Consider the elements $\mathrm{Ne}, \mathrm{Na}, \mathrm{Mg}$. The element with highest first ionization…
Consider the elements $\mathrm{Ne}, \mathrm{Na}, \mathrm{Mg}$. The element with highest first ionization enthalpy and element with lowest second ionization enthalpy respectively are
$\mathrm{Na}, \mathrm{Ne}$
$\mathrm{Ne}, \mathrm{Mg}$
$\mathrm{Na}, \mathrm{Na}$
$\mathrm{Mg}, \mathrm{Na}$
Solution
$\begin{aligned} & { }_{10} \mathrm{Ne}=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6,{ }_{11} \mathrm{Na}=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^1, \\ & { }_{12} \mathrm{Mg}=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2\end{aligned}$
$\Rightarrow$ Highest first ionization enthalpy $=\mathrm{Ne}$, due to stable fully-filled configuration.
$\Rightarrow$ Lowest second ionization enthalpy $=\mathrm{Mg}$, due to stable noble - gas configuration obtained after losing two electrons.