A configuration with the lowest ionization enthalpy among the following is:
1. \(1 s^2 2 s^2 2 p^5\)$\mathrm{}$
2. \(1 s^2 2 s^2 2 p^3\)
3. \(1 s^2 2 s^2 2 p^6 3 s^1\)
4. \(1 s^2 2 s^2 2 p^6\)

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

The incorrect order of electronegativity is :

Among the following, the most characteristic oxidation states for lead and tin are, respectively:

1. +4, +2

2. +2, +4

3. +4, +4

4. +2, +2

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

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

1. $C{a}^{2+}>K^{+}>S^{2-}>C{l}^{-}$

2. $C{l}^{-}>S^{2-}>C{a}^{2+}>K^{+}$

3. $S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

4. $K^{+}>C{a}^{2+}>C{l}^{-}>S^{2-}$

The correct order of ionic radii is:

1. $H^{-}>H^{+}>H$ $$
2. $N{a}^{+}>F^{-}>O^{2-}$
3. $F^{-}>O^{2-}>N{a}^{+}$
4. $N^{3-}>M{g}^{2+}>A{l}^{3+}$

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:

Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$, the simplest formula for this compound is: 

1. ${\mathrm{Mg}}_2{\mathrm{X}}_3$

2. ${\mathrm{MgX}}_2$

3. ${\mathrm{Mg}}_2\mathrm{X}$

4. ${\mathrm{Mg}}_3{\mathrm{X}}_2$

Element of the third period that is expected to exhibit positive electron gain enthalpy is: