If first ionization enthalpies of element X a | vedclass

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Question

(A) For element $X$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $3069 - 419 = 2650 \ kJ \ mol^{-1}$,which is very large.

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

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(A) For element $X$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $3069 - 419 = 2650 \ kJ \ mol^{-1}$,which is very large. This indicates that the removal of the second electron requires breaking a stable noble gas configuration,characteristic of an alkali metal.For element $Y$: The difference between the $2^{nd}$ and $1^{st}$ ionization enthalpy is $1145 - 590 = 555 \ kJ \ mol^{-1}$,which is relatively small. This indicates that the second electron is removed from the same valence shell,characteristic of an alkaline earth metal.Therefore,$X$ is an alkali metal and $Y$ is an alkaline earth metal.

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

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