. 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,
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Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
The hydronium ion concentration at pH = 5.
..............(1)
The hydronium ion concentration at pH = 3.
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.