The correct sequence of elements in decreasin | vedclass

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(D) The first ionisation energy depends on the electronic configuration and effective nuclear charge.Electronic configurations are: $Na (Z=11): [Ne] 3

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$Mg > Al > Na$

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(D) The first ionisation energy depends on the electronic configuration and effective nuclear charge.Electronic configurations are: $Na (Z=11): [Ne] 3s^1$,$Mg (Z=12): [Ne] 3s^2$,$Al (Z=13): [Ne] 3s^2 3p^1$.$Mg$ has a fully filled $3s$ orbital,which is more stable and requires more energy to remove an electron compared to $Al$ and $Na$.$Al$ has a higher effective nuclear charge than $Na$,making its first ionisation energy higher than $Na$.Thus,the decreasing order is $Mg > Al > Na$.

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

The correct sequence of elements in decreasing order of first ionisation energy is

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

$X$ $495$ $4563$

$Y$ $731$ $1450$

Which of the following electronic configurations will have the least $I.P.$ value?

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