Periodic Table

Understanding Electron Gain Enthalpy
Electron gain enthalpy is the energy change when an electron is added to a neutral atom in its gaseous state. The trend of electron gain enthalpy is influenced by:
- Atomic size (smaller atoms have higher negative values).
- Effective nuclear charge (more nuclear attraction increases the enthalpy).
- Electronic configuration stability (half-filled and fully-filled orbitals resist electron gain).
Analyzing the Given Configurations
- Element I ([He] 2s 2p ): This corresponds to Fluorine (F), which has a very high electron gain enthalpy due to its high nuclear attraction and small size.
- Element II ([Ne] 3s 3p ): This corresponds to Chlorine (Cl), which has slightly lower but still high electron gain enthalpy compared to Fluorine.
- Element III ([He] 2s 2p ): This corresponds to Oxygen (O), which has lower electron gain enthalpy than Fluorine due to higher repulsion in a smaller size.
- Element IV ([Ne] 3s 3p ): This corresponds to Sulfur (S), which has a lower electron gain enthalpy compared to Fluorine and Chlorine.
Determining the Correct Order From known values: Thus, the correct order is:

To understand the ionic radii of N^3-, O^2-, and F^-, let us consider the following points:

The ionic radius is the effective size of an ion in a crystal lattice.

The ionic radius depends on the charge of the ion: the more negative the charge, the larger the ionic radius, as additional electrons increase electron-electron repulsion.

For the same element in different ionic states, the more negatively charged ion will generally have a larger ionic radius because of the greater repulsion between the added electrons.

Now, let's examine each ion:

N^3- (Nitride ion) has the highest negative charge (3^-), which means it has gained three electrons. The added electrons cause the electron-electron repulsion to be high, which leads to an increase in size. This results in the largest ionic radius among the three ions. The ionic radius of N^3- is 1.71 Å.

O^2- (Oxide ion) has a charge of 2^-, meaning it has gained two electrons. It has less repulsion compared to N^3- but still exhibits a larger size than F^-. The ionic radius of O^2- is 1.40 Å.

F^- (Fluoride ion) has the smallest negative charge (1^-), which means it has only gained one electron. Since it has the least repulsion between electrons, it will have the smallest ionic radius among the three. The ionic radius of F^- is 1.36 Å.

Thus, the correct order of ionic radii is:

N^3- > O^2- > F^-

So, the correct answer is option (2) 1.71, 1.40, and 1.36.