Question

Arrange the following in order of increasing radius and increasing ionization energy. a. \(\mathrm{N}^{+}, \mathrm{N}, \mathrm{N}^{-}\) b. Se, \(\mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\) c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Short answer

The short answer based on the step-by-step solution is: a. Increasing radius: \(N^+ < N < N^-\) Increasing ionization energy: \(N^- < N < N^+\) b. Increasing radius: \(Cl^+ < Cl < Se < Se^-\) Increasing ionization energy: \(Se^- < Se < Cl < Cl^+\) c. Increasing radius: \(Sr^{2+} < Rb^+ < Br^-\) Increasing ionization energy: \(Br^- < Rb^+ < Sr^{2+}\)

Step-by-step solution

a. Atomic Radius Order

For N⁺, N, and N⁻ ions: When we go from N⁺ to N, we are adding an electron which increases the repulsion between electrons, causing the atom to expand, and the radius to increase. When going from N to N⁻, another electron is added, causing even more repulsion and an even bigger expansion. So, the order of increasing radius is: \[N^+ < N < N^-\]

a. Ionization Energy Order

Ionization energy is the energy required to remove an electron from an atom or ion. It's generally easier to remove an electron from a negatively charged species than a neutral one, and even more difficult to remove one from a positively charged species. Therefore, the order of increasing ionization energy is: \[N^- < N < N^+\]

b. Atomic Radius Order

For Se, Se⁻, Cl, and Cl⁺: As we go from Se to Se⁻, an electron is added, increasing the atomic radius. Comparing Cl and Cl⁺, removing an electron will reduce electron-electron repulsion and decrease the atomic radius. Comparing Se⁻ with Cl, selenium lies below chlorine in the periodic table, meaning it has a larger atomic radius since the number of energy levels increases. So, the order of increasing radius is: \[Cl^+ < Cl < Se < Se^-\]

b. Ionization Energy Order

As discussed earlier, positive ions generally have higher ionization energies than neutral atoms, and neutral atoms have higher ionization energies than negative ions. Also, ionization energy generally increases across a period from left to right. Therefore, the order of increasing ionization energy is: \[Se^- < Se < Cl < Cl^+\]

c. Atomic Radius Order

For Br⁻, Rb⁺, and Sr²⁺: Comparing Br⁻ and Rb⁺, bromine has gained an electron and rubidium has lost one, making Br⁻ larger than Rb⁺. Between Rb⁺ and Sr²⁺, strontium has lost two electrons, further reducing its electron-electron repulsion and making it even smaller. So, the order of increasing radius is: \[Sr^{2+} < Rb^+ < Br^-\]

c. Ionization Energy Order

Again, we know negative ions have lower ionization energies than neutral atoms or positive ions. We can also consider that the second ionization energy of strontium is much higher than the first ionization energy of rubidium, since it is harder to remove an additional electron. Thus, the order of increasing ionization energy is: \[Br^- < Rb^+ < Sr^{2+}\]

Key concepts

Periodic Trends

Understanding the periodic trends is crucial when arranging elements or ions by atomic radius and ionization energy. In general, atomic radius decreases across a period from left to right and increases down a group. This is due to the increasing number of protons in the nucleus, which pulls the electrons closer, reducing the size of the atom across a period. Conversely, as one moves down a group, additional energy levels are added, making the atom larger.

Ionization energy, the energy required to remove an electron from an atom in its gaseous state, shows an opposite trend. It generally increases across a period due to the stronger attraction between the nucleus and the electrons as the atomic number increases. Contrarily, ionization energy decreases down a group as the outermost electrons are farther from the nucleus and shielded by inner electrons, making them easier to remove.

Atomic Structure

Atomic structure plays a fundamental role in determining the atomic radius and ionization energy of elements. An atom is composed of a nucleus, containing protons and neutrons, surrounded by electrons in regions called electron shells or orbitals. The number of protons in the nucleus determines the atomic number and therefore, the identity of the element.

Size of Ions

An element can exist as an ion by gaining or losing electrons, which significantly affects its size. For example, when an atom loses an electron to form a cation, the loss reduces electron-electron repulsion, and the remaining electrons are pulled closer to the nucleus, resulting in a smaller radius. In contrast, an anion is formed by the addition of electrons, increasing repulsion within the electron cloud, and causing an increase in the atomic radius.

Electron Configuration

Electron configuration describes the distribution of electrons in an atom’s orbitals. It provides invaluable insights into chemical reactivity, bonding, and periodic trends, including atomic radius and ionization energy. Electrons occupy subshells in an order described by the Aufbau principle, starting from the lowest energy level.

For instance, the ionization energy can be influenced by the stability of the electron configuration. Atoms with a full or half-full valence shell require more energy to remove an electron due to their relative stability. Conversely, ions formed by removing an electron from a stable configuration will have a high ionization energy because the resulting configuration will be more stable.