Electronic Arrangement & Ionization Enthalpy

To determine the electronic arrangement with the lowest ionization enthalpy among the given options, we need to understand the concept of ionization enthalpy and how it is influenced by electronic configurations.

Ionization Enthalpy:
Ionization enthalpy is the energy required to remove an electron from an atom or ion in its gaseous state. It is a measure of the stability of the atom and its tendency to lose electrons. The lower the ionization enthalpy, the easier it is to remove an electron.

Electronic Configurations:
The electronic configuration of an atom describes the distribution of electrons in its various atomic orbitals. Each electronic configuration is unique and determines the atom's chemical behavior and properties.

Now, let's analyze each option to determine which one has the lowest ionization enthalpy.

(a) 1s2 2s2 2p3:
This electronic configuration corresponds to nitrogen (N). It has 5 valence electrons in the 2p orbital. The removal of an electron would result in a stable half-filled 2p orbital, which requires a moderate amount of energy. Nitrogen has a higher ionization enthalpy compared to other elements in this set.

(b) 1s2 2s2 2p6 3s1:
This electronic configuration corresponds to sodium (Na). It has 1 valence electron in the 3s orbital. The removal of this electron results in a stable noble gas configuration of neon (1s2 2s2 2p6). Sodium has a relatively low ionization enthalpy due to the electron's position in the outermost orbital.

(c) 1s2 2s2 2p6:
This electronic configuration corresponds to the noble gas neon (Ne). It has a completely filled valence shell, and removing an electron requires a significant amount of energy. Neon has a high ionization enthalpy.

(d) 1s2 2s2 2p5:
This electronic configuration corresponds to fluorine (F). It has 7 valence electrons in the 2p orbital. The removal of an electron results in a stable noble gas configuration of neon. Fluorine has a relatively high ionization enthalpy due to the electron's position in the outermost orbital and the strong electron-electron repulsion in the 2p subshell.

Conclusion:
Among the given options, the electronic arrangement with the lowest ionization enthalpy is option (b) - 1s2 2s2 2p6 3s1, corresponding to sodium (Na). Sodium's electron configuration and the position of the valence electron in the outermost orbital make it easier to remove compared to the other options.