Question

67. What is the oxidation number of Sulphur in \( Na _{2} S _{4} O _{6} \) ? (a) \( \frac{2}{3} \) (b) \( \frac{3}{2} \) (c) \( \frac{3}{5} \) (d) \( \frac{5}{2} \)

Answer 1

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×+1+(4X)+(6)×−(2)
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.