Would You Expect the Second Electron Gain Enthalpy of O as Positive, More Negative Or Less Negative than the First? Justify Your Answer. - Chemistry

Question

Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Answer in Brief

Solution 1

When an electron is added to O atom to form O^– ion, energy is released. Thus, the first electron gain enthalpy of O is negative.

$O_{(g)} + e^{-} ⟶ O^{-}_{(g)}$

On the other hand, when an electron is added to O^- ion to form O^2- ion, energy has to be given out in order to overcome the strong electronic repulsions. Thus, the second electron gain enthalpy of O is positive.

$O_{(g)} + e^{-} ⟶ O^{-}_{(g)}$

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Solution 2

$O_{(g)} + e^{-} ⟶ O^{-}_{(g)}$ (∆ eg H = – 141 kJ mol ^-1)

$O^{-}_{(g)} + e^{-} ⟶ O^{2 -}_{(g)}$ (∆ eg H = + 780 kJ mol ^-1)

The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and second incoming electron.

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Physical Properties - Electron Gain Enthalpy

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Q 3.21 Q 3.20 (ii)Q 3.22

Chapter 3: Classification of Elements and Periodicity in Properties - EXERCISES [Page 97]

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Chapter 3 Classification of Elements and Periodicity in Properties
EXERCISES | Q 3.21 | Page 97

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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

The formation of the oxide ion, $O_{2}^{-} (g)$ , from oxygen atom requires first an exothermic and then an endothermic step as shown below:

$O (g) + e^{-} ⟶ O^{-} (g); ∆ H^{Θ} = - 14 kJ mol^{- 1}$

$O^{-} (g) + e^{-} ⟶ O^{2 -} (g); ∆ H^{Θ} = + 780 kJ mol^{- 1}$

Thus process of formation of $O^{2 -}$ in gas phase is unfavourable even though $O^{2 -}$ 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) $S (g)$

(ii) $Na (g)$

(iii) $O (g)$

(iv) $Cl (g)$

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) $Al^{3 +} < Mg^{2 +} < Na^{+} < F^{-}$ (increasing ionic size)

(ii) $B < C < N < O$ (increasing first ionisation enthalpy)

(iii) $I < Br < Cl < F$ (increasing electron gain enthalpy)

(iv) $Li < Na < K < Rb$ (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

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) 1s²2s² sp⁶	(A) – 53
(ii) 1s² 2s² 2p⁶ 3s¹	(B) – 328
(iii) 1s² 2s² 2p⁵	(C) – 141
(iv) 1s² 2s² 2p⁴	(D) + 48

Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.

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Would You Expect the Second Electron Gain Enthalpy of O as Positive, More Negative Or Less Negative than the First? Justify Your Answer. - Chemistry

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