Q. The oxidation number of S in ${Na}_{2}S_{4}O_{6}$ 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 other two have zero)
Explanation
(d)
${Na}_{2}S_{4}O_{6}$ is
[image:3_image5.emf]
The two S atoms which are linked to each other have 0 oxidation number. The oxidation number of other two S-atoms can be calculated as
$2x + 2 \times 0 + 6 \times - 2 = - 2$
$2x = 12 - 2 = 10$
$x = + 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.