Advanced Bonding: Resonance, Formal Charge, and Orbital Theory

At the introductory level, chemical bonding often feels like connecting dots. You count valence electrons, draw lines, and ensure everyone has eight electrons. However, reality is more nuanced. Why do some molecules break the octet rule? Why are some bonds stronger than others? Why does oxygen attract electrons more aggressively than nitrogen?

To master university-level chemistry, you must look beyond simple Lewis structures and understand the quantum mechanical nature of electrons. This guide breaks down advanced bonding concepts—from the anatomy of sigma and pi bonds to the predictive power of Molecular Orbital (MO) theory.

The Nature of Covalent Bonds: Sigma ($\sigma$) and Pi ($\pi$) Bonds

Not all covalent bonds are created equal. While a single line in a drawing represents a bond, the physical shape of the electron clouds differs depending on whether it is a single, double, or triple bond.

1. Sigma ($\sigma$) Bonds: The Foundation

A sigma bond is the strongest type of covalent bond. It is formed by the head-on overlap of atomic orbitals directly along the internuclear axis (the invisible line connecting two nuclei).

Orbital Overlap: Can occur between two s-orbitals, one s and one p, or two p-orbitals facing each other.
Rotation: Because the overlap is cylindrical and centered on the axis, atoms can rotate freely around a sigma bond without breaking it.
Occurrence: Every single bond is a sigma bond. The first bond formed between any two atoms is always a sigma bond.

2. Pi ($\pi$) Bonds: The Reinforcement

A pi bond is formed by the sideways (lateral) overlap of parallel p-orbitals. These orbitals interact above and below the internuclear axis, not directly between the nuclei.

Orbital Overlap: strictly between two parallel p-orbitals.
Rotation: Pi bonds lock the molecule in place. Rotation would break the parallel alignment of the p-orbitals, destroying the bond.
Occurrence: Pi bonds only exist in double and triple bonds.

Case Study: The Carbon-Carbon Triple Bond

Consider the triple bond in an alkyne ($C \equiv C$). Many students mistakenly think this consists of three identical bonds. In reality, the orbital composition is distinct:

The First Bond: A $\sigma$ bond formed by the head-on overlap of $sp$ hybrid orbitals.
The Second & Third Bonds: Two distinct $\pi$ bonds formed by the sideways overlap of unhybridized $p$ orbitals (one pair on the y-axis, one pair on the z-axis).

Therefore, a triple bond is composed of one $\sigma$ bond and two $\pi$ bonds. This accumulation of electron density makes triple bonds shorter and stronger than single bonds, though the individual $\pi$ bonds are weaker than the $\sigma$ bond.

Resonance, Delocalization, and Calculating Formal Charge

Sometimes, a single Lewis structure cannot accurately depict a molecule. When electrons can move (delocalize) across adjacent parallel p-orbitals, we use resonance structures to represent the molecule.

But how do you know which resonance structure contributes most to the actual hybrid? You use Formal Charge.

The Formal Charge Formula

Formal charge (FC) allows us to determine the distribution of charge within a molecule, assuming all electrons in bonds are shared equally.

$$FC = V - (L + \frac{1}{2}S)$$

$V$ = Valence electrons (from the periodic table).
$L$ = Lone pair electrons (count individual dots).
$S$ = Shared bonding electrons (count lines as 2).

Rules for Stability

The "best" or major resonance structure follows these priorities:

Minimizing Charge: Structures where atoms have an FC of 0 are preferred over those with charges.
Electronegativity: If a negative FC is necessary, it should reside on the most electronegative atom (e.g., Oxygen rather than Carbon).
Octet Rule: Structures that satisfy the octet rule are generally preferred (though exceptions exist, see next section).

Practice Problem: Isobutyryl Chloride

Consider the resonance forms for an acyl chloride ($R-COCl$). You might draw a structure where the C=O double bond breaks, putting a negative charge on oxygen and a positive charge on the carbon, which then pulls electrons from Chlorine to form a double bond ($C=Cl$).

Compare the two forms:

Structure A (Neutral): $C=O$ double bond, $C-Cl$ single bond. All FCs are 0.
Structure B (Charged): $C-O^-$ and $C=Cl^+$. Chlorine has a positive formal charge, and Oxygen has a negative one.

Conclusion: Structure A is the major contributor because it minimizes charge separation and avoids placing a positive charge on the electronegative chlorine atom.

