. 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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Explanation:
Atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. To attain noble gas configuration it is necessary for the magnesium atom to lose two valence electrons and therefore, it forms ions.
On the other hand, atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. To attain noble gas configuration it needs to gain two electrons. And, on gaining the electrons it forms ions.
Hence, when both magnesium and oxygen ions chemically combine with each other then it forms the compound magnesium oxide ().
This is because magnesium transfer its two valence electrons to the oxygen atom and due to the formation of opposite charges on these atoms they get attracted towards each other.
