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प्रश्न
The formation of the oxide ion, \[\ce{O2- (g)}\], from oxygen atom requires first an exothermic and then an endothermic step as shown below:
\[\ce{O (g) + e- -> O- (g) ; ∆H^Θ = - 14 kJ mol^{-1}}\]
\[\ce{O- (g) + e- -> O^{2-} (g) ; ∆H^Θ = + 780 kJ mol^{-1}}\]
Thus process of formation of \[\ce{O^{2-}}\] in gas phase is unfavourable even though \[\ce{O^{2-}}\] is isoelectronic with neon. It is due to the fact that,
पर्याय
oxygen is more electronegative.
addition of electron in oxygen results in larger size of the ion.
electron repulsion outweighs the stability gained by achieving noble gas configuration.
\[\ce{O-}\] ion has comparatively smaller size than oxygen atom.
उत्तर
electron repulsion outweighs the stability gained by achieving noble gas configuration.
Explanation:
There is a lot of repulsion when similar charges approach each other as \[\ce{O- (g)}\], and electron are both negatively charged. To add an electron under such situation, the force of repulsion is to be overcome by applying external energy.
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संबंधित प्रश्न
Which of the following pair of elements would have a more negative electron gain enthalpy?
F or Cl
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Which of the following pair of elements would have a more negative electron gain enthalpy?
O or F
Which of the following statements are correct?
(i) Helium has the highest first ionisation enthalpy in the periodic table.
(ii) Chlorine has less negative electron gain enthalpy than fluorine.
(iii) Mercury and bromine are liquids at room temperature.
(iv) In any period, atomic radius of alkali metal is the highest.
In which of the following options the order of arrangement does not agree with the variation of the property indicated against it?
(i) \[\ce{Al^{3+} < Mg^{2+} < Na+ < F-}\] (increasing ionic size)
(ii) \[\ce{B < C < N < O}\] (increasing first ionisation enthalpy)
(iii) \[\ce{I < Br < Cl < F}\] (increasing electron gain enthalpy)
(iv) \[\ce{Li < Na < K < Rb}\] (increasing metallic radius)
Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron hence 2p electron is more shielded by the inner core of electrons than the 2s electrons.
Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
The correct order of electron gain enthalpy (−ve value) is ______.
