Question

For each pair, identify the larger atom or ion. (a) \(\mathrm{F}\) or \(\mathrm{F}^{-}\) (b) \(\mathrm{Sr}\) or \(\mathrm{Sr}^{2+}\)

Short answer

The larger particles are \(\mathrm{F}^{-}\) in comparison to \(\mathrm{F}\) and \(\mathrm{Sr}\) in comparison to \(\mathrm{Sr}^{2+}\)

Step-by-step solution

Identify size for each pair

(a) For \(\mathrm{F}\) or \(\mathrm{F}^{-}\), the ion \(\mathrm{F}^{-}\) is larger than its parent atom \(\mathrm{F}\). This is because \(\mathrm{F}^{-}\) has one more electron than \(\mathrm{F}\), causing a greater electron-electron repulsion, which in turn increases the size of the electron cloud. The atomic radius is generally considered to be the total distance from the atomic nucleus to the outermost electron orbital.(b) For \(\mathrm{Sr}\) or \(\mathrm{Sr}^{2+}\), \(\mathrm{Sr}^{2+}\) is the cation which is smaller than its parent atom \(\mathrm{Sr}\), because when two electrons are lost, there's less repulsion among the remaining electrons, allowing them to move closer to the nucleus. The atom has more electron-electron repulsion and therefore a larger radius.

Key concepts

Understanding Atomic Radius

The term 'atomic radius' refers to the size of an atom, which is typically measured from the nucleus to the outermost layer of electrons (electron cloud). Imagine each atom as a tiny, mostly empty sphere, with a tiny, dense nucleus at its center. Surrounding the nucleus, electrons move about in regions called orbitals, defining the space the atom occupies. In simple terms, the atomic radius is like the radius of a circle, but for a three-dimensional sphere representing the atom.

Imagine you could inflate a balloon to increase its size. Similarly, the atomic radius expands due to forces within the atom. One key force is the balance between the positively charged nucleus attracting negatively charged electrons and the electrons naturally repelling each other because they have the same charge. The atomic radius can differ depending on numerous factors such as the number of electron shells, the energy levels the electrons occupy, and even the presence of nearby atoms.

Ion Size Comparison

When talking about ions, the comparison of size involves understanding how adding or removing electrons from an atom will affect its radius. Simply put, negative ions, also known as anions, are usually larger than their parent atoms. This is the case because anions have extra electrons, leading to increased electron-electron repulsion and a subsequent expansion of the electron cloud.

Conversely, positive ions, or cations, end up smaller than the atoms they derive from. Upon losing electrons, there is a reduction in repulsion across the electron cloud so it contracts, bringing electrons closer to the nucleus. This effect can be readily seen in transition metals where the loss of electrons can lead to a significant decrease in size due to the collapse of electron shells.

Electron-Electron Repulsion

In the atomic world, electron-electron repulsion is a critical factor in determining atomic and ionic size. Since electrons all share a negative charge, they repel each other. Think of it like trying to hold two negatively charged magnets close to one another—they push away. Within an atom, these repulsive forces are balanced by the attraction to the positively charged nucleus.

The more electrons there are, the more they push apart from one another, generally increasing the size of the atom. In ions, the balance shifts. In anions with extra electrons, the electron-electron repulsion is greater, leading to a larger radius. In cations, there are fewer electrons, less repulsion, and thus a smaller radius. This principle is beautifully demonstrated in the differences between neutral atoms and their ionized counterparts, as shown in the textbook exercise.