Question

Write electron configurations for a. the cations \(\mathrm{Mg}^{2+}, \mathrm{K}^{+},\) and \(\mathrm{Al}^{3+}\) . b. the anions \(\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-},\) and \(\mathrm{Te}^{2-}\)

Short answer

The electron configurations for the given cations and anions: Cations: 1. \(\mathrm{Mg}^{2+}\): \(1s^2 2s^2 2p^6\) 2. \(\mathrm{K}^{+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6\) 3. \(\mathrm{Al}^{3+}\): \(1s^2 2s^2 2p^6\) Anions: 1. \(\mathrm{N}^{3-}\): \(1s^2 2s^2 2p^6\) 2. \(\mathrm{O}^{2-}\): \(1s^2 2s^2 2p^6\) 3. \(\mathrm{F}^{-}\): \(1s^2 2s^2 2p^6\) 4. \(\mathrm{Te}^{2-}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

Step-by-step solution

Write Electron Configurations for Cations

To write the electron configurations for the cations \(\mathrm{Mg}^{2+}, \mathrm{K}^{+}, \mathrm{Al}^{3+}\), we start by finding the electron configurations for neutral Mg, K, and Al atoms, and then remove the necessary electrons to account for the charge. a. Mg has 12 protons and 12 electrons in a neutral atom: \[1s^2 2s^2 2p^6 3s^2\] Since \(\mathrm{Mg}^{2+}\) has lost two electrons, the electron configuration for \(\mathrm{Mg}^{2+}\) will be: \[1s^2 2s^2 2p^6\] b. K has 19 protons and 19 electrons in a neutral atom: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^1\] Since \(\mathrm{K}^{+}\) has lost one electron, the electron configuration for \(\mathrm{K}^{+}\) will be: \[1s^2 2s^2 2p^6 3s^2 3p^6\] c. Al has 13 protons and 13 electrons in a neutral atom: \[1s^2 2s^2 2p^6 3s^2 3p^1\] Since \(\mathrm{Al}^{3+}\) has lost three electrons, the electron configuration for \(\mathrm{Al}^{3+}\) will be: \[1s^2 2s^2 2p^6\]

Write Electron Configurations for Anions

To write the electron configurations for the anions \(\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Te}^{2-}\), we start by finding the electron configurations for neutral N, O, F, and Te atoms, and then add the necessary electrons to account for the charge. a. N has 7 protons and 7 electrons in a neutral atom: \[1s^2 2s^2 2p^3\] Since \(\mathrm{N}^{3-}\) has gained three electrons, the electron configuration for \(\mathrm{N}^{3-}\) will be: \[1s^2 2s^2 2p^6\] b. O has 8 protons and 8 electrons in a neutral atom: \[1s^2 2s^2 2p^4\] Since \(\mathrm{O}^{2-}\) has gained two electrons, the electron configuration for \(\mathrm{O}^{2-}\) will be: \[1s^2 2s^2 2p^6\] c. F has 9 protons and 9 electrons in a neutral atom: \[1s^2 2s^2 2p^5\] Since \(\mathrm{F}^{-}\) has gained one electron, the electron configuration for \(\mathrm{F}^{-}\) will be: \[1s^2 2s^2 2p^6\] d. Te has 52 protons and 52 electrons in a neutral atom: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4\] Since \(\mathrm{Te}^{2-}\) has gained two electrons, the electron configuration for \(\mathrm{Te}^{2-}\) will be: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\]

Key concepts

Cations and Their Electron Configurations

Cations are positively charged ions formed when an atom loses one or more electrons. Losing electrons is often a result of the atom achieving a more stable, usually a noble gas-like, electron configuration. Cations are often metals, which tend to lose electrons easily. For example, in the periodic table:

• Magnesium (\(\mathrm{Mg}\)) loses two electrons to form the cation \(\mathrm{Mg}^{2+}\).
• Potassium (\(\mathrm{K}\)) loses one electron to form \(\mathrm{K}^{+}\).
• Aluminum (\(\mathrm{Al}\)) loses three electrons to form \(\mathrm{Al}^{3+}\).

These transformations towards cations involve removing electrons, which decreases the size of the electron cloud around the nucleus, making cations smaller than their neutral atom counterparts.

Anions and Their Electron Configurations

Anions are negatively charged ions formed when an atom gains one or more electrons. Like cations, the goal is often achieving a more stable, often noble gas-like, electron configuration. However, anions are non-metals, which tend to gain electrons rather than lose them. For instance, in the periodic table:

• Nitrogen (\(\mathrm{N}\)) gains three electrons to form the anion \(\mathrm{N}^{3-}\).
• Oxygen (\(\mathrm{O}\)) gains two electrons to form \(\mathrm{O}^{2-}\).
• Fluorine (\(\mathrm{F}\)) gains one electron to form \(\mathrm{F}^{-}\).
• Tellurium (\(\mathrm{Te}\)) gains two electrons to form \(\mathrm{Te}^{2-}\).

When electrons are added, the size of the electron cloud around the nucleus increases, making anions larger than their neutral atoms.

The Process of Electron Removal for Cations

Electron removal is a process that occurs when an atom transitions into a cation by losing electrons. This process is crucial in shaping the element’s chemical characteristics. When atoms lose electrons:

• The remaining electrons are held more tightly to the nucleus, which usually increases the ionization energy.
• The atomic radius decreases, as the reduced electron cloud means electrons experience less repulsion from each other.

When pairing electron removal and electron configuration, it's important to remember that:

• Electrons are typically removed from the highest energy levels first.
• For \(\mathrm{Mg}^{2+}\), \(\mathrm{K}^{+}\), and \(\mathrm{Al}^{3+}\), electrons were removed from the outermost shells.

The Process of Electron Addition for Anions

Electron addition is an essential concept in understanding how atoms achieve stability by forming anions through gaining electrons. The addition of electrons typically happens in non-metals, allowing them to complete their valence shells. When electrons are added:

• Atoms often achieve a complete valence shell, resembling the electron configuration of noble gases, which is energetically favorable.
• The added electrons increase the electron-electron repulsions, generally increasing the size of the atom's electron cloud.

When considering electron configurations for anions, remember:

• Electrons are added to the next available energy level in the atom.
• For \(\mathrm{N}^{3-}\), \(\mathrm{O}^{2-}\), \(\mathrm{F}^{-}\), and \(\mathrm{Te}^{2-}\), electrons were added to fill their outermost orbits, achieving stable configurations.