To determine the number of valence electrons for an atom with the given electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^2\), let's follow these steps:
1. Identify the electron configuration:
- The given electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^2\).
2. Determine the outermost shell or energy level:
- The outermost shell is the one with the highest principal quantum number \(n\).
- In this configuration, the highest \(n\) is 3 (associated with 3s and 3p orbitals).
3. Count the electrons in the outermost shell:
- The 3s subshell has 2 electrons (\(3s^2\)).
- The 3p subshell has 2 electrons (\(3p^2\)).
4. Sum the electrons in the outermost shell to find the valence electrons:
- In the 3rd shell: electrons in 3s subshell + electrons in 3p subshell.
- \(3s^2\) contributes 2 electrons.
- \(3p^2\) contributes 2 electrons.
5. Add these numbers together:
- \(2 \text{ (from } 3s^2\text{) } + 2 \text{ (from } 3p^2\text{) } = 4\) valence electrons.
Therefore, the atom has [tex]\(\boxed{4}\)[/tex] valence electrons.
