Advertisements
Advertisements
Question
Give reasons: SO2 is reducing while TeO2 is an oxidising agent.
Advertisements
Solution
In case of sulphur, because of the presence of empty d-orbital, it can expand its oxidation state from the + 4 to the +6 oxidation state. Hence, it acts as a reducing agent.
Te is a heavy element and so because of the inert pair effect, the lower oxidation state is more stable. Hence, it acts as an oxidising agent.
APPEARS IN
RELATED QUESTIONS
Account for the following: Oxygen shows catenation behavior less than sulphur.
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
List the important sources of sulphur.
Give reasons Thermal stability decreases from H2O to H2Te.
Draw the structures of `H_3PO_2`
Match the items of Columns I and II and mark the correct option.
| Column I | Column II |
| (A) \[\ce{H2SO4}\] | (1) Highest electron gain enthalpy |
| (B) \[\ce{CCl3NO2}\] | (2) Chalcogen |
| (C) \[\ce{Cl2}\] | (3) Tear gas |
| (D) Sulphur | (4) Storage batteries |
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
Write a balanced chemical equation for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen.
These are physical properties of an elements.
- Sublimation enthalpy
- Ionisation enthalpy
- Hydration enthalpy
- Electron gain enthalpy
The total number of above properties that affect the reduction potential is ______. (Integer answer)
What is the basicity of \[\ce{H3PO4}\]?
