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If first ionization enthalpies of elements 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 correct statement is:
Detailed Solution
X has a low first IE and a very high second IE: after losing one electron it attains a noble gas configuration, so X is an alkali metal. Y has a higher first IE (stable $ns^2$) but a comparatively low second IE, so Y is an alkaline earth metal.