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(B) The difference $(IP_2 - IP_1)$ is related to the energy required to remove the second electron compared to the first. For $Mg$ $(Z=12)$,the electr

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$Mg$

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(B) The difference $(IP_2 - IP_1)$ is related to the energy required to remove the second electron compared to the first. For $Mg$ $(Z=12)$,the electronic configuration is $[Ne] 3s^2$. $IP_1$ involves removing an electron from the stable $3s^2$ orbital,while $IP_2$ involves removing an electron from the $3s^1$ orbital. For $Al$ $(Z=13)$,the configuration is $[Ne] 3s^2 3p^1$. $IP_1$ involves removing the $3p^1$ electron,and $IP_2$ involves removing a $3s^2$ electron. Generally,the jump $(IP_2 - IP_1)$ is significantly higher when the second electron is removed from a more stable or inner shell. For $Mg$,$(IP_2 - IP_1)$ is approximately $15.03 \ eV - 7.64 \ eV = 7.39 \ eV$,which is $For $Al$,$(IP_2 - IP_1)$ is approximately $18.82 \ eV - 5.98 \ eV = 12.84 \ eV$,which is $> 11 \ eV$. Thus,the condition holds for $Mg$.

Element	$IE_1 + IE_2$ $(kJ/mol)$	$IE_3 + IE_4$ $(kJ/mol)$
$P$	$2.45$	$8.82$
$Q$	$2.85$	$6.11$

For which element is the difference between the second and first ionization potential $(IP_2 - IP_1 < 11 \ eV)$ found?

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The sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:

Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$

$P$ $2.45$ $8.82$

$Q$ $2.85$ $6.11$

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