Question

Among the second period elements the actual ionization enthalpies are in the order 
Li < B < Be < C < O < N < F < Ne. 
Explain why 

1. Be has higher Δ_iH than B
2. O has lower Δ_iH than N and F?

Answer 1

Actual order (second period): \[ \text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne} \]

(a) Why Be has higher Δ_iH than B

Electronic configurations: \[ \text{Be}: 1s^{2} 2s^{2} \] \[ \text{B}: 1s^{2} 2s^{2} 2p^{1} \]

• In Be, the electron to be removed is from a completely filled \(2s\) subshell, which is relatively stable and more strongly held by the nucleus.
• In B, the electron to be removed is a single \(2p\) electron, which is higher in energy and less penetrating than \(2s\); it is held less tightly.
• Therefore, it requires more energy to remove an electron from Be than from B, so Be has higher ionization enthalpy.

(b) Why O has lower Δ_iH than N and F

Electronic configurations: \[ \text{N}: 1s^{2} 2s^{2} 2p^{3} \quad (\text{half‑filled, stable}) \] \[ \text{O}: 1s^{2} 2s^{2} 2p^{4} \] \[ \text{F}: 1s^{2} 2s^{2} 2p^{5} \]

• N has a half‑filled \(2p^{3}\) configuration, which is extra stable; removing an electron disturbs this stability, so Δ_iH of N is relatively high.
• In O, the \(2p^{4}\) configuration has one pair of electrons in a \(2p\) orbital; the inter‑electronic repulsion within this pair makes it easier to remove one of these electrons, so Δ_iH of O is lower than that of N.
• F has higher effective nuclear charge and smaller size than O; its valence electrons are held more strongly, so its ionization enthalpy is higher than that of O.

Key Idea (LaTeX Summary)

\[ \text{Stable (filled/half‑filled subshell) } \Rightarrow \text{higher ionization enthalpy} \] \[ \text{Paired electrons in same orbital} \Rightarrow \text{extra repulsion} \Rightarrow \text{lower ionization enthalpy} \]