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Question
Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is ______.
Options
\[\ce{F > Cl > Br > I}\]
\[\ce{F < Cl < Br < I}\]
\[\ce{F < Cl > Br > I}\]
\[\ce{F < Cl < Br < I}\]
Solution
Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is \[\ce{F < Cl > Br > I}\].
Explanation:
Chlorine has higher electron gain enthalpy than fluorine. This is due to small size of fluorine atom, i.e., the electron density is high which resists the addition of an electron \[\ce{(F < Cl)}\].
In general, electron gain enthalpy decreases as atomic size increases. Thus, electron gain enthalpy follows the order \[\ce{Cl > Br > I}\]
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RELATED QUESTIONS
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.
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F or Cl
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\[\ce{O (g) + e- -> O- (g) ; ∆H^Θ = - 14 kJ mol^{-1}}\]
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Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy.
| Column (I) | Column (II) |
| Electronic configuration | Electron gain enthalpy/kJ mol–1 |
| (i) 1s2 2s2 sp6 | (A) – 53 |
| (ii) 1s2 2s2 2p6 3s1 | (B) – 328 |
| (iii) 1s2 2s2 2p5 | (C) – 141 |
| (iv) 1s2 2s2 2p4 | (D) + 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.
Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
Assertion: The most electronegative element in the periodic table is F.
Reason: Fluorine has the highest negative electron gain enthalpy.
The correct order of electron gain enthalpy (−ve value) is ______.
