Question

The oxidation state of sulphur in $Na_2S_4O_6$ is

• A. + 6
• B. $\frac{+3}{2}$
• C. $\frac{+5}{2}$
• D. -2

Answer 1

Answer: (C)

Oxidation state of $Na =+1$
Oxidation state of $O =-2$
The average oxidation state of Sulphur
$\Rightarrow 2\left( Na ^{+}\right)+4( S )+6\left( O ^{2-}\right)=0$
$\Rightarrow 2(+1)+4( x )+6(-2)=0$
$\Rightarrow x =\frac{+5}{2}$