Covalent Bond

Covalent bonds hold together every amino acid, nucleotide, lipid, and drug on the MCAT. Reasoning at the level of electrons, geometry, polarity, and stereochemistry lets you handle unfamiliar molecules on test day. This guide builds from Lewis structures up through stereochemistry, each layer extending the one before.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Lewis Electron Dot Formulas

The Core Idea

Must know

Every covalent bond is a shared pair of electrons — each atom contributes one (or one donates both in a coordinate covalent bond). A Lewis structure shows all valence electrons explicitly: lines (bonding pairs) or dots (lone pairs). The goal is a full valence shell on every atom: 2 for H, 8 for most second-row atoms (the octet rule), with exceptions.

How to Draw Lewis Structures

Must know

1. Count total valence electrons (subtract 1 per positive charge, add 1 per negative charge).
2. Arrange atoms — least electronegative in the center (never H).
3. Connect with single bonds.
4. Distribute remaining electrons as lone pairs, completing terminal octets first, then the center.
5. Form multiple bonds if the central atom lacks an octet (convert adjacent lone pairs into bonds).

Octet Rule Exceptions

Must know

There are three categories: electron-deficient/incomplete octet ($\ce{BF3}$, B has 6 e⁻), odd-electron radicals ($\ce{NO}$, 11 e⁻), and expanded octet (period 3+ atoms like $\ce{PCl5}$, $\ce{SF6}$).

Worked Example — $\ce{NO2-}$: Total = N(5) + 2×O(6) + 1 = 18 e⁻. N central; single bonds to both O (4 e⁻); complete O octets (12 e⁻); remaining lone pair on N. N has only 6 e⁻, so convert one O lone pair to a double bond → $\ce{O=N-O-}$ with a lone pair on N.

Quick check: How many total valence electrons does $\ce{SO4^{2-}}$ have?

Answer: S(6) + 4×O(6) + 2 = 32 e⁻

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Resonance Structures

The Concept

Must know

When more than one valid Lewis structure can be drawn (differing only in electron placement, not atom position), these are resonance structures. The molecule does NOT flip between them — the real electron distribution is a weighted resonance hybrid, with properties intermediate between contributors. E.g., the C–O bonds in $\ce{CO3^{2-}}$ are all equal length, between a C=O double and C–O single bond.

Rules for Drawing

Must know

• Move electrons (lone pairs or π bonds), never atoms.
• All structures have the same electron count.
• Major contributors: more bonds, fewer charges; negative charge on the more electronegative atom.

Delocalized Electrons

Must know

Electrons in resonance are delocalized — spread over multiple atoms, which lowers molecular energy (resonance stabilization). Benzene is canonical: six π electrons delocalized over all six carbons give exceptional stability. Carboxylate ions ($\ce{COO-}$), amide bonds (N–C=O in peptides), and aromatic rings are key biological examples.

Quick check: In the resonance structures of benzene, what moves between structures?

Answer: The π electrons (double bonds) shift; no atoms move. All C–C bonds are equal (~140 pm), intermediate between single and double.

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Formal Charge

Why It Matters

Must know

Formal charge compares a molecule's electron distribution to the isolated neutral atoms. It identifies the most stable resonance structure and shows where charge density sits — relevant to reactivity and acid-base behavior.

The Formula

Must know

$\text{Formal Charge} = \text{(Valence } \mathrm{e}^-\text{)} - \text{(Lone pair } \mathrm{e}^-\text{)} - \frac{1}{2}\text{(Bonding } \mathrm{e}^-\text{)}$

The best Lewis structure minimizes formal charges, places negatives on the most electronegative atoms, and avoids adjacent same-sign charges.

Worked Example — $\ce{CO2}$ ($\ce{O=C=O}$): C: 4 − 0 − ½(8) = 0; each O: 6 − 4 − ½(4) = 0. All zero → best structure. A triple/single-bond alternative gives a +1 and −1, so it is less stable.

Quick check: In which resonance structure of $\ce{CO3^{2-}}$ would carbon have a formal charge of zero?

Answer: In the conventional structures with one C=O and two C–O single bonds, C has formal charge 0, and the two single-bonded O's each carry −1 (negatives on the most electronegative atom) — consistent with the −2 ion.

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Lewis Acids and Bases

The Concept

Must know

A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. This is the broadest framework — it does NOT require proton transfer and subsumes Arrhenius and Brønsted–Lowry. The bond formed is a coordinate covalent (dative) bond: both electrons come from the base.

Typical Lewis acids: electron-deficient atoms ($\ce{BF3}$, $\ce{AlCl3}$), metal cations ($\ce{Fe^{3+}}$, $\ce{Mg^{2+}}$), $\ce{H+}$. Typical Lewis bases: atoms with lone pairs ($\ce{NH3}$, $\ce{H2O}$, $\ce{F-}$), anions, π systems.

Passage-level

Metal ions in enzyme active sites (e.g., $\ce{Zn^{2+}}$ in carbonic anhydrase) act as Lewis acids accepting electron pairs from substrate.

Quick check: In $\ce{BF3 + NH3 -> F3B-NH3}$, identify the Lewis acid and base.

Answer: $\ce{BF3}$ is the acid (B is electron-deficient, accepts the lone pair); $\ce{NH3}$ is the base (N donates). A new N→B coordinate covalent bond forms.

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Partial Ionic Character and the Role of Electronegativity

Electronegativity and Bond Character

Must know

Electronegativity (EN) is an atom's ability to attract bonding electrons. Must know the relative order: F > O > N ≈ Cl > C > H (F ~4.0, O ~3.5, N ~3.0, C ~2.5, H ~2.1; memorize the trend, not exact numbers). Unequal sharing gives partial charges: the more electronegative atom is $\delta-$, the other $\delta+$.

Rules of thumb: ΔEN < 0.5 → nonpolar covalent; 0.5–1.7 → polar covalent; > ~1.7 → largely ionic (boundary is gradual).

Charge Distribution in Biomolecules

Must know

O–H bonds are highly polar (O δ−, H δ+), explaining the hydrogen-bond donor ability of alcohols, water, and carboxylic acids. C–H bonds (ΔEN ~0.4) are essentially nonpolar — why alkyl groups are hydrophobic.

Quick check: In an amide bond (N–C=O), which atom bears the greater electron density, and why?

Answer: Oxygen (most electronegative) draws the greatest density. N's lone pair delocalizes into the carbonyl, giving N partial positive character and O extra negative density. This resonance makes the amide planar with partial double-bond character — relevant to protein backbone rigidity.

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Dipole Moment

Vector Logic

Must know

A dipole moment ($\mu$) is a vector. The MCAT emphasizes the net molecular dipole moment = the vector sum of all bond dipoles.

$\mu = q \times d$

where $q$ is partial charge and $d$ is the separation; units are debyes (D). (You won't compute debye values quantitatively.)

Predicting Molecular Polarity

Know the logic

Even with polar bonds, a molecule is nonpolar if symmetric geometry cancels the dipoles. Nonpolar by symmetry: $\ce{CO2}$ (linear), $\ce{BF3}$ (trigonal planar), $\ce{CH4}$/$\ce{CCl4}$ (tetrahedral). Polar (no cancellation): $\ce{H2O}$ (bent), $\ce{NH3}$ (pyramidal), $\ce{CHCl3}$ (asymmetric).

Quick check: Is $\ce{CH2Cl2}$ polar?

Answer: Yes. Replacing two Cl with two H breaks the symmetry of $\ce{CCl4}$, so the bond dipoles no longer cancel — net dipole toward the Cl side (DCM is a polar aprotic solvent).

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σ and π Bonds

Physical Picture

Must know

A sigma (σ) bond comes from head-on orbital overlap along the internuclear axis; every single bond is a σ bond and σ bonds allow free rotation. A pi (π) bond comes from side-by-side overlap of parallel p orbitals above and below the axis. A double bond = 1σ + 1π; a triple bond = 1σ + 2π. π bonds do not allow rotation — rotating would break the parallel overlap.

MCAT relevance: π bonds are individually weaker than σ but restrict rotation — the basis of cis-trans isomerism, double-bond planarity, and ring/peptide rigidity.

Effect on Bond Length and Energy

Must know

More bonds between the same two atoms = shorter, stronger bond: single < double < triple in strength; the reverse in length. Bond energy does not simply scale with order — each π adds less than the σ (less efficient overlap). Reason from the trend; don't memorize values.

Quick check: Which has the shorter bond, $\ce{N2}$ or $\ce{O2}$?

Answer: $\ce{N2}$ (triple bond) is shorter and stronger than $\ce{O2}$ (double bond) — why atmospheric N₂ is so inert.

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Hybrid Orbitals and Molecular Geometry

Why Hybridization Exists

Must know

Pure s and p orbitals have the wrong geometry for observed bond angles. Hybridization mixes atomic orbitals on the same atom into equal-energy orbitals with optimal bonding geometry. E.g., C in $\ce{CH4}$ makes four equivalent 109.5° bonds — mixing 2s + three 2p gives four sp³ orbitals pointing to tetrahedron corners.

The Three Hybridizations

Must know

Assign by counting electron groups (σ bonds + lone pairs) on the central atom:

Hybrid	Orbitals mixed	Leftover p (for π)	Electron geometry	Examples
sp³	1s + 3p	0	Tetrahedral (109.5°)	$\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$
sp²	1s + 2p	1	Trigonal planar (120°)	$\ce{BF3}$, alkenes, carbonyl C, benzene
sp	1s + 1p	2	Linear (180°)	$\ce{CO2}$, alkynes, $\ce{HCN}$

The leftover p orbital on sp²/sp atoms is what forms π bonds and resonance delocalization.

VSEPR and Shape Prediction

Must know

VSEPR: electron groups (bonding + lone pairs) arrange to minimize repulsion. The key distinction:

• Electron geometry: counts all groups (including lone pairs) — sets the hybridization.
• Molecular geometry: describes only atom positions.

Lone pairs repel more than bonding pairs, compressing bond angles (e.g., $\ce{NH3}$ ~107°, $\ce{H2O}$ ~104.5° from 109.5°).

Reference (common AXE cases):

Electron groups	Lone pairs	Molecular geometry	Hybridization
2	0	Linear	sp
3	0	Trigonal planar	sp²
3	1	Bent	sp²
4	0	Tetrahedral	sp³
4	1	Trigonal pyramidal	sp³
4	2	Bent	sp³

Quick check: What is the hybridization of nitrogen in $\ce{NH3}$, and electron vs. molecular geometry?

Answer: N has 4 electron groups (3 bonds + 1 lone pair) → sp³ → tetrahedral electron geometry → trigonal pyramidal molecular geometry (~107°).

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Structural Formulas

Notation Conventions

Must know

For the tested atoms (H, C, N, O, F, S, P, Si, Cl), be able to read/draw: Lewis structures (all electrons), structural formulas (bonds as lines), condensed formulas ($\ce{CH3CH2OH}$), and skeletal (line-angle) formulas (carbons at vertices, C–H implied).

Normal valences: H = 1, C = 4, N = 3 (4 if charged), O = 2, halogens (F, Cl) = 1, Si = 4; S and P are variable (2/4/6 and 3/5).

Quick check: In a line-angle benzene (hexagon, alternating double bonds), how many hydrogens are on each carbon?

Answer: Each C is sp², forms 2 C–C bonds + 1 C–H → 1 H per carbon, 6 total ($\ce{C6H6}$).

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Multiple Bonding: Rigidity in Molecular Structure

Must know

Multiple bonds create rigidity because π bonds need parallel p orbitals — rotating breaks the overlap. This is the origin of geometric (cis-trans) isomerism and double-bond planarity.

Passage-level

Biological examples:

1. Peptide bond: resonance gives the N–C=O ~40% double-bond character → planar, restricted rotation (rotation remains at Cα via φ/ψ angles).
2. Aromatic side chains (Phe, Tyr, Trp): rigid planar π systems affect folding and stacking.
3. Unsaturated fatty acids: cis double bonds kink the chain, preventing tight packing → liquid at room temperature, increasing membrane fluidity.

Quick check: Why can't cis- and *trans-*2-butene interconvert at room temperature without a catalyst?

Answer: Interconversion requires breaking the π bond (rotating about C=C), which needs far more energy than is available thermally; a catalyst, heat, or light is required.

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Stereochemistry of Covalently Bonded Molecules

Isomers: Overview

Must know

Isomers = same molecular formula, different arrangement. The hierarchy:

• Constitutional (structural) isomers — different connectivity.
• Stereoisomers — same connectivity, different spatial arrangement.
  • Conformational — differ by rotation about single bonds.
  • Configurational — enantiomers (non-superimposable mirror images) and diastereomers (stereoisomers that are NOT mirror images, including cis-trans and epimers).

Structural (Constitutional) Isomers

Must know

Same formula, different connectivity. E.g., $\ce{C4H10}$: butane vs. isobutane — distinct compounds with different physical properties.

