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(C) Ionization enthalpy $(IE)$ depends on atomic size and electronic stability.Smaller atomic size leads to higher $IE$ because the valence electrons

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$[Ne]\, 3s^2\, 3p^3$

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(C) Ionization enthalpy $(IE)$ depends on atomic size and electronic stability.Smaller atomic size leads to higher $IE$ because the valence electrons are more strongly attracted by the nucleus.Additionally,half-filled and fully-filled orbitals possess extra stability,requiring more energy to remove an electron.Comparing the given configurations: $[Ne]\, 3s^2\, 3p^1$,$[Ne]\, 3s^2\, 3p^2$,and $[Ne]\, 3s^2\, 3p^3$ belong to the $3^{rd}$ period,while $[Ar]\, 3d^{10}\, 4s^2\, 4p^3$ belongs to the $4^{th}$ period.Elements in the $3^{rd}$ period have smaller atomic radii than those in the $4^{th}$ period,resulting in higher $IE$ for $3^{rd}$ period elements.Among the $3^{rd}$ period elements,$[Ne]\, 3s^2\, 3p^3$ has a half-filled $p$-orbital,which provides extra stability,making its ionization enthalpy the highest.

The electronic configuration with the highest ionization enthalpy is

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