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प्रश्न
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
उत्तर
Oxygen is more electronegative than sulphur, the bond angle of \[\ce{H2O}\] is greater, and the body pair electron of an \[\ce{OH}\] bond will be closer to oxygen, causing bond-pair bond-pair repulsion between bond pairs of two \[\ce{OH}\] bonds.
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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)
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
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.
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 O → O− and O → O2− 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.
The boiling points of hydrides of group 16 are in the order:
Which of the following statement is incorrect?
Strong reducing behaviour of \[\ce{H3PO2}\] is due to ______.
