Question

Which of the following relation is correct with respect to $I^{st}$ and $2^{nd}$ Ionization energy of Sodium and Magnesium?

• 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.
$\therefore$ $I{I}_{Na}>I{I}_{Mg}$