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(B) The second ionisation energy $(IE_2)$ refers to the energy required to remove an electron from a unipositive ion.$1$. For $Te$ $(5s^2 5p^4)$ and $

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$In > Sr$

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(B) The second ionisation energy $(IE_2)$ refers to the energy required to remove an electron from a unipositive ion.$1$. For $Te$ $(5s^2 5p^4)$ and $Sb$ $(5s^2 5p^3)$: The $IE_2$ of $Te$ is higher than $Sb$ because $Sb^+$ has a stable half-filled $p^2$ configuration,but $Te^+$ has a $p^3$ configuration. Actually,$Sb$ has a stable half-filled $p^3$ configuration,so removing the second electron from $Sb^+$ is harder than from $Te^+$. Thus,$Te > Sb$ is correct.$2$. For $In$ $(5s^2 5p^1)$ and $Sr$ $(5s^2)$: $Sr^+$ has a stable $s^1$ configuration,while $In^+$ has a $s^2$ configuration. Removing an electron from $Sr^+$ is harder,so $Sr > In$. The comparison $In > Sr$ is incorrect.$3$. $He$ has a very high $IE_2$ due to its noble gas configuration.$4$. $Fe < Fe^+$ is a comparison of ionisation energy,which is conceptually incorrect as $IE$ refers to neutral atoms,but $Fe^+$ is already an ion. The question asks for the incorrect comparison among the options provided regarding $IE_2$ trends.

Find the incorrect comparison of second ionisation energy.

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