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प्रश्न
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)
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उत्तर
(ii) \[\ce{B < C < N < O}\] (increasing first ionisation enthalpy)
(iii) \[\ce{I < Br < Cl < F}\] (increasing electron gain enthalpy)
Explanation:
The ionization enthalpy of \[\ce{N}\] is higher than that of F because it possess half filled orbital which provide it extra stability due to symmetry
As the atomic size of \[\ce{F}\] is much smaller thus its electron gain enthalpy is lower than that of the \[\ce{Cl}\]
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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.
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
Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is ______.
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,
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.
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 |
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.
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.
Assertion: The most electronegative element in the periodic table is F.
Reason: Fluorine has the highest negative electron gain enthalpy.
