If first ionization enthalpies of element $X$ and $Y$ are $419 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $590 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively and second ionization enthalpies of $\mathrm{X}$ and $\mathrm{Y}$ are 3069 $\mathrm{kJ} \mathrm{mol}^{-1}$ and $1145 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. Then correct statement is :

• Correct answer
  A

  X is an alkali metal and $Y$ is an alkaline earth metal.
• B

  X is an alkaline earth metal and $Y$ is an alkali metal.
• C

  Both $X$ and $Y$ are alkali metals.
• D

  Both $X$ and $Y$ are alkaline earth metals.

Solution

The highest jump ion successive ionisation energy indicates a stable noble gas configuration.
The highest jump is observed in the $\mathrm{IE}_1 \& \mathrm{IE}_2$. It shows the attainment of stable noble gas configuration of $X$ after losing one electron. Hence, $\mathrm{X}$ is an alkali metal. The $\mathrm{IE}_1$ of $Y$ is $+590 \mathrm{~kJ} \mathrm{~mol}^{-1}$ which is greater than $\mathrm{IE}_1$ of $X$ due to $n s^2$ configuration. $\therefore \quad X=$ Alkali metal and $Y=$ Alkaline Earth metal

Hence, the correct answer is option: 'A'.