. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?
1 answer:
Answer:
<u>Four</u>
Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,
Ca → Ca²⁺ + 2 e⁻
These two electrons are accepte by Sulfur as,
S + 2 e⁻ → S²⁻
So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,
<u>Four</u>
Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,
Ca → Ca²⁺ + 2 e⁻
These two electrons are accepte by Sulfur as,
S + 2 e⁻ → S²⁻
So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,
You might be interested in
Answer:
See explanation below
Explanation:
You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.
Now let's analyze what is happening in the reaction so we can predict the final product.
We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.
Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.
Hope this helps
