Question

question 15 6 pts what is the abbreviated electron configuration, and, how many valence electrons does the element sulfur (s) have? periodic table image ○ abbreviated electron configuration = he 2s² 2p⁶ 3s² 3p⁴ and valence electrons = 4 ○ abbreviated electron configuration = ne 3s² 3p⁴ and valence electrons = 4 ○ abbreviated electron configuration = he 2s² 2p⁶ 3s² 3p⁴ and valence electrons = 6 ○ abbreviated electron configuration = ne 3s² 3p⁴ and valence electrons = 6

Explanation:

Step1: Determine the noble gas core for sulfur

Sulfur (S) has an atomic number of 16. The noble gas before sulfur is neon (Ne) with an atomic number of 10. So the abbreviated electron configuration starts with [Ne].

Step2: Determine the remaining electron configuration

After the neon core, the remaining electrons are \(16 - 10 = 6\). These fill the \(3s\) and \(3p\) orbitals. The electron configuration for the remaining electrons is \(3s^2 3p^4\). So the abbreviated electron configuration is \([Ne] 3s^2 3p^4\).

Step3: Determine the number of valence electrons

Valence electrons are the electrons in the outermost shell. For sulfur, the outermost shell is the 3rd shell, with electrons in \(3s^2\) and \(3p^4\). So the number of valence electrons is \(2 + 4 = 6\).

Answer:

abbreviated electron configuration = \([Ne] 3s^2 3p^4\) and valence electrons = 6 (the last option: "abbreviated electron configuration = [Ne] \(3s^2\) \(3p^4\) and valence electrons = 6")