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Question

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

Vedclass · Answer

(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

Similar Questions

Which of the following sets has the highest sum of $IE$ values compared to all other sets mentioned?

Among the second period elements,the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$. Explain why:
$(i)$ $Be$ has higher $\Delta_{i}H$ than $B$
$(ii)$ $O$ has lower $\Delta_{i}H$ than $N$ and $F$?

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 $C$?

Both nitrogen and oxygen have electrons in the $2p$ orbital. In which of these is the electron-electron repulsion greater,and what is its effect?

For which of the following reactions is the $\Delta H^o$ value equal to the first ionization energy of $Ca$?

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