Question

the lewis diagram below represents an aluminum ion. alⁿ what is the value of n in the diagram? choose 1 answer: a 3− b 1− c 1+ d 3+

Explanation:

Step1: Recall aluminum's electron configuration

Aluminum (Al) has an atomic number of 13, so its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^1\). To achieve a stable octet (like a noble gas, e.g., Ne with \(1s^2 2s^2 2p^6\)), aluminum tends to lose its 3 valence electrons (2 from \(3s\) and 1 from \(3p\)).

Step2: Determine the charge of the aluminum ion

When an atom loses electrons, it forms a positive ion (cation). Since aluminum loses 3 electrons, the charge of the aluminum ion (\(Al^{n}\)) is \(+3\) (because the number of protons (13) now exceeds the number of electrons (13 - 3 = 10) by 3). So \(n = 3+\).

Answer:

D. \(3+\)