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 $Mg > Al > Na$.

List-$I$ (Electronic configuration)	List-$II$ ($\Delta_{i}H$ in $kJ\ mol^{-1}$)
$(a)$ $1s^{2} 2s^{2}$	$(i)$ $801$
$(b)$ $1s^{2} 2s^{2} 2p^{4}$	$(ii)$ $899$
$(c)$ $1s^{2} 2s^{2} 2p^{3}$	$(iii)$ $1314$
$(d)$ $1s^{2} 2s^{2} 2p^{1}$	$(iv)$ $1402$

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

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