Question

which shows the electron configuration for silicon (si)?
○ ar4s²3d¹⁰4p⁴
○ ar3d¹⁴4s²
○ ne3s²3p⁴
○ ne3s²3p²

Explanation:

1. Determine the atomic number of Si: Silicon (Si) has an atomic number of 14.
2. Recall the electron configuration rules: Electrons fill orbitals in order of increasing energy. The noble gas before Si is Neon (Ne, atomic number 10), so we use [Ne] as the core.
3. Calculate the remaining electrons: $14 - 10 = 4$ electrons to distribute in the $n = 3$ shell.
4. Fill the orbitals: The 3s orbital holds 2 electrons, leaving $4 - 2 = 2$ electrons for the 3p orbital. So the electron configuration is $[\text{Ne}]3s^23p^2$.
5. Eliminate other options:

• $[\text{Ar}]4s^23d^{10}4p^4$ corresponds to a higher - atomic - number element (e.g., Selenium, atomic number 34), not Si.
• $[\text{Ar}]3d^{14}4s^2$ is incorrect as the 3d orbital can hold a maximum of 10 electrons, and this configuration does not match Si.
• $[\text{Ne}]3s^23p^4$ has 10 + 2+4 = 16 electrons, corresponding to Sulfur (S), not Si.

Answer:

D. $[\text{Ne}]3s^23p^2$ (assuming the last option is D, with the option text $[\text{Ne}]3s^23p^2$)