Which of the following relations is correct w | vedclass

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(D) The electronic configuration of $Na$ $(Z=11)$ is $[Ne] 3s^1$. After losing one electron,it achieves the stable noble gas configuration of $Ne$ $(1

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$II_{Na} > II_{Mg}$

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(D) The electronic configuration of $Na$ $(Z=11)$ is $[Ne] 3s^1$. After losing one electron,it achieves the stable noble gas configuration of $Ne$ $(1s^2 2s^2 2p^6)$.Removing the second electron from this stable configuration requires a very high amount of energy.The electronic configuration of $Mg$ $(Z=12)$ is $[Ne] 3s^2$. After losing one electron,it becomes $[Ne] 3s^1$,which is relatively easier to ionize further compared to the stable $Ne$ core of $Na^+$.Therefore,the second ionization potential of sodium is significantly higher than that of magnesium,i.e.,$II_{Na} > II_{Mg}$.

$B$	$801$
$Al$	$577$
$Ga$	$579$
$In$	$558$
$Tl$	$589$

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

Which of the following relations is correct with respect to the first $(I)$ and second $(II)$ ionization potentials of sodium $(Na)$ and magnesium $(Mg)$?

Similar Questions

Which of the following electronic configurations has the highest ionization enthalpy?

In which of the following,elements are correctly arranged in the decreasing order of their second ionisation enthalpies?

Which one of the following has the lowest ionization energy?

The first ionization enthalpy values (in $kJ\,mol^{-1}$) of group $13$ elements are :

$B$ $801$

$Al$ $577$

$Ga$ $579$

$In$ $558$

$Tl$ $589$

How would you explain this deviation from the general trend?

The ionization potential $(IP)$ values for an element are given below. Identify the element:
$(IP)_1$ $7.1 \ eV$
$(IP)_2$ $10.3 \ eV$
$(IP)_3$ $36.2 \ eV$
$(IP)_4$ $39.1 \ eV$

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