Chemistry - PERIODIC TABLE Question with Solution | TestHub

The correct statement is $I.P_3$ of Mg $>$$I.P_2$ of Na. Ionization energy generally increases across a period. Magnesium has a higher effective nuclear charge than sodium. Removing a third electron from Mg²⁺ requires significantly more energy than removing a second electron from Na⁺ due to noble gas configuration.

Explanation:

Ionization Potential (I.P.) is the energy required to remove an electron from a gaseous atom or ion.

Option 1: $I.P_1$ of Mg < $I.P_1$ of Al. This is incorrect. Generally, $I.P_1$ increases across a period. However, due to the stable half-filled p-orbital configuration of Al after removing the first electron, and the full s-orbital of Mg, $I.P_1$ of Mg ($[Ne]3s^2$) is actually higher than $I.P_1$ of Al ($[Ne]3s^{2}3p^1$). This is an exception to the general trend.

Option 2: $I.P_3$ of Mg > $I.P_2$ of Na. This is correct.

- $I.P_2$ of Na: Removing an electron from Na⁺ ($[Ne]$). This involves breaking a stable noble gas configuration.

- $I.P_3$ of Mg: Removing an electron from Mg²⁺ ($[Ne]$). This also involves breaking a stable noble gas configuration.

- However, Mg²⁺ has a higher nuclear charge (+12) than Na⁺ (+11) for the same electron configuration. Therefore, it requires more energy to remove an electron from Mg²⁺ than from Na⁺.

Option 3: $I.P_1$ of Mg > $I.P_2$ of Na. This is incorrect.

- $I.P_1$ of Mg: Removing the first electron from Mg ($[Ne]3s^2$).

- $I.P_2$ of Na: Removing the second electron from Na ($[Ne]3s^1\to [Ne]$). This is a very high energy process as it involves removing an electron from a stable noble gas configuration. Thus, $I.P_2$ of Na is much greater than $I.P_1$ of Mg.

Option 4: $I.P_2$ of Mg > $I.P_2$ of Al. This is incorrect.

- $I.P_2$ of Mg: Removing the second electron from Mg⁺ ($[Ne]3s^1$).

- $I.P_2$ of Al: Removing the second electron from Al⁺ ($[Ne]3s^2$).

- Mg⁺ has a $3s^1$ configuration, while Al⁺ has a $3s^2$ configuration. Removing an electron from a filled s-orbital (Al⁺) is generally harder than from a single s-electron (Mg⁺). Also, Al has a higher nuclear charge. Therefore, $I.P_2$ of Al is greater than $I.P_2$ of Mg.