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प्रश्न
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
विकल्प
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)
Why is H2O a liquid and H2S a gas?
Why are halogens strong oxidising agents?
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
Give a reason for the following:
Fluorine gives only one oxide but chlorine gives a series of oxides.
Arrange the following in order of the property indicated set.
HF, HCl, HBr, HI - decreasing bond enthalpy.
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 |
Given below are two statements labelled as Assertion (A) and Reason (R).
Assertion (A): Electron gain enthalpy of oxygen is less than that of Flourine but greater than Nitrogen.
Reason (R): Ionisation enthalpies of the elements follow the order Nitrogen > Oxygen > Fluorine.
Select the most appropriate answer from the options given below:
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:
