Chapter 2 — Structure and Bonding: Lewis Structures, Hybridization, and Molecular Orbitals

"It is one of the most beautiful results of modern physics that from a few simple rules about electrons, we can deduce the architecture of every organic molecule." — Linus Pauling, The Nature of the Chemical Bond (1939)

Chapter 1 argued that organic chemistry is built on a single element — carbon — and claimed that the properties of carbon (tetravalency, strong $\sigma$ and $\pi$ bonds, moderate electronegativity, catenation) explain the character of the whole discipline. This chapter backs up that claim by showing why carbon has those properties. The answer lies in the behavior of electrons: specifically, in how carbon's six electrons arrange themselves around its nucleus and how those electrons reorganize when carbon forms bonds to other atoms.

By the end of the chapter you should be able to:

• Draw a Lewis structure for any simple organic molecule, showing every bond, every lone pair, and every formal charge correctly.
• Predict the geometry of any atom in any such molecule using VSEPR.
• Assign the hybridization ($sp^3$, $sp^2$, or $sp$) of every atom.
• Identify every $\sigma$ bond and every $\pi$ bond in the molecule.
• Draw the molecular orbital diagram of simple diatomic molecules ($H_2$, $O_2$, $CO$) and identify the HOMO and LUMO.
• Recognize when a molecule requires resonance structures for an accurate description and draw them.
• Predict the polarity of a bond from electronegativity differences and the overall dipole moment of a molecule.

These are the drawing skills on which the rest of the book depends. Every mechanism you will meet in Part III and beyond is drawn on top of an accurate Lewis structure. If you draw the Lewis structure wrong, the mechanism will be wrong. So this chapter matters.

2.1 A short review of atomic structure

You have met this material in general chemistry. We review only the parts needed for the rest of the chapter.

An atom consists of a dense, positively charged nucleus (protons and neutrons) surrounded by electrons in a much larger volume. The number of protons defines the element; the number of electrons in the neutral atom equals the number of protons. The number of neutrons determines the isotope but not the chemistry (except in mass spectrometry, Chapter 6).

Electrons do not orbit the nucleus in classical trajectories. They occupy orbitals — regions of space where the probability of finding the electron is high. Each orbital has a characteristic shape and a characteristic energy. The orbitals are organized into shells and subshells:

• The first shell contains one $s$ orbital (labeled $1s$).
• The second shell contains one $s$ orbital ($2s$) and three $p$ orbitals ($2p_x$, $2p_y$, $2p_z$).
• The third shell contains one $s$, three $p$, and five $d$ orbitals.
• And so on.

For the light atoms that appear in organic molecules (mostly hydrogen through chlorine), only the $s$ and $p$ orbitals matter. You will almost never think about $d$ orbitals in this book, except briefly when transition metals appear in Chapter 37.

The shapes of orbitals

An $s$ orbital is spherical. A $p$ orbital is a dumbbell with two lobes of opposite phase along a single axis, with a node (zero probability of finding the electron) at the nucleus. The three $p$ orbitals in a given shell lie along three perpendicular axes ($x$, $y$, $z$).

Figure 2.1 — The shapes of the atomic orbitals you will use constantly in organic chemistry. The $1s$ orbital is a small sphere; the $2s$ orbital is a larger sphere with a spherical node inside; each $2p$ orbital is a dumbbell with two lobes of opposite phase along one of the three axes. Phase (shown here as shading) will matter when we combine orbitals to form bonds.

The "phase" of an orbital — the fact that a $p$ orbital has a positive lobe on one side and a negative lobe on the other — does not refer to charge. The electron is still negative on both sides. It refers to the sign of the wavefunction, which is a mathematical property of the orbital derived from quantum mechanics. The reason we care: when two orbitals combine to form a bond, they must have the same phase at the region of overlap to produce a bonding interaction. Opposite phases produce an antibonding interaction. We will return to this in Section 2.8.

The Aufbau principle and valence electrons

Electrons fill orbitals from lowest energy to highest, two electrons per orbital, with the Pauli exclusion principle requiring opposite spins in the same orbital and Hund's rule requiring electrons to occupy separate orbitals of equal energy (degenerate orbitals) before pairing.

