The correct order of second ionisation potent | vedclass

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Question

(A) The electronic configurations are: $K (Z=19): [Ar] 4s^1$,$Ca (Z=20): [Ar] 4s^2$,$Ba (Z=56): [Xe] 6s^2$.After the first ionization,the configuratio

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$K > Ca > Ba$

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(A) The electronic configurations are: $K (Z=19): [Ar] 4s^1$,$Ca (Z=20): [Ar] 4s^2$,$Ba (Z=56): [Xe] 6s^2$.After the first ionization,the configurations become: $K^+: [Ar]$,$Ca^+: [Ar] 4s^1$,$Ba^+: [Xe] 6s^1$.The second ionization potential involves removing an electron from these ions.For $K^+$,the electron is removed from the stable noble gas configuration $([Ar])$,which requires a very high amount of energy.For $Ca^+$ and $Ba^+$,the second electron is removed from the $ns^1$ orbital.Comparing $Ca^+$ and $Ba^+$,$Ca$ is above $Ba$ in the group,so $Ca^+$ has a smaller size and higher effective nuclear charge,making its second $IP$ higher than that of $Ba^+$.Therefore,the order is $K > Ca > Ba$.

The correct order of second ionisation potentials $(IP)$ of $Ca$,$Ba$ and $K$ is

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