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Molecules & structure

The Lewis structure is an electron ledger — the exact counterpart of the species ledger, but counting valence electrons instead of atoms. For each molecule the enginederives and checks the accounting: the valence-electron total, that every atom completes an octet (a duet for H), each atom's formal charge, and that the formal charges sum to the molecular charge. It refuses any structure that fails. The shape then follows from the electron-domain count (VSEPR, sourced), and the shape decides polarity — a disclosed model conclusion. Polarity in turn fixes the dominant intermolecular force (with the H-on-N/O/F test — machine-derived from the structure; the boiling point is sourced evidence). Bond ΔEN comes from the Valence Table's electronegativities.

NH3\mathrm{NH_{3}}

Ammonia

polar
Electron ledgermachine-checked

H 3×1 = 3 + N 1×5 = 58 valence electrons

Placed: 6 bonding (3 shared pairs) + 2 nonbonding (1 lone pair) = 8 ✓

AtomLone pairsBonds (Σ order)Formal charge
NN1130
HH1010
HH2010
HH3010
Shape (VSEPR)geometry: openstax-chemistry-2e

4 electron domains(3 bonding + 1 lone) → electron geometry tetrahedral, shape trigonal pyramidal.

ideal angle 109.5° — the lone pair compresses the actual angle to about 107°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • H–N ×3ΔEN 0.84 · polar covalent
Molecular polaritymodel-assumed

polar The molecule is trigonal pyramidal (not planar), so the three N–H bond dipoles and the lone pair do not cancel — a net dipole points through the nitrogen toward the lone pair.

Intermolecular forcesfrom the structure

Dominant: hydrogen bonding · present: London dispersion, dipole–dipole, hydrogen bonding · H-bond donor

boils at -33.3 °Cnist-webbook-boiling-points

Eight valence electrons: three N–H bonding pairs and one lone pair on nitrogen. Four electron domains make the electron geometry tetrahedral, but with one lone pair the molecular shape is trigonal pyramidal — a tripod with the lone pair on top. The pyramid is not symmetric top-to-bottom, so the three polar N–H dipoles plus the lone pair give a net dipole: ammonia is polar. Contrast the ammonium ion, where a fourth N–H bond replaces the lone pair and the shape becomes a symmetric tetrahedron.

NH4+\mathrm{NH_{4}}^{+}

Ammonium ion

charge +1

Also: ammonium ion

Electron ledgermachine-checked

H 4×1 = 4 + N 1×5 = 5 − 1 (charge)8 valence electrons

Placed: 8 bonding (4 shared pairs) + 0 nonbonding (0 lone pairs) = 8 ✓

AtomLone pairsBonds (Σ order)Formal charge
NN104+1
HH1010
HH2010
HH3010
HH4010
Shape (VSEPR)geometry: openstax-chemistry-2e

4 electron domains(4 bonding + 0 lone) → electron geometry tetrahedral, shape tetrahedral.

ideal angle 109.5°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • H–N ×4ΔEN 0.84 · polar covalent

A polyatomic cation, eight valence electrons: nitrogen contributes five and four hydrogens one each, minus one for the +1 charge. Four N–H bonds, no lone pairs — a symmetric tetrahedron. The formal charge tells the story: nitrogen owns half of four bonding pairs (4 electrons) against its 5 group electrons, so its formal charge is 5 − 4 = +1, and every hydrogen is 0 — the ion's +1 sits on nitrogen. This is ammonia with its lone pair donated to a proton, turning the pyramid into a tetrahedron.

  • The +1 charge is a property of the whole ion, so molecular polarity (a neutral-molecule idea) is not stated here — the formal-charge sum, +1, equals the ion charge, and that is the machine-checked claim.
CO2\mathrm{CO_{2}}

Carbon dioxide

nonpolar
Electron ledgermachine-checked

C 1×4 = 4 + O 2×6 = 1216 valence electrons

Placed: 8 bonding (4 shared pairs) + 8 nonbonding (4 lone pairs) = 16 ✓

AtomLone pairsBonds (Σ order)Formal charge
CC1040
OO1220
OO2220
Shape (VSEPR)geometry: openstax-chemistry-2e

2 electron domains(2 bonding + 0 lone) → electron geometry linear, shape linear.

ideal angle 180°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • C=O ×2ΔEN 0.89 · polar covalent
Molecular polaritymodel-assumed

nonpolar The molecule is linear, so the two equal, opposite C=O bond dipoles cancel exactly — polar bonds, but no net molecular dipole.

Intermolecular forcesfrom the structure

Dominant: London dispersion · present: London dispersion

sublimes at -78.5 °Cnist-webbook-boiling-points

Sixteen valence electrons: two C=O double bonds and two lone pairs on each oxygen. Carbon has two electron domains (the two double bonds) and no lone pairs, so the molecule is linear. Each C=O bond is polar (ΔEN 0.89), but the two bond dipoles point in exactly opposite directions along the linear axis and cancel — so CO₂ has polar bonds yet is a nonpolar molecule. That is the lesson water inverts: shape decides.

