Question

For each of the following statements, provide an explanation (a) \(\mathrm{O}^{2-}\) is larger than \(\mathrm{O} ;\) (b) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{O}^{2-}\) : (c) \(S^{2-}\) is larger than \(K^{+}\); (d) \(\mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}\).

Short answer

a) O²⁻ is larger than O because O²⁻ has gained two extra electrons, increasing electron-electron repulsion and causing the outer electron cloud to expand. b) S²⁻ is larger than O²⁻ because S²⁻ has one extra electron shell compared to O²⁻, resulting in a larger atomic size. c) S²⁻ is larger than K⁺ due to the gain of electrons in S²⁻ increasing electron-electron repulsion, while the loss of an electron in K⁺ and a higher effective nuclear charge in K⁺ results in a smaller atomic size. d) K⁺ is larger than Ca²⁺ because Ca²⁺ has lost more electrons, decreasing electron-electron repulsion, and has a higher effective nuclear charge, attracting the remaining electrons more strongly and resulting in a smaller atomic size.

Step-by-step solution

a) O²⁻ is larger than O

This can be explained by the fact that \(\mathrm{O}^{2-}\) has gained two extra electrons, which increases electron-electron repulsion in the atom, causing the outer electron cloud to expand and hence the atomic size of \(\mathrm{O}^{2-}\) is larger than that of \(\mathrm{O}\).

b) S²⁻ is larger than O²⁻

Both ions \(\mathrm{S}^{2-}\) and \(\mathrm{O}^{2-}\) have the same number of electrons but the \(\mathrm{S}^{2-}\) ion has one extra electron shell as sulfur (atomic number 16) has 3 electron shells whereas oxygen (atomic number 8) has only 2 electron shells. Therefore, the atomic size of \(\mathrm{S}^{2-}\) is larger than that of \(\mathrm{O}^{2-}\).

c) S²⁻ is larger than K⁺

In the case of \(\mathrm{S}^{2-}\) and \(\mathrm{K}^{+}\) ions, \(\mathrm{S}^{2-}\) has gained two electrons, while \(\mathrm{K}^{+}\) has lost one electron. The gain of electrons in \(\mathrm{S}^{2-}\) leads to an increase in electron-electron repulsion, while the loss of one electron in \(\mathrm{K}^{+}\) reduces electron-electron repulsion. Moreover, \(\mathrm{K}^{+}\) has more protons compared to \(\mathrm{S}^{2-}\) resulting in a higher effective nuclear charge, which attracts the remaining electrons more strongly. Hence, the atomic size of \(\mathrm{S}^{2-}\) is larger than that of \(\mathrm{K}^{+}\).

d) K⁺ is larger than Ca²⁺

Both ions \(\mathrm{K}^{+}\) and \(\mathrm{Ca}^{2+}\) have lost electrons, but \(\mathrm{Ca}^{2+}\) has lost two while \(\mathrm{K}^{+}\) has only lost one electron. The loss of electrons leads to a decrease in electron-electron repulsion. Additionally, \(\mathrm{Ca}^{2+}\) ion has more protons compared to \(\mathrm{K}^{+}\) ions resulting in a higher effective nuclear charge, which leads to stronger attraction to the remaining electrons in the ion. Consequently, the atomic size of \(\mathrm{K}^{+}\) is larger than that of \(\mathrm{Ca}^{2+}\).

Key concepts

Electron-Electron Repulsion

In an atom, electrons are negatively charged particles that repel each other due to their like charges. This principle, known as electron-electron repulsion, plays a crucial role in determining the size of an atom. When additional electrons are added to an atom, as in the case of anion formation (e.g., \(\mathrm{O}^{2-}\)), the increased number of electrons enhances repulsion among them. This repulsion causes the electron cloud to expand outward, making the anion larger than the neutral atom.

Let's consider \(\mathrm{O}^{2-}\) and \(\mathrm{O}\). The oxygen anion has two extra electrons compared to the neutral oxygen atom, leading to greater repulsion and a subsequent increase in atomic size. Similar effects occur in other ions, and understanding how electron-electron repulsion influences atomic size is essential in predicting and explaining the relative sizes of atoms and ions in different states.

Ionic Radius

The ionic radius refers to the size of an atom's ion and can vary greatly depending on whether an electron is added (anions) or removed (cations). In general, anions are larger than their neutral counterparts because the addition of electrons increases repulsion, as discussed previously. In contrast, cations are typically smaller due to the loss of one or more electrons, which reduces repulsion and allows the remaining electrons to be drawn closer to the nucleus.

For example, comparing \(\mathrm{S}^{2-}\) and \(\mathrm{O}^{2-}\), we see that the sulfur anion is larger because sulfur has an additional electron shell compared to oxygen. Therefore, with more shells, the outermost electrons are further from the nucleus, resulting in a larger ionic radius. Additionally, comparing cations like \(\mathrm{K}^{+}\) and \(\mathrm{Ca}^{2+}\), despite both having lost electrons, the calcium cation is smaller because it has lost more electrons and has a higher charge, leading to a stronger attraction of the electrons to the nucleus and a smaller ionic radius.

Effective Nuclear Charge

The effective nuclear charge (ENC) is the net positive charge experienced by electrons in an atom and is critical in determining atomic and ionic sizes. It is influenced by the number of protons in the nucleus (which attracts electrons) and the inner electrons that shield the outer electrons from the full nuclear charge.

An increase in the number of protons (and thus a higher atomic number) generally translates to a larger ENC, causing electrons to be attracted more strongly to the nucleus. This attraction can cause a decrease in atomic size, as seen when comparing \(\mathrm{K}^{+}\) to \(\mathrm{Ca}^{2+}\); although both are cations, \(\mathrm{Ca}^{2+}\) experiences a higher ENC due to its greater number of protons, pulling its electrons closer to the nucleus and resulting in a smaller ionic radius. Understanding the balance between the number of protons, the degree of electron shielding, and the resulting effective nuclear charge is essential for explaining trends in atomic sizes across different elements and ions.