Problem 20 - Chemistry The Central Science

Question

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\) , \((\mathbf{b}) \mathrm{P}^{3-},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ru}^{3+},(\mathbf{e}) \mathrm{As}^{3-},(\mathbf{f}) \mathrm{Ag}^{+}\)

Step-by-Step Solution

Answer

The electron configurations for the given ions are as follows: \(Cd^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8\) \(P^{3-}: 1s^2 2s^2 2p^6 3s^2 3p^6\) \(Zr^{4+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) \(Ru^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\) \(As^{3-}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) \(Ag^{+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^{10}\) Only the \(P^{3-}\) and \(As^{3-}\) ions have noble-gas configurations (Ne and Ar, respectively).

Step 1: Identify the atomic numbers of the elements

Find the atomic numbers of the elements based on their symbols in the periodic table: Cd: 48 P: 15 Zr: 40 Ru: 44 As: 33 Ag: 47

Step 2: Write the ground-state electron configurations of the neutral atoms

Write the electron configurations for the neutral atoms using the periodic table and the standard method for writing electron configurations: Cd: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) P: \(1s^2 2s^2 2p^6 3s^2 3p^3\) Zr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2\) Ru: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7\) As: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3\) Ag: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^{10}\)

Step 3: Adjust the electron configurations for given ion charges

Cd²⁺: Remove two electrons from the highest-energy orbital, which is the 4d orbital: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8\) P³⁻: Add three electrons to the highest-energy orbital, which is the 3p orbital: \(1s^2 2s^2 2p^6 3s^2 3p^6\) Zr⁴⁺: Remove four electrons from the highest-energy orbitals, starting with the 4d and then the 5s orbitals: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^0\) Ru³⁺: Remove three electrons from the highest-energy orbitals, starting with the 4d and then the 5s orbitals: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^4\) As³⁻: Add three electrons to the highest-energy orbital, which is the 4p orbital: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) Ag⁺: Remove one electron from the highest-energy orbital, which is the 5s orbital: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^{10}\)

Step 4: Compare the electron configurations with noble gas configurations

The electron configurations of noble gases are complete with full energy levels, and their general form is (\(n-1\)d^{10})np^6, where \(n\) is the principal quantum number of the noble gases. In our adjusted electron configurations from Step 3, the following ions match the noble gas configurations: - P³⁻ matches the configuration of Ne (Z=10): \(1s^2 2s^2 2p^6\) - As³⁻ matches the configuration of Ar (Z=18): \(1s^2 2s^2 2p^6 3s^2 3p^6\) Conclusion: Only P³⁻ and As³⁻ ions have noble-gas configurations.

Key Concepts

Noble Gas Configurations

Noble gas configurations make atoms or ions more stable by having completely filled electron shells. These configurations are noted for their complete subshells, which result in minimal chemical reactivity. Noble gases, like helium, neon, and argon, have fully filled electron configurations such as:

• Neon: \[1s^2 2s^2 2p^6\]
• Argon: \[1s^2 2s^2 2p^6 3s^2 3p^6\]

Atoms often gain or lose electrons to achieve a noble gas configuration, which results in ion formation. This process brings them to a lower-energy, more stable state. For instance, in our exercise, the ions \(\mathrm{P}^{3-}\) and \(\mathrm{As}^{3-}\) have electron configurations that match those of neon and argon respectively, reaching stable noble gas states.

Ion Charges

Ion charges arise when atoms gain or lose electrons, altering their electron count relative to protons. This imbalance results in an overall positive or negative charge on the ion. Ions with a positive charge, known as cations, form from losing electrons, whereas negatively charged ions, or anions, result from gaining electrons.
To determine the charge, consider the number of electrons transferred compared to the neutral atom:

• \(\mathrm{Cd}^{2+}\): Two electrons are lost, resulting in a +2 charge.
• \(\mathrm{P}^{3-}\): Three electrons are gained, leading to a -3 charge.
• \(\mathrm{Zr}^{4+}\): Four electrons are removed, giving a +4 charge.
• \(\mathrm{Ru}^{3+}\): Three electrons are lost, resulting in a +3 charge.
• \(\mathrm{As}^{3-}\): Three electrons added, leading to a -3 charge.
• \(\mathrm{Ag}^{+}\): One electron is removed, leading to a +1 charge.

Identifying these charges helps predict ion interactions and formation of chemical compounds.

Electron Configuration of Ions

Understanding electron configurations of ions provides insight into the structure and reactivity of atoms. By identifying which electrons to remove or add based on the ion charge, we can pinpoint the resulting configuration.
For cations, electrons are removed from the highest energy orbitals first:

• For \(\mathrm{Cd}^{2+}\), the \(5s\) electrons are removed, resulting in: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^8\]
• For \(\mathrm{Ag}^{+}\), the \(5s\) electron is removed, leading to: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^{10}\]

For anions, electrons are added to the lowest energy orbitals available:

• For \(\mathrm{P}^{3-}\), three electrons are added to the \(3p\), resulting in: \[1s^2 2s^2 2p^6 3s^2 3p^6\]
• For \(\mathrm{As}^{3-}\), three electrons complete the \(4p\), leading to: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\]

Mastering this concept allows one to predict how ions behave in various chemical reactions and their roles in forming ionic compounds.