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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
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)
Write the order of thermal stability of the hydrides of Group 16 elements.
Arrange the following in the order of property indicated for the given set:
F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
Arrange the following in order of the property indicated set.
HF, HCl, HBr, HI - decreasing bond enthalpy.
The formation of \[\ce{O^+_2[PtF6]^-}\] is the basis for the formation of first xenon compound. This is because ____________.
Which of the following statement is incorrect?
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:
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
What is the basicity of \[\ce{H3PO4}\]?
