Atomic Structure & Bonding

Atomic Orbitals & Electron Configuration

Part 1 of 7

Every reaction you will draw in Organic Chemistry is, underneath the curved arrows, a story about electrons moving between orbitals. Before you can predict where a nucleophile attacks or why a carbocation rearranges, you need a physically honest picture of where electrons actually live around an atom. That picture is the atomic orbital.

An orbital is not a tiny planetary orbit. It is a region of space — described by a wavefunction $\psi$ — where there is roughly a 90% probability of finding an electron. The square of the wavefunction, $\psi^2$, gives the electron probability density. This matters for orgo because bonding is orbital overlap: a $\sigma$ bond forms where two orbitals share electron density between two nuclei, and the shape of the orbital dictates the direction a bond can point.

The orbitals you will use constantly are:

• $s$ orbitals — spherical, no directional preference. One per shell.
• $p$ orbitals — two-lobed (dumbbell-shaped), aligned along the $x$, $y$, and $z$ axes. Three per shell from $n = 2$ upward.

For carbon and its neighbors (N, O, F), the $1s$, $2s$, and three $2p$ orbitals are the entire toolkit. Everything else in this course is built from how those five orbitals are filled, mixed, and overlapped.

Quantum Numbers: The Orbital's Address

Each orbital is specified by a set of quantum numbers, and three rules govern how electrons fill them. You do not need to solve the Schrödinger equation, but you must be fluent with the bookkeeping, because it tells you the number of valence electrons — and that number is the single most important fact about an atom in orgo.

Three filling rules:

1. Aufbau principle — fill the lowest-energy orbital available first ($1s$ before $2s$ before $2p$).
2. Pauli exclusion principle — no two electrons share all four quantum numbers, so any orbital holds at most two electrons, and they must have opposite spin.
3. Hund's rule — within a set of degenerate (equal-energy) orbitals like the three $2p$, electrons occupy them singly with parallel spins before any orbital is doubled up.

Hund's rule is why carbon, in its ground state, has two unpaired electrons in separate $2p$ orbitals — a fact we will need to confront the moment we ask how carbon makes four bonds.

Worked Example: The Ground-State Configuration of Carbon

Carbon has 6 electrons. Applying Aufbau, Pauli, and Hund in order:

• $1s^2$ — the first two electrons pair up in the lowest orbital (these are core electrons, chemically inert).
• $2s^2$ — the next two fill and pair the $2s$.
• $2p^2$ — the final two electrons enter the $2p$ set. By Hund's rule they go into two different $2p$ orbitals with parallel spin, not into the same one.

So the full configuration is $1s^2\,2s^2\,2p^2$, often written $[\text{He}]\,2s^2\,2p^2$.

The valence shell is $n = 2$: that is $2s^2\,2p^2$, a total of 4 valence electrons. Those four electrons are what carbon uses to bond.

Here is the puzzle that motivates the rest of this unit. The ground-state orbital diagram shows a filled $2s$ pair and only two unpaired $2p$ electrons:

$2s\!\uparrow\!\downarrow \quad 2p\!\uparrow \;\; 2p\!\uparrow \;\; 2p\;\;\;$

Reading this literally, carbon should form only 2 bonds (one per unpaired electron) — but methane is $\text{CH}_4$, with 4 identical bonds. Resolving that contradiction is exactly what promotion and hybridization (Part 3) accomplish. For now, hold onto the key count: carbon brings 4 valence electrons to the table.

Why carbon is special: With 4 valence electrons, carbon sits exactly halfway to a full octet. It has no strong drive to either lose electrons (like a metal) or grab them (like a halogen), so it overwhelmingly shares them in covalent bonds — and it can do so with up to four other atoms, including other carbons. That is the structural root of all of organic chemistry's diversity.

Checkpoint — Orbitals & Filling Rules

The Octet Rule and Why It Works

Second-row elements (C, N, O, F) are most stable when surrounded by eight valence electrons — a filled $2s$ plus filled $2p$ set, the configuration of neon. This is the octet rule, and it is the reason Lewis structures (Part 2) are drawn the way they are.

