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) ; ∆ Hᶱ = – 141 kJ mol^–1

O^– (g) + e^– → O^2– (g); ∆ Hᶱ = + 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,

(i) Oxygen is more electronegative.

(ii) Addition of electron in oxygen results in larger size of the ion.

(iii) Electron repulsion outweighs the stability gained by achieving noble gas configuration.

(iv) O^– ion has comparatively smaller size than oxygen atom.

The element expected to form largest ion to achieve the nearest noble gas configuration is:
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The order of electron gain enthalpy in kJ/mol of fluorine, chlorine, bromine and iodine, respectively are
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The correct order of the ionic radii of O²⁻, N³⁻, F⁻, Mg²⁺, Na⁺ and Al³⁺ is:

3.31. The first (ΔiH₁) and the second (ΔiH₂) ionization enthalpies (in kJ mol^–1) and the (Δ_egH) electron gain enthalpy (in kJ mol^–1) of a few elements are given below:

Elements      ΔH1      ΔH2       Δ_egH

I                    520       7300      –60

II                  419        3051     –48

III               1681        3374     –328

IV               1008        1846     –295

V                 2372       5251       +48

VI                 738       1451        –40

Which of the above elements is likely to be:

(a) The least reactive element

(b) The most reactive metal

(c) The most reactive non-metal

(d) The least reactive non-metal

(e) The metal which can form a stable binary halide of the formula MX₂ (X = halogen)

(f) The metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?

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.