Arrange the ions \(\mathrm{Cl}^{-}, \mathrm{Br}^{-}, \mathrm{N}^{3-}\), and \(\mathrm{O}^{2-}\) in order of increasing polarizability, giving reasons for your decisions.

In the context of polarizability, the term 'electron cloud distortion' refers to how the negative charge surrounding an ion can be deformed under the influence of an external electric field.

Imagine the electron cloud like a sponge surrounding the ion's nucleus; when an electric field is present, it's as though this sponge is squeezed, causing it to change shape. The ease with which an ion's electron cloud can be distorted is its polarizability. Large ions with many electrons in their outer shells have 'squishier' sponges, making them more polarizable. In the given exercise, we see that ions with more electrons and larger sizes are more susceptible to this distortion, hence, bromide (Br-) with more electrons and a larger size, distorts more easily than others.

The polarizability of an ion is intimately linked to its size. Larger ions have more electron shells, which means their outer electrons are further from the nucleus and experience a weaker pull from the positively charged nucleus. This allows the outer electrons to be more easily influenced by external electric fields.

In the provided exercise, the ionic size helps us understand why bromide (Br-) is more polarizable than chloride (Cl-), which in turn is more polarizable than oxide (O2-) and nitride (N3-) ions. Larger ionic size translates to a higher ability of the electron cloud to distort, thus yielding more polarizable ions.

Charge density is a crucial factor affecting polarizability. It is the ratio of an ion's charge to its volume. Higher charge density means the electrical charge is more concentrated and the electrostatic forces holding the electron cloud in place are stronger.

For ions of similar sizes, those with higher charges will have a higher charge density and thus be less polarizable, because their electron cloud is held more tightly by the nucleus. This concept helps explain why, despite being small, ions like O2- and N3- are less polarizable: they have high charge densities due to their extra electrons creating a significant negative charge in a relatively small volume.

When analyzing polarizability, we can use the organization of the periodic table to predict trends. Elements are arranged into groups and periods, which represent columns and rows, respectively.

Looking vertically down a group, such as the halogens which include chlorine and bromine, atoms get larger and the additional electron shells make them more polarizable. Horizontally, across a period, atoms generally get smaller and more tightly bound, diminishing polarizability. This hierarchy becomes evident when comparing ions from different periods and groups, as in the example with the halide ions (Br- and Cl-) versus the smaller period 2 anions (O2- and N3-).

Comparing anions is central to understanding trends in polarizability. When anions vary in size, charge, and position on the periodic table, comparing them involves examining these factors together.

For instance, looking at the ions in our exercise, the N3- is the smallest and holds the most negative charge among them, which makes it the least polarizable. Conversely, Br- is the largest and carries a single negative charge, resulting in a lower charge density and higher polarizability.

This direct comparison of anions highlights that both the atomic number and the charge must be considered to determine an ion's polarizability, thus explaining the observed trend of N3- < O2- < Cl- < Br- in the given exercise.