Comprehensive Study Notes on Lewis Dot Structures, Electronegativity, and Chemical Bonding

Course Scope and Content Boundary

• Content coverage ends after slide 75 of Chapter 4; students are responsible for all material in Chapter 4 up through slide 75.

• Material beginning after slide 75 in Chapter 4 is excluded from assessment responsibility.

• Chapter 5 focuses on drawing chemical structures using Lewis structures following the study of compound nomenclature.

Fundamentals of Lewis Dot Structures

• A Lewis dot structure is a structural representation that displays the valence electrons of an atom as dots drawn around its chemical symbol.

• Valence electrons are the electrons located in the outermost shell of an atom and are the sole electrons that participate in forming covalent or ionic chemical bonds.

• The number of valence electrons for a main-group element corresponds directly to its group number on the periodic table:

  • Chlorine ($Cl$): Located in Group 7; possesses 7 valence electrons in its outer shell. Represented by drawing 7 dots around the $Cl$ symbol.

  • Sodium ($Na$): Located in Group 1; possesses 1 valence electron in its outer shell. Represented by drawing 1 dot around the $Na$ symbol.

Electronegativity and Bond Polarity

• Electronegativity is the measure of an atom's ability to pull shared electrons toward its own nucleus when involved in a covalent bond.

• Periodic Trends in Electronegativity:

  • Increases moving left-to-right across a period on the periodic table. Elements such as Fluorine ($F$) and Chlorine ($Cl$) have 7 valence electrons and strongly attract shared electrons to complete their valence octet.

  • Decreases moving down a group on the periodic table.

  • Noble Gases Exception: Although noble gases have 8 valence electrons in their outer shell, they do not exhibit significant electronegativity values because they are inherently stable and generally do not form chemical bonds.

• Unequal Electron Sharing and Partial Charges:

  • In a covalent bond between two different atoms, electrons can be shared equally or unequally.

  • Unequal sharing occurs when one atom has a higher electronegativity than the other, pulling the shared electron density closer to its nucleus.

  • This asymmetric charge distribution creates bond polarity, resulting in partial positive ($\delta^+$) and partial negative ($\delta^-$) charges.

  • Delta symbols ($\delta$) denote partial electric charges that are strictly less than whole integer values (less than $+1$ or $-1$).

• Example: Hydrochloric Acid ($HCl$):

  • The bond line between $H$ and $Cl$ represents 2 shared electrons (1 contributed by Hydrogen, 1 by Chlorine).

  • Chlorine has a high electronegativity and pulls the shared electron pair toward its nucleus.

  • The electron pair spends more time near the Chlorine nucleus, resulting in a partial negative charge ($\delta^-$) on Chlorine and leaving Hydrogen with a partial positive charge ($\delta^+$).

Quantifying Bond Polarity from Electronegativity Differences

• Method of Calculation:

  • Calculate the difference in electronegativity ($\Delta \text{EN}$) for a single bond by subtracting the smaller electronegativity value from the larger electronegativity value:     $\Delta \text{EN} = \text{EN}_{\text{larger}} - \text{EN}_{\text{smaller}}$

  • Calculations are performed individually per single bond; values are not multiplied by stoichiometric coefficients or subscripts in the molecular formula.

• Example: Water ($H_2O$):

  • Oxygen ($O$) electronegativity = $3.5$

  • Hydrogen ($H$) electronegativity = $2.1$

  • Electronegativity difference per $O-H$ bond: $3.5 - 2.1 = 1.4$

  • Each $O-H$ bond is evaluated individually as having a difference of $1.4$.

• Classification Ranges for Bond Types:

  • $0.0$ to $0.4$: Nonpolar Covalent (Pure Covalent). Electrons are shared equally between atoms. No partial charges exist. Example: Chlorine gas ($Cl_2$), where $3.0 - 3.0 = 0.0$

  • $0.5$ to $1.9$: Polar Covalent. Shared electrons are pulled closer to the more electronegative atom, generating partial positive and partial negative charges.

  • Greater than $1.9$ (Values of $2.0$ and above): Ionic Bond. Electron sharing ceases; complete transfer of one or more electrons occurs from one atom to another, forming discrete ions.

Molecular Dipoles and the Octet Rule

• Molecular Dipole:

  • A dipole exists when a molecule exhibits net separated positive and negative regions across the overall molecular structure ("two poles").

  • Determining whether a molecule acts as a dipole requires drawing its complete Lewis structure and geometry.

  • Example: Water ($H_2O$) has a partial negative charge at the top (Oxygen atom) and partial positive charges at the bottom (Hydrogen atoms).

• The Octet Rule:

  • Main-group atoms react, share, or transfer electrons to achieve a stable outer shell of 8 valence electrons, matching the electron configuration of a noble gas.

Exceptions to the Octet Rule

• Hydrogen ($H$):

  • Requires only 2 valence electrons to complete its outer shell (attaining the stable configuration of Helium, $He$).

  • Hydrogen forms exactly 1 single bond (2 shared electrons) and never forms double bonds, triple bonds, or unshared lone pairs.

• Beryllium ($Be$):

  • Exceptions permit Beryllium to form 2 single bonds, surrounding itself with only 4 valence electrons.

• Boron ($B$):

  • Exceptions permit Boron to form 3 single bonds, surrounding itself with only 6 valence electrons.

• Expanded Octets:

  • Elements located in Row 3 or lower of the periodic table can accommodate more than 8 valence electrons around the central atom.

  • Expansion is possible because period 3 and higher elements have accessible $d$-orbitals (such as the $3d$ sublevel present starting in period 3 / row 4) to hold additional electrons.

Types of Covalent Bonds

• Single Covalent Bond:

  • Represented by 1 straight line between atoms.

  • Involves 2 shared electrons (1 pair).

• Double Bond:

  • Represented by 2 parallel lines between atoms.

  • Involves 4 shared electrons (2 pairs; 2 electrons contributed by each atom).

  • Example: Oxygen gas ($O_2$), where two Oxygen atoms share 4 electrons to give each atom an octet of 8 electrons.

• Triple Bond:

  • Represented by 3 parallel lines between atoms.

  • Involves 6 shared electrons (3 pairs; 3 electrons contributed by each atom).

  • Example: Nitrogen gas ($N_2$), where two Nitrogen atoms share 6 electrons to give each atom an octet of 8 electrons.

• Dual-Counting Rule: All shared bonding electrons count toward the total octet requirement for both participating atoms simultaneously.

Step-by-Step Procedure for Drawing Lewis Structures

• Step 1: Identify the Central Atom.

  • The central atom is typically the first element written in the molecular chemical formula (excluding Hydrogen, which is always terminal).

• Step 2: Sum Total Valence Electrons.

  • Calculate the exact number of available valence electrons by summing the valence electrons contributed by every individual atom in the formula.

• Step 3: Connect Atoms with Single Bonds.

  • Draw single bond lines connecting the central atom to each outer terminal atom. Deduct 2 valence electrons per single bond from the available total pool.

• Step 4: Complete Outer Atom Octets.

  • Place remaining available electrons as lone pair dots around terminal outer atoms until each outer atom achieves its stable valence configuration (2 electrons for Hydrogen, 8 electrons for main group atoms).

• Step 5: Complete Central Atom Octet and Form Multiple Bonds if Needed.

  • Check the outer atoms first to ensure stability, then check the central atom.

  • Place remaining electrons on the central atom as lone pairs.

  • If the total valence electron pool is completely exhausted and the central atom still lacks an octet (fewer than 8 electrons), take a lone pair from an outer atom and convert it into a double or triple bond between that outer atom and the central atom.

Step-by-Step Worked Examples

• Example 1: Formaldehyde ($CH_2O$)

  • Central Atom Determination: Carbon ($C$) is written first and serves as the central atom.

  • Total Valence Electron Calculation:

  • Carbon (Group 4): 4 valence electrons

  • Hydrogen (Group 1): $1 \times 2 = 2$ valence electrons

  • Oxygen (Group 6): 6 valence electrons

  • Total available valence electrons: $4 + 2 + 6 = 12$ electrons.

  • Connect Atoms: Draw single bonds from central $C$ to two $H$ atoms and one $O$ atom ($H-C$, $H-C$, $C-O$). Uses $3 \times 2 = 6$ electrons. Remaining electrons: $12 - 6 = 6$ electrons.

  • Satisfy Outer Atoms: Hydrogens are satisfied with 1 single bond each. Place all remaining 6 electrons on Oxygen as 3 lone pairs. Remaining unassigned electrons: $6 - 6 = 0$ electrons.

  • Evaluate Central Atom Octet: Carbon currently has 3 single bonds around it ($3 \times 2 = 6$ electrons), failing the octet requirement.

  • Form Multiple Bonds: Convert 1 lone pair on the Oxygen atom into a $C=O$ double bond.

  • Final Structure Check:

  • Oxygen has 2 single/double bond connections (4 electrons) + 2 lone pairs (4 electrons) = 8 valence electrons.

  • Carbon has 4 bonding pairs (1 double bond + 2 single bonds = 8 electrons) = 8 valence electrons.

  • Each Hydrogen has 1 single bond = 2 valence electrons.

• Example 2: Ammonia ($NH_3$)

  • Central Atom Determination: Nitrogen ($N$) is the central atom.

  • Total Valence Electron Calculation:

  • Nitrogen (Group 5): 5 valence electrons

  • Hydrogen (Group 1): $1 \times 3 = 3$ valence electrons

  • Total available valence electrons: $5 + 3 = 8$ electrons.

  • Connect Atoms: Draw single bonds from central $N$ to three $H$ atoms ($N-H$). Uses $3 \times 2 = 6$ electrons. Remaining electrons: $8 - 6 = 2$ electrons.

  • Satisfy Outer Atoms: All three Hydrogens are satisfied with 1 bond each.

  • Satisfy Central Atom: Place the remaining 2 electrons onto the central Nitrogen atom as 1 lone pair. Remaining unassigned electrons: $2 - 2 = 0$ electrons.

  • Final Structure Check: Nitrogen has 3 bonding pairs (6 electrons) + 1 lone pair (2 electrons) = 8 valence electrons. All Hydrogens have 2 electrons.

• Example 3: Methane ($CH_4$)

  • Central Atom Determination: Carbon ($C$) is the central atom.

  • Total Valence Electron Calculation:

  • Carbon (Group 4): 4 valence electrons

  • Hydrogen (Group 1): $1 \times 4 = 4$ valence electrons

  • Total available valence electrons: $4 + 4 = 8$ electrons.

  • Connect Atoms: Draw single bonds from central $C$ to four $H$ atoms ($C-H$). Uses $4 \times 2 = 8$ electrons. Remaining electrons: $8 - 8 = 0$ electrons.

  • Final Structure Check: Carbon has 4 bonding pairs = 8 valence electrons. All four Hydrogens have 1 bond each = 2 valence electrons. Structure is complete.