Which of the following electronic configurati | vedclass

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(B) The tendency to form a univalent gaseous cation is determined by the first ionization energy $(IE_1)$.Lower ionization energy corresponds to a hig

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$1s^2 2s^2 2p^6 3s^2 3p^1$

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(B) The tendency to form a univalent gaseous cation is determined by the first ionization energy $(IE_1)$.Lower ionization energy corresponds to a higher tendency to form a cation.Comparing the given configurations:$A$: $1s^2 2s^2 2p^6 3s^2$ (Fully filled $s$-orbital,high $IE_1$)$B$: $1s^2 2s^2 2p^6 3s^2 3p^1$ (Valence electron in $3p$ orbital,lowest $IE_1$ among the options)$C$: $1s^2 2s^2 2p^6 3s^2 3p^2$ (Higher $IE_1$ than $B$ due to increased effective nuclear charge)$D$: $1s^2 2s^2 2p^6 3s^2 3p^3$ (Half-filled $p$-orbital,high $IE_1$)Therefore,the configuration $1s^2 2s^2 2p^6 3s^2 3p^1$ has the lowest ionization energy and the highest tendency to form a univalent cation.

Which of the following electronic configurations will have the highest tendency to form a univalent gaseous cation?

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