Question

Write electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

Short answer

The electron configurations for the most stable ions are: Al^3+: \(1s^2 2s^2 2p^6\) Ba^2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) Se^2-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) I^-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

Step-by-step solution

1. Identify the most stable ion for each element

: Al (Aluminum) belongs to Group 13, meaning it has 3 valence electrons. To achieve a full outer shell similar to noble gases, it will lose 3 electrons, forming Al^3+ ion. Ba (Barium) belongs to Group 2, meaning it has 2 valence electrons. To achieve a full outer shell similar to noble gases, it will lose 2 electrons, forming Ba^2+ ion. Se (Selenium) belongs to Group 16, meaning it has 6 valence electrons. To achieve a full outer shell similar to noble gases, it will gain 2 electrons, forming Se^2- ion. I (Iodine) belongs to Group 17, meaning it has 7 valence electrons. To achieve a full outer shell similar to noble gases, it will gain 1 electron, forming I^- ion.

2. Write the electron configuration for each most stable ion

: For Al^3+ ion, the atomic number of Al is 13; losing 3 electrons will result in 10 electrons. Electron configuration for Al^3+: \(1s^2 2s^2 2p^6\) For Ba^2+ ion, the atomic number of Ba is 56; losing 2 electrons will result in 54 electrons. Electron configuration for Ba^2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) For Se^2- ion, the atomic number of Se is 34; gaining 2 electrons will result in 36 electrons. Electron configuration for Se^2-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) For I^- ion, the atomic number of I is 53; gaining 1 electron will result in 54 electrons. Electron configuration for I^-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are paramount in dictating an element's chemical properties, especially its bonding behavior. These electrons reside in the outermost shell, or energy level, of an atom.

For example, aluminum (Al) has three valence electrons, which it can lose to form a positively charged ion (Al³⁺), resulting in a configuration that mimics the stable noble gas nearest to Al in the periodic table. This tendency to lose or gain electrons is a crux for understanding the formation of ionic compounds.

Understanding valence electrons is critical for predicting how atoms will interact:

Ionic Compounds

Ionic compounds form when atoms with distinct differences in electronegativity transfer electrons, creating ions that bond due to electrostatic forces. The electron donors become positively charged cations, while the acceptors become negatively charged anions.

By examining the exercise provided, we see examples of such transfer: Aluminum loses three electrons to form Al³⁺, while iodine gains one electron to become I^-. In a crystal lattice, these oppositely charged ions attract each other, resulting in a stable ionic compound.

Properties of ionic compounds include:

Noble Gas Electron Configuration

Noble gases are the elements in Group 18 of the periodic table, renowned for their chemical inertness. They possess a complete valence shell, which makes them particularly stable. Other elements strive to achieve a similar 'noble gas electron configuration', which lends to stability.

For instance, when barium (Ba) loses two electrons to form Ba²⁺, it attains an electron configuration similar to xenon (Xe), its nearest noble gas neighbor. This drive toward noble gas-like electron configurations underpins the formation of ions from elements, as seen through the textbook examples.