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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
Account for the following: Oxygen shows catenation behavior less than sulphur.
a. Explain the trends in the following properties with reference to group 16:
1 Atomic radii and ionic radii
2 Density
3 ionisation enthalpy
4 Electronegativity
b. In the electolysis of AgNO3 solution 0.7g of Ag is deposited after a certain period of time. Calulate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9g mol-1)
Give reasons: SO2 is reducing while TeO2 is an oxidising agent.
List the important sources of sulphur.
Why does NH3 form hydrogen bond but PH3 does not?
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].
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.
The boiling points of hydrides of group 16 are in the order:
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?