Practice Problem: Phosphate Ion ($PO_4^{3-}$)

In the phosphate ion, drawing four single $P-O$ bonds results in a phosphorus atom with a +1 formal charge. While this obeys the octet rule, it is not the lowest energy state.

By moving a lone pair from one oxygen to form a double bond ($P=O$), phosphorus expands its octet (10 electrons). The result?

Phosphorus FC: $5 - (0 + \frac{1}{2}(10)) = 0$.
Double-bonded Oxygen FC: $6 - (4 + \frac{1}{2}(4)) = 0$.

This structure minimizes formal charge on the central atom, making it a significant resonance contributor.

Exceptions to the Octet Rule: Expanded and Incomplete Octets

The "Rule of Eight" is strictly obeyed by Period 2 elements (C, N, O, F). However, elements in Period 3 and below (like P, S, Cl) have access to empty d-orbitals, allowing them to accommodate more than 8 valence electrons. This is called an expanded octet or hypervalency.

Identifying Hypervalent Molecules

To determine if a molecule has an expanded octet, count the electron domains around the central atom. If the total shared electrons + lone pair electrons > 8, it is hypervalent.

Comparative Examples:

$H_2O$ (Water): Oxygen (Period 2) has 2 bonds + 2 lone pairs = 8 electrons. Follows Octet.
$SF_6$ (Sulfur Hexafluoride): Sulfur (Period 3) is bonded to 6 Fluorine atoms. $6 \times 2 = 12$ electrons. Expanded Octet.
$SO_3$ (Sulfur Trioxide):
- Strict Octet structure: One double bond, two single bonds. Sulfur has +2 charge.
- Expanded structure: Three double bonds. Sulfur has 12 electrons but 0 Formal Charge. The expanded version is the dominant contributor because it neutralizes the charge on the central atom.

Note on Incomplete Octets: Elements like Boron often form stable compounds with fewer than 8 electrons. For example, in $BH_3$, Boron has only 6 valence electrons and is stable (though highly reactive).

Introduction to Molecular Orbital (MO) Theory

Lewis structures and Valence Bond Theory explain molecular geometry well, but they fail to explain magnetic properties or bond strength trends in depth. Molecular Orbital (MO) theory solves this by visualizing orbitals not as belonging to individual atoms, but to the entire molecule.

Bonding vs. Antibonding Orbitals

When atomic orbitals combine, they conserve the number of orbitals. Two atomic orbitals combine to create:

Bonding Molecular Orbital ($\sigma$ or $\pi$): Lower energy, constructive interference. Electrons here stabilize the molecule.
Antibonding Molecular Orbital ($\sigma^*$ or $\pi^*$): Higher energy, destructive interference. Electrons here destabilize the molecule.

Calculating Bond Order

Bond order tells us the strength of the bond. A bond order of 0 means the molecule is unstable and will not exist.

$$Bond \ Order = \frac{1}{2} (N_{bonding} - N_{antibonding})$$

Step-by-Step: Ranking Bond Strength in Diatomic Hydrogen

Let's use MO diagrams to rank the bond energy of $H_2$, $H_2^-$, and $H_2^{2-}$ from weakest to strongest.

Determine Electron Counts:
- $H_2$: 2 valence electrons.
- $H_2^-$: 3 valence electrons.
- $H_2^{2-}$: 4 valence electrons.
Fill the MO Diagram: (Order: $\sigma_{1s}$ then $\sigma^*_{1s}$)
- $H_2$: 2 e- in bonding. Bond Order = $0.5(2-0) = 1$.
- $H_2^-$: 2 e- in bonding, 1 e- in antibonding. Bond Order = $0.5(2-1) = 0.5$.
- $H_2^{2-}$: 2 e- in bonding, 2 e- in antibonding. Bond Order = $0.5(2-2) = 0$.
Conclusion: Higher bond order equates to higher bond energy.
Ranking (Weakest to Strongest): $H_2^{2-}$ (unstable) < $H_2^-$ < $H_2$.

Key Takeaways

Bond Composition: Single bonds are always $\sigma$. Multiple bonds are composed of one $\sigma$ and remaining $\pi$ bonds.
Formal Charge: The most stable Lewis structure minimizes formal charge. Use $FC = V - (L + \frac{1}{2}S)$.
Expanded Octets: Atoms in Period 3 and below (like P, S, Cl) can exceed 8 valence electrons to minimize formal charge.
MO Theory: Provides a more accurate picture of stability. Electrons in antibonding orbitals lower the bond order and weaken the bond.