The oxidation number of `S` in `Na_(2)S_(4)O_(6)` is

Options

A. `+2.5`
B. `+2 and +3` (two S have +2 and other two 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 have 0)

Answer

D

Explanation

The structural formula of `Na_(2)S_(4)O_(6)` (sodium tetrathionate) is as follows
`{:(" O O"),(" || ||"),(Na^(+)O^(-)-S-S-S-S-O^(-)Na^(+)),(" || ||"),(" O O"):}`
The O.N of the two central sulphur atoms which are linked only to each other i.e., -S-S- is zero. Let the O.N of the other two sulphur atoms be x each. Here, O.N. of O is -2 and O.N. of tetrathionate ion = -2
`:. 6xx(-2)+2xx0+2xx x=-2`
`-12+2x=-2`
`x=5`
Thus, two S atoms in `Na_(2)S_(4)O_(6)` have O.N. of zero each, while the other two S atoms have O.N of 5 each.