The octet rule has a hard physical ceiling for the second row: there are only four valence orbitals available ($2s$ + three $2p$), and each holds two electrons, so $4 \times 2 = 8$ is the maximum. This is why carbon, nitrogen, oxygen, and fluorine cannot exceed an octet — they have no $2d$ orbitals to expand into. Memorize this: an "expanded octet" on carbon is always a mistake.

Counting toward the octet, every bond and lone pair contributes:

• Each shared (bonding) pair counts as 2 electrons toward the octet of both atoms it connects.
• Each lone pair counts as 2 electrons toward only the atom that owns it.

Carbon reaching an octet with four single bonds (as in $\text{CH}_4$) shares $4 \times 2 = 8$ electrons — a complete octet, no lone pairs. Nitrogen, with 5 valence electrons, typically forms 3 bonds plus 1 lone pair ($3\times2 + 2 = 8$); oxygen, with 6, forms 2 bonds plus 2 lone pairs ($2\times2 + 4 = 8$). These "default" bonding patterns — C makes 4, N makes 3, O makes 2, H and halogens make 1 — are the scaffolding you will lean on for every structure in this course.

Checkpoint — The Octet Rule

Exit Ticket — Part 1 Synthesis

Lewis Structures & Formal Charge

Part 2 of 7

A Lewis structure is the working diagram of organic chemistry. It shows every bond, every lone pair, and — once you add formal charges — exactly where a molecule is electron-rich (nucleophilic) and electron-poor (electrophilic). You will draw thousands of these. Drawing them correctly and quickly is a non-negotiable skill, because a wrong Lewis structure produces a wrong reaction mechanism every single time.

A reliable five-step procedure:

1. Count total valence electrons. Sum the valence electrons of every atom. For an anion, add one electron per negative charge; for a cation, subtract one per positive charge.
2. Place the skeleton. Carbon is almost always central; hydrogen and halogens are always terminal (one bond each). The least electronegative atom (other than H) tends to be central.
3. Connect with single bonds, then distribute remaining electrons as lone pairs, completing octets on the outer atoms first.
4. Form multiple bonds by converting lone pairs into bonding pairs wherever a central atom is short of an octet.
5. Assign formal charges to check that you have the best structure.

Steps 1 and 5 are where students lose points, so we will drill the electron count and formal charge hardest.

Formal Charge: The Bookkeeping That Reveals Reactivity

Formal charge answers the question: relative to a free, neutral atom, did this atom gain or lose ownership of electrons when it joined the molecule? The rule for counting "owned" electrons is the heart of it: an atom owns all of its lone-pair electrons but only half of each bonding pair (the other half belongs to its bonding partner).

$\text{FC} = (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons})$

A faster, equivalent form many people prefer:

$\text{FC} = (\text{valence electrons}) - (\text{lone pairs} \times 2) - (\text{number of bonds})$

The second form works because half of the bonding electrons equals exactly the number of bonds (each bond = 2 shared electrons, half of which is 1).

Three anchor cases to memorize cold, because they recur constantly:

• Oxygen with 3 bonds + 1 lone pair (as in $\text{H}_3\text{O}^+$): $\text{FC} = 6 - 2 - 3 = +1$.
• Oxygen with 1 bond + 3 lone pairs (as in hydroxide $\text{OH}^-$): $\text{FC} = 6 - 6 - 1 = -1$.
• Nitrogen with 4 bonds + 0 lone pairs (as in ammonium $\text{NH}_4^+$): $\text{FC} = 5 - 0 - 4 = +1$.

Note the pattern: an atom with more bonds than its neutral default tends to be positive; an atom with fewer bonds (and extra lone pairs) tends to be negative. The sum of all formal charges must equal the overall charge on the species.

Worked Example: Formal Charges in the Nitrate Ion ($\text{NO}_3^-$)

Step 1 — Count electrons. N contributes 5, each O contributes 6, and the $-1$ charge adds 1: $5 + (3 \times 6) + 1 = 24$ valence electrons.

Step 2-4 — Build it. Nitrogen is central, bonded to three oxygens. To give nitrogen an octet we use one $\text{N=O}$ double bond and two $\text{N-O}$ single bonds. The two single-bonded oxygens each carry 3 lone pairs; the double-bonded oxygen carries 2 lone pairs.

