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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 :
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The correct answer is:
X is an alkali metal and $Y$ is an alkaline earth metal.
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
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
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