Question

write the condensed (noble - gas) electron configuration of aluminum. for multi - digit superscripts or coefficients, use each number in succession.

Explanation:

Step1: Determine the atomic number of aluminum

Aluminum has an atomic number of 13.

Step2: Identify the noble - gas core

The noble - gas core for aluminum is neon ([Ne]) which has 10 electrons.

Step3: Determine the remaining electrons

Since aluminum has 13 electrons and neon has 10, there are \(13 - 10=3\) remaining electrons.

Step4: Write the electron configuration for the remaining electrons

The remaining 3 electrons are in the \(3s\) and \(3p\) orbitals. The \(3s\) orbital can hold 2 electrons and the \(3p\) orbital can hold 6 electrons. For 3 remaining electrons, the configuration is \(3s^{2}3p^{1}\)

Answer:

\([Ne]3s^{2}3p^{1}\)