Ionisation energy Questions in English

(C) Statement $I$: The electronic configuration of the ions after the first ionization are $B^+ (2s^2)$, $Al^+ (3s^2)$, and $Ga^+ (4s^2)$. As the principal quantum number increases from $B$ to $Ga$, the size of the ion increases, leading to a decrease in the ionization enthalpy. Thus, the order $B > Al > Ga$ is correct.
Statement $II$: The general trend for the first ionization enthalpy in group $14$ is $C > Si > Ge > Sn < Pb$. Due to the poor shielding effect of $d$ and $f$ electrons in $Pb$, the effective nuclear charge increases, making the first ionization enthalpy of $Pb$ higher than that of $Sn$. The given order $Si < Ge < Pb < Sn$ is incorrect.

(B) The total ionization energy required to convert $1 \text{ mole}$ of $Li(g)$ to $Li^{2+}(g)$ is the sum of the first and second ionization enthalpies: $IE_{total} = 520 + 7297 = 7817 \text{ kJ mol}^{-1}$.
Given mass of $Li = 3.5 \text{ mg} = 3.5 \times 10^{-3} \text{ g}$.
Molar mass of $Li = 7 \text{ g mol}^{-1}$.
Number of moles of $Li = \frac{3.5 \times 10^{-3} \text{ g}}{7 \text{ g mol}^{-1}} = 0.5 \times 10^{-3} \text{ mol}$.
Total energy required = $\text{Moles} \times IE_{total} = 0.5 \times 10^{-3} \text{ mol} \times 7817 \text{ kJ mol}^{-1} = 3.9085 \text{ kJ}$.
Rounding to the nearest integer, we get $4 \text{ kJ}$.

(C) $1$. Electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$. It has a stable half-filled $p$-orbital, which makes its first ionization enthalpy higher than that of $O$ ($Z=8$, $1s^2 2s^2 2p^4$).
$2$. $F$ ($Z=9$, $1s^2 2s^2 2p^5$) has a higher effective nuclear charge than $O$, leading to a higher first ionization enthalpy.
$3$. Thus, Assertion $A$ is correct.
$4$. Reason $R$ states that the loss of an electron from $O$ leads to a stable half-filled $p$-orbital. The configuration of $O^+$ is $1s^2 2s^2 2p^3$, which is indeed a stable half-filled $p$-orbital. However, this explains the second ionization enthalpy of $O$, not the first. Therefore, Reason $R$ is incorrect as an explanation for the first ionization enthalpy trend.
$5$. Hence, $A$ is correct but $R$ is not correct.