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Q3.16(ii):
Among the second period elements the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$. Explain why (ii) $O$ has lower $\Delta_{i} H$ than $N$ and $F$?

Solution :

Step 1: Electronic Configuration Analysis

To determine the relative ionization enthalpies ($\Delta_{i}H$) of Nitrogen ($N$) and Oxygen ($O$), we must first analyze their ground-state electronic configurations in the second period of the periodic table.

  • Nitrogen ($N$, $Z=7$): $1s^{2} 2s^{2} 2p^{3}$
  • Oxygen ($O$, $Z=8$): $1s^{2} 2s^{2} 2p^{4}$

[Per the Aufbau Principle and Hund’s Rule of Maximum Multiplicity]

Step 2: Analysis of Subshell Stability

The stability of an atom is significantly influenced by the occupancy of its valence subshells. Specifically, half-filled and fully-filled orbitals possess extra stability due to symmetry and exchange energy.

The $2p$ subshell consists of three degenerate orbitals ($p_{x}, p_{y}, p_{z}$).

  • In Nitrogen, the $2p^{3}$ configuration implies one electron in each of the three $p$-orbitals. This is a symmetrical, half-filled configuration, which provides additional stability. [Per Hund's Rule]
  • In Oxygen, the $2p^{4}$ configuration implies one orbital is doubly occupied ($p_{x}^{2} p_{y}^{1} p_{z}^{1}$). This introduces inter-electronic repulsion between the two electrons sharing the same $p$-orbital.

Step 3: Energetic Considerations for Ionization

Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. The process can be represented as:

$N(g) \rightarrow N^{+}(g) + e^{-}$; $\Delta_{i}H = 1402 \text{ kJ/mol}$

$O(g) \rightarrow O^{+}(g) + e^{-}$; $\Delta_{i}H = 1314 \text{ kJ/mol}$

Although Oxygen has a higher effective nuclear charge ($Z_{eff}$) than Nitrogen—which would normally pull electrons closer and increase $\Delta_{i}H$—the stability factor dominates here.

  • Removing an electron from Nitrogen requires breaking the stable, half-filled $2p$ configuration.
  • Removing an electron from Oxygen actually relieves the inter-electronic repulsion present in the doubly occupied $2p$ orbital, making the removal energetically more favorable than expected.

Step 4: Comparative Orbital Visualization

Nitrogen (2p^3) Oxygen (2p^4)

Step 5: Conclusion on Periodic Trends

Regarding Fluorine ($F$), it possesses a $2p^{5}$ configuration. Despite the inter-electronic repulsions, the significantly higher $Z_{eff}$ of Fluorine compared to Nitrogen creates a stronger electrostatic attraction between the nucleus and the valence electrons, which overrides the stability factor of Nitrogen. Therefore, the $\Delta_{i}H$ of $N$ is lower than $F$ due to the atomic size and $Z_{eff}$ increase, but Oxygen is lower than Nitrogen due to the destabilization caused by electron-electron repulsion in the $2p^{4}$ orbital.

Final Solution: Oxygen exhibits a lower ionization enthalpy than Nitrogen because Nitrogen's half-filled $2p^{3}$ subshell configuration provides extra electronic stability, whereas the removal of an electron from Oxygen's $2p^{4}$ configuration reduces the inter-electronic repulsions existing in the doubly occupied $p$-orbital.


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