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Q3.17:
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Solution :

Step 1: Electronic Configuration and First Ionization Enthalpy ($IE_1$)

To analyze the ionization enthalpies, we must first establish the ground state electronic configurations of Sodium ($Na$, $Z=11$) and Magnesium ($Mg$, $Z=12$):

  • $Na (11): 1s^2 2s^2 2p^6 3s^1$
  • $Mg (12): 1s^2 2s^2 2p^6 3s^2$

The first ionization enthalpy ($IE_1$) is the energy required to remove the most loosely bound electron from a neutral gaseous atom:

  • $Na(g) \rightarrow Na^+(g) + e^-$
  • $Mg(g) \rightarrow Mg^+(g) + e^-$

Analysis: Although Magnesium has a higher nuclear charge ($Z_{eff}$) which increases attraction, the $3s^2$ orbital of Magnesium is completely filled, providing extra stability [Per Hund’s Rule and Pauli Exclusion Principle]. However, Sodium's electron is in the $3s^1$ configuration. The effective nuclear charge experienced by the valence electron in Sodium is significantly lower than that of Magnesium, and removing an electron from $Na$ achieves a stable noble gas configuration ($[Ne]$). Despite this, the higher nuclear charge of $Mg$ makes its first electron harder to remove than the first electron of $Na$. Therefore, $IE_1(Na) < IE_1(Mg)$.

Step 2: Analysis of Second Ionization Enthalpy ($IE_2$)

The second ionization enthalpy ($IE_2$) involves removing an electron from the unipositive ion:

  • $Na^+(g) \rightarrow Na^{2+}(g) + e^-$ (Configuration: $1s^2 2s^2 2p^5$)
  • $Mg^+(g) \rightarrow Mg^{2+}(g) + e^-$ (Configuration: $1s^2 2s^2 2p^6$)

Analysis: When the first electron is removed, Sodium achieves the stable neon configuration ($1s^2 2s^2 2p^6$). Removing a second electron from $Na^+$ requires breaking this highly stable octet (noble gas core), which demands a massive amount of energy. Conversely, $Mg^+$ has the configuration $1s^2 2s^2 2p^6 3s^1$. Removing this $3s^1$ electron allows Magnesium to achieve the stable noble gas configuration ($Mg^{2+}$, $1s^2 2s^2 2p^6$). [By the Stability of Noble Gas Configurations], the energy required to disturb the stable octet of $Na^+$ is significantly higher than the energy required to reach the stable configuration of $Mg^{2+}$.

Step 3: Comparative Data Summary

Element Electronic Configuration ($Ion^+$) Stability Status
$Na^+$ $[He] 2s^2 2p^6$ Stable Noble Gas Core
$Mg^+$ $[Ne] 3s^1$ Unstable / Readily loses electron

Step 4: Visualization of Atomic Stability

Na+ (Stable) Config: 1s2 2s2 2p6 Mg+ (Unstable) Config: [Ne] 3s1 IE2 Path

Final Solution: The first ionization enthalpy of Sodium is lower because it is easier to remove an electron from the $3s^1$ orbital of Sodium than the $3s^2$ orbital of Magnesium. However, the second ionization enthalpy of Sodium is significantly higher because it involves the removal of an electron from a stable, noble-gas configuration ($2p^6$), whereas the second ionization of Magnesium results in the stable noble-gas configuration ($2p^6$).


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