Knowing the electron gain enthalpy values for O → O^- and O → O^2- as -141 and 702 kJ mol^-1 respectively
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).
Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2- ion is much more than the oxide involving O- ion. Hence, the oxide having O2- ions are more stable than oxides having O-. Hence, we can say that formation of O2- is energetically more favourable than formation of O-.
-
List the uses of Neon and argon gases.
3 years ago
-
Why do noble gases have comparatively large atomic sizes?
3 years ago
-
Which one of the following does not exist? (i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6
3 years ago
-
Arrange the following in the order of property indicated for each set: (i) F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy. (ii) HF, HCl, HBr, HI - increasing acid strength.
3 years ago
-
How are XeO3 and XeOF4 prepared?
3 years ago
- 321 Forums
- 27.3 K Topics
- 53.8 K Posts
- 0 Online
- 12.4 K Members