Oxidation state of sulphur atoms in sodium tetrathionate(Na2S4O6) are

Oxidation state of sulphur atoms in sodium tetrathionate( ${Na}_{2} S_{4} O_{6}$ ) are

A
$+ 6, - 2, - 2, + 6$
B
$+ 5, 0, 0, + 5$
C
$\frac{+ 5}{2}, \frac{+ 5}{2}, \frac{+ 5}{2}, \frac{+ 5}{2}$
D
$+ 4, + 2, + 2, + 2$

Solution:

Sodium tetra thionate is ${Na}_{2} S_{4} O_{6}$

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