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# What is the electron configuration for S^(2-) ion?

Mar 24, 2016

${\text{S}}^{2 -} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$

#### Explanation:

A good starting point when looking for the electron configuration of an ion is the electron configuration of the neutral atom.

In your case, the neutral atom is sulfur, $\text{S}$, which is located in period 3, group 16 of the periodic table. Sulfur's has an atomic number equal to $16$, which means that a neutral sulfur atom has a total of $16$ electrons surrounding its nucleus.

The electron configuration of a neutral sulfur atom will thus be

$\text{S: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{4}$

Now, the sulfide anion, ${\text{S}}^{2 -}$, is formed when two electrons are added to a neutral sulfur atom.

As you can see in the configuration of the neutral atom, these two electrons will be added to the 3p-orbitals, which can hold a maximum of six electrons between them.

The electron configuration of the sulfide anion will thus be

${\text{S}}^{2 -} : \textcolor{w h i t e}{a} 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$

The noble gas shorthand notation for the sulfide anion will use the electron configuration of neon, the noble gas that comes immediately before sulfur in the periodic table.

"S"^(2-): color(white)(a)["Ne"] 3s^2 3p^6