Step 5 — Assign formal charges:

• Nitrogen: 4 bonds (one double counts as 2 bonds), 0 lone pairs. $\text{FC} = 5 - 0 - 4 = +1$.
• Double-bonded O: 2 bonds, 2 lone pairs. $\text{FC} = 6 - 4 - 2 = 0$.
• Each single-bonded O: 1 bond, 3 lone pairs. $\text{FC} = 6 - 6 - 1 = -1$.

Check: $(+1) + (0) + (-1) + (-1) = -1$. This matches the overall charge of the ion — the structure is valid.

Now the deeper point. We chose which oxygen got the double bond, but that choice is arbitrary: the three oxygens are chemically identical. The single Lewis structure is therefore a lie of convenience — the real ion has the negative charge and the double-bond character spread evenly across all three oxygens. That delocalization is resonance, and the "best structure" rules below are how we reason about which contributing structures matter most.

Checkpoint — Calculating Formal Charge

Choosing the "Best" Lewis Structure

When more than one valid Lewis structure can be drawn for a species (different placements of bonds and charges), they are resonance contributors, and they are not equally important. The real molecule resembles the lowest-energy, most stable contributors most closely. Rank contributors by these criteria, in order of priority:

1. Complete octets win. A structure where every second-row atom (especially carbon) has a full octet is far better than one with an electron-deficient atom. This rule outranks all others.
2. Minimize formal charges. Fewer atoms bearing nonzero formal charge means a more stable contributor. The ideal is zero formal charge everywhere.
3. Negative charge on the most electronegative atom. If a negative charge must exist, it is most stable on the atom best able to hold it (O over N over C). Conversely, positive charge prefers the less electronegative atom.
4. Avoid like charges on adjacent atoms and avoid large charge separation.

These same rules let you reject impossible structures (e.g., a carbon with only 6 electrons when a better octet structure exists) and identify the dominant contributor that controls a molecule's behavior.

Crucial distinction: Resonance contributors are not different molecules in equilibrium, and the molecule does not flicker between them. They are simply incomplete sketches of one real, delocalized structure. The molecule is a single resonance hybrid — a weighted average — at all times. Part 7 returns to resonance in depth; here, just internalize that "best structure" means "most stable contributor," ranked by octets first, then formal charge.

Checkpoint — Best Structure & Resonance

Exit Ticket — Part 2 Synthesis

Hybridization

Part 3 of 7

In Part 1 we hit a contradiction: ground-state carbon ($1s^2\,2s^2\,2p^2$) has only two unpaired electrons, yet methane has four identical bonds in a perfect tetrahedron. Hybridization is the model that resolves this — and it is arguably the single most useful concept in all of orgo, because hybridization at a carbon predicts its geometry, its bond angles, its bond strengths, and even its acidity.

The idea, in two moves:

1. Promotion. One $2s$ electron is promoted into the empty $2p$ orbital, giving carbon four singly-occupied valence orbitals: $2s^1\,2p^1\,2p^1\,2p^1$. The small energy cost of promotion is repaid many times over by forming two extra bonds.
2. Mixing. The pure $s$ and $p$ orbitals — which point in incompatible directions and have different energies — are mathematically combined (hybridized) into a new set of equivalent orbitals that point toward the corners of a regular shape and are perfectly suited for bonding.

The number of atomic orbitals you mix always equals the number of hybrid orbitals you get out (orbitals are conserved). Mix 1 $s$ + 3 $p$ and you get four $sp^3$ orbitals; mix 1 $s$ + 2 $p$ and you get three $sp^2$ (leaving one pure $p$ untouched); mix 1 $s$ + 1 $p$ and you get two $sp$ (leaving two pure $p$).

