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

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources


Write a one sentence answer...



Explain in detail...


I want someone to double check my answer

Describe your changes (optional) 200

Dec 12, 2016


#"S"^(2-):1s^2 2s^2 2p^6 3s^2 3p^6#


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, #"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

#"S: " 1s^2 2s^2 2p^6 3s^2 3p^4#

Now, the sulfide anion, #"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

#"S"^(2-):color(white)(a) 1s^2 2s^2 2p^6 3s^2 3p^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#


Was this helpful? Let the contributor know!
Impact of this question
32810 views around the world
You can reuse this answer
Creative Commons License