The electronic configuration of elements A, B | vedclass

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Question

(A) The elements $A, B$ and $C$ belong to Group $1$ (alkali metals) with configurations $[He] 2s^1$ $(Li)$,$[Ne] 3s^1$ $(Na)$ and $[Ar] 4s^1$ $(K)$.As

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$A > B > C$

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(A) The elements $A, B$ and $C$ belong to Group $1$ (alkali metals) with configurations $[He] 2s^1$ $(Li)$,$[Ne] 3s^1$ $(Na)$ and $[Ar] 4s^1$ $(K)$.As we move down the group,the atomic size increases due to the addition of new shells.An increase in atomic size leads to a decrease in the effective nuclear attraction on the valence electron.Therefore,the energy required to remove the valence electron (first ionization potential) decreases as we move down the group.The order of ionization potential is $A > B > C$.

Atomic number	$I_1$ $(kJ/mol)$	$I_2$ $(kJ/mol)$	$I_3$ $(kJ/mol)$
$n$	$I_1$	$I_2$	$I_3$
$n+1$	$1681$	$3374$	$6050$
$n+2$	$2081$	$3952$	$6122$
$n+3$	$496$	$4562$	$6910$
$n+4$	$738$	$1451$	$7733$

The electronic configuration of elements $A, B$ and $C$ are $[He] 2s^1, [Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?

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