For the second-row elements of greatest interest in organic chemistry:

The valence electrons are those in the outermost shell — the only ones that participate in bonding. Chemistry is valence-electron chemistry. Inner-shell (core) electrons are spectators.

2.2 The covalent bond

A covalent bond is a pair of electrons shared between two atoms. The attraction that holds the bond together is electrostatic: both electrons are attracted to both positively charged nuclei simultaneously. The shared-pair picture, first proposed by Gilbert N. Lewis in 1916, remains the working picture of chemical bonding for most purposes in organic chemistry.

Two key quantitative features:

• Bond length — the equilibrium distance between the two nuclei, typically 1.0 to 1.5 Å for common organic bonds. (1 Å = $10^{-10}$ m = 100 picometers.)
• Bond dissociation energy (BDE) — the energy required to break the bond homolytically (each atom gets one electron), typically 60 to 150 kcal/mol.

Representative values for bonds that will appear constantly in this book:

Three patterns to notice:

1. Multiple bonds are shorter than single bonds. A $C=C$ is about 0.2 Å shorter than a $C-C$. A $C \equiv C$ is even shorter. The reason: multiple bonds include both $\sigma$ and $\pi$ contributions, and the additional $\pi$ bonds pull the nuclei closer.
2. Multiple bonds are stronger than single bonds, but not by a factor of two or three. The BDE of a $C=C$ (146) is not twice the BDE of a $C-C$ (83). It is about 63 kcal/mol more — which is approximately the strength of a single $\pi$ bond by itself. A $C=C$ is, roughly, a $C-C$ $\sigma$ bond plus a $\pi$ bond of about 63 kcal/mol.
3. Bonds to hydrogen are relatively strong. The $C-H$ BDE of 105 is higher than the $C-C$ BDE of 83. This is because hydrogen has no core electrons and its $1s$ orbital overlaps well with the partner orbital on carbon.

These bond energies will return constantly. When you want to know whether a reaction is thermodynamically favorable, you will often compute $\Delta H$ of the reaction by summing the bonds broken and the bonds formed. Chapter 5 does this in detail.

2.3 Lewis structures

A Lewis structure is a two-dimensional drawing that shows every atom in a molecule, every bond, every lone pair, and every formal charge. Drawing a Lewis structure correctly is the first skill of organic chemistry and the prerequisite for everything else in this book.

The procedure has four steps.

Step 1: Count the total valence electrons

Add up the valence electrons contributed by all atoms. For an ion, add electrons for each negative charge and subtract electrons for each positive charge.

Example: $H_2O$. - 2 hydrogens × 1 valence electron = 2 - 1 oxygen × 6 valence electrons = 6 - Total: 8 valence electrons.

Example: $NH_4^+$. - 1 nitrogen × 5 = 5 - 4 hydrogens × 1 = 4 - Adjust for $+1$ charge: subtract 1 - Total: 8 valence electrons.

Step 2: Arrange the atoms

Carbon atoms are almost always in the interior; hydrogens are always on the exterior (hydrogen forms only one bond). Heteroatoms (O, N, halogens) are usually between carbon and hydrogen. The overall skeleton is often suggested by the name or the formula; when it is ambiguous, standard heuristics apply: central atoms are usually the less electronegative ones (except hydrogen), and the structure is usually as symmetric as possible.

Step 3: Place the electrons

Place two electrons (one bonding pair) between each pair of bonded atoms. Then distribute the remaining electrons as lone pairs, starting with the most electronegative atoms (which prefer lone pairs), until every non-hydrogen atom has its octet filled — eight valence electrons surrounding it, counting shared pairs as contributing to both atoms.

If there are not enough electrons to give everyone an octet with single bonds, make double or triple bonds until everyone is satisfied.

Step 4: Check the formal charges

The formal charge on an atom in a Lewis structure is:

$$\text{formal charge} = (\text{valence electrons in the free atom}) - (\text{non-bonding electrons}) - \frac{1}{2}(\text{bonding electrons})$$

Equivalently: what is the "normal" number of bonds for this atom, and does it have that many?

