The graph shows the variation of Ionization P | vedclass

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(A) In a group,the Ionization Potential $(I.P.)$ generally decreases from top to bottom due to an increase in atomic size and shielding effect.However

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$13$ group

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(A) In a group,the Ionization Potential $(I.P.)$ generally decreases from top to bottom due to an increase in atomic size and shielding effect.However,the graph shows an irregular trend $(A > B  D This specific zig-zag pattern of $I.P.$ values is characteristic of elements in group $13$ (e.g.,$B, Al, Ga, In, Tl$).In group $13$,the $I.P.$ decreases from $B$ to $Al$,then increases from $Al$ to $Ga$ due to poor shielding of $d$-electrons,decreases from $Ga$ to $In$,and increases from $In$ to $Tl$ due to poor shielding of $f$-electrons.Therefore,the group is $13$.

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

Element	Ionization energy $1^{st}$ $(kJ/mol)$	Ionization energy $2^{nd}$ $(kJ/mol)$
$X$	$495$	$4563$
$Y$	$731$	$1450$

The graph shows the variation of Ionization Potential $(I.P.)$ with Atomic Number $(At. No.)$ for elements $A$ to $E$. If these elements belong to the same group in the periodic table,identify the group.

Similar Questions

The first and second ionisation enthalpies of a metal are $496$ and $4560 \; kJ \; mol^{-1},$ respectively. How many moles of $HCl$ and $H_{2}SO_{4}$ respectively,will be needed to react completely with $1$ mole of the metal hydroxide?

In each period,the alkali metals...

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

$A$ large difference between the fourth and fifth ionization energies indicates the presence of

Identify the elements $X$ and $Y$ using the ionisation energy values given below :

Element Ionization energy $1^{st}$ $(kJ/mol)$ Ionization energy $2^{nd}$ $(kJ/mol)$

$X$ $495$ $4563$

$Y$ $731$ $1450$

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