The Three Hybridization States of Carbon

The fastest way to assign hybridization in practice is to count regions of electron density — that is, the number of $\sigma$ bonds plus lone pairs (a multiple bond counts as just one region, because its $\pi$ component does not change the geometry). This steric number maps directly onto hybridization:

What the leftover pure $p$ orbitals do is the whole point for orgo:

• $sp^3$ carbon (e.g., methane, the carbon in an alkane): four $\sigma$ bonds, no leftover $p$, so no $\pi$ bonds are possible. This is a saturated, tetrahedral carbon.
• $sp^2$ carbon (e.g., an alkene carbon, a carbonyl carbon): three $\sigma$ bonds in a plane, plus one perpendicular $p$ orbital that forms a $\pi$ bond (the second bond of a C=C or C=O double bond).
• $sp$ carbon (e.g., an alkyne carbon, a nitrile carbon): two $\sigma$ bonds at $180^\circ$, plus two perpendicular $p$ orbitals that form two $\pi$ bonds (the extra two bonds of a triple bond).

So the rule of thumb that ties it together: single bond region $\rightarrow$ all $\sigma$; a double bond hides one $\pi$; a triple bond hides two $\pi$ — and every $\pi$ bond demands a leftover pure $p$ orbital, which only $sp^2$ and $sp$ carbons have.

Worked Example: %s-Character, Bond Length, and Acidity

Hybridization is not just geometry — it changes the energy of the electrons in the hybrid orbital, with real chemical consequences. The key parameter is %s-character, the fraction of the hybrid that comes from the low-lying, tightly-held $s$ orbital:

• $sp^3$: 1 part $s$ out of 4 total $\rightarrow$ 25% s-character
• $sp^2$: 1 part $s$ out of 3 $\rightarrow$ 33% s-character
• $sp$: 1 part $s$ out of 2 $\rightarrow$ 50% s-character

More s-character means the orbital is lower in energy and held closer to the nucleus. Two consequences you must be able to reason through:

1. Bond length / strength. A bond using an $sp$ orbital is shorter and stronger than one using $sp^3$, because the electrons sit closer to the nucleus. This is why a $\equiv\text{C-H}$ bond is shorter than a $=\text{C-H}$, which is shorter than a $-\text{C-H}$.

2. Acidity of terminal alkynes. Compare the C-H bonds of ethane ($sp^3$), ethene ($sp^2$), and ethyne ($sp$). When that C-H is deprotonated, the lone pair left behind sits in the carbon hybrid orbital. An $sp$ orbital (50% s) holds that lone pair closest to the nucleus and most stably, so the conjugate base (the carbanion) is most stable for the $sp$ case.

$\text{p}K_a:\quad \text{ethane }(sp^3)\approx 50 \;>\; \text{ethene }(sp^2)\approx 44 \;>\; \text{ethyne }(sp)\approx 25$

A lower $\text{p}K_a$ means a stronger acid. Terminal alkynes are dramatically more acidic than alkenes or alkanes entirely because of the s-character of the carbon orbital holding the resulting lone pair. This single line of reasoning — more s-character $\rightarrow$ more stable carbanion $\rightarrow$ stronger acid — is a favorite exam target.

Checkpoint — Assigning Hybridization

Checkpoint — s-Character & Its Consequences

Exit Ticket — Part 3 Synthesis

Sigma & Pi Bonds: Orbital Overlap and Molecular Orbitals

Part 4 of 7

Hybridization (Part 3) told us the shape of a carbon's bonding orbitals. This part tells us what happens when those orbitals overlap to actually form a bond — and why the two flavors of covalent bond, $\sigma$ (sigma) and $\pi$ (pi), behave so differently. The distinction between $\sigma$ and $\pi$ is the engine behind almost every reaction of alkenes, alkynes, and carbonyls.

Molecular orbital (MO) theory is the underlying framework. When two atomic orbitals overlap, they combine into two molecular orbitals (orbitals are conserved — two in, two out):

• A bonding MO, formed by constructive (in-phase) overlap. It concentrates electron density between the nuclei, is lower in energy than the original atomic orbitals, and holds the bond together.
• An antibonding MO (marked with an asterisk, e.g. $\sigma^*$), formed by destructive (out-of-phase) overlap. It has a node between the nuclei, is higher in energy, and weakens or breaks the bond when occupied.

Electrons fill the bonding MO first. A stable bond exists only when more electrons occupy bonding MOs than antibonding MOs.

