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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 understand the anomaly in the ionization enthalpy ($\Delta_{i}H$) of Oxygen ($Z=8$) relative to Nitrogen ($Z=7$) and Fluorine ($Z=9$), we must first write the ground-state electronic configurations according to the Aufbau Principle, Pauli Exclusion Principle, and Hund’s Rule of Maximum Multiplicity.

Element Atomic Number ($Z$) Electronic Configuration
Nitrogen ($N$) 7 $1s^2 2s^2 2p^3$
Oxygen ($O$) 8 $1s^2 2s^2 2p^4$
Fluorine ($F$) 9 $1s^2 2s^2 2p^5$

Step 2: Theoretical Justification of Stability

Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from an isolated gaseous atom. The magnitude of this energy is governed by:

  • Effective Nuclear Charge ($Z_{eff}$): As we move left to right across a period, $Z_{eff}$ increases due to the addition of protons, which generally increases $\Delta_{i}H$ [Per Periodic Trend Principles].
  • Exchange Energy and Stability: Subshells that are half-filled ($p^3$) or completely filled ($p^6$) possess additional stability due to the symmetric distribution of electron density and maximum exchange energy [Per Hund’s Rule of Maximum Multiplicity].

Step 3: Comparative Analysis of N and O

Nitrogen ($2p^3$) possesses a half-filled p-subshell. In this configuration, each of the three $p$-orbitals contains one electron with parallel spins, which minimizes inter-electronic repulsion and maximizes exchange energy. Removing an electron from Nitrogen requires overcoming this inherent stability.

Conversely, Oxygen ($2p^4$) has one paired set of electrons in the $2p$ subshell ($p_x^2, p_y^1, p_z^1$). The presence of two electrons in the same orbital introduces inter-electronic repulsion. This repulsion makes it energetically easier to remove an electron from the $2p^4$ configuration of Oxygen compared to the stable, half-filled $2p^3$ configuration of Nitrogen [Per Electronic Stability Principles].

Step 4: Comparative Analysis of O and F

While Oxygen has a lower $\Delta_{i}H$ than Nitrogen, it follows the trend relative to Fluorine. Fluorine ($2p^5$) has a higher $Z_{eff}$ than Oxygen. Because Fluorine is closer to the stable noble gas configuration (Neon, $2p^6$), the valence electrons in Fluorine are held more tightly by the nucleus. Therefore, Oxygen possesses a lower $\Delta_{i}H$ than Fluorine due to both lower effective nuclear charge and the configuration-based stability gap [Per Coulomb’s Law].

Step 5: Visual Representation of Valence Orbitals

Nitrogen (2p³) ↑ ↑ ↑ Oxygen (2p⁴) ↑↓ ↑ ↑

Final Solution: Oxygen has a lower ionization enthalpy than Nitrogen because the removal of an electron from Nitrogen requires disrupting a stable, half-filled $2p^3$ configuration, whereas in Oxygen, the removal of an electron is facilitated by inter-electronic repulsion present in the $2p^4$ configuration. Oxygen has a lower ionization enthalpy than Fluorine due to the smaller effective nuclear charge of Oxygen compared to Fluorine.


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