From the ground state electronic configuratio | vedclass

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Question

(B) The second ionization energy is the energy required to remove an electron from a unipositive ion $(M^+)$.$1$. For option $A$ $(Mg)$: $Mg^+ (1s^2,

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The second ionization energy is the energy required to remove an electron from a unipositive ion $(M^+)$.$1$. For option $A$ $(Mg)$: $Mg^+ (1s^2, 2s^2, 2p^6, 3s^1) \rightarrow Mg^{2+} + e^-$. The electron is removed from the $3s$ orbital.$2$. For option $B$ $(Na)$: $Na^+ (1s^2, 2s^2, 2p^6) \rightarrow Na^{2+} + e^-$. The electron is removed from the stable noble gas configuration $(2p^6)$,which requires very high energy.$3$. For option $C$ $(Ne)$: $Ne^+ (1s^2, 2s^2, 2p^5) \rightarrow Ne^{2+} + e^-$. The electron is removed from the $2p$ orbital.$4$. For option $D$ $(F)$: $F^+ (1s^2, 2s^2, 2p^4) \rightarrow F^{2+} + e^-$. The electron is removed from the $2p$ orbital.Comparing these,removing an electron from the stable octet configuration of $Na^+$ requires the highest energy.

From the ground state electronic configuration of the elements given below,pick up the one with the highest value of second ionization energy.

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