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प्रश्न
Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
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उत्तर
As the size of the elements increases in the order O < S < Se < Te, the X−H bond strength decreases from H2O to H2Te and therefore, the bond dissociation enthalpy decreases. Hence, due to the increase in the tendency to release proton, the element's reducing tendency also increases. Therefore, H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
संबंधित प्रश्न
Give reasons: SO2 is reducing while TeO2 is an oxidising agent.
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
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 |
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
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The correct order of ΔiHs among the following elements is
Which of the following compound is a peroxide?
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)
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follows the order:
H2O > H2Te > H2Se > H2S
Statement II: On the basis of molecular mass, H2O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has a higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
