Question

The oxidation state of $S$ in $N{a}_2{S}_4{O}_6$ is:

• A. 2.5
• B. +2 and +3 (two $S$ have +2 and two $S$ have +3)
• C. + 2 and +3 (three $S$ have +2 and one $S$ has +3)
• D. + 5 and 0 (two $S$ have +5 and the other two $S$ have 0)

Answer 1

Answer: (A)

O.N of $S$ is $X$
O.N of $Na$ is $+1$
O.N of $O$ is -2
Sum of all O.N is $2\times +1+\left(4X\right)+\left(6\right)\times -\left(2\right)$
$4X-10$ is equal to 0
$X$ equals to 10/4 which equals to 2.5

his compound must have sulfur atoms with mixed oxidation states. Since it is normal for sulfur to have oxidation states of -2, 0, +2, +4, and +6, it is most likely that there are three sulfurs with a +2 oxidation state and one sulfur that is +4.
Hence option $A$ is correct.