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प्रश्न
Explain the following properties of group 16 elements :
1) Electro negativity
2) Melting and boiling points
3) Metallic character
4) Allotropy
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उत्तर
1) Electronegativity :
The elements of group 16 have higher values of electronegativity than the corresponding elements of group 15. Oxygen is the second most electronegative element, the first being fluorin. The electronegativity decreases on going down the group. The decrease in electronegativity down the group is due to increase in size of the atoms.
2) Melting and boiling points :
The melting and boiling points increases with the increase in atomic number as we go down the group.
3) Metallic character :
The first four elements namely oxygen, sulphur, selenium and tellurium are non-metals. The non-metallic character is stronger in O and S are weaker in Se an Te. On the other hand, last element is markedly metallic. However, it is radioactive and is only short-lived.
4) Allotropy :
All the elements of the group exhibit allotropy. For example, oxygen exists as O2 and O3 (ozone.) Sulphur exists in a number of allotropic forms of which yellow ortho-rhombic, α and β -monoclinic forms are most important. All these allotropic forms of sulphur are nonmetallic. Selenium exists in eight allotropic forms of which three are red monoclinic forms containing Se8 rings.
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संबंधित प्रश्न
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
List the important sources of sulphur.
Why is H2O a liquid and H2S a gas?
Why does NH3 form hydrogen bond but PH3 does not?
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.
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).
Why are halogens strong oxidising agents?
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Arrange the following in the order of property indicated for the given set:
F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
Give reasons Thermal stability decreases from H2O to H2Te.
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.
The boiling points of hydrides of group 16 are in the order:
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?
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) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
Which of the following compound is a peroxide?
What is the basicity of \[\ce{H3PO4}\]?
______ is a gaseous element of group 16.
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follows the order:
H2O > H2Te > H2Se > H2S
Statement II: On the basis of molecular mass, H2O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has a higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
