Difficulty: Introductory | Prerequisites: General chemistry (atomic orbitals, electron configuration, periodic trends).
Electronic structure is the foundation of organic chemistry. Before you can understand why molecules react, you need to know how atoms share electrons, what shape that sharing produces, and how shape determines polarity. This material bridges general chemistry (where you learnt about orbitals and electronegativity) and the organic reactions that come next. If you cannot assign hybridization or predict geometry on sight, every later topic, from resonance to conformational analysis, will feel harder than it needs to be.
Count an atom's bonding groups plus lone pairs to get its hybridization (sp, sp², sp³) and predict its geometry and bond angles. Sigma bonds form from head-on orbital overlap; pi bonds form from side-on p-orbital overlap. Molecular polarity depends on both bond polarity and molecular shape: symmetric molecules can be non-polar even with polar bonds.
Hybridization
The mixing of atomic orbitals on a single atom to form new, equivalent hybrid orbitals that better describe bonding geometry. In simple terms, it is the way an atom rearranges its orbitals to make the right number of bonds and lone pairs.
sp³ hybridization
Four hybrid orbitals arranged tetrahedrally (109.5° bond angles), formed from one s and three p orbitals. Think of it as the shape you get when an atom needs four groups around it, like carbon in methane.
sp² hybridization
Three hybrid orbitals arranged in a trigonal plane (120° bond angles), with one unhybridized p orbital remaining. This is what you see at any atom involved in a double bond, like the carbons in ethylene.
sp hybridization
Two hybrid orbitals arranged linearly (180° bond angles), with two unhybridized p orbitals remaining. This appears at atoms with two groups total (bonds plus lone pairs counted as groups), such as the carbon in a nitrile (C≡N).
Sigma (σ) bond
A bond formed by head-on overlap of orbitals along the internuclear axis. Think of it as the first bond between any two atoms, the strong backbone that holds them together.
Pi (π) bond
A bond formed by side-on overlap of unhybridized p orbitals above and below the internuclear axis. This is the second (and third) bond in double and triple bonds. It cannot exist without a sigma bond already in place.
Molecular polarity (dipole moment)
The overall measure of charge separation in a molecule, determined by both the polarity of individual bonds and the three-dimensional shape. In simple terms, a molecule is polar if its bond dipoles do not cancel out when you add them as vectors.
Intermolecular forces (IMF)
Forces of attraction between molecules (not within them). Stronger IMF means higher boiling points. The hierarchy runs: ionic > hydrogen bonding > dipole-dipole > London dispersion.
Count the steric number: the total number of atoms bonded to the atom plus lone pairs on that atom.
Steric number 4 → sp³ (tetrahedral, 109.5°)
Steric number 3 → sp² (trigonal planar, 120°)
Steric number 2 → sp (linear, 180°)
Example from the exam: in sulforaphane, sulfur S1 has three bonds and one lone pair (steric number 4), so it is sp³. The nitrogen has one lone pair and two bonds (steric number 3), so it is sp², giving a C-N-C bond angle of approximately 120°.
A single bond = one σ bond (head-on overlap of hybrid orbitals).
A double bond = one σ bond + one π bond.
A triple bond = one σ bond + two π bonds.
The σ bond uses each atom's hybrid orbitals. The π bond uses each atom's leftover unhybridized p orbitals.
Example from the exam: in the C=N bond of sulforaphane, the σ bond is sp (from C) overlapping with sp² (from N). The π bond is p (from C) overlapping with p (from N). Carbon is sp because it has two groups (double bond to N and single bond to S). Nitrogen is sp² because it has three groups (double bond to C, single bond to the chain, and one lone pair).
A molecule is polar if it has polar bonds and those bond dipoles do not cancel.
Symmetry is the deciding factor. Linear symmetric molecules like CO₂ (O=C=O) and CS₂ (S=C=S) have polar bonds, but the dipoles point in opposite directions and cancel. The molecule is non-polar overall.
Bent molecules like SO₂ have polar bonds that do not cancel, because the geometry is asymmetric. SO₂ is the most polar of the set {O₂, CO₂, CS₂, SO₂}.
O₂ has a non-polar bond to begin with (identical atoms), so it has no dipole at all.
Boiling point depends on the strength of intermolecular forces.
The hierarchy: ionic compounds > hydrogen bonding > dipole-dipole > London dispersion forces.
Example from the exam: among CF₄, NF₃, OF₂, and LiF, the answer is LiF. It is an ionic compound, so it has the strongest intermolecular forces (electrostatic attraction between ions) and therefore the highest boiling point.