Learning to spot formal charges by glance (by looking at bond counts) is faster than computing them with the formula every time. Make it a habit.

Worked examples of Lewis structures

Worked Problem 2.1 — Methane, $CH_4$

1. Valence electrons: $4 + 4(1) = 8$.
2. Carbon is the central atom; four hydrogens around it.
3. Four $C-H$ bonds use all 8 electrons.
4. Carbon has 4 bonds (normal for C); each H has 1 bond (normal for H). No formal charges.

Lewis structure:

H | H - C - H | H

Worked Problem 2.2 — Formaldehyde, $H_2CO$

1. Valence electrons: $4 + 2(1) + 6 = 12$.
2. Carbon is central, with two hydrogens and an oxygen around it.
3. If we connect with single bonds — two $C-H$ and one $C-O$ — we use $3 \times 2 = 6$ electrons. We have 6 left. We could give the oxygen 3 lone pairs (6 electrons), but then carbon has only 6 electrons (three bonds, no lone pair) and lacks its octet.
4. Make the $C-O$ into a $C=O$ double bond. Now: two $C-H$ single bonds (2 pairs), one $C=O$ double bond (2 pairs = 4 electrons). Total used: $2 + 2 + 4 = 8$. We have 4 electrons left — give them to oxygen as two lone pairs.
5. Now check: carbon has 4 bonds (normal). Oxygen has 2 bonds and 2 lone pairs (normal). No formal charges.

Lewis structure:

H H \ / C = O: ..

(The two dots on oxygen represent its two lone pairs.)

Worked Problem 2.3 — Hydroxide anion, $OH^-$

1. Valence electrons: $6 + 1 + 1 = 8$ (the extra 1 is for the negative charge).
2. Oxygen bonded to one hydrogen.
3. One $O-H$ bond uses 2 electrons. The remaining 6 electrons are three lone pairs on oxygen.
4. Oxygen has 1 bond and 3 lone pairs. Formula: formal charge = $6 - 6 - \frac{1}{2}(2) = -1$. So the oxygen bears a formal negative charge, consistent with the overall charge of the ion.

Lewis structure:

.. :: O — H (–)

Skeletal (line-angle) formulas — the drawings chemists actually use

Writing out every hydrogen and every lone pair gets tedious fast. Chemists abbreviate by drawing only the carbon skeleton as a zig-zag of lines and implicitly filling in the hydrogens:

• Each vertex and each line endpoint is a carbon.
• Each carbon has enough hydrogens attached to give it four bonds total.
• Heteroatoms (O, N, halogens) are drawn explicitly with their attached hydrogens.
• Lone pairs are usually omitted (implied by formal charges and by the normal bond count of the element).

Figure 2.2 — How to read a skeletal formula. The structure on the left is butane ($C_4H_{10}$); each vertex is a carbon, and the hydrogens are implied. The middle is 2-butanol ($C_4H_{10}O$), with the OH explicit. The right is acetone ($C_3H_6O$), with the carbonyl oxygen explicit.

Learn to read skeletal formulas fluently. Every structure in the rest of this book is drawn this way. Spending a few minutes with Figure 2.2 and copying the examples until the conversion feels natural is time very well spent.

Common Mistake 2.1

A student draws cyclohexane as a hexagon and counts: six carbons, zero hydrogens. Wrong. Every vertex is a carbon with enough hydrogens to reach four bonds. For cyclohexane ($C_6H_{12}$), each vertex has two hydrogens (because each carbon is bonded to two other carbons and to two hydrogens). For benzene ($C_6H_6$), each vertex has one hydrogen (each carbon has two bonds to neighbors in the ring plus one double bond, plus one hydrogen).

Always count hydrogens implicitly. Do not trust the drawing to show them.

2.4 Hybridization

Here is a problem you might have spotted already. Carbon's ground-state configuration is $1s^{2}2s^{2}2p^{2}$. Of its four valence electrons, two are in the $2s$ orbital (paired) and two are in the $2p$ orbitals (unpaired, one each in two different $2p$ orbitals by Hund's rule). So carbon has two unpaired electrons — and should therefore form two bonds, not four.

But carbon famously forms four bonds. What is going on?

The answer is hybridization: when carbon forms bonds, its valence orbitals rearrange. The $2s$ and three $2p$ orbitals mix to form new orbitals that are neither purely $s$ nor purely $p$ but something in between. The number of hybrid orbitals produced equals the number of atomic orbitals that went into the mixing. Three different hybridization schemes are possible for carbon, corresponding to three different bonding situations.

$sp^3$: four bonds, all single

When carbon forms four single bonds (as in methane), all four of its valence orbitals — the $2s$ and the three $2p$ orbitals — hybridize. The result is four equivalent $sp^3$ hybrid orbitals, each one-quarter $s$ character and three-quarters $p$ character. The four $sp^3$ orbitals point toward the corners of a regular tetrahedron, making angles of $109.5°$ with each other. Each one contains one electron.

These are the four bonding orbitals of $sp^3$ carbon. In methane, each of them overlaps with the $1s$ orbital of one hydrogen to form a $C-H$ bond. In ethane, three of them form $C-H$ bonds and the fourth forms a $C-C$ bond (overlapping with an $sp^3$ orbital on the other carbon).

$sp^3$ carbon is saturated, tetrahedral, and forms only $\sigma$ bonds. It has no $\pi$ bonds.

$sp^2$: three $\sigma$ bonds plus one $\pi$ bond

When carbon forms one double bond (as in ethylene, $CH_2=CH_2$), it hybridizes differently. Only the $2s$ and two of the three $2p$ orbitals mix, giving three $sp^2$ hybrid orbitals. The remaining $2p$ orbital (conventionally called $2p_z$) is left unhybridized.

The three $sp^2$ hybrid orbitals lie in a plane at $120°$ angles to each other. They form $\sigma$ bonds to three neighboring atoms. The unhybridized $2p_z$ orbital is perpendicular to that plane, with lobes above and below. When two $sp^2$ carbons are next to each other, their $2p_z$ orbitals overlap sideways to form a $\pi$ bond.

So in ethylene: - Three $\sigma$ bonds per carbon: two $C-H$ and one $C-C$ $\sigma$ bond (both $\sigma$ bonds from $sp^2$-$s$ or $sp^2$-$sp^2$ overlap). - One $\pi$ bond between the two carbons, formed by side-by-side overlap of the two $2p_z$ orbitals.

The $C=C$ double bond is therefore one $\sigma$ bond plus one $\pi$ bond. Not two independent bonds; one of each kind.

Figure 2.3 — Hybridization in carbon. Top: $sp^3$ carbon with four equivalent tetrahedral hybrid orbitals (left: methane, right: ethane). Middle: $sp^2$ carbon with three planar hybrid orbitals at 120° and one unhybridized $2p_z$ perpendicular to the plane; two $sp^2$ carbons form a $\sigma$ bond through their hybrids and a $\pi$ bond through side-by-side overlap of their $2p_z$ orbitals (ethylene). Bottom: $sp$ carbon with two linear hybrid orbitals at 180° and two unhybridized p orbitals; two $sp$ carbons form a $\sigma$ plus two $\pi$ bonds (acetylene).

$sp$: two $\sigma$ bonds plus two $\pi$ bonds

When carbon forms a triple bond (as in acetylene, $HC \equiv CH$), only the $2s$ and one $2p$ orbital hybridize. The two $sp$ hybrid orbitals point in opposite directions at $180°$ (linear geometry). The two unhybridized $p$ orbitals ($2p_y$ and $2p_z$) are perpendicular to the $sp$ axis and to each other.

In acetylene: - Two $\sigma$ bonds per carbon: one $C-H$ and one $C-C$ $\sigma$ bond. - Two $\pi$ bonds between the two carbons, formed by side-by-side overlap of the $2p_y$ and the $2p_z$ pairs.

The $C \equiv C$ triple bond is one $\sigma$ bond plus two $\pi$ bonds.

How to assign hybridization by inspection

You do not need to re-derive the hybridization every time. There is a fast rule:

• Count the steric number of the atom = (number of $\sigma$-bonded neighbors) + (number of lone pairs).
• Steric number 4 → $sp^3$ (tetrahedral).
• Steric number 3 → $sp^2$ (trigonal planar).
• Steric number 2 → $sp$ (linear).

