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

The key idea is that ionization enthalpy depends on electronic configuration and stability of the subshell — half-filled and fully-filled configurations are extra stable.

1. Be vs B: Be has the configuration $1s^2 2s^2$ — the $2s$ subshell is completely filled, making it very stable. Removing an electron from a filled $2s$ orbital requires extra energy. B has $1s^2 2s^2 2p^1$ — the single $2p$ electron is less tightly bound and easier to remove. Hence, $\Delta_i H$ of Be is higher than that of B.
2. O vs N and F: N has $1s^2 2s^2 2p^3$ — the $2p$ subshell is half-filled, giving extra stability. Removing an electron from a half-filled $p$ subshell is harder. O has $1s^2 2s^2 2p^4$ — the fourth $p$ electron must pair up in an orbital, causing electron-electron repulsion. This makes it easier to remove than from N. …

The ionization enthalpy ($\Delta_i H$) of an element is the energy required to remove the most loosely bound electron from a gaseous atom. Across a period, we generally expect $\Delta_i H$ to increase as nuclear charge increases and atomic radius decreases. But period 2 shows a famous zigzag pattern: Li < B < Be < C < O < N < F < Ne. Two specific anomalies stand out, and both are rooted in the stability conferred by particular electronic configurations.

Why Be has higher $\Delta_i H$ than B

1. Electronic configurations matter.

   Beryllium (atomic number 4) has the configuration $1s^2 2s^2$. Boron (atomic number 5) is $1s^2 2s^2 2p^1$. The electron to be removed from Be comes from the filled $2s$ subshell; from B, it comes from the singly occupied $2p$ orbital.

2. Subshell stability.

   A filled $2s^2$ subshell is particularly stable — it has spherical symmetry and no unpaired electrons. Removing an electron from a filled subshell disrupts this stable arrangement, requiring extra energy. In contrast, the single $2p$ electron in boron is in a higher‑energy orbital (the $2p$ level is slightly above $2s$) and is also less tightly bound because it experiences less effective nuclear charge (the $2s$ electrons partially shield it). So removing that $2p$ electron is easier.

3. Penetration and shielding.

   The $2s$ electron penetrates closer to the nucleus than a $2p$ electron, so it feels a stronger effective nuclear charge ($Z_{\text{eff}}$). This makes the $2s$ electron harder to remove. For Be, $Z_{\text{eff}}$ for the $2s$ electron is about 3.0; for B, the $2p$ electron feels $Z_{\text{eff}} \approx 2.4$ (the $2s$ electrons shield it poorly, but the $2p$ orbital is less penetrating). The net effect: Be’s ionization enthalpy (899 kJ/mol) is higher than B’s (801 kJ/mol).

Why O has lower $\Delta_i H$ than N and F

1. Half‑filled vs. electron‑pair repulsion.

   Nitrogen (atomic number 7) has the configuration $1s^2 2s^2 2p^3$. The three $2p$ electrons occupy three different orbitals (Hund’s rule), each with one electron — a half‑filled $p$ subshell. This arrangement has extra stability due to exchange energy (all spins parallel) and symmetry. Removing an electron from N means breaking this stable half‑filled configuration, which requires more energy.

   Oxygen (atomic number 8) is $1s^2 2s^2 2p^4$. Now one of the $2p$ orbitals is doubly occupied. The two electrons in the same orbital experience strong electron‑electron repulsion. This repulsion makes one of those paired electrons easier to remove — it’s already “pushed away” by its partner. So O’s ionization enthalpy (1314 kJ/mol) is actually lower than N’s (1402 kJ/mol).

2. Why is O lower than F? …