What is the correct order of the oxidation state of sulfur in the anions $SO_3^{2-}$,${S_2}O_4^{2-}$,and ${S_2}O_6^{2-}$?

  • A
    ${S_2}O_6^{2-} < {S_2}O_4^{2-} < SO_3^{2-}$
  • B
    ${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$
  • C
    $SO_3^{2-} < {S_2}O_4^{2-} < {S_2}O_6^{2-}$
  • D
    ${S_2}O_4^{2-} < {S_2}O_6^{2-} < SO_3^{2-}$

Explore More

Similar Questions

The oxidation number of $H$ in $KH$,$MgH_2$,and $NaOH$ are respectively

Maximum oxidation state is present in

The correct order of $N$ compounds in their decreasing order of oxidation states is:

What is the change in oxidation number of selenium in the following redox reaction?
$SeO_{3(aq)}^{2-} + Cl_{2(g)} + 2OH^- \longrightarrow SeO_{4(aq)}^{2-} + 2Cl_{(aq)}^- + H_2O$

Carbon is present in the highest oxidation number in

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo