Polar Covalent Bonds, Electronegativity, and Lewis Structures - Chapter 10 - Part 2

Polar Covalent Bonds and Electronegativity

Definition of Polar Covalent Bonds

A polar covalent bond is a covalent bond in which the bonding electrons spend more time near one atom than the other. This unequal sharing leads to partial charges on the atoms involved.

• Electronegativity (X): A measure of the ability of an atom in a molecule to attract bonding electrons to itself.

• Electronegativity is related to ionization energy and electron affinity.

Trends in Electronegativity

• Electronegativity increases from left to right across a period and from bottom to top within a group in the periodic table.

• Fluorine (F), Oxygen (O), Nitrogen (N), and Chlorine (Cl) have the highest electronegativity values.

Bond Polarity and Electronegativity Difference

• During bond formation, electrons are pulled toward the more electronegative atom.

• The direction of electron shift can be predicted using the electronegativity scale.

• Example: In H–Cl, electrons are pulled toward Cl (X = 3.0) rather than H (X = 2.1), making Cl partially negative and H partially positive. Thus, HCl is a polar molecule.

• The difference in electronegativity between two atoms is a rough measure of bond polarity:

  • Very large difference: Ionic bond

  • Large difference: Polar covalent bond

  • Small difference: Nonpolar covalent bond

Example: Ranking Bond Polarity

• Arrange C–N, Na–F, and O–H by increasing polarity using electronegativity differences:

• C–N: 3.0 (N) – 2.5 (C) = 0.5

• Na–F: 4.0 (F) – 0.9 (Na) = 3.1

• O–H: 3.5 (O) – 2.1 (H) = 1.4

• Order: C–N < O–H < Na–F

Writing Lewis Electron-Dot Formulas

Steps for Drawing Lewis Structures

1. Calculate the number of valence electrons.

2. Write the skeleton structure of the molecule or ion.

3. Distribute electrons to the atoms surrounding the central atom(s) to satisfy the octet rule.

4. Distribute the remaining electrons as pairs to the central atom(s).

5. If the central atom still needs electrons, use nonbonding electrons from adjacent atoms to form double or triple bonds.

6. Always make sure all atoms have an octet (except for exceptions noted below).

Examples of Lewis Structures

• OF2: 20 valence electrons. O is the central atom. Distribute electrons to F atoms first, then remaining electrons to O.

• NF3: 26 valence electrons. N is the central atom. Distribute electrons to F atoms, then remaining electrons to N.

• NH2OH (hydroxylamine): 14 valence electrons. N is the central atom. Distribute electrons to H and O, then remaining electrons to N.

• CO2: 16 valence electrons. C is the central atom. Distribute electrons to O atoms. If C lacks an octet, convert lone pairs on O to double bonds.

• HCN: 10 valence electrons. C is the central atom. Distribute electrons to N. If C lacks an octet, convert lone pairs on N to triple bond.

• PCl4+: 32 valence electrons. P is the central atom. Distribute electrons to Cl atoms. Enclose structure in brackets with charge.

Delocalized Bonding: Resonance

Definition and Application

• Delocalized bonding occurs when a bonding pair of electrons is spread over several atoms, not just two.

• Resonance is used when a single Lewis structure cannot adequately represent the bonding. All possible structures are drawn and connected by double-headed arrows.

• Example: The acetate ion (CH3COO–) has two resonance structures, each with a double bond between C and one O atom.

Exceptions to the Octet Rule

Types of Exceptions

• Too few electrons: Elements in Groups IIA and IIIA (e.g., Be, B, Al) may have fewer than eight electrons around the central atom.

• Odd number of electrons: Some molecules (e.g., NO) have an odd number of electrons and cannot satisfy the octet rule for all atoms.

• Expanded octet: Elements in period 3 or beyond (e.g., S, P, Cl) can have more than eight electrons due to available d orbitals.

• Elements in the second period (only s and p subshells) cannot expand their octet.

• Example: Phosphorus pentafluoride (PF5) and iodine pentafluoride (IF5) have more than eight valence electrons on the central atom.

Formal Charge and Lewis Formulas

Definition and Calculation

• The formal charge on an atom in a Lewis structure is the hypothetical charge assuming equal sharing of bonding electrons and that lone pair electrons belong entirely to one atom.

• Formula:

$\text{Formal charge} = \text{valence electrons on free atom} - \frac{1}{2}(\text{number of electrons in bonds}) - (\text{number of lone-pair electrons})$

• The sum of formal charges in a molecule or ion equals the overall charge.

Rules for Choosing the Best Lewis Structure

1. Choose the structure with the lowest magnitudes of formal charges.

2. If two structures have the same formal charge magnitudes, choose the one with the negative formal charge on the more electronegative atom.

3. Avoid like charges on adjacent atoms.

Example: CO2 Formal Charges

Structure	C	O (double bond)	O (double bond)
O=C=O	0	0	0
O–C≡O	0	–1	+1

The left structure (O=C=O) is preferred because all formal charges are zero.

Example: H2SO4 (Sulfuric Acid)

• Calculate formal charges for each atom to determine the best Lewis structure.

• Structures with zero formal charges are preferred, often requiring double bonds to oxygen atoms.

Bond Length and Bond Order

Definitions

• Bond length is the distance between the nuclei of two bonded atoms.

• Bond order is the number of electron pairs shared between two atoms (single = 1, double = 2, triple = 3).

• Bond length decreases as bond order increases.

Example: Propylene Molecule

• Double bonds are shorter (134 pm) than single bonds (150 pm).

Example: Nitrogen Bonds

• N2H4 (single bond), N2 (triple bond), N2F2 (double bond).

• The triple bond (N2) is shortest; the single bond (N2H4) is longest.

Bond Enthalpy

Definition and Application

• Bond enthalpy is the average enthalpy change for breaking a specific bond in a molecule in the gas phase.

• Bond enthalpy is a measure of bond strength: higher bond enthalpy means a stronger bond.

Estimating Enthalpy Change (ΔH) for Reactions

• ΔH can be estimated by considering the energy required to break bonds and the energy released when new bonds form:

$\Delta H = \text{sum of bond energies for bonds broken} - \text{sum of bond energies for bonds formed}$

• If ΔH is negative, heat is released (exothermic reaction).

• If ΔH is positive, heat is absorbed (endothermic reaction).

Example Calculation

• For a reaction breaking 1 C=C (602 kJ) and 1 C–Cl (240 kJ), and forming 1 C–C (346 kJ) and 2 C–Cl (654 kJ):

$\Delta H = (602 + 240)\ \text{kJ} - (346 + 654)\ \text{kJ} = 842\ \text{kJ} - 1000\ \text{kJ} = -158\ \text{kJ}$

• The reaction is exothermic.