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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,
विकल्प
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.
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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
What is the significance of the terms - ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Hint: Requirements for comparison purposes.
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.
Describe the theory associated with the radius of an atom as it gains an electron.
Describe the theory associated with the radius of an atom as it loses an electron.
Which of the following pair of elements would have a more negative electron gain enthalpy?
O or F
Which of the following elements will gain one electron more readily in comparison to other elements of their group?
(i) \[\ce{S (g)}\]
(ii) \[\ce{Na (g)}\]
(iii) \[\ce{O (g)}\]
(iv) \[\ce{Cl (g)}\]
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.
Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.
| Elements | ∆H1 | ∆H2 | ∆egH | |
| (i) Most reactive non-metal | A. | 419 | 3051 | – 48 |
| (ii) Most reactive metal | B. | 1681 | 3374 | – 328 |
| (iii) Least reactive element e | C. | 738 | 1451 | – 40 |
| (iv) Metal forming binary halide | D. | 2372 | 5251 | + 48 |
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.
Assertion: The most electronegative element in the periodic table is F.
Reason: Fluorine has the highest negative electron gain enthalpy.
