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(C) The Ionization Potential $(IP)$ decreases down a group as the atomic size increases and the valence electrons are further from the nucleus.For nob

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$He : Xe$

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(C) The Ionization Potential $(IP)$ decreases down a group as the atomic size increases and the valence electrons are further from the nucleus.For noble gases,the $IP$ values are: $He (2372 \ kJ/mol)$,$Ne (2080 \ kJ/mol)$,$Ar (1520 \ kJ/mol)$,$Kr (1351 \ kJ/mol)$,$Xe (1170 \ kJ/mol)$.Calculating the ratios:$A) He/Ne = 2372/2080 \approx 1.14$$B) Ne/Ar = 2080/1520 \approx 1.37$$C) He/Xe = 2372/1170 \approx 2.03$$D) Kr/Xe = 1351/1170 \approx 1.15$The highest ratio is obtained for the pair $He : Xe$.

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

Find the highest ratio of $IP$ values of the given pair of elements.

Similar Questions

The element having the greatest difference between its first and second ionization energies is:

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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