The oxidation state of sulfur in the anions $SO_3^{2-}$,$S_2O_4^{2-}$,and $S_2O_6^{2-}$ increases in the order:
- A$S_2O_6^{2-} < S_2O_4^{2-} < SO_3^{2-}$
- B$SO_3^{2-} < S_2O_4^{2-} < S_2O_6^{2-}$
- C$S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$
- D$S_2O_4^{2-} < S_2O_6^{2-} < SO_3^{2-}$
Explore More
Similar Questions
Consider the following molecules: $Br_3O_8$,$F_2O$,$H_2S_4O_6$,$H_2S_5O_6$,and $C_3O_2$. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is . . . . .
MediumIIT 2023
View SolutionWhich of the following sequences is correct with reference to the oxidation number of iodine?
Medium
View SolutionOxidation number of Nitrogen in $NH_4NO_3$ is
Medium
View SolutionWhat is the oxidation number of $Cl$ in $HOClO_2$?
The correct order of the oxidation states of nitrogen in $NO$,$N_2O$,$NO_2$,and $N_2O_3$ is:
DifficultJEE MAIN 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo