Source: Exam 1 Key, Organic Chemistry I, University of Minnesota Twin Cities
Difficulty: Introductory to Intermediate | Prerequisites: General chemistry (atomic orbitals, electron configuration, periodic trends)
This material covers the foundational language of organic chemistry: how atoms bond, what shape those bonds take, and how electrons distribute across a molecule. If you cannot assign hybridisation, draw a Lewis structure, or push electrons through a resonance form, every later topic (reactions, stereochemistry, spectroscopy) will feel arbitrary. These three skills are tested early and assumed for the rest of the course. You should already be comfortable with electron configuration, electronegativity trends, and the octet rule from general chemistry.
Hybridisation tells you an atom's geometry and what orbitals form its bonds. Lewis structures assign every valence electron a home and let you calculate formal charges. Resonance structures show how electrons can be delocalised across a molecule, and the best contributor is the one with the most bonds, fewest charges, and negative charge on the most electronegative atom.
Tags: hybridisation, sp2, sp3, sp, orbital overlap, sigma bond, pi bond, formal charge, Lewis structure, resonance, resonance contributor, delocalisation, conjugation
Hybridisation
The mixing of atomic orbitals (s, p, and sometimes d) on a single atom to form new, equivalent hybrid orbitals that determine molecular geometry.
Think of it as: the atom blending its orbitals into a set of identical ones so it can point bonds in the right directions.
Sigma (σ) bond
A covalent bond formed by head-on (end-to-end) overlap of orbitals along the internuclear axis. Every single bond is a σ bond; every double and triple bond contains exactly one σ bond.
In simple terms, this means: the first bond between any two atoms is always a σ bond.
Pi (π) bond
A covalent bond formed by side-by-side (lateral) overlap of unhybridised p orbitals above and below the internuclear axis.
Think of it as: the "second" bond in a double bond, or the second and third in a triple bond. It cannot exist without a σ bond already in place.
Formal charge
The charge assigned to an atom in a Lewis structure, calculated as: valence electrons minus lone-pair electrons minus half of bonding electrons.
In simple terms, this means: how many electrons the atom "owns" in the structure compared to how many it normally has.
Resonance structures
Two or more valid Lewis structures for the same molecule that differ only in the placement of electrons (not atoms). The true structure is a weighted average (hybrid) of all contributors.
Think of it as: different ways of drawing the same molecule's electrons. The real molecule does not flip between them; it sits somewhere in between.
Resonance contributor (major vs. minor)
The individual Lewis structure that contributes most to the resonance hybrid. The best contributor has the most covalent bonds, the fewest formal charges, and any negative charge on the most electronegative atom.
In simple terms, this means: draw all the options, then rank them. The one that looks most "normal" (full octets, minimal charges, charges in sensible places) wins.
Count the number of electron domains (bonding pairs + lone pairs) around the atom.
4 domains → sp3, tetrahedral geometry, ~109.5° angles
3 domains → sp2, trigonal planar geometry, ~120° angles
2 domains → sp, linear geometry, 180° angles
A double bond counts as one domain. A triple bond counts as one domain. Only lone pairs and bond connections matter, not how many bonds are in each connection.
In the structure containing a C=O bond with nitrogen in the ring and a C-H bond:
Oxygen in the C=O is sp2 hybridised (three electron domains: one double bond to C, two lone pairs).
The C-N-C bond angle is 120°, because nitrogen is sp2 hybridised in this conjugated system (three bonding domains, and the lone pair is part of the π system).
σ bond in a C=O double bond: formed by overlap of an sp2 orbital on carbon with an sp2 orbital on oxygen.
π bond in a C=O double bond: formed by side-by-side overlap of a p orbital on carbon with a p orbital on oxygen (these are the unhybridised p orbitals perpendicular to the plane).
C-H single bond: formed by overlap of an sp2 orbital on carbon with a 1s orbital on hydrogen.
The σ bond always uses the hybrid orbital of each atom.
The π bond always uses the leftover unhybridised p orbital(s).
Hydrogen only has a 1s orbital to contribute.
The hybridisation of each atom tells you which orbital it uses for its σ bonds.
Count total valence electrons for the molecule (adjust for charges).
Place the least electronegative atom in the centre (hydrogen is always terminal).
Connect atoms with single bonds first, then distribute remaining electrons as lone pairs to satisfy octets.
If octets are not satisfied, convert lone pairs to bonding pairs (forming double or triple bonds).
Formal charge = (valence electrons of the free atom) - (lone pair electrons) - (half of bonding electrons)
Alternatively: formal charge = (valence electrons) - (dots) - (lines), where dots are lone pair electrons and lines are bonds.
The sum of all formal charges must equal the overall charge of the molecule or ion.
For the structure containing Li, N, B, F, O, and H:
Li: 0 (one valence electron, one bond, no lone pairs → 1 - 0 - 1 = 0)
H: 0 (one valence electron, one bond → 1 - 0 - 1 = 0)
N: +1 (five valence electrons, four bonds, no lone pairs → 5 - 0 - 4 = +1)
B: -1 (three valence electrons, four bonds, no lone pairs → 3 - 0 - 4 = -1)
F: 0 (seven valence electrons, one bond, three lone pairs → 7 - 6 - 1 = 0)
O: -1 (six valence electrons, one bond, three lone pairs → 6 - 6 - 1 = -1, or two bonds and two lone pairs depending on the structure)
Nitrogen with four bonds and no lone pairs → +1 formal charge
Oxygen with one bond and three lone pairs → -1 formal charge
Boron with four bonds and no lone pairs → -1 formal charge
Carbon with three bonds and one lone pair → -1 formal charge
Move only electrons (lone pairs or π bonds), never atoms.