Sigma vs. Pi: It Is All About How the Orbitals Meet

The difference between $\sigma$ and $\pi$ bonds is purely geometric — the direction of overlap:

Sigma ($\sigma$) bond — head-on overlap. Orbitals overlap end-to-end, directly along the internuclear axis. The electron density is cylindrically symmetric around that axis. Every single bond is a $\sigma$ bond, and it is always the first bond between any two atoms. Sigma bonds form from $s$-$s$, $s$-$p$, hybrid-hybrid, or hybrid-$s$ overlap.

Because $\sigma$ density is symmetric about the bond axis, a $\sigma$ bond permits free rotation: you can twist one end relative to the other without breaking any overlap. This is why single bonds rotate freely at room temperature and alkanes adopt many conformations.

Pi ($\pi$) bond — side-on overlap. Two parallel, unhybridized $p$ orbitals overlap sideways, above and below the internuclear axis. The electron density sits in two lobes, one on each face of the bond, with a node in the plane of the nuclei. A $\pi$ bond is always the second (and third) bond of a multiple bond — never the first.

Because $\pi$ overlap depends on the two $p$ orbitals staying parallel, a $\pi$ bond locks rotation: twisting one end by $90^\circ$ destroys the overlap and breaks the bond. This rotational rigidity is why alkenes have cis/trans (E/Z) isomers — the double bond cannot rotate to interconvert them.

Worked Example: Counting $\sigma$ and $\pi$ Bonds, and Bond Order

A reliable counting rule for any structure:

• A single bond = 1 $\sigma$, 0 $\pi$.
• A double bond = 1 $\sigma$ + 1 $\pi$.
• A triple bond = 1 $\sigma$ + 2 $\pi$.

The first bond between two atoms is always the $\sigma$; any additional bonds are $\pi$.

Example — propyne, $\text{CH}_3\!-\!\text{C}\!\equiv\!\text{CH}$. Count every connection:

• 3 C-H bonds on the methyl group: $3\,\sigma$.
• 1 C-C single bond (methyl to the alkyne): $1\,\sigma$.
• The C≡C triple bond: $1\,\sigma + 2\,\pi$.
• 1 terminal ≡C-H bond: $1\,\sigma$.

Total: 6 $\sigma$ bonds and 2 $\pi$ bonds. Notice the triple bond contributed only one $\sigma$ (its first bond) and two $\pi$.

Bond order quantifies how many net bonding electron pairs hold two atoms together. From MO theory:

$\text{Bond order} = \frac{(\text{bonding electrons}) - (\text{antibonding electrons})}{2}$

For a triple bond like C≡C, bond order = 3 (one $\sigma$ + two $\pi$ = three shared pairs, none antibonding). Higher bond order means a shorter, stronger bond: a C≡C triple bond is shorter and stronger than a C=C double bond, which is shorter and stronger than a C-C single bond. The same MO formula explains why $\text{O}_2$ is paramagnetic — its MO diagram forces two electrons into separate antibonding $\pi^*$ orbitals with parallel spins, an outcome the simple Lewis dot structure completely misses.

Checkpoint — Sigma vs. Pi

Checkpoint — Counting & Bond Order

Exit Ticket — Part 4 Synthesis

Electronegativity, Bond Polarity & Dipole Moments

Part 5 of 7

A covalent bond shares electrons — but rarely equally. Electronegativity is an atom's pull on the shared electrons of a bond. When two bonded atoms differ in electronegativity, the electron density shifts toward the greedier atom, creating a polar bond: a separation of partial charge, written $\delta^-$ (partial negative, on the more electronegative atom) and $\delta^+$ (partial positive). This polarity is the origin of almost every intermolecular force and every "where does the nucleophile attack" question in orgo.

The trend on the periodic table, which you should know without a chart:

• Electronegativity increases left $\rightarrow$ right across a period (more nuclear charge pulling on the same shell).
• Electronegativity increases bottom $\rightarrow$ top up a group (valence electrons closer to the nucleus).
• The orgo "all-stars," roughly: $\text{F} > \text{O} > \text{N} \approx \text{Cl} > \text{C} \approx \text{H}$.

The key reference point for organic chemistry: carbon (2.5) and hydrogen (2.2) are nearly equal, so C-H bonds are essentially nonpolar. That is why hydrocarbon chains are greasy and hydrophobic. But the moment carbon bonds to O, N, or a halogen, a polar bond appears with carbon as the $\delta^+$ end — and that electron-poor carbon becomes the electrophilic site that nucleophiles attack.

