Question

The oxidation state of S atom in $[HS{O}_4{]}^{-}$ is

• A. +2
• B. +4
• C. +6
• D. -2

Answer 1

The correct option is C

+6

Since this is not a neutral compound the sum of oxidation numbers of all the elements will not be equal to 0 here. Instead, it will be equal to the charge on the ion.
We know that

O.N. of (O) = -2

O.N. of (H) =+1
Let the oxidation number for sulphur be $x$
So we can formulate the equation of oxidation numbers as given below.

$[HS{O}_4{]}^{-}⟹1+x+4(-2)=-1$
$x=+6$
Hence, the oxidation number of sulphur in $[HS{O}_4{]}^{-}$ ion is +6