Question

For each of the following pairs, indicate which species is smaller. Explain your reasoning in terms of the electron structure of each species. a. Na or \(\mathrm{Na}^{+}\) c. \(\mathrm{Mg}\) or \(\mathrm{Mg}^{2+}\) b. F or \(\mathrm{F}^{-}\) d. \(\mathrm{S}\) or \(\mathrm{S}^{2-}\)

Short answer

In summary, based on the electron structure of each species: a. \(\mathrm{Na}^{+}\) is smaller than Na due to greater charge density. b. F is smaller than \(\mathrm{F}^{-}\) due to increased electron-electron repulsion in the ion. c. \(\mathrm{Mg}^{2+}\) is smaller than Mg because of stronger electrostatic attraction between remaining electrons and protons. d. S is smaller than \(\mathrm{S}^{2-}\) as the additional electrons lead to increased electron-electron repulsion and a larger ion.

Step-by-step solution

Introduction

To determine the size of these atomic species, it is important to understand that as we add or subtract electrons to/from neutral atoms, the electrostatic forces between the protons and the electrons will change. The more electrons are added or subtracted, the stronger or weaker this force will be, affecting the size of these species (ions).

Compare Na and \(\mathrm{Na}^{+}\)

The Na atom has an electron configuration of [Ne]3s1, while \(\mathrm{Na}^{+}\) is formed by ionizing the neutral atom, leaving it with an electron configuration of [Ne]. Since the sodium cation, \(\mathrm{Na}^{+}\), has one fewer electron, its charge density is greater than that of the neutral sodium atom. This greater charge density causes the remaining electrons to be more strongly attracted to the nucleus, making \(\mathrm{Na}^{+}\) smaller than Na.

Compare \(\mathrm{Mg}\) and \(\mathrm{Mg}^{2+}\)

The Mg atom has an electron configuration of [Ne]3s2, while \(\mathrm{Mg}^{2+}\) has an electron configuration of [Ne] after losing two electrons. The magnesium cation, \(\mathrm{Mg}^{2+}\), has lost two electrons but its nuclear charge remains the same, hence it will be significantly smaller compared to the neutral Mg atom due to stronger electrostatic attraction between protons and remaining electrons.

Compare F and \(\mathrm{F}^{-}\)

The F atom has an electron configuration of [He]2s2 2p5, while \(\mathrm{F}^{-}\) gains one electron, resulting in an electron configuration of [He]2s2 2p6. The additional electron in \(\mathrm{F}^{-}\) causes a reduced electrostatic attraction between the protons and electrons, as they are somewhat further out and they experience stronger electron-electron repulsions. As a result, the added electron pushes the others outward, increasing the size of the ion. Thus, \(\mathrm{F}^{-}\) is larger than F.

Compare \(\mathrm{S}\) and \(\mathrm{S}^{2-}\)

The S atom has an electron configuration of [Ne]3s2 3p4, while the \(\mathrm{S}^{2-}\) gains two electrons, resulting in an electron configuration of [Ne]3s2 3p6. The addition of two electrons to the sulfur atom increases electron-electron repulsion, pushing the electrons further away from the nucleus, thus expanding the ion. Consequently, \(\mathrm{S}^{2-}\) is larger than S.

Summary

In summary, the sizes of the atomic species are as follows: a. \(\mathrm{Na}^{+}\) is smaller than Na b. F is smaller than \(\mathrm{F}^{-}\) c. \(\mathrm{Mg}^{2+}\) is smaller than Mg d. S is smaller than \(\mathrm{S}^{2-}\)

Key concepts

Electron Configuration

Understanding electron configuration is crucial when examining atomic sizes. The electron configuration tells us how electrons are distributed among the various atomic orbitals. For example, the sodium atom (Na) has the electron configuration of [Ne]3s1, meaning it has one electron in its outer 3s orbital beyond the stable neon core. When sodium loses an electron to become \( \text{Na}^+ \), its configuration becomes [Ne], filling all the orbitals up to neon. This loss of an outer electron stabilizes the atom and impacts its radius.

Similarly, magnesium (Mg) has an electron configuration of [Ne]3s2 but loses two electrons to form \( \text{Mg}^{2+} \), reverting to [Ne]. This change indicates a return to the stable configuration of neon, significantly decreasing its size. Understanding these patterns assists in predicting how the removal or addition of electrons affects the size of an ion.

Cation vs Anion Sizes

Cations and anions have different sizes due to their electron configurations. Cations, like \( \text{Na}^+ \) and \( \text{Mg}^{2+} \), are positively charged ions that form when an atom loses electrons. Losing electrons results in a stronger electrostatic attraction between the remaining electrons and the nucleus, making the atom more compact and smaller. That's why \( \text{Na}^+ \) is smaller than Na and \( \text{Mg}^{2+} \) is smaller than Mg.

Anions, however, such as \( \text{F}^- \) and \( \text{S}^{2-} \), increase in size compared to their parent atoms. This increase happens because anions gain electrons, introducing added electron-electron repulsion which pushes the electrons further apart. This explains why \( \text{F}^- \) is larger than F and \( \text{S}^{2-} \) is larger than S. The increased size is directly linked to the filled outer shells and additional repulsive forces at play among the electrons.

Electrostatic Forces

Electrostatic forces play a major role in determining the size of ions. These forces arise due to the attraction and repulsion between charged particles—protons and electrons, in an atom or ion. When an atom loses electrons and forms a cation, such as \( \text{Na}^+ \) or \( \text{Mg}^{2+} \), the reduced electron count results in a stronger pull by the nucleus on the remaining electrons. This contraction results in a smaller radius for the ion.

Conversely, when an atom gains electrons to form an anion, like \( \text{F}^- \) or \( \text{S}^{2-} \), the added electrons increase repulsion among the electrons. This pushes them further apart, thereby increasing the size of the ion. Understanding these interactions helps explain the differences in atomic and ionic radii observed across periods and groups in the periodic table.

Ionization

Ionization involves the process of gaining or losing electrons by an atom to form ions, significantly affecting their size. The ionization of sodium Na to \( \text{Na}^+ \), for instance, involves the loss of an electron from its outermost shell. This results in a positive charge and a smaller ionic radius due to the reduction in electron cloud size around the nucleus.

Similarly, for magnesium, forming \( \text{Mg}^{2+} \) through the loss of two electrons leads to an even smaller size compared to its neutral atom. On the other hand, elements like fluorine \( \text{F} \) and sulfur \( \text{S} \) gain electrons to form anions, \( \text{F}^- \) and \( \text{S}^{2-} \) respectively. The process of gaining electrons results in a larger size due to increased electron-electron repulsion, which affects the overall electron distribution. Understanding ionization helps in predicting and explaining the periodic trends in sizes and reactivity of different elements and ions.