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प्रश्न
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
पर्याय
Low oxidation state of phosphorus
Presence of two –OH groups and one P–H bond
Presence of one –OH group and two P–H bonds
High electron gain enthalpy of phosphorus
उत्तर
Strong reducing behaviour of \[\ce{H3PO2}\] is due to presence of one –OH group and two P–H bonds.
Explanation:
\[\ce{H3PO2}\] has one O – H group and two P – H bonds.
The existence of a P – H link gives phosphorus oxyacids their reducing characteristics. It has a strong proclivity for releasing protons. As a result, it demonstrates a decreasing nature.
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संबंधित प्रश्न
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)
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
List the important sources of sulphur.
Write the order of thermal stability of the hydrides of Group 16 elements.
Knowing the electron gain enthalpy values for O → O− and O → O2− as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−? (Hint: Consider lattice energy factor in the formation of compounds).
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
Which of the following statement is incorrect?
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)
______ is a gaseous element of group 16.
