Question

Write electron configurations for the following ions. a. \(\mathrm{Ni}^{2+}\) b. \(\mathrm{Cd}^{2+}\) c. \(\mathrm{Zr}^{3+}\) and \(\mathrm{Zr}^{4+}\) d. \(\mathrm{Os}^{2+}\) and \(\mathrm{Os}^{3+}\)

Short answer

The electron configurations for the given ions are as follows: a. \(\mathrm{Ni}^{2+}\): \([\mathrm{Ar}] \, 3\mathrm{d}^8\) b. \(\mathrm{Cd}^{2+}\): \([\mathrm{Kr}] \, 4\mathrm{d}^{10}\) c. \(\mathrm{Zr}^{3+}\): \([\mathrm{Kr}]\); \(\mathrm{Zr}^{4+}\): \([\mathrm{Kr}]\) d. \(\mathrm{Os}^{2+}\): \([\mathrm{Xe}] \, 4\mathrm{f}^{14} \, 5\mathrm{d}^6\); \(\mathrm{Os}^{3+}\): \([\mathrm{Xe}] \, 4\mathrm{f}^{14} \, 5\mathrm{d}^5\)

Step-by-step solution

Determine atomic numbers

We can find the atomic numbers of each element by referring to the periodic table. The atomic number represents the number of protons (and electrons in a neutral atom) in the element: a. \(\mathrm{Ni}\): Atomic number is 28 b. \(\mathrm{Cd}\): Atomic number is 48 c. \(\mathrm{Zr}\): Atomic number is 40 d. \(\mathrm{Os}\): Atomic number is 76

Write ground-state electron configurations for neutral elements

Next, use the atomic number to determine the ground-state electron configuration for the neutral atoms of each element. The electron configuration will be in the form of orbital symbols and superscripts representing the number of electrons in each orbital: a. \(\mathrm{Ni}\): \([\mathrm{Ar}] \, 4\mathrm{s}^2 \, 3\mathrm{d}^8 \) b. \(\mathrm{Cd}\): \([\mathrm{Kr}] \, 5\mathrm{s}^2 \, 4\mathrm{d}^{10} \) c. \(\mathrm{Zr}\): \([\mathrm{Kr}] \, 5\mathrm{s}^2 \, 4\mathrm{d}^2 \, 4\mathrm{f}^0 \,\) (\(4\mathrm{f}\) is included to show that \(\mathrm{Zr}\) is in the \(\mathrm{d}\)-block) d. \(\mathrm{Os}\): \([\mathrm{Xe}] \, 6\mathrm{s}^2 \, 4\mathrm{f}^{14} \, 5\mathrm{d}^6\)

Adjust electron configurations for each ion

Now, we need to adjust each electron configuration for the ions provided by removing the appropriate number of electrons from the valence shell. Valence electrons are the outermost electrons that are removed first when forming cations: a. \(\mathrm{Ni}^{2+}\): \([\mathrm{Ar}] \, 4\mathrm{s}^0 \,3\mathrm{d}^8 \rightarrow [\mathrm{Ar}] \, 3\mathrm{d}^8\) b. \(\mathrm{Cd}^{2+}\): \([\mathrm{Kr}] \, 5\mathrm{s}^0 \,4\mathrm{d}^{10} \rightarrow [\mathrm{Kr}] \, 4\mathrm{d}^{10}\) c. \(\mathrm{Zr}^{3+}\): \([\mathrm{Kr}] \, 5\mathrm{s}^0 \,4\mathrm{d}^0 \, 4\mathrm{f}^0 \rightarrow [\mathrm{Kr}]\) \(\mathrm{Zr}^{4+}\): \([\mathrm{Kr}] \,5\mathrm{s}^0 \, 4\mathrm{d}^0 \, 4\mathrm{f}^0 \rightarrow [\mathrm{Kr}]\) d. \(\mathrm{Os}^{2+}\): \([\mathrm{Xe}] \, 6\mathrm{s}^0 \,4\mathrm{f}^{14} \, 5\mathrm{d}^6 \rightarrow [\mathrm{Xe}] \, 4\mathrm{f}^{14} \, 5\mathrm{d}^6\) \(\mathrm{Os}^{3+}\): \([\mathrm{Xe}] \, 6\mathrm{s}^0 \,4\mathrm{f}^{14} \, 5\mathrm{d}^5 \rightarrow [\mathrm{Xe}] \, 4\mathrm{f}^{14} \, 5\mathrm{d}^5\)

Key concepts

Ions

Ions are atoms or molecules that have gained or lost one or more electrons, resulting in a net electrical charge. When an atom loses electrons, it forms a positively charged ion, also known as a cation. When an atom gains electrons, it becomes negatively charged and is called an anion. For example, in the exercise above, we deal with positive ions (cations) such as

• Ni²⁺
• Cd²⁺
• Zr³⁺ and Zr⁴⁺
• Os²⁺ and Os³⁺

Removing electrons means the atom is losing negatively charged particles, making the overall charge positive.
When writing electron configurations for ions, we adjust the configurations by removing electrons from the neutral atom's outermost orbitals, until the charge of the ion is satisfied. Understanding ions is fundamental in chemistry as they are integral to the formation of compounds and chemical reactions.

Atomic Number

The atomic number is a key concept in chemistry that indicates the number of protons found in the nucleus of an atom. This number also tells us the number of electrons in a neutral atom because atoms are electrically neutral when the number of protons equals the number of electrons.
Using the periodic table, we can easily find the atomic number for any element to determine its electron configuration. For example, in this exercise:

• Nickel (Ni) has an atomic number of 28.
• Cadmium (Cd) has an atomic number of 48.
• Zirconium (Zr) has an atomic number of 40.
• Osmium (Os) has an atomic number of 76.

By knowing an element's atomic number, chemists can determine not only the electron configuration but also understand its chemical properties and behaviors. Understanding the atomic number helps predict how an atom will interact with other atoms during chemical reactions.

Electron Orbitals

Electron orbitals are regions within an atom where there is a high probability of finding electrons. Electrons surround the nucleus in atomic orbitals arranged in a specific order of energy levels: s, p, d, and f orbitals.
Each type of orbital holds a different number of electrons:

• s orbital: Can hold up to 2 electrons.
• p orbitals: Can hold up to 6 electrons.
• d orbitals: Can hold up to 10 electrons.
• f orbitals: Can hold up to 14 electrons.

In the given examples, electron configurations were used to represent the number of electrons in each orbital of the atom. Accurate knowledge of how electrons are distributed among orbitals is crucial for understanding an element's reactivity, its placement in the periodic table, and the types of bonds it can form.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in determining how an element will react with others. They are typically found in the s and p orbitals of an atom's highest energy level, or shell.
These electrons play a vital role in chemical bonding because they are the ones involved in forming bonds. When atoms form ions, such as cations, they lose valence electrons. This makes understanding changes in electron configuration crucial. For example, the removal of these electrons in the exercise led to the formation of ions:

• In Ni²⁺, removing 2 electrons from the 4s orbital results in [Ar] 3d⁸.
• In Cd²⁺, removing 2 electrons from the 5s orbital results in [Kr] 4d¹⁰.

Mastering the concept of valence electrons aids in predicting how different elements will interact or bond, which is fundamental in both chemistry and material science.