Question

Select the larger member of each pair. (a) \(\mathrm{Ca}\) and \(\mathrm{Ca}^{2+}\) (b) \(\mathrm{Sr}\) and \(\mathrm{Sr}^{2+}\) (c) \(\mathrm{F}\) and \(\mathrm{F}^{-}\) (d) \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}^{2+}\)

Short answer

Answer: (a) Ca (b) Sr (c) F⁻ (d) Cu⁺

Step-by-step solution

Understand Atomic and Ionic Radius

In a neutral atom, the atomic radius is the distance between the nucleus and the outermost electron shell. When an electron is removed from an atom (forming a positive ion or cation) or added to an atom (forming a negative ion or anion), the atomic radius changes forming an ionic radius. Cations: When an electron is removed, the remaining electrons will be more attracted to the nucleus, decreasing the ionic radius compared to the atomic radius of the neutral atom. Anions: When an electron is added, repulsion among electrons increases, increasing the ionic radius compared to the atomic radius of the neutral atom.

Apply the concept to each pair

Compare each pair of ions based on the number of electrons they have. (a) \(\mathrm{Ca}\) and \(\mathrm{Ca}^{2+}\): Ca has 20 electrons, while Ca²⁺ loses 2 electrons and has only 18 electrons, so the cation is smaller. The larger member of this pair is: \(\boxed{\mathrm{Ca}}\) (b) \(\mathrm{Sr}\) and \(\mathrm{Sr}^{2+}\): Sr has 38 electrons, while Sr²⁺ loses 2 electrons and has only 36 electrons, so the cation is smaller. The larger member of this pair is: \(\boxed{\mathrm{Sr}}\) (c) \(\mathrm{F}\) and \(\mathrm{F}^{-}\): F has 9 electrons, while F⁻ gains one electron and now has 10 electrons, so the anion is larger. The larger member of this pair is: \(\boxed{\mathrm{F}^{-}}\) (d) \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}^{2+}\): Cu⁺ has 28 electrons, while Cu²⁺ has 27 electrons, so the ion with lesser charge is larger. The larger member of this pair is: \(\boxed{\mathrm{Cu}^{+}}\)

Key concepts

Cation and Anion Size Comparison

Understanding how ions form is crucial when comparing the sizes of cations and anions. A cation is a positively charged ion, created when one or more electrons are removed from a neutral atom. This removal leads to a decreased atomic radius, as the lesser number of electrons are pulled closer to the nucleus due to the unbalanced positive charge.

On the other hand, an anion, which is a negatively charged ion, is formed by the addition of one or more electrons to a neutral atom. This increase in electron count leads to greater electron-electron repulsion, causing the ion to expand and the atomic radius to increase. Consequently, anions are usually larger than the parent atoms from which they are derived, while cations are smaller.

Electron Removal and Atomic Size

When an electron is removed from an atom, the balance between the protons in the nucleus and the electrons in the orbitals shifts. This results in a greater effective nuclear charge, meaning that the nucleus holds the remaining electrons more strongly. As a consequence, the electrons are pulled closer to the nucleus, reducing the atomic size.

An important consideration in atomic size is the electron configuration, specifically in relation to the energy level of the electron that is being removed. If the removal of an electron results in the loss of an entire electron shell, as seen with the transition from Ca to Ca²⁺, the reduction in size is particularly significant due to the loss of a whole electron energy level.

Electron Addition and Ionic Size

The process of electron addition impacts ionic size significantly. As electrons are added to an atom to form an anion, the additional negative charges repel each other and spread out to reduce repulsion, a concept known as electron-electron repulsion. This spreading out causes the ion to enlarge, and the ionic radius increases.

A particular case of interest is the addition of an electron to a halogen atom, such as fluorine, which results in the formation of a halide ion. Taking the example from the exercise, fluoride ions F^- are larger than neutral fluorine atoms because the added electron increases repulsion among the now more numerous electrons, forcing them to occupy a larger space around the nucleus.