Question

Choose the larger particle in each pair: (a) \(S\) or \(S^{2-}\); (b) \(\mathrm{Al}^{3+}\) or \(\mathrm{Al} ;\) (c) \(\mathrm{Au}^{+}\) or \(\mathrm{Au}^{3+}\).

Short answer

(a) S2- is larger than S; (b) Al atom is larger than Al3+; (c) Au+ is larger than Au3+.

Step-by-step solution

Understanding Atomic and Ionic Size

Recognize that the size of an atom or ion is affected by the number of electrons and the effective nuclear charge. Generally, the more electrons an atom or ion has, the larger it is, due to increased electron-electron repulsion. Conversely, the more positive the charge (the more electrons are removed), the smaller the ion becomes, due to a greater effective nuclear charge pulling the remaining electrons closer.

Comparing sulfur (S) and sulfide (S2-)

Decide which is larger between a sulfur atom (S) and a sulfide ion (S2-). Since the sulfide ion has gained two extra electrons compared to the neutral sulfur atom, it will be larger due to increased electron-electron repulsion.

Comparing aluminum ion (Al3+) and aluminum atom (Al)

Decide which is larger between an aluminum ion (Al3+) and an aluminum atom (Al). The aluminum ion has lost three electrons and will have a smaller radius than the neutral aluminum atom because the effective nuclear charge is stronger on the remaining electrons, pulling them closer to the nucleus.

Comparing gold(I) ion (Au+) and gold(III) ion (Au3+)

Decide which is larger between a gold(I) ion (Au+) and a gold(III) ion (Au3+). The Au3+ ion has lost more electrons than Au+, and therefore, the effective nuclear charge experienced by the remaining electrons is higher. This means that Au+ is larger than Au3+ as it has more electrons.

Key concepts

Electron-Electron Repulsion

When exploring the realm of atomic and ionic sizes, the concept of electron-electron repulsion plays a pivotal role. Simply put, electrons all have the same negative charge and, as a result, they repel each other. Picture them as a group of people all trying to hold onto magnets with the same pole facing each other; they naturally want to push apart.

Now, consider the case of a neutral atom versus an ion with additional electrons. The more electrons present, the more they will repel each other, leading to an increase in the overall size of the ion. As mentioned in the textbook solution, a sulfide ion (\(S^{2-}\)) carries two extra electrons compared to a neutral sulfur atom (\(S\)), and because of this, the repulsions between these additional electrons force the electron cloud to expand, making the sulfide ion noticeably larger than the sulfur atom.

This concept elucidates why, as atoms turn into anions (negatively charged ions), their sizes tend to increase. Understanding electron-electron repulsion provides a foundational explanation for the variance in size between different ions and atoms.

Effective Nuclear Charge

Diving deeper into the fabric of atoms, we encounter the term effective nuclear charge (ENC). This concept is vital when determining why certain ions are smaller than others. The ENC is the net positive charge exerted by the nucleus onto the electron cloud, accounting for the fact that inner-shell electrons partially shield outer-shell electrons from the full nuclear charge.

In our textbook example concerning aluminum, the aluminum ion (\(Al^{3+}\)) is smaller than the neutral aluminum atom (\(Al\) itself), because when it loses three electrons to become \(Al^{3+}\), the reduction in electron-electron repulsion means that the remaining electrons feel a stronger pull from the nucleus. The result is a higher ENC, which draws the electrons closer, reducing the radius of the ion.

The ENC is a crucial concept in predicting the behavior of electrons within an atom or ion. A higher ENC creates tighter, closer shells of electrons, which shrinks the atomic or ionic radius, as clearly seen by comparing neutral atoms to their positively charged ion counterparts.

Comparison of Atomic Radii

Understanding the Variance in Atomic Sizes

Finally, let's explore the comparison of atomic radii. Atoms are the basic units of chemical elements and are unique in their sizes, which are often measured in terms of atomic radii. Atomic radius is the measure from the center of the nucleus to the boundary of the surrounding cloud of electrons.

In the textbook solution, we see an example of such comparison between gold ions: \(Au^+\) and \(Au^{3+}\). Due to the loss of electrons, the \(Au^{3+}\) ion has a higher effective nuclear charge on its remaining electrons than the \(Au^+\) ion, which causes the radius of \(Au^{3+}\) to be smaller than that of \(Au^+\) despite both being gold ions.

When comparing atomic radii across the periodic table, we find patterns such as radii increasing down a group as additional electron shells are added, and decreasing across a period as electrons are added to the same shell but the nucleus gains more protons, thus increasing the effective nuclear charge. These principles allow chemists and students alike to predict and rationalize the relative sizes of atoms and ions, enhancing their understanding of chemical reactions and the periodic trends.