The Bonding Spectrum and the Dipole Moment

Bond character is a continuum, indexed by the electronegativity difference $\Delta\text{EN}$:

These cutoffs are guidelines, not hard walls — the point is the trend: the bigger the $\Delta\text{EN}$, the more the electrons localize on the electronegative atom, sliding from equal sharing toward full transfer.

The dipole moment ($\mu$) quantifies a bond's (or a molecule's) polarity. It is the product of the magnitude of the partial charge and the distance separating the charges:

$\mu = q \times d$

where $q$ is the partial charge and $d$ is the separation distance. Dipole moment is a vector — it has direction, conventionally drawn as an arrow pointing from $\delta^+$ toward $\delta^-$ (some textbooks use a crossed arrow with the cross on the positive end). Its unit is the debye (D).

The vector nature is the crux of the next idea: a molecule's overall dipole is the vector sum of its individual bond dipoles. Individual bonds can be very polar, yet if their dipole vectors point in opposing directions and cancel, the molecule has no net dipole at all.

Worked Example: Why Geometry Decides Molecular Polarity

To predict whether a molecule is polar, you need its shape, which comes from VSEPR (Valence Shell Electron Pair Repulsion): electron-density regions around a central atom arrange themselves as far apart as possible, giving exactly the geometries from hybridization — linear ($sp$, $180^\circ$), trigonal planar ($sp^2$, $120^\circ$), tetrahedral ($sp^3$, $109.5^\circ$). Then you add up the bond dipoles as vectors.

Case 1 — carbon dioxide, $\text{O=C=O}$. Each C=O bond is strongly polar (O is much more electronegative than C). But the central carbon is $sp$, so the molecule is linear: the two C=O dipoles point in exactly opposite directions and cancel. Net dipole = 0. $\text{CO}_2$ is nonpolar despite two very polar bonds.

Case 2 — water, $\text{H-O-H}$. Each O-H bond is polar. Oxygen is $sp^3$ with two lone pairs, so the molecule is bent ($\sim 104.5^\circ$), not linear. The two O-H dipoles do not oppose each other; they add to a large net dipole pointing toward the oxygen. Water is strongly polar — the reason it is an excellent solvent.

Case 3 — carbon tetrachloride, $\text{CCl}_4$. Each C-Cl bond is polar, but the carbon is $sp^3$ tetrahedral and all four substituents are identical chlorines arranged symmetrically. The four dipoles point toward the four corners and sum to zero. $\text{CCl}_4$ is nonpolar. Contrast with chloroform, $\text{CHCl}_3$: replacing one Cl with H breaks the symmetry, the dipoles no longer cancel, and the molecule becomes polar.

The lesson, in one line: polar bonds + symmetric geometry $\rightarrow$ nonpolar molecule; polar bonds + asymmetric geometry (or lone pairs that break symmetry) $\rightarrow$ polar molecule. You cannot judge molecular polarity from bonds alone — you must consider the 3-D shape.

Checkpoint — Bond Polarity & Electronegativity

Checkpoint — Molecular Polarity & Geometry

Exit Ticket — Part 5 Synthesis

Problem-Solving Workshop

Part 6 of 7

You now have all the tools: Lewis structures and formal charge (Part 2), hybridization and s-character (Part 3), $\sigma$/$\pi$ bonds and bond order (Part 4), and electronegativity, dipoles, and VSEPR geometry (Part 5). Real exam questions rarely test these in isolation — they hand you a structure and ask you to chain the ideas together. This workshop drills that integrated workflow on real organic molecules.

A master procedure for analyzing any structure handed to you:

1. Draw / verify the Lewis structure — correct electron count, octets satisfied.
2. Assign formal charges — locate any charged atoms; these flag reactive sites.
3. Count regions of electron density per central atom ($\sigma$ bonds + lone pairs) to assign hybridization and VSEPR geometry.
4. Identify $\sigma$ vs $\pi$ bonds — singles are $\sigma$; doubles add one $\pi$; triples add two $\pi$.
5. Evaluate polarity — bond dipoles from $\Delta\text{EN}$, then the vector sum given the geometry.

