An electron can be added to a halogen atom to form a halide ion with?

Analysis

The question asks about the electronic configuration of a halide ion formed when an electron is added to a neutral halogen atom.

Concept Explanation

• Halogens: These are elements in Group 7 (or Group 17) of the Periodic Table. They include Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).
• Valence Electrons: A neutral halogen atom has 7 valence electrons in its outermost shell. This makes them highly reactive because they are one electron short of a stable "octet" (a full outer shell of 8 electrons).
• Ion Formation: To achieve stability, a halogen atom gains one electron from another atom to complete its outer shell. When it gains this electron, it becomes a negatively charged ion called a halide ion.
• The Octet Rule: Atoms generally react to achieve a stable electron configuration, which for most elements is having 8 electrons in their outermost shell (like the noble gases).

Step-by-Step Solution

1. Identify the starting state: A neutral halogen atom has 7 valence electrons ($ns^2 np^5$).
2. Identify the process: The atom gains 1 electron to become a halide ion ($X^-$).
3. Calculate the final state: $7 \text{ (original valence electrons)} + 1 \text{ (gained electron)} = 8 \text{ valence electrons}$.
4. Conclusion: The resulting halide ion now has a stable octet configuration, meaning it has 8 electrons in its outermost shell.

Identifying the Answer

Comparing our result to the given options:

• A. 8 valence electrons (Matches our calculation)
• B. 7 valence electrons (This is the neutral atom, not the ion)
• C. 2 valence electrons (Incorrect)
• D. 3 valence electrons (Incorrect)

Exam Tip

Remember the "Octet Rule": Almost all main-group elements (except Hydrogen, Helium, Lithium, and Beryllium) "want" to have 8 electrons in their outer shell to be stable. If you see a question about an ion forming to become stable, the answer is almost always 8!