Problem 18 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Account for the following trends in ionization energy. (Section \(3.7)\) (a) The increase from oxygen to fluorine. (b) The decrease from nitrogen to phosphorus. (c) The decrease from phosphorus to sulfur.

Step-by-Step Solution

Answer

Ionization energy increases from oxygen to fluorine due to increased nuclear charge. It decreases from nitrogen to phosphorus due to increased electron shielding. It also decreases from phosphorus to sulfur due to electron repulsion in the p-orbital.

Step 1: Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It generally increases across a period (from left to right) on the periodic table and decreases down a group (from top to bottom). These trends occur due to the effective nuclear charge and electron shielding.

Step (a): Analyzing the Increase from Oxygen to Fluorine

Oxygen and fluorine are both in the same period on the periodic table. As you move from oxygen to fluorine, the nuclear charge increases because you're adding more protons to the nucleus. This higher nuclear charge strongly attracts the electrons, resulting in a higher ionization energy for fluorine compared to oxygen.

Step (b): Analyzing the Decrease from Nitrogen to Phosphorus

Nitrogen and phosphorus are in the same group, with phosphorus below nitrogen. As you move down a group from nitrogen to phosphorus, the principal quantum number increases, which means electrons occupy shells further from the nucleus. The increased distance and electron shielding reduce the effective nuclear charge, thereby decreasing the ionization energy for phosphorus compared to nitrogen.

Step (c): Analyzing the Decrease from Phosphorus to Sulfur

Phosphorus and sulfur are in the same period. However, sulfur has an electron pairing in the p-orbital, which slightly increases the electron-electron repulsion within the orbital. This repulsion makes it easier to remove an electron from sulfur than phosphorus, resulting in a decrease in ionization energy from phosphorus to sulfur.

Key Concepts

Periodic Table Trends

In order to understand ionization energy, it's useful to begin by looking at trends on the periodic table. Ionization energy generally increases from left to right across a period. This is due to the increase in nuclear charge. More protons mean a stronger attraction between the nucleus and electrons. As a result, more energy is needed to remove an electron. Conversely, ionization energy tends to decrease as you move down a group. Moving down a group increases the distance between the nucleus and the electrons because additional electron shells are added. This larger distance weakens the nucleus’s hold on the electrons, requiring less energy for their removal.

Effective Nuclear Charge

The term "effective nuclear charge" refers to the net positive charge experienced by an electron in a multi-electron atom.

• While nuclear charge refers to the total positive charge from the protons, effective nuclear charge considers the fact that inner electrons shield outer electrons from the full effect of this charge.
• Effective nuclear charge can be approximated using the formula: \( Z_{eff} = Z - S \), where \( Z \) is the actual nuclear charge and \( S \) is the shielding constant, often related to the number of inner-shell electrons.

Across a period, the effective nuclear charge felt by valence electrons increases because protons are added while the shielding effect remains relatively constant. Thus, the electrons are pulled closer to the nucleus, increasing the ionization energy.

Electron Shielding

Electron shielding is the phenomenon where inner electrons in an atom repel outer ones, reducing the full effect of the nuclear charge on these outer electrons. This is especially significant when analyzing changes in ionization energy.

• When moving down a group, additional electron shells mean more inner electrons can effectively shield the outermost electrons from the nucleus.
• This increased shielding results in a decrease in the effective nuclear charge felt by the outer electrons, making it easier to remove them.

For instance, even though elements like nitrogen and phosphorus belong to the same group, increased electron shielding in phosphorus due to its larger atomic size leads to a decrease in ionization energy.

Electron Repulsion

Electron repulsion plays a critical role when analyzing ionization energies, particularly in elements where electrons begin to pair in orbitals.
When electrons pair up in an orbital, they repel each other because like charges repel. This repulsion slightly lowers the energy required to remove one of these paired electrons, leading to decreases in ionization energy.
For example, as sulfur begins to pair its electrons in the p-orbitals, the repulsion among these paired electrons results in a lower ionization energy relative to phosphorus, which has unpaired electrons in its p-orbitals.