Question

identify the electron configuration for aluminum (al)
1s² 2s² 2p⁶ 3s² 3p³
1s² 2s² 2p⁶ 3s² 4p¹
1s² 2s² 2p⁶ 3s² 3p¹

Explanation:

Step1: Determine the atomic number of aluminum

Aluminum (\(Al\)) has an atomic number of \(13\). This means it has \(13\) electrons.

Step2: Fill the electron shells

• The \(1s\) orbital can hold \(2\) electrons: \(1s^{2}\)
• The \(2s\) orbital can hold \(2\) electrons: \(2s^{2}\)
• The \(2p\) orbital can hold \(6\) electrons: \(2p^{6}\)
• The \(3s\) orbital can hold \(2\) electrons: \(3s^{2}\)
• After filling \(1s\), \(2s\), \(2p\), and \(3s\) (total \(2 + 2+6 + 2=12\) electrons), we have \(1\) electron left for the \(3p\) orbital: \(3p^{1}\)

Combining these, the electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)

Answer:

C. \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)