Question

Arrange each of the following sets of atoms and ions, in order of increasing size: (a) \(\mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se} ;\) (b) \(\mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\); (c) \(\mathrm{Ca}^{2 \mathrm{Ti}^{4+}}, \mathrm{Sc}^{3+}\); (d) \(\mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}\).

Short answer

The short answer is: (a) \(\mathrm{Te}^{2-} < \mathrm{Se}^{2-} < \mathrm{Se}\) (b) \(\mathrm{Fe}^{3+} < \mathrm{Fe}^{2+} < \mathrm{Co}^{3+}\) (c) \(\mathrm{Ti}^{4+} < \mathrm{Ca}^{2+} < \mathrm{Sc}^{3+}\) (d) \(\mathrm{Be}^{2+} < \mathrm{Na}^{+} < \mathrm{Ne}\)

Step-by-step solution

Determine the number of protons and electrons for each species

To determine the number of protons, look up the atomic number of each element. To find the number of electrons, take into account the charge of the ion. (a) \(\mathrm{Se}^{2-}: Z = 34, \text{ e- } = 36\) \(\mathrm{Te}^{2-}: Z = 52, \text{ e- } = 54\) \(\mathrm{Se}: Z = 34, \text{ e- } = 34\) (b) \(\mathrm{Co}^{3+}: Z = 27, \text{ e- } = 24\) \(\mathrm{Fe}^{2+}: Z = 26, \text{ e- } = 24\) \(\mathrm{Fe}^{3+}: Z = 26, \text{ e- } = 23\) (c) \(\mathrm{Ca}^{2+}: Z = 20, \text{ e- } = 18\) \(\mathrm{Ti}^{4+}: Z = 22, \text{ e- } = 18\) \(\mathrm{Sc}^{3+}: Z = 21, \text{ e- } = 18\) (d) \(\mathrm{Be}^{2+}: Z = 4, \text{ e- } = 2\) \(\mathrm{Na}^{+}: Z = 11, \text{ e- } = 10\) \(\mathrm{Ne}: Z = 10, \text{ e- } = 10\)

Arrange species in each set in order of increasing size

Based on periodic table trends and considering the number of protons and electrons in each species, we can now arrange them in order of increasing size. (a) For isoelectronic species with the same number of electrons, the size of the ions decreases as the nuclear charge (number of protons) increases. So the order is: \(\mathrm{Te}^{2-} < \mathrm{Se}^{2-} < \mathrm{Se}\) (b) The same rule applies here: \(\mathrm{Fe}^{3+} < \mathrm{Fe}^{2+} < \mathrm{Co}^{3+}\) (c) Here, \(\mathrm{Ca}^{2+}\) and \(\mathrm{Ti}^{4+}\) are isoelectronic and have the same number of electrons, so they would be compared based on the nuclear charge. \(\mathrm{Sc}^{3+}\) has more electrons than either \(\mathrm{Ca}^{2+}\) or \(\mathrm{Ti}^{4+}\), so it will be larger. Thus, the order is: \(\mathrm{Ti}^{4+} < \mathrm{Ca}^{2+} < \mathrm{Sc}^{3+}\) (d) \(\mathrm{Be}^{2+}\) has fewer electrons than both \(\mathrm{Na}^{+}\) and \(\mathrm{Ne}\), therefore it is the smallest. The next smallest species is \(\mathrm{Na}^{+}\) since it has fewer protons than \(\mathrm{Ne}\). The order is: \(\mathrm{Be}^{2+} < \mathrm{Na}^{+} < \mathrm{Ne}\)

Key concepts

Ionic Radius

The ionic radius is an important concept in chemistry, as it helps to explain how the size of an ion influences various properties. Surrounding electrons create an electron cloud around the nucleus of an atom. Ions are atoms that have lost or gained electrons, which affects the size of this cloud.
The ionic radius can change due to the gain or loss of electrons. When an atom gains electrons, forming an anion, the additional negative charge causes the electron cloud to expand, making the ion larger. Conversely, when an atom loses electrons to form a cation, there is a reduction in repulsion among the remaining electrons, leading to a smaller ionic radius.
For example, in \( \text{Se}^{2-} \), gaining two electrons increases the radius, whereas in \( \text{Co}^{3+} \), losing electrons reduces the radius. The ionic radius is not a fixed property and varies depending on the atomic structure and electron configuration, making it essential to consider the specific conditions of each ion when comparing their sizes.

Nuclear Charge

Nuclear charge plays a crucial role in determining the size and behavior of atoms and ions. The nuclear charge is essentially the total positive charge of the protons in an atom's nucleus.
As more protons are added to a nucleus, the atomic nucleus becomes more positively charged. This increase in positive charge pulls the negatively charged electrons closer to the nucleus, often resulting in a smaller ionic radius.
In isoelectronic series, species have the same number of electrons but different nuclear charges. For instance, \( \text{Ti}^{4+} \), \( \text{Ca}^{2+} \), and \( \text{Sc}^{3+} \) all share the same amount of electrons, yet \( \text{Ti}^{4+} \) has the highest nuclear charge and smallest ionic radius.

• Higher nuclear charge = smaller ion (due to greater attraction)
• Lower nuclear charge = larger ion (due to lesser attraction)

Understanding nuclear charge also aids in predicting the reactivity and bonding characteristics of elements within the periodic table.

Isoelectronic Species

Isoelectronic species are atoms and ions that have the same number of electrons. These species are particularly intriguing because, while they have a shared electron count, their sizes vary due to differences in nuclear charge.
For instance, consider \( \text{Ca}^{2+} \), \( \text{Ti}^{4+} \), and \( \text{Sc}^{3+} \). Despite all having 18 electrons, they differ in size due to their individual nuclear charges. A higher nuclear charge results in a tighter pull on the electrons, decreasing the size of the ion.

• Isoelectronic species with more protons exhibit a smaller ionic radius.
• Species with fewer protons are larger, even though they share the same electron count.

Being able to identify and understand isoelectronic species aids students in predicting and explaining differences in physical and chemical properties across different elements and compounds.

Periodic Trends

Periodic trends are patterns observed in the periodic table that relate to atomic structure, including atomic size, ionic radius, and reactivity. These trends help predict the behavior of elements.
Across a period, the atomic size generally decreases from left to right. This occurs because the nuclear charge increases, pulling the electrons inwards and reducing the atomic radius. Down a group, the atomic size increases due to the addition of electron shells, which outweighs the increase in nuclear charge.
For ionic radius:

• Across a period, cations become smaller, and anions become larger.
• Down a group, the ionic radius increases due to additional electron shells.

Understanding periodic trends gives insights into elemental properties, helping to predict and justify how ions will arrange by size as demonstrated in exercises with species like \( \text{Be}^{2+} \), \( \text{Na}^{+} \), and \( \text{Ne} \). Recognizing these trends is essential for mastering topics in chemistry.