Solution:
$\text{Hint: In the period, elements having half-filled and fully-filled orbitals need more ionization energy than elements having partial filled atomic orbitals}$ $1. \text{The metallic radius increases down the group because the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.}$ $\text{This results in a larger metallic radius. Thus, Li < Na < K < Rb as increasing order of metallic radius is correct.}$ $\text{Li= 152 pm; Na=186 pm; K=227 pm; Rb=248 pm}$ $2. \text{Electron gain enthalpy become less negative as we go down a group because the size of the atom increases and the added electron would be farther from the nucleus. But Cl has high negative electron gain enthalpy than F.}$ $\text{Fluorine due to its small size interelectronic repulsion takes place when an extra electron is added, and because of it, F has less electron gain enthalpy than Cl.}$ $\text{The given Increasing order of electron gain enthalpy (with negative sign) that is, I < Br < F < Cl is correct.}$ $\text{F=-328 kJ/mol ; Cl=- 349 kJ/mol; Br=- 325 kJ/mol; I=- 295 kJ/mol}$ $3. \text{In the case of an isoelectronic ion, as the negative charge increases the size of the ion increases, and as the positive charge on the ion increases the size of the ion decreases. Hence, Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} \text{ in Increasing ionic size is correct.}$ $4. \text{The given order of first ionization enthalpy is wrong. From left to right across the period the ionization energy increases because the effective nuclear charge increases and the size of the element decreases.}$ $\text{But the ionization energy of nitrogen is more than oxygen because nitrogen has a stable half-filled orbital from which removing electron is difficult. Thus, the correct order is B < C < O < N.}$ $\text{B=801 kJ/mol; C=1086 kJ/mol; O=1314 kJ/mol; N=1402 kJ/mol}$