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प्रश्न
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.
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उत्तर
- Electronic Configuration: All the elements, oxygen (O), sulphur (S), selenium (Se), tellurium (Te), and polonium (Po), belong to Group 16, known as chalcogens. They have six valence electrons, with the general outer electronic configuration ns2np4. where n varies from 2 to 6. This similarity in valence shell configuration justifies their placement in the same group.
- Oxidation state: Due to the presence of six valence electrons, these elements commonly exhibit an oxidation state of −2.
- Oxygen shows −2 oxidation state predominantly, due to its small size and high electronegativity. It also shows −1 (as in H2O2), 0 (as in O2), and +2 (in OF2) states.
- The −2 oxidation state becomes less stable down the group due to decreasing electronegativity.
- Heavier elements like S, Se, Te, and Po also exhibit +2, +4, and +6 oxidation states due to the availability of vacant d-orbitals for bonding.
- Formation of hydrides: All Group 16 elements form binary hydrides of the general formula H2E, where E = O, S, Se, Te, or Po, These hydrides are volatile and covalent in nature.
- Oxygen and sulphur also form peroxides (e.g. H2O2, H2S2).
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Volatility decreases down the group, while thermal stability and boiling point increase due to increasing molecular weight and decreasing hydrogen bonding.
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संबंधित प्रश्न
Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
List the important sources of sulphur.
Write the order of thermal stability of the hydrides of Group 16 elements.
Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe
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).
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Arrange the following in the order of the property indicated against set :
H2O, H2S, H2Se, H2Te − increasing acidic character.
Explain the following properties of group 16 elements :
1) Electro negativity
2) Melting and boiling points
3) Metallic character
4) Allotropy
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
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?
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:
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
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:
______ is a radioactive element in group 16 elements.
______ is a gaseous element of group 16.
