In the chemical reaction $C l_{2} + H_{2} S ⟶ 2 H C l + S$ , the oxidation number of sulphur changes from to .

-2

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

-1

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

0.0

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

Open in App

Solution

The correct options are
A -2
D 0.0
The oxidation state of sulphur in $H_{2} S i s - 2$ , while it is zero in 'S' i.e. in this reaction oxidation of sulphur and reduction of chlorine takes place.

Suggest Corrections

Similar questions

C l_{2} + H_{2} S ⟶ 2 H C l + S

C l_{2} + H_{2} S \to 2 H C l + S

H_{2} S + C l_{2} \to 2 H C l + S

In the reaction, H₂S + Cl₂ → 2HCl + S; hydrogen sulphide is:

The oxidation number of sulphur in $S_{8}$ , $S_{2}$ $F_{2}$ , $H_{2} S$ respectively, are

Join BYJU'S Learning Program