To determine the number of pairs of electrons in the valence shell of the Lewis structure of calcium sulfide (CaS), we first need to understand how the atoms in this compound combine.
- Calcium (Ca) has 2 valence electrons. It typically loses these two electrons to form a Ca²⁺ ion.
- Sulfur (S) has 6 valence electrons. It typically gains 2 electrons to fill its valence shell and form an S²⁻ ion.
When calcium and sulfur combine to form calcium sulfide, the calcium loses its two valence electrons, and sulfur gains these two electrons.
Now, let's consider the number of pairs of electrons:
- Calcium's 2 valence electrons are lost and do not form pairs in the Lewis structure.
- Sulfur's 6 valence electrons, after gaining 2 from calcium, results in a total of 8 electrons around the sulfur atom.
In the Lewis structure:
- The 6 original valence electrons of sulfur can be visualized as 3 pairs (4 electrons) and 2 single electrons.
- After gaining 2 electrons to achieve a full octet, sulfur will have a total of 8 electrons, which constitutes 4 pairs of electrons (8 electrons ÷ 2 electrons per pair).
Therefore, in the Lewis structure of calcium sulfide, there are four pairs of electrons present in the valence shell of sulfur.
The correct answer is D. four.
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