Count bonded neighbors, not bonds: a $C=O$ counts as one neighbor (the oxygen), not two. A $C \equiv N$ counts as one neighbor (the nitrogen). The double and triple bonds contribute $\pi$ bonds that do not change the steric number.

This rule also extends to heteroatoms. Oxygen in water has steric number $2 + 2 = 4$ (2 hydrogens bonded, 2 lone pairs) → $sp^3$. Nitrogen in ammonia has steric number $3 + 1 = 4$ → $sp^3$. Oxygen in a carbonyl has steric number $1 + 2 = 3$ → $sp^2$. Nitrogen in a nitrile has steric number $1 + 1 = 2$ → $sp$.

Worked Problem 2.4 — Hybridization of each atom in acetone

Acetone is $(CH_3)_2C=O$.

• The two methyl carbons each have four $\sigma$-bonded neighbors (one carbon, three hydrogens), no lone pairs. Steric number 4 → $sp^3$.
• The central (carbonyl) carbon has three $\sigma$-bonded neighbors (two carbons and one oxygen via the $\sigma$ component of the $C=O$), no lone pairs. Steric number 3 → $sp^2$.
• The oxygen has one $\sigma$-bonded neighbor (the carbonyl carbon) and two lone pairs. Steric number 3 → $sp^2$.

So acetone has two $sp^3$ carbons, one $sp^2$ carbon, and one $sp^2$ oxygen. The three $sp^2$ atoms form a trigonal-planar arrangement; the two $sp^3$ methyl carbons are tetrahedral.

2.5 VSEPR and molecular geometry

Valence shell electron-pair repulsion (VSEPR) theory says: electron pairs around a central atom — bonding pairs or lone pairs — repel each other and arrange themselves to be as far apart as possible. Combined with hybridization, this gives a procedure for predicting the 3D shape of any molecule.

The predictions match the hybridization assignments directly:

However — and this is important — the molecular geometry (the shape made by the atoms) differs from the electron geometry (the shape made by all electron pairs) when there are lone pairs. Lone pairs are invisible to someone looking at the atoms, but they still take up space.

• In ammonia ($NH_3$), the steric number is 4 ($sp^3$, electron geometry tetrahedral), but only three positions are occupied by atoms (three hydrogens). The fourth is a lone pair. The molecular geometry is trigonal pyramidal, and the H-N-H angles are about $107°$ — slightly less than the tetrahedral $109.5°$, because the lone pair takes up slightly more space than a bonded hydrogen.
• In water ($H_2O$), steric number is 4, but only two positions are occupied by atoms (two hydrogens), and two are lone pairs. The molecular geometry is bent (V-shaped), and the H-O-H angle is about $104.5°$.

Memorizing the nine or so common geometries (linear, bent, trigonal planar, trigonal pyramidal, tetrahedral, trigonal bipyramidal, seesaw, T-shaped, octahedral) is easier than it sounds. The first five are the ones that appear constantly in organic chemistry. The last four (all with steric number 5 or 6) require expanded-octet considerations that rarely come up for second-row atoms.

2.6 Resonance

Some molecules cannot be adequately described by a single Lewis structure. Consider the nitrate anion, $NO_3^-$. If you follow the procedure of Section 2.3, you end up with nitrogen double-bonded to one oxygen and single-bonded (with a formal negative charge) to the other two:

          O    (-)
          ‖
      (-) O — N — O (-)

But this picture makes a specific prediction: one of the N-O bonds should be shorter (it's the double bond) than the other two (single bonds). Experimentally, all three N-O bonds in nitrate are identical, with bond length 1.24 Å — exactly intermediate between a typical $N-O$ single (1.44 Å) and a typical $N=O$ double bond (1.20 Å).

The resolution is that nitrate is not accurately described by any single Lewis structure. It is described by a set of three resonance structures, each showing the double bond to a different oxygen. The actual molecule is a hybrid that averages over all three. No single structure is correct. The hybrid is.