Conformational Isomers

Must know

Conformers interconvert by rotation about single bonds (no bonds broken) and are NOT isolable at room temperature. Represented with Newman projections (front atom = lines from center; rear = lines from a circle). Energy order: anti (groups 180°, lowest) < gauche (60°) < eclipsed (0°, highest, torsional strain). For cyclohexane, chairs are most stable; large groups prefer equatorial (less 1,3-diaxial strain); a ring flip swaps the two chairs.

Quick check: In the Newman projection of ethane, which conformation is most stable and why?

Answer: The staggered conformation (60° between H's) minimizes torsional strain — C–H electron clouds are as far apart as possible.

Cis-Trans (Geometric) Isomers

Must know

Arise when rotation is restricted (a double bond or a ring) and each carbon bears two different substituents. Cis: same-priority groups on the same side; trans: opposite sides (for rings, same vs. opposite face).

E/Z Nomenclature (CIP for alkenes): Assign priority to the two groups on each alkene carbon by atomic number. If the two highest-priority groups are on the same side → Z (zusammen); opposite sides → E (entgegen).

CIP priority rules: higher atomic number → higher priority (I > Br > Cl > S > O > N > C > H); at ties, go to the next atom; treat a double bond as duplicated atoms (C=O → C bonded to (O,O)).

Quick check: Assign E or Z to $\ce{ClCH=CHBr}$ with Cl and Br on the same side.

Answer: On each carbon the halogen outranks H, and Cl and Br are on the same side → Z.

Enantiomers and Chirality

Must know

A molecule is chiral if non-superimposable on its mirror image — most often from a chiral center (stereocenter): a carbon bonded to four different groups. Enantiomers are non-superimposable mirror images with identical physical properties EXCEPT they rotate plane-polarized light oppositely. A molecule with no stereocenter, or with an internal mirror plane (meso compound), is achiral.

R and S Configuration (CIP System)

Must know

1. Assign priorities 1–4 by CIP (atomic number, then first point of difference).
2. Orient with the lowest priority (4) pointing away.
3. Trace 1 → 2 → 3: clockwise = R, counterclockwise = S.
   Trick: if group 4 points toward you, trace and then reverse the answer.

Fischer projection convention: chain vertical, most oxidized carbon at top; horizontal bonds point toward the viewer, vertical bonds point away. Misreading this inverts the configuration — central to sugar chemistry.

Worked Example — L-alanine: chiral carbon bears $\ce{-NH2}$, $\ce{-COOH}$, $\ce{-CH3}$, $\ce{-H}$. Priorities: $\ce{NH2}$ (N) > $\ce{COOH}$ (C bonded to O,O) > $\ce{CH3}$ > H. In the Fischer projection H is horizontal (toward viewer), so trace 1→2→3 (counterclockwise) and reverse → S. So L-alanine = (S)-alanine. (All natural amino acids except glycine are L, and almost all are S — cysteine is the exception, R, because its sulfur raises priority.)

Quick check: A chiral center with priorities $\ce{-OH}$(1) > $\ce{-COOH}$(2) > $\ce{-CH2CH3}$(3) > $\ce{-H}$(4); with H pointing away, 1→2→3 is clockwise. Configuration?

Answer: H (4) already points away, so no correction. Clockwise = R.

Diastereomers

Must know

Diastereomers are stereoisomers that are NOT mirror images. For $n$ stereocenters, the maximum number of stereoisomers is $2^n$ (fewer if meso forms exist). Unlike enantiomers, diastereomers have different physical properties, so they can be separated by ordinary techniques. Epimers differ at exactly one stereocenter (glucose vs. galactose at C-4). A meso compound has stereocenters but an internal mirror plane → achiral.

Quick check: Tartaric acid has two stereocenters. List its stereoisomers.

Answer: (2R,3R) and (2S,3S) (an enantiomer pair), plus meso-tartaric acid (2R,3S), which is achiral — 3 total, not 4, because the meso form's mirror image is superimposable on itself.

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Polarization of Light and Specific Rotation

Plane-Polarized Light

Must know

A polarimeter passes light through a filter to make plane-polarized light; a chiral solution rotates the plane. Dextrorotatory (+) = clockwise; levorotatory (−) = counterclockwise.

Crucially: R/S does NOT predict +/−; the sign must be measured. (E.g., L-alanine is actually (+).)

Specific Rotation

Must know

$[\alpha]_\lambda^T = \frac{\alpha_{\text{observed}}}{l \times c}$

with $l$ = path length in decimeters (dm) and $c$ = concentration in g/mL (λ usually the 589 nm sodium D line, T usually 25°C). Specific rotation is an intrinsic property, like melting point.

