Use the electron-dot symbols to write the oquation for the formation of the icmic compounds from sach of the following pairs
- Kand F
b. Na and O
Ca and P
d. Al and S
The Correct Answer and Explanation is :
Here are the electron-dot symbols and equations for the formation of ionic compounds for the following pairs:
a) K and F (Potassium and Fluorine)
Potassium (K) has an electron configuration of [Ar] 4s¹, meaning it has one electron in its outermost shell. Fluorine (F), on the other hand, has an electron configuration of [He] 2s² 2p⁵, with seven valence electrons.
- Potassium loses one electron to achieve a stable configuration, becoming K⁺ (cation).
- Fluorine gains one electron to complete its valence shell, becoming F⁻ (anion).
The ionic bond is formed as K⁺ and F⁻ are attracted by electrostatic forces.
Electron-dot symbols:
- K: ·K → K⁺ + e⁻
- F: F + e⁻ → F⁻
Equation for ionic compound formation:
[ K + F \rightarrow K^+ + F^- ]
Resulting ionic compound: KF (Potassium Fluoride).
b) Na and O (Sodium and Oxygen)
Sodium (Na) has an electron configuration of [Ne] 3s¹, with one valence electron, while oxygen (O) has an electron configuration of [He] 2s² 2p⁴, with six valence electrons.
- Sodium loses one electron to form Na⁺.
- Oxygen gains two electrons to form O²⁻.
Electron-dot symbols:
- Na: ·Na → Na⁺ + e⁻
- O: O + 2e⁻ → O²⁻
Equation for ionic compound formation:
[ 2Na + O \rightarrow 2Na^+ + O^{2-} ]
Resulting ionic compound: Na₂O (Sodium Oxide).
c) Ca and P (Calcium and Phosphorus)
Calcium (Ca) has an electron configuration of [Ar] 4s², with two valence electrons. Phosphorus (P) has an electron configuration of [Ne] 3s² 3p³, with five valence electrons.
- Calcium loses two electrons to form Ca²⁺.
- Phosphorus gains three electrons to form P³⁻.
Electron-dot symbols:
- Ca: ·Ca → Ca²⁺ + 2e⁻
- P: P + 3e⁻ → P³⁻
Equation for ionic compound formation:
[ 3Ca + 2P \rightarrow 3Ca^{2+} + 2P^{3-} ]
Resulting ionic compound: Ca₃P₂ (Calcium Phosphide).
d) Al and S (Aluminum and Sulfur)
Aluminum (Al) has an electron configuration of [Ne] 3s² 3p¹, with three valence electrons. Sulfur (S) has an electron configuration of [Ne] 3s² 3p⁴, with six valence electrons.
- Aluminum loses three electrons to form Al³⁺.
- Sulfur gains two electrons to form S²⁻.
Electron-dot symbols:
- Al: ··Al → Al³⁺ + 3e⁻
- S: S + 2e⁻ → S²⁻
Equation for ionic compound formation:
[ 2Al + 3S \rightarrow 2Al^{3+} + 3S^{2-} ]
Resulting ionic compound: Al₂S₃ (Aluminum Sulfide).
Explanation:
The formation of ionic compounds involves the transfer of electrons between metals and nonmetals. Metals (like K, Na, Ca, Al) lose electrons to achieve a stable electron configuration, becoming positively charged cations (K⁺, Na⁺, Ca²⁺, Al³⁺). Nonmetals (like F, O, P, S) gain electrons to complete their valence shells, becoming negatively charged anions (F⁻, O²⁻, P³⁻, S²⁻). The electrostatic attraction between these oppositely charged ions forms an ionic bond, resulting in the creation of ionic compounds like KF, Na₂O, Ca₃P₂, and Al₂S₃.
I will now generate an image to represent this.
Here is the illustration showing the formation of ionic compounds with electron-dot symbols for each pair of elements: K and F, Na and O, Ca and P, and Al and S. It visually represents how the electrons are transferred, resulting in the formation of cations and anions that attract each other, forming ionic bonds.
