Consider the elements Ne,Na,and Mg. The eleme | vedclass

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(B) The electronic configurations are: $_{10}Ne = 1s^2 2s^2 2p^6$ $_{11}Na = 1s^2 2s^2 2p^6 3s^1$ $_{12}Mg = 1s^2 2s^2 2p^6 3s^2$ $1$. Highest first i

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$Ne$,$Mg$

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(B) The electronic configurations are: $_{10}Ne = 1s^2 2s^2 2p^6$ $_{11}Na = 1s^2 2s^2 2p^6 3s^1$ $_{12}Mg = 1s^2 2s^2 2p^6 3s^2$ $1$. Highest first ionization enthalpy: $Ne$ has the highest first ionization enthalpy because it is a noble gas with a stable,fully-filled valence shell $(2s^2 2p^6)$. $2$. Lowest second ionization enthalpy: The second ionization enthalpy involves removing an electron from a unipositive ion. For $Na^+$,the configuration becomes $1s^2 2s^2 2p^6$ (stable noble gas configuration),making the second ionization enthalpy very high. For $Mg^+$,the configuration is $1s^2 2s^2 2p^6 3s^1$. Removing the second electron from $Mg^+$ leads to the stable $Mg^{2+}$ ion $(1s^2 2s^2 2p^6)$,which is energetically favorable. Thus,$Mg$ has the lowest second ionization enthalpy among the given elements.

Consider the elements $Ne$,$Na$,and $Mg$. The element with the highest first ionization enthalpy and the element with the lowest second ionization enthalpy,respectively,are:

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