Figure 2.4 — The three resonance structures of the nitrate anion. Each shows nitrogen with a formal positive charge and one oxygen double-bonded. The actual nitrate ion is a resonance hybrid in which the $\pi$ electrons are delocalized equally over all three N-O bonds. Each bond has a bond order of roughly $\frac{4}{3}$.

Rules for resonance structures

1. Resonance structures differ only in the placement of electrons, never in the placement of atoms. If you move an atom to draw a "resonance structure," you have drawn a constitutional isomer, not a resonance structure.

2. The actual molecule is a hybrid of the resonance structures, not a rapidly interconverting mixture. A molecule does not flip between its resonance structures. The electrons are simultaneously delocalized across all of them. The resonance structures are tools we use because we cannot easily draw a delocalized electron distribution with Lewis structures; they are not snapshots of the molecule at different times.

3. Resonance structures are drawn with a double-headed arrow ($\leftrightarrow$), not an equilibrium arrow ($\rightleftharpoons$). The distinction matters: $\leftrightarrow$ means "the same molecule, drawn in different ways"; $\rightleftharpoons$ means "two different molecules in equilibrium."

4. Not all resonance structures contribute equally. The most-stable resonance structures contribute most to the hybrid. Stability rules (in roughly decreasing importance): - Maximum number of octets filled. - Minimum number of formal charges. - Any formal negative charge on the most electronegative atom; any formal positive charge on the least electronegative atom. - Adjacent charges of the same sign are bad.

5. Curved arrows are used to go from one resonance structure to the next. Each curved arrow represents the movement of an electron pair (two electrons). We will use curved-arrow notation constantly from Chapter 10 onward.

Where resonance matters in organic chemistry

Resonance is crucial in several specific contexts:

• Carboxylate anions ($RCOO^-$) — the negative charge is distributed equally over both oxygens via two equivalent resonance structures. This delocalization stabilizes the anion relative to an alkoxide anion, which is why carboxylic acids are much more acidic than alcohols. We will use this fact dozens of times in Chapter 3 onward.
• Enolate anions (the deprotonated form of a carbonyl $\alpha$-carbon) — the negative charge is delocalized between the $\alpha$-carbon and the carbonyl oxygen. Chapter 27 is entirely about the consequences.
• Aromatic rings like benzene — six $\pi$ electrons delocalized over a six-membered ring. Chapter 20.
• Conjugated $\pi$ systems (alternating single and double bonds) — electrons flow over the whole conjugated region. Chapter 19.
• Amides — the nitrogen lone pair delocalizes into the carbonyl, giving amides partial double-bond character at the $C-N$ bond (and explaining why amides do not behave like typical amines in nucleophilicity).

Resonance will return in almost every chapter of the book. Get comfortable with the arrow-drawing conventions and the stability rules now.

2.7 Electronegativity and bond polarity

Two atoms bonded covalently rarely share their electrons equally. The atom with higher electronegativity — a measure of how strongly an atom attracts electron density in a bond — pulls the shared electrons toward itself, making the other atom partially positive.

The Pauling electronegativity scale (selected values for atoms common in organic chemistry):

A bond is called polar if the electronegativity difference between the two atoms is substantial (conventionally, more than about 0.4 on the Pauling scale). A bond with a large difference (more than about 2.0) is usually treated as ionic rather than covalent.

Polar bonds have a dipole moment — a vector pointing from the $\delta^+$ end to the $\delta^-$ end, with magnitude reflecting the size of the partial charges. The symbols $\delta^+$ and $\delta^-$ (Greek letter delta) denote partial positive and negative charges — not full unit charges.

The overall dipole moment of a molecule is the vector sum of all its individual bond dipoles plus any lone-pair contributions. A symmetric molecule like $CO_2$ (linear, $O=C=O$) has two $C=O$ bond dipoles pointing in opposite directions that cancel exactly — the molecule has no net dipole moment even though both bonds are highly polar. A molecule like water ($H_2O$, bent) has two $O-H$ dipoles that do not cancel — they add to give water its large net dipole moment of 1.85 D (debye).

Why bond polarity matters for reactivity

This is the key insight of Chapter 2 for the rest of the book: polar bonds are reactive. Polar bonds are where electrons go and come from.

