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Chemical Bonding - Lewis Dot Structures

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Valence Electrons: These are the electrons in the outermost shell of an atom. In Lewis Dot Structures, they are represented as dots around the chemical symbol (e.g., Carbon is shown with 44 dots: ⋅C˙⋅\cdot \dot{C} \cdot).

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The Octet Rule: Atoms tend to gain, lose, or share electrons to achieve a full outer shell of 88 electrons, resembling the electron configuration of a Noble Gas. Exception: Hydrogen (HH) follows the Duet Rule, seeking only 22 electrons.

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Covalent Bonding: Occurs when two non-metal atoms share pairs of electrons. A single bond involves 22 electrons, a double bond involves 44 electrons, and a triple bond involves 66 electrons.

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Lone Pairs: These are pairs of valence electrons that are not involved in bonding (non-bonding electrons), such as the two pairs remaining on Oxygen in H2OH_2O.

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Ionic Bonding: Represented by showing the transfer of electrons. The metal atom becomes a cation (e.g., [Na]+[Na]^{+}) and the non-metal becomes an anion (e.g., [:Cl¨:]−[:\ddot{Cl}:]^{-}), both enclosed in square brackets with their respective charges.

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Electronegativity: The ability of an atom to attract shared electrons. In Lewis structures, the least electronegative atom (excluding HH) is usually placed in the center.

📐Formulae

TVE=∑(Valence Electrons of each atom)−(Ionic Charge)TVE = \sum (\text{Valence Electrons of each atom}) - (\text{Ionic Charge})

Number of Bonds=(Electrons needed for full octets)−(Total Valence Electrons)2\text{Number of Bonds} = \frac{(\text{Electrons needed for full octets}) - (\text{Total Valence Electrons})}{2}

Formal Charge=V−L−S2\text{Formal Charge} = V - L - \frac{S}{2}

💡Examples

Problem 1:

Draw the Lewis Dot Structure for Methane (CH4CH_4).

Solution:

The central Carbon atom is surrounded by four Hydrogen atoms, each connected by a single line representing a shared pair of electrons (C−HC-H).

Explanation:

Carbon has 44 valence electrons and each of the 44 Hydrogen atoms provides 11. Total valence electrons = 4+(4×1)=84 + (4 \times 1) = 8. Carbon forms 44 single bonds to satisfy its octet, and each Hydrogen achieves a duet.

Problem 2:

Draw the Lewis Dot Structure for Carbon Dioxide (CO2CO_2).

Solution:

O¨=C=O¨\ddot{O} = C = \ddot{O}

Explanation:

Total valence electrons = 4 (from C)+2×6 (from O)=164 \text{ (from } C) + 2 \times 6 \text{ (from } O) = 16. To satisfy the octet rule for all atoms with only 1616 electrons, Carbon must form double bonds with each Oxygen atom. Each Oxygen retains 22 lone pairs.

Problem 3:

Represent the ionic bond in Magnesium Oxide (MgOMgO) using Lewis notation.

Solution:

[Mg]2+[:O¨:]2−[Mg]^{2+} [:\ddot{O}:]^{2-}

Explanation:

Magnesium (MgMg) has 22 valence electrons and loses them to achieve a stable core, becoming [Mg]2+[Mg]^{2+}. Oxygen (OO) has 66 valence electrons and gains those 22 electrons to complete its octet, becoming [O]2−[O]^{2-} with 88 dots around it.

Lewis Dot Structures - Revision Notes & Key Formulas | IB Grade 9 Science