Optical Purity and Enantiomeric Excess

Must know

A racemic mixture (equal enantiomers) has net rotation = 0 (optically inactive). Enantiomeric excess:

$\text{ee} = \frac{[\alpha]_{\text{mixture}}}{[\alpha]_{\text{pure enantiomer}}} \times 100\% = \%(\text{major}) - \%(\text{minor})$

Worked Example: observed $[\alpha] = +9.2°$, pure (R) $[\alpha] = +23.1°$ → ee = 9.2/23.1 ≈ 40%. With %R + %S = 100 and %R − %S = 40 → %R = 70, %S = 30.

Quick check: If equal amounts of two enantiomers are mixed, what is the observed specific rotation?

Answer: Zero — equal and opposite rotations cancel (a racemic mixture).

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Absolute and Relative Configuration

Must know

Absolute configuration is the actual 3-D arrangement at a stereocenter (R/S or D/L). Relative configuration describes one stereocenter relative to another or to a standard, without the actual 3-D arrangement.

The D/L system (referenced to D-glyceraldehyde) assigns by the orientation in a Fischer projection: L = the reference group (amino in amino acids, OH in sugars) on the left; D = on the right.

Important: D/L, +/−, and R/S are three independent systems. Most natural L-amino acids are S (cysteine is the exception).

Quick check: L-cysteine has the S or R configuration?

Answer: R. The $\ce{-CH2SH}$ side chain's sulfur (atomic number 16) outranks the carboxyl carbon, raising the side chain's CIP priority and reversing the label to R, even though the spatial arrangement is the same "L-type."

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Common Confusions & Tricks

1. Electron geometry vs. molecular geometry. Lone pairs count for hybridization and electron geometry, but the shape name describes atom positions only. $\ce{H2O}$ is sp³ (tetrahedral electron geometry) but bent.

2. Formal charge vs. partial charge. Formal charge is bookkeeping (integers, equal sharing assumed); partial charge ($\delta+$/$\delta-$) reflects actual electron density from EN differences. They can disagree — in CO, the formal −1 sits on carbon despite oxygen being more electronegative.

3. Lewis acid ≠ Brønsted acid. A Lewis acid accepts an electron pair. $\ce{H+}$ is both, but $\ce{BF3}$ and metal cations are Lewis acids only — look for an incomplete octet or positive charge.

4. R/S with group 4 pointing toward you. Trace 1→2→3, then flip. (Or: swapping any two groups inverts the configuration — two swaps return the original.)

5. E/Z ≠ trans/cis. For simple disubstituted alkenes, cis = Z and trans = E, but with three or four different substituents you MUST use CIP priorities — old "cis" can be E.

6. Meso compounds are achiral despite stereocenters. An internal mirror plane makes them superimposable on their mirror image — not optically active.

7. Specific rotation: path length in decimeters. A 10 cm tube = 1 dm. Using cm gives answers off by 10×.

8. Conformational vs. configurational isomers. Conformers interconvert by rotation (not isolable); configurational isomers require bond breaking and ARE isolable. Cis-trans alkene isomers are NOT conformers.

9. Bond energy rises with bond order but NOT proportionally. C=C is ~1.8× C–C, not 2× — why π bonds are reactive targets.

10. Hybridization "if you see X, think Y": planar/trigonal → sp²; linear → sp; tetrahedral/pyramidal/bent (no double bond) → sp³. Double bond → each atom sp²; triple bond → each atom sp.

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Key Equations

Equation	Variables & When to Use
$\text{FC} = V - N - \frac{B}{2}$	Formal charge: $V$ = valence e⁻, $N$ = nonbonding e⁻, $B$ = bonding e⁻. Find the most stable Lewis structure.
$\mu = q \times d$	Dipole moment: $q$ = partial charge, $d$ = separation. Molecular dipole = vector sum of bond dipoles. Units: debyes (D).
$[\alpha]_\lambda^T = \dfrac{\alpha_{\text{obs}}}{l \times c}$	Specific rotation: $l$ = path length (dm), $c$ = concentration (g/mL). Intrinsic property of a chiral compound.
$\text{ee} = \dfrac{[\alpha]_{\text{mixture}}}{[\alpha]_{\text{pure}}} \times 100\% = \%(\text{major}) - \%(\text{minor})$	Enantiomeric excess.
Max stereoisomers $= 2^n$	$n$ = number of stereocenters; fewer if meso forms exist.