Q.Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However, oxygen has lower ionisation enthalpy than nitrogen. Explain.

Concept: Electronic Configuration and Stability

The apparent contradiction resolves when we examine the electronic configurations and the energy required to remove versus add an electron.

Nitrogen ($1s^2 2s^2 2p^3$) has a half-filled $2p$ subshell, which is exceptionally stable due to exchange energy and symmetry. Adding an electron disrupts this stability, requiring energy input (positive electron gain enthalpy). Removing an electron from this stable configuration demands considerable energy, giving nitrogen a high ionisation enthalpy.

Oxygen ($1s^2 2s^2 2p^4$) has one electron beyond the half-filled state. Adding an electron to form $O^-$ ($2p^5$) moves toward a stable noble-gas configuration, releasing energy (negative electron gain enthalpy). However, removing an electron from oxygen's $2p^4$ configuration actually relieves some electron-electron repulsion in the doubly-occupied orbital, making ionisation easier than from nitrogen's stable half-filled state. …

Why electronic configuration governs both properties

Electron gain enthalpy and ionisation enthalpy measure opposite processes—adding versus removing an electron—but both depend critically on the stability of the atom's electronic configuration. The key insight is that stability is not a single number; an atom can be stable against one perturbation yet vulnerable to another, depending on which orbitals are involved and how electrons are arranged within them.

Nitrogen ($1s^2 \, 2s^2 \, 2p^3$) and oxygen ($1s^2 \, 2s^2 \, 2p^4$) sit at a turning point in the $2p$ subshell. Nitrogen achieves a half-filled $p$ subshell with one electron in each of the three $2p$ orbitals, all with parallel spins. Oxygen has one orbital doubly occupied, forcing two electrons into the same spatial region.

Step-by-step explanation

1. Nitrogen's positive electron gain enthalpy

When nitrogen attempts to gain an electron, the incoming electron must enter an already singly-occupied $2p$ orbital, creating a pair. This costs energy for two reasons:

• Loss of exchange energy: The half-filled configuration maximises exchange stabilisation (electrons with parallel spins can exchange positions, lowering energy). Pairing destroys this symmetry.
• Electron–electron repulsion: Forcing two electrons into the same orbital increases Coulombic repulsion.

The energy required to overcome these penalties exceeds the energy released by adding an electron to the nuclear attraction. The process is endothermic, so $\Delta_{\text{eg}}H > 0$.

2. Oxygen's negative electron gain enthalpy

Oxygen already has one paired orbital ($2p_x^2 \, 2p_y^1 \, 2p_z^1$, for instance). Adding an electron completes a second pair in a previously singly-occupied orbital:

The incoming electron experiences strong nuclear attraction ($Z_{\text{eff}}$ is high because oxygen is further right in the period) and does not disrupt an especially stable arrangement—one pair already exists. The energy released exceeds the pairing cost, making $\Delta_{\text{eg}}H < 0$.

3. Nitrogen's high ionisation enthalpy

Removing an electron from nitrogen means breaking up the half-filled $2p^3$ configuration:

This destroys the exchange stabilisation and the symmetric distribution of electrons. The resulting $2p^2$ configuration is significantly less stable. Nitrogen holds its electrons tightly, requiring substantial energy to ionise.

4. Oxygen's lower ionisation enthalpy

Removing an electron from oxygen eliminates the pairing repulsion in the doubly-occupied orbital:

The product $\text{O}^+$ now has the coveted half-filled $2p^3$ configuration. The system actually gains stability by ionisation, partially offsetting the energy cost of removing an electron. Consequently, oxygen's ionisation enthalpy is lower than nitrogen's, despite oxygen having a higher nuclear charge.

--- …