The oxidation number of S in Na2S4O6 is

Question

The oxidation number of S in Na2S4O6 is (A) + 2.5 for each S atom (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).

Tardigrade Admin · Accepted Answer

: The structure of Na2S4O6 is The two S atoms which are linked to each other have O.N. of zero. The O.N. of other two S- atoms can be calculated as follows: 2x + 2 × 0 + 6 × -2 = -2 or 2 x = 12 — 2 = 10 or x = + 5

Q. The oxidation number of S in $N a_{2} S_{4} O_{6}$ is

+ 2.5 for each S atom

+ 2 and + 3 (two S have + 2 and other two have + 3)

+ 2 and + 3 (three S have + 2 and one S has + 3)

+ 5 and 0 (two S have + 5 and the other two have 0).

: The structure of $N a_{2} S_{4} O_{6}$ is The two $S$ atoms which are linked to each other have O.N. of zero. The O.N. of other two $S$ - atoms can be calculated as follows : $2 x + 2 \times 0 + 6 \times - 2 = - 2$ or $2 x = 12—2 = 10$ or $x = + 5$

Q. The oxidation number of S in Na2​S4​O6​ is