The formation of the oxide ion O^2- (g), from 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, process of formation of O^2- in gas phase is unfavorable even though O^2- is isoelectronic with neon. It is due to the fact that

1. Electron repulsion outweighs the stability gained by achieving  noble gas Configuration 

2. $O^{-}$ ion has comparatively smaller size than oxygen atom

3. Oxygen is more electronegative

4. Addition of electron in oxygen result in large size of the ion 

For the second period elements the correct increasing order of first ionisation enthalpy is:

1.  Li<Be<B<C<O<N<F<Ne

2.  Li<Be<B<C<N<O<F<Ne

3.  Li<B<Be<C<O<N<F<Ne

4.  Li<B<Be<C<N<O<F<Ne

The order of electron gain enthalpy in the case of halogens is:

1. F > Cl > Br > I

2. I > Br > Cl > F

3. Cl > F > Br > I

4. I > Br > F > Cl

Which pair has largest size difference?

1. Na and Na⁺

2. Na and K

3. Cl and Cl^–

4. Mg and Mg⁺

The correct order of 3rd ionisation potential of Na, Mg, Al and Si is

1. Si > Al > Mg > Na

2. Mg > Si > Al > Na

3. Mg > Na > Si > Al

4. Mg > Na > Al > Si

The correct order of second ionisation energies of B , C , N & O is: 

1. O > N > C > B

2. O > C >N >B

3. O > B > C > N

4. O > N > B > C

The correct order of the decreasing ionic radii among the following isoelectronic species is:

1. Ca²⁺ > K⁺ >S^2–> Cl^–

2. Cl^– > S^2– > Ca^2– > K⁺

3. S^2– > Cl^– > K⁺ > Ca²⁺

4. K⁺ > Ca²⁺ > Cl^– > S^2–

Which of the following has the highest electron affinity?

1. F^–

2. O^–

3. O

4. Na

The electronegativity of the following elements increases in the order:

1. C, N, Si, P

2. N, Si, C, P

3. Si, P, C, N

4. P, Si, N, C