Q3 / 160 · AP EAMCET 2023 (16 May Shift 2) · Chemistry

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

  • A
    $\mathrm{Na}, \mathrm{Ne}$
  • Correct answer
    B
    $\mathrm{Ne}, \mathrm{Mg}$
  • C
    $\mathrm{Na}, \mathrm{Na}$
  • D
    $\mathrm{Mg}, \mathrm{Na}$

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.

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

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