Difficulty: Introductory to Intermediate | Prerequisites: General chemistry (Lewis structures, electron configuration)
This material covers the first half of CHEM 225 (Organic Chemistry II) at UMich, focusing on the language of bonding: how to assign formal charges, count and classify bonds as sigma or pi, determine hybridisation states, and predict relative bond lengths and strengths. These are the ground-level tools you will use for every mechanism, spectroscopy problem, and synthesis question for the rest of the course. If you are comfortable drawing Lewis structures and assigning electron configurations from general chemistry, you are ready for this.
Organic chemistry runs on a small set of bonding rules. Formal charge tells you where electrons "belong" on a Lewis structure. Sigma bonds form from head-on orbital overlap, pi bonds from side-on overlap, and the hybridisation of an atom (sp, sp2, sp3) determines bond angles, geometry, and relative bond lengths. Triple bonds are shortest and strongest; single bonds are longest and weakest.
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, it tells you whether an atom "owns" more or fewer electrons than it would as a free, neutral atom.
Sigma (σ) bond
A bond formed by head-on (end-to-end) overlap of orbitals along the internuclear axis. Think of it as the first bond that forms between any two atoms. Every single bond is a sigma bond.
Pi (π) bond
A bond formed by side-on (lateral) overlap of parallel p orbitals. In simple terms, this is the "second" bond in a double bond and the "second and third" bonds in a triple bond. Pi bonds require unhybridised p orbitals, so they only form between sp2 or sp hybridised atoms.
Hybridisation
The mixing of atomic orbitals (s and p) to form new, equivalent hybrid orbitals. Three key states in organic chemistry:
sp3: four hybrid orbitals, tetrahedral geometry (109.5°), no pi bonds
sp2: three hybrid orbitals + one unhybridised p orbital, trigonal planar geometry (120°), one pi bond possible
sp: two hybrid orbitals + two unhybridised p orbitals, linear geometry (180°), up to two pi bonds possible
Bond order
The number of bonding electron pairs shared between two atoms. Single bond = 1, double bond = 2, triple bond = 3. Higher bond order means shorter and stronger bonds.
Bond length
The equilibrium distance between two bonded nuclei. Shorter bonds are stronger. Triple bonds are shorter than double bonds, which are shorter than single bonds between the same two atoms.
Lewis acid
A species that accepts an electron pair. In simple terms, anything with an empty or electron-deficient orbital. Examples: BeCl2, Al(OH)3, CH3+, BF3.
Lewis base
A species that donates an electron pair. Any molecule or ion with a lone pair available for donation.
The formula is straightforward:
\text{Formal Charge} = \text{Valence electrons} - \text{Lone pair electrons} - \frac{1}{2}(\text{Bonding electrons})Alternatively, count bonds instead of bonding electrons: FC = valence electrons - lone pair electrons - number of bonds.
Worked example from the exam (Q1)
The question gives four nitrogen-containing species and asks which has N with a formal charge of zero.
Compound I (pyridine-type ring): nitrogen has a lone pair and three bonds. FC = 5 - 2 - 3 = 0. Neutral nitrogen.
Compound II (tertiary amine ring with exocyclic bond): nitrogen has three bonds and a lone pair. FC = 5 - 2 - 3 = 0. Also neutral, but check the structure carefully for any additional bonds.
Compound III (N flanked by two oxygens in a ring, as in a morpholine-type system): if N has two bonds to O and one other bond plus a lone pair, FC = 0. But if N is doubly bonded, count again.
Compound IV (NaNH2): the nitrogen in the amide ion (NH2-) has one lone pair and two bonds. FC = 5 - 2 - 2 = +1? No. NH2 minus has two bonds and two lone pairs: FC = 5 - 4 - 2 = -1. This nitrogen carries a negative charge, so it is not zero.
The key insight: count bonds and lone pairs carefully. A neutral nitrogen in an organic molecule typically has three bonds and one lone pair, or (in a positively charged species) four bonds and no lone pair.
Quick rule of thumb
Neutral N: 3 bonds + 1 lone pair
Neutral O: 2 bonds + 2 lone pairs
Neutral C: 4 bonds + 0 lone pairs
If an atom has more bonds than its neutral count, it carries a positive formal charge
If it has fewer bonds (more lone pairs), it carries a negative formal charge
How sigma bonds form
Sigma bonds result from head-on overlap of any two orbitals along the bond axis. This includes:
s + s overlap (e.g. H-H)
s + sp3 overlap (e.g. C-H in methane)
sp3 + sp3 overlap (e.g. C-C in ethane)
sp2 + sp2 overlap (e.g. C-C sigma component in ethene)
sp + sp overlap (e.g. C-C sigma component in ethyne)
Every covalent bond has exactly one sigma bond. A double bond = 1 sigma + 1 pi. A triple bond = 1 sigma + 2 pi.
How pi bonds form
Pi bonds form from lateral overlap of unhybridised p orbitals. The p orbitals must be parallel to each other and perpendicular to the bond axis.
From the exam (Q12): pi bonds form between py and py orbitals (or px and px, as long as they are the same type and parallel). They do not form between sp2 and p, or between sp2 and sp2, because hybridised orbitals overlap head-on to make sigma bonds.
Counting sigma and pi bonds (Q10)
For C2H4 (ethylene):
The C=C bond contributes 1 sigma + 1 pi
Each C-H bond is 1 sigma (there are four C-H bonds)
Total sigma bonds: 1 (C-C) + 4 (C-H) = 5
Total pi bonds: 1
Answer: 5 sigma bonds
Counting pi bonds in line structures (Q3)
To identify pi bonds, look for double and triple bonds:
HCCCH2CH3 (structure I): the CC triple bond has 2 pi bonds. This one has two pi bonds.
H2CCCHCH3 (structure II): this contains C=C=C (an allene), which has 2 pi bonds. Also two.
H2CCHCHCH2 (structure III): two separate C=C double bonds, each contributing 1 pi bond. Also two.
Answer: all three have two pi bonds (answer D).
The relationship
Higher s-character in a hybrid orbital means electrons are held closer to the nucleus, which means shorter and stronger bonds.
sp3 has 25% s-character (1 part s, 3 parts p)
sp2 has 33% s-character (1 part s, 2 parts p)
sp has 50% s-character (1 part s, 1 part p)
So bond lengths follow this order (longest to shortest): sp3-sp3 > sp2-sp2 > sp-sp
Predicting shortest and longest bonds (Q6 and Q7)
The exam shows a conjugated chain with numbered bonds. To identify the shortest bond, look for the highest bond order (triple > double > single) and the most s-character in the hybridised atoms. To identify the longest bond, look for a single bond between two sp3 carbons.
For a structure like: (ring)-CH=CH-C(triple bond)CH
A C-C triple bond will be the shortest (bond order 3, sp hybridised carbons)
A C-C single bond between two sp3 carbons will be the longest
Among single bonds, one between sp3-sp3 carbons is longer than one between sp2-sp3, which is longer than sp2-sp2
Geometry from hybridisation (Q5 in Problem II)
sp3: tetrahedral, 109.5°. Example: the oxygen in an alcohol (two bonds + two lone pairs)
sp2: trigonal planar, 120°. Example: the carbonyl carbon (C=O), the nitrogen in an amide if its lone pair is delocalised
sp: linear, 180°. Example: a carbon in a triple bond or an allene central carbon
Lewis acids and hybridisation (Q11)
A Lewis acid accepts an electron pair, so it needs an empty or partially empty orbital.
BeCl2: Be has only two bonds, sp hybridised, with an empty p orbital. Lewis acid.
Al(OH)3: Al can expand its coordination. Lewis acid.
CH3+: carbocation, sp2 with an empty p orbital. Lewis acid.
NaBH4: the B in BH4 minus has a full octet (four bonds, no empty orbital). It donates hydride (H minus), making it a Lewis base, not a Lewis acid. Answer: NaBH4 is not a Lewis acid.
