Match List-I with List-II:List-I (Electronic | vedclass

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Question

(A) The electronic configuration $ns^2$ corresponds to alkaline earth metals (e.g.,$Be$,$Mg$),which have relatively high ionization energy due to the

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-I$

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(A) The electronic configuration $ns^2$ corresponds to alkaline earth metals (e.g.,$Be$,$Mg$),which have relatively high ionization energy due to the stable fully filled $s$-orbital.$ns^2np^1$ corresponds to Group $13$ elements (e.g.,$B$,$Al$),which have lower ionization energy than Group $2$ due to the removal of a $p$-electron.$ns^2np^3$ corresponds to Group $15$ elements (e.g.,$N$,$P$),which have high ionization energy due to the stable half-filled $p$-orbital.$ns^2np^6$ corresponds to Group $18$ elements (Noble gases,e.g.,$Ne$,$Ar$),which have the highest ionization energy due to their stable octet configuration.Comparing the given values:$A (ns^2) = II (899 	ext{ kJ/mol})$$B (ns^2np^1) = III (800 	ext{ kJ/mol})$$C (ns^2np^3) = IV (1402 	ext{ kJ/mol})$$D (ns^2np^6) = I (2080 	ext{ kJ/mol})$Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

List-$I$ (Electronic configuration of neutral atom where $n=2$)	List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$	$I. 2080$
$B. ns^2np^1$	$II. 899$
$C. ns^2np^3$	$III. 800$
$D. ns^2np^6$	$IV. 1402$

Match List-$I$ with List-$II$:
List-$I$ (Electronic configuration of neutral atom where $n=2$) List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$ $I. 2080$
$B. ns^2np^1$ $II. 899$
$C. ns^2np^3$ $III. 800$
$D. ns^2np^6$ $IV. 1402$

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Match List-$I$ with List-$II$:List-$I$ (Electronic configuration of neutral atom where $n=2$)List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)$A. ns^2$$I. 2080$$B. ns^2np^1$$II. 899$$C. ns^2np^3$$III. 800$$D. ns^2np^6$$IV. 1402$