Question

which is the electron configuration for an oxygen atom? 1s²2s²2p⁴ 1s²2s²2p⁶3s²3p⁶4s² 1s²2s²2p⁶ 1s²2s²2p⁶3s²3p⁶

Explanation:

Step1: Determine the atomic number of oxygen

Oxygen has an atomic number of \(8\).

Step2: Write the electron configuration

Electrons fill orbitals in the order \(1s\), \(2s\), \(2p\), etc. The first shell (\(n = 1\)) has one sub - shell (\(1s\)) which can hold \(2\) electrons. The second shell (\(n=2\)) has two sub - shells (\(2s\) and \(2p\)). The \(2s\) sub - shell can hold \(2\) electrons and the \(2p\) sub - shell can hold \(6\) electrons.
For \(8\) electrons:

• \(1s\) gets \(2\) electrons: \(1s^{2}\)
• \(2s\) gets \(2\) electrons: \(2s^{2}\)
• Remaining \(8-(2 + 2)=4\) electrons go to \(2p\): \(2p^{4}\)

Answer:

\(1s^{2}2s^{2}2p^{4}\)