When we eventually start writing mechanisms in Chapter 10, we will be tracking where electrons move. Electrons move from electron-rich regions (nucleophiles, lone pairs, $\pi$ bonds) to electron-poor regions (electrophiles, atoms bearing a $\delta^+$ charge, empty orbitals). The $\delta^+$ and $\delta^-$ labels on a polar bond tell us at a glance where the reactivity is.

• A $C-H$ bond is only slightly polar ($\Delta$EN = 0.35) and is mostly unreactive — this is why alkanes are famously inert.
• A $C=O$ bond is strongly polar ($\Delta$EN = 0.89) and is highly reactive — the carbon is strongly $\delta^+$ and is the target of every nucleophilic attack in Part VI.
• A $C-Cl$ bond is moderately polar ($\Delta$EN = 0.61) and is the site of the substitution and elimination reactions of Part III.

Spectroscopy Clue 2.1

Bond polarity shows up directly in infrared spectroscopy (Chapter 6). A polar bond absorbs infrared radiation strongly; a symmetric non-polar bond absorbs weakly or not at all. The $C=O$ stretch in a carbonyl is the most intense IR absorption in most organic spectra — a consequence of the very large $C=O$ dipole. A $C=C$ stretch in an alkene is much weaker, because alkenes are much less polar.

When you meet IR spectroscopy in Chapter 6, you will see that the intensity of each absorption is a clue to bond polarity. You are reading polarity from an experimental observable.

2.8 Molecular orbital theory — a short introduction

Lewis structures and hybridization give us a reliable picture for most everyday purposes. But they do not describe everything. Three questions they struggle with:

1. Why is $O_2$ a diradical? (Two unpaired electrons in separate orbitals.)
2. Why is the bond in $H_2^+$ (a one-electron bond) real?
3. Why are aromatic rings particularly stable?

The answer to all three is molecular orbital theory, which treats the electrons of a molecule as occupying molecular orbitals that extend over the whole molecule, not just between two specific atoms.

The core idea is simple. When two atomic orbitals (AOs) on adjacent atoms overlap, they combine to form two molecular orbitals (MOs): one bonding MO (lower in energy than the original AOs, with electron density concentrated between the atoms) and one antibonding MO (higher in energy, with a node between the atoms).

Two AOs in → two MOs out. Always.

Figure 2.5 — The molecular orbital diagram for $H_2$. Two $1s$ atomic orbitals (one on each hydrogen) combine to give two molecular orbitals: a low-energy bonding orbital ($\sigma_{1s}$) with electron density concentrated between the nuclei, and a high-energy antibonding orbital ($\sigma_{1s}^*$) with a node between them. The two electrons of $H_2$ fill the bonding orbital, giving a stable molecule with a net bond order of 1.

For the diatomic molecules of greatest interest:

• $H_2$: 2 electrons, both in $\sigma_{1s}$. Bond order = (bonding electrons − antibonding electrons)/2 = 1. Stable molecule.
• $He_2$: 4 electrons — 2 in $\sigma_{1s}$, 2 in $\sigma_{1s}^*$. Bond order = 0. No stable $He_2$ at ordinary temperatures.
• $O_2$: 16 electrons. The full MO treatment is more complex, but the result is that two electrons end up in a pair of degenerate $\pi^*$ orbitals, each singly occupied (by Hund's rule). This is why $O_2$ is paramagnetic (a diradical) — a fact that Lewis structures cannot easily explain.

HOMO and LUMO

Two MOs deserve special names:

• The HOMO (Highest Occupied Molecular Orbital) is the highest-energy MO that contains electrons. Its electrons are the most loosely held and the most available to donate to other molecules. The HOMO is where a molecule's nucleophilicity lives.
• The LUMO (Lowest Unoccupied Molecular Orbital) is the lowest-energy empty MO. It is the MO best able to accept electrons from another molecule. The LUMO is where a molecule's electrophilicity lives.

When two molecules react, the key interaction is usually HOMO-of-one with LUMO-of-the-other. This framework, called frontier molecular orbital theory, was elaborated by Kenichi Fukui (Nobel Prize, 1981). It will return in Chapter 19 (Diels-Alder), Chapter 35 (drug binding), and Chapter 39 (pericyclic reactions).