CF₄ is non-polar (symmetric tetrahedral), so it has only London dispersion forces. NF₃ and OF₂ are polar, giving them dipole-dipole interactions, but neither matches ionic bonding.
Steric number = (number of bonded atoms) + (number of lone pairs)
Steric number 4 → sp³ → tetrahedral → 109.5°
Steric number 3 → sp² → trigonal planar → 120°
Steric number 2 → sp → linear → 180°
Formal charge = (valence electrons) - (lone pair electrons) - (½ × bonding electrons)
Hybridization and molecular geometry explain why water is bent (sp³ oxygen with two lone pairs), which is why ice floats and life on Earth works the way it does. Molecular polarity determines solubility ("like dissolves like"), which is central to drug design: a drug must be polar enough to dissolve in blood but non-polar enough to cross cell membranes.
Students often assume that any molecule with polar bonds must be polar overall. It need not be. If the geometry is symmetric (linear, tetrahedral with identical substituents), the bond dipoles cancel and the molecule is non-polar. CO₂ and CF₄ are the classic examples.
Students sometimes forget to count lone pairs when determining hybridization. A lone pair occupies a hybrid orbital and counts toward the steric number, even though it does not show up as a bond line in a structural drawing.
A common mix-up: thinking that double bonds use two sigma bonds. They do not. A double bond is always one sigma plus one pi. A triple bond is one sigma plus two pi.
Students confuse bond polarity with molecular polarity. Bond polarity is about the electronegativity difference between two atoms. Molecular polarity is about whether all the bond dipoles in the molecule add up to a net dipole.
⚠️ Hybridization questions are nearly guaranteed. Be ready to assign hybridization to any atom in a structure, including atoms with lone pairs (like nitrogen and oxygen).
⚠️ Orbital overlap questions pair with hybridization. If an atom is sp², its σ bonds use sp² orbitals and its π bond uses the leftover p orbital. Know which orbital comes from which atom.
⚠️ Molecular polarity questions test whether you can connect geometry to dipole cancellation. Practise with the common pairs: CO₂ vs SO₂, BF₃ vs NF₃, CCl₄ vs CHCl₃.
⚠️ Boiling point questions test your grasp of intermolecular force hierarchy. Ionic beats everything; hydrogen bonding beats dipole-dipole; London dispersion is the weakest but always present.
True or false: an atom with three bonds and one lone pair is sp² hybridized. (False, it is sp³ because the steric number is 4.)
Fill in the blank: a double bond consists of one ____ bond and one ____ bond. (sigma; pi)
True or false: CO₂ is a polar molecule. (False, it is linear and the dipoles cancel.)
True or false: LiF has a higher boiling point than NF₃ because LiF is ionic. (True.)
Fill in the blank: the expected bond angle around an sp² atom is ____°. (120°)
Q: What is the hybridization of a carbon atom that forms one double bond and two single bonds?
A: sp². It has three groups around it (steric number 3), which requires three hybrid orbitals.
Q: In sulforaphane, the C=N bond has a σ component and a π component. What orbitals does each atom contribute to the σ bond?
A: Carbon contributes an sp orbital; nitrogen contributes an sp² orbital. (Carbon is sp because it has two groups total; nitrogen is sp² because it has three groups: the double bond, a single bond, and a lone pair.)
Q: Why is SO₂ polar but CO₂ is not, even though both contain polar bonds?
A: CO₂ is linear, so the two C=O dipoles point in exactly opposite directions and cancel. SO₂ is bent (the lone pair on sulfur forces a non-linear geometry), so the two S=O dipoles do not cancel.
Q: Rank the following in order of increasing boiling point: CF₄, NF₃, LiF.
A: CF₄ < NF₃ < LiF. CF₄ is non-polar (London dispersion only). NF₃ is polar (dipole-dipole). LiF is ionic (strongest forces).
Hybridization feeds directly into resonance: you need to know that a p orbital is available for π overlap before you can draw a valid resonance structure. It also underpins conformational analysis, because the geometry around sp³ carbons (tetrahedral) is what creates staggered and eclipsed conformations. Molecular polarity and intermolecular forces come back when you study solubility, chromatography, and reaction solvents later in the course.
Hybridization, sp3, sp2, sp, orbital overlap, sigma bond, pi bond, bond angle, tetrahedral, trigonal planar, linear, steric number, VSEPR, molecular geometry, molecular polarity, dipole moment, bond dipole, vector sum, intermolecular forces, IMF, ionic bonding, hydrogen bonding, dipole-dipole, London dispersion, van der Waals, boiling point, electronegativity, polar molecule, non-polar molecule, sulforaphane, SO2, CO2, CF4, LiF