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

Solution :

Step 1: Electronic Configuration Analysis

To determine the relative ionization enthalpies ($\Delta_{i}H$) of Beryllium ($Z=4$) and Boron ($Z=5$), we must first establish their ground-state electronic configurations per the Aufbau principle [Per the Aufbau Principle, electrons fill orbitals in order of increasing energy].

  • Beryllium ($Be$): $1s^2 2s^2$
  • Boron ($B$): $1s^2 2s^2 2p^1$

Step 2: Analysis of the Ionization Process

The first ionization enthalpy represents the energy required to remove the most loosely bound electron from an isolated gaseous atom:

  • For $Be$: $Be(g) \rightarrow Be^+(g) + e^-$. The electron is removed from the $2s$ orbital.
  • For $B$: $B(g) \rightarrow B^+(g) + e^-$. The electron is removed from the $2p$ orbital.

Step 3: Justification based on Orbital Penetration and Shielding

The discrepancy in the ionization enthalpy arises from the following physical factors:

  1. Orbital Penetration: The $2s$ orbital has a higher penetration power towards the nucleus compared to the $2p$ orbital. Because the $2s$ electrons are closer to the nucleus on average, they experience a higher effective nuclear charge ($Z_{eff}$) and are held more tightly [Per the Shielding Effect and Penetration Power of orbitals].
  2. Symmetry and Stability: The $2s^2$ configuration of Beryllium represents a completely filled subshell. Completely filled subshells possess additional stability due to the symmetrical distribution of electron density, which requires significantly more energy to disrupt compared to the removal of a single electron from an incompletely filled $2p^1$ subshell [Per Hund’s Rule and Orbital Stability criteria].
Be $1s^2 2s^2$ B $1s^2 2s^2 2p^1$ Higher energy to remove 2s

Step 4: Conclusion

Although Boron has a higher atomic number and thus a higher nuclear charge, the removal of an electron from the $2p$ orbital in Boron is easier because it is further from the nucleus and is shielded by the $2s^2$ electrons. Conversely, the $2s$ electrons in Beryllium are held more strongly by the nucleus and belong to a fully-filled, stable configuration.

Final Solution: The ionization enthalpy of Beryllium is higher than that of Boron because Beryllium involves the removal of an electron from a fully-filled $2s$ orbital, which is more stable and closer to the nucleus than the $2p$ electron removed from Boron.


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