Problem 28 - Chemistry: The Central Science in SI Units, Global Edition

Question

Explain the following variations in atomic or ionic radii: (a) \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\) (b) \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

Step-by-Step Solution

Answer

The three cases demonstrate different trends in atomic and ionic radii: (a) \(\mathrm{I}^{-} > \mathrm{I} > \mathrm{I}^{+}\): The iodine anion has gained an electron, causing repulsion between electrons and resulting in an increased atomic radius. The iodine cation has lost an electron, reducing electron-electron repulsion and decreasing its radius. (b) \(\mathrm{Ca}^{2+} > \mathrm{Mg}^{2+} > \mathrm{Be}^{2+}\): As we move down Group 2 elements in the Periodic Table, the atomic radius increases due to the addition of electron shells. All three ions have divalent charges but differ in electron shells. (c) \(\mathrm{Fe} > \mathrm{Fe}^{2+} > \mathrm{Fe}^{3+}\): As neutral iron loses electrons to form cations, electron-electron repulsion decreases and effective nuclear charge felt by the remaining electrons increases, causing the cations to have smaller radii.

Step 1: Understanding Atomic/ Ionic Radius

The radius of an atom/ ion is determined by the number of electron shells it has (the period it belongs to in the Periodic Table), whether it has lost, gained or has not changed the number of electrons compared to the neutral atom (ionization state) and the effective nuclear charge felt by the outermost (valence) electrons.

Step 2: Iodine Anion, Neutral Atom, and Cation

(a) Compare \(\mathrm{I}^{-}\), \(\mathrm{I}\), and \(\mathrm{I}^{+}\). All have the same numbers of protons, but differ in the number of electrons. - \(\mathrm{I}^{-}\) has gained one electron, so the repulsion between the extra electron and the other electrons causes the atomic radius to increase. - \(\mathrm{I}\) is neutral iodine and thus has a smaller radius because it does not have the extra electron of \(\mathrm{I}^{-}\). - \(\mathrm{I}^{+}\) has lost an electron, reducing the electron-electron repulsion, which results in a decrease in radius due to the attractive force from the protons in the nucleus.

Step 3: Divalent Cations of Group 2 Elements

(b) Compare \(\mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}\), and \(\mathrm{Be}^{2+}\). These are all divalent cations, so they have the same charge state, but the atoms from which they are derived differ in their position in the Periodic Table. - Group 2 elements (where Ca, Mg, and Be are located) have a trend of increasing atomic radius as we move down the group due to the addition of electron shells. Hence, \(\mathrm{Ca}^{2+}\) has a larger ionic radius than \(\mathrm{Mg}^{2+}\), which in turn has a larger radius than \(\mathrm{Be}^{2+}\).

Step 4: Iron in Different Oxidation States

(c) Compare \(\mathrm{Fe}\), \(\mathrm{Fe}^{2+}\), and \(\mathrm{Fe}^{3+}\). These all feature the same atomic species, but the number of electrons differs. - \(\mathrm{Fe}\) is neutral iron and has the most electrons of the three. As electrons are lost to form \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\), the cations are smaller because the electron-electron repulsion decreases and the effective nuclear charge felt by the remaining electrons increases. Therefore, \(\mathrm{Fe} > \mathrm{Fe}^{2+} > \mathrm{Fe}^{3+}\).

Key Concepts

Ionization State

The ionization state of an atom refers to its electrical charge resulting from the loss or gain of electrons. When an atom gains electrons, it becomes an anion with a negative charge. Conversely, when it loses electrons, it becomes a cation with a positive charge. This process can significantly affect the atomic or ionic radius based on electron-electron repulsion and the number of electron shells.

• Anions are larger because adding electrons increases repulsion among electrons, causing the outermost shell to expand.
• Cations are smaller since losing electrons reduces repulsion, allowing the nuclei to pull the remaining electrons closer.

Understanding ionization state explains why \( \mathrm{I}^{-} \) is larger than \( \mathrm{I} \), and \( \mathrm{I}^{+} \) is the smallest. Similarly, for metals like iron, from \( \mathrm{Fe} \) to \( \mathrm{Fe}^{3+} \), the decrease in electron count leads to a smaller size due to less repulsion and greater attractive forces towards the nucleus.

Effective Nuclear Charge

The effective nuclear charge (\( Z_{eff} \)) is a crucial concept that determines the overall force exerted by the nucleus on the electrons in an atom. It's the net positive charge experienced by electrons after accounting for shielding effects from other electrons, especially in inner shells.

As electrons are removed, the \( Z_{eff} \) increases because the remaining electrons experience less shielding and thus feel a stronger attraction to the nucleus. For instance:

• In \( \mathrm{Fe}^{2+} \) and \( \mathrm{Fe}^{3+} \), as electrons are stripped away, the remaining electrons are pulled closer, reducing atomic size.
• Even among similar ions like \( \mathrm{Ca}^{2+} \), \( \mathrm{Mg}^{2+} \), and \( \mathrm{Be}^{2+} \), those lower in the periodic table tend to have larger radii due to additional shielding, even as \( Z_{eff} \) acts strongly on outer shells.

Thus, the concept of \( Z_{eff} \) explains the variation in sizes among ions and contributes to understanding why cations are generally smaller than their neutral atoms.

Periodic Table Trends

The periodic table is a powerful tool for predicting and explaining trends in atomic and ionic sizes. As you move across a period from left to right, atomic numbers increase, but the addition of electrons generally does not increase the size due to increased \( Z_{eff} \) pulling electrons closer.

Key trends include:

• Within the same group, atomic and ionic radii increase from top to bottom because of additional electron shells, even as \( Z_{eff} \) might increase slightly.
• In a period, as seen in \( \mathrm{Ca}^{2+} \), \( \mathrm{Mg}^{2+} \), \( \mathrm{Be}^{2+} \), moving left to right generally results in smaller radii due to stronger attraction from increasing protons.

By understanding these periodic trends, one can predict and explain the relative sizes of ions and atoms, providing a clearer insight into chemical properties and behaviors.