Problem 35 - Chemistry The Central Science In Si Units

Question

Provide a brief explanation for each of the following: \((\mathbf{a}) \mathrm{Cl}^{-}\) is larger than Ar. (b) \(\mathrm{P}^{3-}\) is larger than \(\mathrm{S}^{2-}\). (c) \(\mathrm{K}^{+}\) is larger than \(\mathrm{Na}^{+} .(\mathbf{d}) \mathrm{F}^{-}\) is larger than \(\mathrm{F}\).

Step-by-Step Solution

Answer

Ionic size increases when gaining electrons and decreases when losing electrons, with an additional effect from electron repulsion and nuclear charge.

Step 1: Understanding Chloride and Argon

Chloride ion (8 ext{Cl}^{-}9) has gained an extra electron compared to chlorine atom, increasing its electron cloud and causing electron-electron repulsion. As a noble gas, Argon (8 ext{Ar}9) has a stable, full electron shell without this additional electron and repulsion. Thus, 8 ext{Cl}^{-}9 is larger than 8 ext{Ar}9.

Step 2: Comparing Phosphide and Sulfide Ions

The 8 ext{P}^{3-}9 ion has three extra electrons, resulting in a larger ionic radius due to increased electron-electron repulsion and less effective nuclear charge compared to 8 ext{S}^{2-}9. Moreover, 8 ext{P}9 is larger than 8 ext{S}9 in their respective periods, contributing to a larger overall ionic size for 8 ext{P}^{3-}9.

Step 3: Evaluating Potassium and Sodium Ions

Though both ions have lost an electron, 8 ext{K}^{+}9 is in the higher period than 8 ext{Na}^{+}9, meaning 8 ext{K}9 originally had more electron shells even after losing one electron. As a result, 8 ext{K}^{+}9 ends up having a larger radius than 8 ext{Na}^{+}9.

Step 4: Fluoride Ion vs. Fluorine Atom

When Fluorine (8 ext{F}9) gains an electron to become Fluoride (8 ext{F}^{-}9), the increased electron-electron repulsion within the electron cloud leads 8 ext{F}^{-}9 to have a larger radius than the neutral 8 ext{F}9 atom.

Key Concepts

Chloride vs Argon

The size of ions and atoms is primarily influenced by their electron configurations. Chloride ion ( ext{Cl}^{-} ) has gained an electron compared to its neutral atom, resulting in a full outer electron shell like argon. However, because of this extra electron, there is increased electron-electron repulsion, causing the electron cloud to expand. This effect makes ( ext{Cl}^{-} ) larger than the Argon atom ( ext{Ar} ), even though both have the same number of electrons. Argon, being a noble gas, naturally has stable and full electron shells without additional electron gain, leading to a compact size.

Phosphide vs Sulfide

The comparison between phosphide ( ext{P}^{3-} ) and sulfide ( ext{S}^{2-} ) ions highlights how the number of extra electrons influences ionic size. Phosphide ion has three extra electrons, while sulfide ion has two. With more electrons added, phosphide's electron cloud experiences more repulsion, thus it becomes larger than sulfide. Additionally, phosphorus is in an earlier period than sulfur, meaning even their neutral atoms differ in size, with phosphorus being larger. This basic size difference between their atoms is also reflected in their ionic radii when both gain electrons.

Potassium ion vs Sodium ion

Potassium ion ( K^{+} ) and sodium ion ( Na^{+} ) both result from losing one electron, essentially having the same noble gas configuration. However, because potassium is located a period below sodium in the periodic table, it possesses more electron shells initially. Even after losing an electron, the remaining electron cloud is spread over more shells, making K^{+} larger than Na^{+} . The initial presence of additional electron layers in potassium leads to its greater ionic radius.

Fluoride ion vs Fluorine atom

When a fluorine atom ( F ) gains an electron to form a fluoride ion ( F^{-} ), the extra negative charge results in increased repulsion among electrons. This repulsion causes the electron cloud to expand, making the fluoride ion larger than its corresponding neutral atom. The fluorine atom's initial compact size is due to its valency seeking to stabilize by gaining one electron to achieve the noble gas electron configuration similar to neon. Upon gaining this electron, as F^{-} , the corresponding expansion in size due to repulsion is noticeable.