Question

write the electron configuration for a neutral atom of silicon.

Explanation:

Step1: Determine the atomic number of silicon

The atomic number of silicon (\(Si\)) is \(14\). This means a neutral silicon atom has \(14\) electrons.

Step2: Fill the electron shells and sub - shells

• The first shell (\(n = 1\)) has an \(s\) sub - shell. The \(1s\) sub - shell can hold a maximum of \(2\) electrons. So, we have \(1s^{2}\).
• The second shell (\(n=2\)) has \(s\) and \(p\) sub - shells. The \(2s\) sub - shell can hold \(2\) electrons (\(2s^{2}\)), and the \(2p\) sub - shell can hold \(6\) electrons (\(2p^{6}\)).
• The third shell (\(n = 3\)): After filling the first two shells (\(1s^{2}2s^{2}2p^{6}\), which accounts for \(2 + 2+6=10\) electrons), we have \(14 - 10=4\) electrons left. The \(3s\) sub - shell holds \(2\) electrons (\(3s^{2}\)), and the remaining \(2\) electrons go into the \(3p\) sub - shell (\(3p^{2}\)).

Answer:

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