Q. The oxidation state of sulphur in sodium tetrathionate (${Na}_{2}S_{4}O_{6}$) is
A) 2
B) 0
C) 2.5
D) 3.5
Explanation
(c)
Let the oxidation state of sulphur in ${Na}_{2}S_{4}O_{6}$ is $x.$
${Na}_{2}S_{4}O_{6}$
1$\times 2 + 4 \times x + ( - 2) \times 6 = 0$
$2 + 4x - 12 = 0$
$4x - 10 = 0$
$4x = 10$
$x = \frac{10}{4} = 2.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.