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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 ionization enthalpy ($\Delta_{i}H$), we must first establish the ground state electronic configurations of Beryllium ($Be$, $Z=4$) and Boron ($B$, $Z=5$) according to the Aufbau principle.

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

[Justification: The ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.]

Step 2: Analysis of the Ionization Process

For Beryllium, the electron to be removed is from the $2s$ orbital. For Boron, the electron to be removed is from the $2p$ orbital.

  • $Be \rightarrow Be^+ + e^-$: Electron is removed from a fully filled $2s^2$ subshell.
  • $B \rightarrow B^+ + e^-$: Electron is removed from a $2p^1$ subshell.

Step 3: Justification based on Orbital Penetration and Stability

The higher ionization enthalpy of Beryllium compared to Boron is attributed to two primary quantum mechanical factors:

  1. Penetration Effect: The $2s$ orbital has a higher electron density near the nucleus than the $2p$ orbital. Consequently, the $2s$ electron is more strongly attracted by the nuclear charge ($Z_{eff}$) compared to the $2p$ electron, which experiences more effective shielding from the inner shell electrons [Per the Radial Probability Distribution function].
  2. Electronic Stability: The $2s^2$ configuration is a completely filled subshell. Completely filled and half-filled subshells possess extra stability due to symmetry and the exchange energy [Per Hund’s Rule and the Pauli Exclusion Principle]. Removing an electron from a stable $2s^2$ configuration requires significantly more energy than removing the single $2p^1$ electron from Boron.

Step 4: Visualizing Orbital Energy

Orbital Energy Levels Be (2s) B (2p) Energy (eV)

The diagram illustrates that the $2p$ electron in Boron sits at a higher energy state (closer to the ionization threshold) than the $2s$ electrons in Beryllium, making the $2p$ electron easier to ionize.

Final Solution: The ionization enthalpy of Beryllium is higher than that of Boron because the $2s^2$ orbital of Beryllium is completely filled and experiences a higher effective nuclear charge due to the greater penetration of $2s$ electrons, rendering the removal of an electron energetically unfavorable compared to the removal of the shielded $2p^1$ electron in Boron.


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