Problem 83 - General Chemistry: Principles and Modern Applications

Question

Write electron configurations to show the first two ionizations for Cs. Explain why the second ionization energy is much greater than the first.

Step-by-Step Solution

Answer

The electron configurations for Cs, Cs+ and Cs++ are [Kr] 5s^1, [Kr], and [Ar] 4d^10 4p^6 respectively. The second ionization energy is higher than the first due to the second electron being removed from a lower energy level (4p), which is closer to and more strongly attracted to the atomic nucleus.

Step 1: Electron Configuration of Cs

Cesium (Cs) has 55 electrons. It's electron configuration can be written as \([Kr] 5s^1\), where [Kr] represents core electrons identical to the noble gas krypton.

Step 2: First Ionization of Cs

The first ionization of Cs involves removing one electron from the outermost shell. The electron configuration becomes \([Kr]\). The outermost electron is in the 5s orbital, which is relatively easy to remove, thus the lower ionization energy.

Step 3: Second Ionization of Cs

The second ionization of Cs involves removing another electron. However, this time, the electron is being removed from a complete shell, closer to the nucleus (the 4p orbital), represented as [Ar] 4d^10 4p^6. Removing an electron from this shell requires much more energy because these electrons are much closer to the nucleus and thus more strongly attracted to the positive charge of the protons.

Step 4: Reason for Higher Second Ionization Energy

The second ionization energy is much greater than the first because the second electron being removed is in a lower energy level, closer to the nucleus. The electrons in lower energy levels are held more tightly by the protons in the nucleus, thus requiring more energy to remove.

Key Concepts

Electron Configuration

Electron configuration is the arrangement of electrons in the orbitals of an atom. It provides insight into the structure of an atom and its chemical properties. For cesium (Cs), which has 55 electrons, we describe its electron configuration to understand how these electrons are distributed. The electron configuration for Cs is written as \([Kr] 5s^1\). Here, [Kr] represents the electron configuration of the noble gas krypton, which accounts for the first 54 electrons, also known as core electrons. The remaining electron is in the 5s orbital, making Cs part of the alkali metals group.

• Understanding electron configurations helps predict how atoms will interact in chemical reactions.
• The outermost electrons, also known as valence electrons, play a key role in determining an element's chemical behavior.

This foundational concept in chemistry is essential when discussing ionization energies, as it explains why electrons in various orbitals require different amounts of energy to be removed.

First Ionization Energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The **first ionization energy** specifically refers to the energy needed to remove the outermost electron from a neutral atom. In the case of cesium, removing the single electron in the 5s orbital is relatively easy due to its distance from the nucleus. Because this outermost electron in cesium is in a higher energy level (5s), it experiences less nuclear attraction. This makes it easier to remove, resulting in a lower first ionization energy.

• First ionization energy is influenced by the distance of the valence electron from the nucleus. The greater the distance, the lower the energy required.
• This property typically decreases as you move down a group in the periodic table, as electrons are further from the nucleus.

Thus, the first ionization energy provides crucial information on how elements might form ions and engage in chemical bonds.

Second Ionization Energy

The **second ionization energy** refers to the energy required to remove a second electron, after the first has been removed. For cesium, this becomes significantly more challenging. After removing the first electron, the next electron must be taken from a stable and complete shell closer to the nucleus, specifically the 4p orbital.
The second ionization energy is much higher than the first because these remaining electrons are in lower energy levels, closer to the nucleus, and are thus more strongly attracted to the nuclear charge.

• Electrons closer to the nucleus are more tightly bound, requiring more energy to dislodge.
• The increase in second ionization energy illustrates the increased stability of a complete inner shell.

This concept underscores why ionization energies tend to increase dramatically from the first to second, as removing electrons from inner shells involves overcoming greater electrostatic forces from a now positively charged ion.