Advertisements
Advertisements
प्रश्न
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
Advertisements
उत्तर
The oxygen exists as a diatomic molecule, O2, while sulphur exists as a polyatomic molecule, S8. Hence, there is a large difference between the melting point and the boiling point of oxygen and sulphur.
APPEARS IN
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?
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].
Knowing the electron gain enthalpy values for \[\ce{O -> O-}\] and \[\ce{O -> O^{2-}}\] 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 reasons Thermal stability decreases from H2O to H2Te.
Arrange the following in the order of the property indicated against set :
H2O, H2S, H2Se, H2Te − increasing acidic character.
The boiling points of hydrides of group 16 are in the order:
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 ______.
Which of the following statements are correct?
(i) \[\ce{CaF2 + H2SO4 -> CaSO4 + 2HF}\]
(ii) \[\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}\]
(iii) \[\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}\]
(iv) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