Work the examples below with paper and pencil before reading each resolution. Speed and accuracy here are what separate a confident orgo student from a struggling one.

Worked Problem 1: Full Analysis of Acetonitrile ($\text{CH}_3\text{CN}$)

Structure: $\text{CH}_3\!-\!\text{C}\!\equiv\!\text{N}\!:$ (a methyl group, then a carbon triple-bonded to nitrogen, with a lone pair on N).

Formal charges. Methyl carbon: 4 bonds, 0 LP $\rightarrow 4 - 0 - 4 = 0$. Nitrile carbon: 4 bonds (1 single + triple), 0 LP $\rightarrow 4 - 0 - 4 = 0$. Nitrogen: 3 bonds + 1 LP $\rightarrow 5 - 2 - 3 = 0$. All neutral; net charge 0. Good.

Hybridization (count regions).

• Methyl carbon: 4 $\sigma$ bonds (three C-H + one C-C), 0 LP = 4 regions $\rightarrow sp^3$, tetrahedral, $109.5^\circ$.
• Nitrile carbon: 2 regions (one $\sigma$ to methyl, one $\sigma$ to N — the triple bond is one region) = 2 regions $\rightarrow sp$, linear, $180^\circ$.
• Nitrogen: 2 regions (one $\sigma$ to C + one lone pair; the two $\pi$ bonds add no region) = 2 regions $\rightarrow sp$, linear.

Sigma vs pi. Three C-H ($\sigma$), one C-C ($\sigma$), and the C≡N (1 $\sigma$ + 2 $\pi$). Total: 5 $\sigma$ and 2 $\pi$.

Polarity. The C≡N bond is strongly polar (N more electronegative), and the linear nitrile end does not cancel it; acetonitrile has a large net dipole toward nitrogen. Polar molecule — indeed a common polar aprotic solvent.

Notice how every tool fed into the next: regions gave hybridization and geometry, bond multiplicity gave the $\sigma$/$\pi$ count, and geometry plus $\Delta\text{EN}$ gave the polarity.

Checkpoint — Integrated Analysis I

Worked Problem 2: The Carbonyl Group and a Formal-Charge Trap

Consider the carbonyl carbon in acetone, $(\text{CH}_3)_2\text{C=O}$ — the central carbon bonded to two methyl groups and double-bonded to oxygen.

Hybridization & geometry. The carbonyl carbon has 3 regions (two C-C $\sigma$ + one C=O, which is one region) = $sp^2$, trigonal planar, $120^\circ$. The leftover pure $p$ orbital on this carbon forms the $\pi$ bond to oxygen.

Bonds. The C=O is 1 $\sigma$ + 1 $\pi$. The oxygen, with 2 bonds + 2 lone pairs, is also part of the picture: regions = 2 (one $\sigma$ to C + 2 lone pairs... wait — count carefully: oxygen has one $\sigma$ bond and two lone pairs = 3 regions $\rightarrow sp^2$).

Polarity and reactivity — the payoff. Oxygen pulls the C=O electrons toward itself, so oxygen is $\delta^-$ and the carbonyl carbon is $\delta^+$. That electron-poor, $sp^2$, trigonal-planar carbon is wide open for a nucleophile to attack from above or below the plane — this is the foundation of every carbonyl reaction you will see (nucleophilic addition, substitution at acyl carbons, and more).

The trap. A tempting resonance contributor puts a full positive charge on carbon and a negative on oxygen ($\text{C}^+\!-\!\text{O}^-$ with a C-O single bond). Is it valid? Check octets: that contributor leaves carbon with only 6 electrons (an incomplete octet). By the "best structure" rules (Part 2), it is a minor contributor — the neutral $\text{C=O}$ double-bond structure dominates. But the minor contributor is not useless: it is exactly the picture that explains the carbon's $\delta^+$ electrophilicity. Knowing which contributor dominates (the octet one) and what the minor one teaches you (the reactivity) is the integrated skill being built.

Checkpoint — Integrated Analysis II

Exit Ticket — Part 6 Synthesis

Resonance, Delocalization & Capstone Review

Part 7 of 7

This final part does two things. First, it gives resonance — touched on in Parts 2 and 6 — the full treatment it deserves, because resonance is the concept students most often misunderstand and the one orgo leans on most heavily (acidity, stability, aromaticity, mechanism all depend on it). Second, it ties the whole unit together: structure determines properties, from the orbital all the way up to boiling point and reactivity.