For now, the key point: MO theory and Lewis structures are complementary. Lewis structures give you a working picture of every molecule. MO theory fills in the quantitative details when Lewis structures fall short.

Computational Exercise 2.1

Install Avogadro if you have not already (Appendix E). Build an ethylene molecule ($CH_2=CH_2$) and optimize the geometry. Then, using the Extensions → Surfaces menu, compute and display the HOMO and LUMO of ethylene.

You should see that the HOMO is the $\pi$ bond — a pair of lobes above and below the plane of the molecule, one pair on each carbon. The LUMO is the $\pi^*$ antibonding orbital — similar lobes but with a node between the two carbons.

This is what a $\pi$ bond looks like. Spin the molecule and look at the HOMO from different angles. Notice that the $\pi$ electron density is above and below the molecular plane, not between the two carbons.

When we do the electrophilic addition of HBr to ethylene in Chapter 15, the first step will be the HOMO (the $\pi$ bond) attacking the LUMO of HCl (the $\sigma^*$ of the H-Cl bond). Seeing the HOMO here prepares you for that reasoning.

2.9 Summary and connections to Chapter 3

What you have seen in this chapter:

1. Atomic orbitals ($s$ and $p$) are probability distributions for electrons, with characteristic shapes and energies. The valence electrons of second-row atoms occupy $2s$ and $2p$ orbitals. Phase matters when orbitals combine.

2. Covalent bonds are shared pairs of electrons between two atoms. Bond strength and bond length depend on the atoms involved and on bond order (single, double, triple).

3. Lewis structures show every atom, every bond, every lone pair, and every formal charge. Drawing a Lewis structure is a four-step procedure: count valence electrons, arrange atoms, place bonds and lone pairs, check formal charges.

4. Hybridization explains how carbon (and other atoms) reorganize their valence orbitals when they form bonds. $sp^3$ for four $\sigma$ bonds (tetrahedral). $sp^2$ for three $\sigma$ bonds plus one $\pi$ bond (trigonal planar). $sp$ for two $\sigma$ bonds plus two $\pi$ bonds (linear). Steric number tells you which.

5. VSEPR predicts molecular geometry from steric number and lone-pair count. Electron geometry and molecular geometry differ when lone pairs are present.

6. Resonance is needed when a single Lewis structure cannot describe a molecule accurately. Resonance structures differ only in electron placement; the actual molecule is a hybrid.

7. Electronegativity drives bond polarity. Polar bonds carry $\delta^+$ and $\delta^-$ and are the sites of reactivity. The polarity pattern of a molecule tells you where electrons want to go when the molecule reacts.

8. Molecular orbital theory describes electrons as occupying orbitals that extend over whole molecules, not just between specific pairs of atoms. HOMO and LUMO are the frontier orbitals that govern most reactivity.

What is next

Chapter 3 is about acids and bases. Building on the polarity and resonance ideas of this chapter, we will develop the $pK_a$ framework — the master framework that predicts acidity, basicity, nucleophilicity, leaving-group ability, and equilibrium position in every chapter of the book. It is the single most important framework in organic chemistry, and it rests entirely on the structural ideas you have just learned.

Before proceeding to Chapter 3, check yourself:

• Can you draw a Lewis structure for any molecule of up to about 10 non-hydrogen atoms?
• Can you assign hybridization to every atom in that molecule?
• Can you predict its molecular geometry?
• Can you spot which bonds are polar?

If yes to all four, you are ready for Chapter 3. If not, practice with the exercises. The $pK_a$ framework we build next will use these skills constantly.

Common Mistake 2.2

"Carbon must be $sp^3$ when it has four bonds and $sp^2$ when it has three bonds, right?"

The rule is about steric number, not bond count. Carbon with four $\sigma$ bonds (and no lone pairs) is $sp^3$. Carbon with three $\sigma$ bonds and one $\pi$ bond (total of 4 bonds but steric number 3) is $sp^2$. The $\pi$ bond does not change hybridization; it uses the unhybridized $2p_z$ orbital.

Always count the steric number (neighbors + lone pairs), not the total bond count.