Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:

The formation of the oxide ion O^2– (g), from the oxygen atom requires first an exothermic and then an endothermic step as shown below, 
$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-};$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=–141$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{2-}\left(\mathrm{g}\right);$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=+780$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

Thus, the process of formation of O^2– in the gas phase is unfavorable even though O^2– is isoelectronic with neon. It is due to the fact that:

The correct order of increasing electron gain enthalpy with a negative sign for the elements O, S, F, and Cl is : 

1. Cl < F < S < O                   

2. O < S < F < Cl

3. F < S < O < Cl                   

4. S < O < Cl < F

The correct order of increasing electron affinity for the elements, O, S, F and Cl is:

The electron gain enthalpies of halogens in kJ mol^-1 are given below.

F =-332, Cl =-349, Br =-325, I =-295.

The lesser negative value for F as compared to that of Cl is due to:
 

1. Strong electron-electron repulsions in the compact 2p-subshell of F.

2. Weak electron-electron repulsions in the bigger 3p-subshell of Cl.

3. Smaller electronegativity value of F than Cl.

4. 1 & 2 both 

The reactivity of alkali metals increases, whereas halogen decreases down the group, because:

The incorrect match among the following is:

1. B < C < N < O (increasing first ionisation enthalpy)

2. I < Br < F < Cl (increasing electron gain enthalpy)

3. Li < Na < K < Rb (increasing metallic radius)

4. Al³⁺ < Mg²⁺ < Na⁺ <F^- (increasing ionic size)

 The correct order for electron affinity of halogens is :

1. Br > F

2. F > Cl

3. Br > Cl

4. F > I

An element with higher negative electron gain enthalpy in the given pair is:

(i) O or F  (ii) F or Cl
1. O, Cl
2. F, F
3. O, F
4. F, Cl