Question

If $I_{Mg}=\text{First ionization energy of Mg}$
$I{I}_{Mg}=\text{Second ionization energy of Mg}$
$I_{Na}=\text{First ionization energy of Na}$
$I{I}_{Na}=\text{Second ionization energy of Mg}$
Choose the correct option:

• A. $I_{Mg}=I{I}_{Na}$
• B. $I_{Na}>I{I}_{Mg}$
• C. $I{I}_{Mg}>I{I}_{Na}$
• D. $I{I}_{Na}>I{I}_{Mg}$

Answer 1

The correct option is D $I{I}_{Na}>I{I}_{Mg}$

Electronic configuratin of $Na=1s^{2}2s^{2}2p^{6}3s^1$
Electronic configuratin of $Mg=1s^{2}2s^{2}2p^{6}3s^2$
$I^{st}$ Ionization energy of sodium will be less than magnesium.
As we move across the period, the number of valence electrons in an atom increases and also nuclear charge increases which tend to decrease the radius and ionization energy increases.
When an electron is removed from Na, it acquires noble gas configuration which is more stable and removing electron requires more energy.
$\therefore$ $I{I}_{Na}>I{I}_{Mg}$