The thesis of the unit: a molecule's behavior is dictated, in a clean chain, by

$\text{electron configuration} \rightarrow \text{Lewis structure / formal charge} \rightarrow \text{hybridization} \rightarrow \text{geometry} \rightarrow \text{polarity} \rightarrow \text{properties and reactivity}$

Resonance is what happens when a single Lewis structure cannot honestly capture the electron distribution — when electrons are delocalized over more than two atoms. Mastering it is the capstone skill of atomic structure and bonding.

Resonance: One Molecule, Several Sketches

Resonance structures (contributors) are two or more valid Lewis structures for the same arrangement of atoms that differ only in the placement of electrons — specifically $\pi$ electrons and lone pairs. The real molecule is none of them individually; it is the resonance hybrid, a single, lower-energy, weighted average of all contributors.

The non-negotiable rules of resonance:

1. Atoms never move. Only electrons (specifically $\pi$ bonds and lone pairs) are redistributed. $\sigma$ bonds and the molecular skeleton are fixed. If you moved an atom, you drew a different molecule, not a resonance structure.
2. The total electron count and overall charge are conserved across every contributor.
3. Contributors are connected by a double-headed arrow ($\leftrightarrow$), which means "these are pictures of one hybrid" — not an equilibrium arrow ($\rightleftharpoons$). The molecule does not interconvert between them.
4. Contributors are weighted by stability (the Part 2 best-structure rules: full octets first, then minimal formal charge, then negative charge on electronegative atoms). The more stable a contributor, the more it resembles the true hybrid.

Delocalization lowers energy. This is the chemical punchline: spreading electrons (and charge) over several atoms is stabilizing. A molecule or ion with significant resonance delocalization is lower in energy — more stable — than any single contributor suggests. This extra stability is called resonance (delocalization) energy, and it explains a huge amount of orgo:

• Carboxylic acids are far more acidic than alcohols because the carboxylate conjugate base spreads its negative charge over two equivalent oxygens (two equal contributors).
• Allylic and benzylic cations/radicals are unusually stable because the charge/radical is delocalized.
• Benzene's exceptional stability ("aromaticity") is resonance delocalization of six $\pi$ electrons over the ring.

Worked Example: Resonance in the Acetate Ion and How to Rank Contributors

Acetate, $\text{CH}_3\text{COO}^-$, is the conjugate base of acetic acid. Draw the carboxylate end: the carbon is bonded to the methyl group, double-bonded to one oxygen, and single-bonded to the other oxygen (which bears the $-1$ charge and three lone pairs).

The two contributors. Push the lone pair on the negative oxygen up to form a new $\pi$ bond, and simultaneously push the existing C=O $\pi$ electrons down onto that oxygen as a new lone pair. The result is a second, equivalent structure with the double bond and the negative charge swapped between the two oxygens. Because the two contributors are identical in stability (each: full octets, one $-1$ on an oxygen), they contribute equally.

The hybrid. The true acetate ion has:

• Two identical C-O bonds, each with a bond order of 1.5 (halfway between single and double).
• The $-1$ charge split evenly, $-\tfrac{1}{2}$ on each oxygen.

This equal, symmetric delocalization is precisely why acetate is so stable, and therefore why acetic acid is far more acidic than, say, ethanol (whose alkoxide conjugate base has no resonance to spread the charge).

Ranking contributors — the checklist. When contributors are not equivalent, rank them with the Part 2 rules, in priority order:

1. More complete octets (especially no electron-deficient carbon) — highest priority.
2. Fewer formal charges.
3. Negative charge on the more electronegative atom (and positive on the less electronegative).
4. Less charge separation.

The contributor highest on this list is the major contributor and dominates the hybrid; ones with incomplete octets or awkward charges are minor but can still illustrate reactivity (recall the carbonyl example in Part 6).

Checkpoint — Resonance Fundamentals

Checkpoint — Ranking Contributors & Delocalization

Exit Ticket — Unit Capstone Synthesis