Use curved arrows to show electron movement: from a lone pair into a bond, or from a bond to become a lone pair on the adjacent atom.
The molecular framework (which atoms are bonded to which) stays identical across all resonance forms.
Lone pair next to a π bond: the lone pair can form a new π bond while the existing π bond breaks, pushing electrons onto the far atom. Example: the amide (C=O next to N with a lone pair) gives a second form where N=C and O carries a negative charge.
Positive charge next to a π bond: the π bond can shift toward the positive centre. Example: a carbocation adjacent to a double bond can redistribute the charge.
Equivalent atoms (symmetry): benzene, carboxylate ions, nitrate ions, etc.
When deciding which resonance form contributes most:
More covalent bonds = better. Structures with more bonds are more stable.
Fewer formal charges = better. A structure with no charges beats one with separated charges.
Negative charge on the more electronegative atom = better. Oxygen bearing a negative charge is more reasonable than carbon bearing one.
Every atom has an octet = better (especially C, N, O). Structures that violate the octet rule on second-row elements are poor contributors.
For the molecule H₂C=C-NH₂ (with various resonance forms labelled A through D):
Form D (no formal charges, all atoms have octets, maximum bonding) is the greatest contributor.
Forms A and B have separated formal charges, making them lesser contributors.
Form C places a positive charge on carbon (a terminal CH₂ group), which is particularly unfavourable.
Students often think a double bond counts as two electron domains when determining hybridisation. It does not. A double bond is one domain, a triple bond is one domain. Only the number of connections and lone pairs matters.
Students often confuse formal charge with oxidation state. Formal charge splits bonding electrons equally; oxidation state gives them to the more electronegative atom. They are different numbers used for different purposes.
Students often think resonance structures represent the molecule flipping between forms. The molecule does not switch. The true structure is a single hybrid, a blend of all contributors, all the time.
Students often move atoms when drawing resonance structures. Only electrons (lone pairs and π bonds) move. If you have relocated an atom, you have drawn a constitutional isomer, not a resonance structure.
⚠️ Hybridisation questions appear on nearly every exam. Be able to assign sp, sp2, or sp3 to any atom in a structure within seconds.
⚠️ Formal charge calculation is a common source of lost marks. Memorise the shortcut patterns (N with four bonds = +1, O with one bond and three lone pairs = -1) so you can assign charges by inspection.
⚠️ Resonance ranking is frequently tested as a multiple-choice question. The exam will show you several resonance forms and ask which contributes most. Apply the rules in order: most bonds, fewest charges, negative charge on electronegative atoms.
⚠️ You must be able to draw a valid resonance structure from a given starting structure. Practise pushing lone pairs into adjacent π bonds and vice versa until the arrow-pushing feels automatic.
True or false: A carbon atom with one double bond and two single bonds is sp3 hybridised. (False, it is sp2.)
Fill in the blank: The π bond in a C=O double bond is formed by the overlap of a ____ orbital on carbon and a ____ orbital on oxygen. (p, p)
True or false: If the sum of formal charges in a Lewis structure does not equal the overall molecular charge, you have made an error. (True.)
Fill in the blank: The resonance form with the most ____ bonds and the fewest ____ charges is the major contributor. (covalent, formal)
True or false: Moving a hydrogen atom from one position to another counts as drawing a resonance structure. (False, that is a constitutional isomer.)
Q: What is the hybridisation of a carbon atom that forms one double bond and two single bonds?
A: sp2. Three electron domains (one double bond + two single bonds) give trigonal planar geometry with ~120° bond angles.
Q: In a C=O bond, what type of orbital overlap forms the σ component and what forms the π component?
A: The σ bond is sp2-sp2 overlap (head-on). The π bond is p-p overlap (side-by-side).
Q: A nitrogen atom has four bonds and no lone pairs. What is its formal charge?
A: +1. Nitrogen has five valence electrons; four bonds use four electrons, leaving a formal charge of 5 - 0 - 4 = +1.
Q: You are given two resonance structures. One has no formal charges and full octets. The other has a +1 on nitrogen and a -1 on carbon. Which is the major contributor?
A: The one with no formal charges. Fewer charges and full octets make it the more stable, more significant contributor.
Q: When drawing a resonance structure for an amide (R-CO-NH2), in which direction do you push the lone pair?
A: Push the nitrogen lone pair into the C-N bond to form a C=N double bond, and simultaneously break the C=O π bond so that oxygen picks up a lone pair and a negative formal charge. The result is a form with C=N (N carries +1) and C-O⁻.
Hybridisation and orbital overlap are the foundation for understanding molecular geometry (VSEPR), which in turn determines polarity, physical properties, and reactivity. Resonance is essential for predicting acid/base strength (a conjugate base stabilised by resonance is weaker, making the parent a stronger acid) and for understanding reaction mechanisms throughout organic chemistry, from electrophilic aromatic substitution to carbonyl chemistry.
Formal charge calculation connects directly to predicting reaction sites: nucleophiles attack atoms with positive or partial-positive character, and electrophiles are drawn to negative or electron-rich centres.
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