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(C) The electronic configurations of the ions formed after the removal of the first electron are: $Na^{+} (1s^{2} 2s^{2} 2p^{6})$ $Ne^{+} (1s^{2} 2s^{

Vedclass · Accepted Answer

$Mg < Al < Ne < Na$

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(C) The electronic configurations of the ions formed after the removal of the first electron are: $Na^{+} (1s^{2} 2s^{2} 2p^{6})$ $Ne^{+} (1s^{2} 2s^{2} 2p^{5})$ $Mg^{+} (1s^{2} 2s^{2} 2p^{6} 3s^{1})$ $Al^{+} (1s^{2} 2s^{2} 2p^{6} 3s^{2})$ The second ionisation potential $(IE_{2})$ is the energy required to remove an electron from these unipositive ions. $Na^{+}$ has a stable noble gas configuration $(2s^{2} 2p^{6})$,making it extremely difficult to remove another electron,thus it has the highest $IE_{2}$. $Mg^{+}$ has a single electron in the $3s$ orbital,which is relatively easy to remove. $Al^{+}$ has a stable $3s^{2}$ configuration,making its $IE_{2}$ higher than $Mg^{+}$. $Ne^{+}$ has a $2p^{5}$ configuration,which is less stable than $Na^{+}$ but requires significant energy. Comparing these,the correct order of $IE_{2}$ is $Mg < Al < Ne < Na$.

Which one of the following is the correct order of the second ionisation potential of $Na, Ne, Mg$ and $Al$?

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