Students often think that a nitrogen atom with three bonds is always neutral. It is not: if that nitrogen also has a formal positive charge (four bonds, no lone pair) or negative charge (two bonds, two lone pairs), the bond count alone does not settle it. Always apply the formal charge formula.
Students often confuse sigma and pi bonds with single and double bonds. A double bond is not "two sigma bonds." It is one sigma plus one pi. A triple bond is one sigma plus two pi.
Students sometimes assume that pi bonds can form between any two orbitals. They cannot. Pi bonds require parallel, unhybridised p orbitals of the same type (both py, or both px). A hybrid orbital (sp, sp2, sp3) forms only sigma bonds.
Students often think that longer bonds are stronger. The opposite is true. Shorter bonds have greater orbital overlap and higher bond dissociation energies.
⚠️ Formal charge calculation appears on nearly every CHEM 225 exam. Expect at least one question asking you to identify the charged atom in a structure.
⚠️ Sigma and pi bond counting is a reliable exam staple. Be able to count both in any structure, including ones with triple bonds and allenes.
⚠️ Know the relationship: more s-character = shorter bond = stronger bond. This is tested through questions asking you to rank bond lengths.
⚠️ Lewis acid identification often trips students up. Remember: NaBH4 is a hydride donor (Lewis base). BH3 (not BH4 minus) would be a Lewis acid.
⚠️ Geometry questions (trigonal planar vs tetrahedral) frequently appear in the context of identifying atoms in complex molecules. You need to determine hybridisation first, then state the geometry.
True or false: A nitrogen atom with four bonds and no lone pairs has a formal charge of +1. (True)
Fill in the blank: A double bond consists of one ______ bond and one ______ bond. (sigma, pi)
True or false: An sp-hybridised carbon forms shorter bonds than an sp3-hybridised carbon. (True)
Fill in the blank: Pi bonds form by lateral overlap of ______ p orbitals. (parallel/unhybridised)
True or false: NaBH4 is a Lewis acid. (False, it is a Lewis base / hydride donor)
Q: How many sigma bonds are in C2H4 (ethylene)?
A: 5. There are four C-H sigma bonds and one C-C sigma bond. The C=C double bond contributes one sigma and one pi, but the question asks for sigma only.
Q: Which of the following can form pi bonds: sp2 + p, sp2 + sp2, py + py, or px + py?
A: py + py. Pi bonds require parallel p orbitals of the same orientation. sp2 orbitals form sigma bonds. px + py are perpendicular and cannot overlap laterally.
Q: A compound has molecular formula C5H14. How many constitutional isomers does it have? (Hint: check the degree of unsaturation.)
A: The maximum number of hydrogens for a five-carbon alkane (CnH2n+2) is C5H12. C5H14 has more hydrogens than a saturated hydrocarbon can accommodate, which is impossible for a structure containing only C and H. This is a trick question testing whether you recognise an impossible formula.
Q: In the structure HC triple-bond C-CH2-CH3, how many pi bonds are present?
A: 2. The triple bond contains one sigma bond and two pi bonds.
Q: Rank the following bonds from shortest to longest: C-C single bond (sp3-sp3), C=C double bond (sp2-sp2), C triple-bond C (sp-sp).
A: C triple-bond C (shortest) < C=C < C-C (longest). Higher bond order and more s-character both contribute to shorter bond length.
Formal charge feeds directly into resonance: you cannot draw valid resonance structures without correctly assigning formal charges to each atom. Hybridisation determines the geometry around each atom, which becomes critical when you study stereochemistry (cis/trans isomerism, chirality) later in the course. Bond strength and length relationships will reappear when you study reaction thermodynamics and kinetics: weaker bonds break more easily, which governs reactivity.
formal charge, formal charge calculation, sigma bond, pi bond, sigma vs pi, bond order, bond length, bond strength, hybridisation, sp3, sp2, sp, orbital overlap, Lewis acid, Lewis base, electron pair acceptor, electron pair donor, trigonal planar, tetrahedral, linear geometry, CHEM 225, organic chemistry II, UMich, constitutional isomers, degree of unsaturation, molecular formula