(O.N. of $H$ ) $+$ (O.N. of $Cl$ ) $+($ O.N. of $O )=0$
Let oxidation number of $Cl =x$
We know that
Oxidation number of $H =+1$
Oxidation number of $O =-2$
Therefore, $+1+x+4(-2)=0$
$ 1+x-8 =0 $
$ x =+7 $
So, the oxidation number of chlorine in $HClO_4$ is $+7$