$C a S_{2} + \frac{3}{2} O_{2} ⟶ C a S_{2} O_{3}$

What is the difference in average oxidation number of sulphur in product and in reactant in the given reaction?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 3
Oxidation state of sulphur in reactant ( $C a S_{2}$ ) is -1. It changes to an oxidation state of +2 in product ( $C a S_{2} O_{3}$ ). Therefore the difference is 3.

Suggest Corrections

Similar questions

N a_{2} S_{4} O_{6}

S

S_{4} O_{6}^{2 -}

+ 2.5

S_{4} O_{6}^{2 -}

+ 2.5

N a_{2} S_{4} O_{6}

Find the difference in oxidation numbers of sulphur in the $(S O_{4})^{2 -}$ ion and $S O_{3}$ compound.

Join BYJU'S Learning Program