CH2O\mathrm{CH_{2}O}

Formaldehyde

polar

Also: methanal

Electron ledgermachine-checked

C 1×4 = 4 + H 2×1 = 2 + O 1×6 = 612 valence electrons

Placed: 8 bonding (4 shared pairs) + 4 nonbonding (2 lone pairs) = 12 ✓

AtomLone pairsBonds (Σ order)Formal charge
CC1040
OO1220
HH1010
HH2010
Shape (VSEPR)geometry: openstax-chemistry-2e

3 electron domains(3 bonding + 0 lone) → electron geometry trigonal planar, shape trigonal planar.

ideal angle 120°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • C=OΔEN 0.89 · polar covalent
  • C–H ×2ΔEN 0.35 · nonpolar covalent
Molecular polaritymodel-assumed

polar Trigonal planar, but the bonds differ: the strong C=O dipole is not cancelled by the two weak C–H dipoles, leaving a net dipole toward oxygen.

Intermolecular forcesfrom the structure

Dominant: dipole–dipole · present: London dispersion, dipole–dipole

Twelve valence electrons: two C–H single bonds, one C=O double bond, and two lone pairs on oxygen. Carbon has three electron domains (two single bonds and one double bond count as three domains) and no lone pairs, so the molecule is trigonal planar, angles near 120°. Unlike carbon dioxide, the three bonds are not identical: the strong C=O dipole (ΔEN 0.89) is not balanced by the two weak C–H dipoles (ΔEN 0.35), so the flat molecule has a net dipole toward oxygen — it is polar. Geometry alone does not settle polarity; the bonds have to match too.

CH4\mathrm{CH_{4}}

Methane

nonpolar
Electron ledgermachine-checked

C 1×4 = 4 + H 4×1 = 48 valence electrons

Placed: 8 bonding (4 shared pairs) + 0 nonbonding (0 lone pairs) = 8 ✓

AtomLone pairsBonds (Σ order)Formal charge
CC1040
HH1010
HH2010
HH3010
HH4010
Shape (VSEPR)geometry: openstax-chemistry-2e

4 electron domains(4 bonding + 0 lone) → electron geometry tetrahedral, shape tetrahedral.

ideal angle 109.5°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • C–H ×4ΔEN 0.35 · nonpolar covalent
Molecular polaritymodel-assumed

nonpolar Four identical C–H bonds point to the corners of a symmetric tetrahedron, so any bond dipoles cancel; the C–H bond is nearly nonpolar to begin with (ΔEN 0.35).

Intermolecular forcesfrom the structure

Dominant: London dispersion · present: London dispersion

boils at -161.5 °Cnist-webbook-boiling-points

Eight valence electrons: four C–H bonding pairs, no lone pairs. Four electron domains and no lone pairs give a symmetric tetrahedron, bond angles 109.5°. The C–H bond is barely polar (ΔEN 0.35, on the nonpolar side of the guideline), and the four identical bonds point to the corners of a tetrahedron, so their small dipoles cancel — methane is nonpolar twice over: nearly nonpolar bonds arranged symmetrically.

H2O\mathrm{H_{2}O}

Water

polar

Also: dihydrogen monoxide

Electron ledgermachine-checked

H 2×1 = 2 + O 1×6 = 68 valence electrons

Placed: 4 bonding (2 shared pairs) + 4 nonbonding (2 lone pairs) = 8 ✓

AtomLone pairsBonds (Σ order)Formal charge
OO1220
HH1010
HH2010
Shape (VSEPR)geometry: openstax-chemistry-2e

4 electron domains(2 bonding + 2 lone) → electron geometry tetrahedral, shape bent.

ideal angle 109.5° — two lone pairs compress the actual angle to about 104.5°

Bonds (ΔEN)EN: allred-1961-electronegativity
  • H–O ×2ΔEN 1.24 · polar covalent
Molecular polaritymodel-assumed

polar The molecule is bent, so the two polar O–H bond dipoles (ΔEN 1.24) add rather than cancel — a net dipole points from the H side toward oxygen.

Intermolecular forcesfrom the structure

Dominant: hydrogen bonding · present: London dispersion, dipole–dipole, hydrogen bonding · H-bond donor

boils at 100.0 °Cnist-webbook-boiling-points

Eight valence electrons: two O–H bonding pairs and two lone pairs on oxygen. Four electron domains around O make the electron geometry tetrahedral, but the two lone pairs push the two bonds together into a bent molecule. Because the shape is bent, the two polar O–H bond dipoles do not cancel — water is the archetypal polar molecule, and that net dipole is why it dissolves salts and boils far higher than its mass predicts.