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Chapter 8 · Free lesson

Bonding and molecular properties

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YOUR CHAPTER PLAN

What you will study

  1. Lewis structures count outer electrons
  2. Formal charge and resonance
  3. Electron regions lead to molecular shape
  4. Bonding models: hybridization, sigma/pi bonds, and bond order
  5. Polarity and attractions between molecules
  6. Alkanes and simple organic molecules

By the end, you can…

  1. Count valence electrons and draw a simple Lewis structure.
  2. Use formal charge and resonance to choose the best Lewis structure.
  3. Find molecular shape from bonds and lone pairs.
  4. Use electron-region counts for introductory hybridization and distinguish sigma and pi bonds.
  5. Use bond polarity and shape to decide molecular polarity.
  6. Choose the strongest relevant intermolecular attraction.
  7. Draw Lewis structures and give bond angles for the first ten alkanes and a few simple organic molecules.

These are the skills to check, not just pages to finish. Use the section goals while studying, then try the chapter checks without hints.

Read the idea first. Then notice the clue, follow the route, and make a choice you can explain. You can return to the worked example if a practice question feels unfamiliar.

8.1 Lewis structures count outer electrons

A Lewis structure is a drawing of a molecule's outer (valence) electrons. Lines show shared pairs; dots show pairs that are not shared. It is the first step toward shape and polarity.

What you’ll use from earlier
  • Main-group valence electrons (Chapter 3); formula atom counts (Chapter 2).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
Reading a Lewis structure
HOHbond = 2 shared electronslone pair (not in a bond)Around O: 2 bonds + 2 lone pairs= 8 electrons (an octet)Each H: 1 bond = 2 electrons
Water: 1 + 1 + 6 = 8 valence electrons, all shown.
Key words
  • Lewis structure: valence electrons drawn as bonds and lone pairs.
  • Lone pair: two electrons not used in a bond.
  • Octet: eight electrons around an atom. Hydrogen needs only two.
  • Some atoms have incomplete or expanded octets; check your course's exceptions.
Main-group columns help count valence electrons. White numbers in gold badges: main-group valence electrons. Gray numbers beneath: group numbers. Helium is the exception to the group 18 badge: He has 2 valence electrons, not 8. Carbon (group 14) contributes 4; nitrogen (15) contributes 5; oxygen (16) contributes 6.
Gold badges: main-group valence electrons. Gray numbers: group numbers. Helium has 2.
Count the electron budget first
electrons = sum of valence electrons − (ion charge)
NH4+: 5 + 4(1) − 1 = 8    OH−: 6 + 1 + 1 = 8
Negative ion: add electrons. Positive ion: subtract them.
The build order
StepWhat you do
1Count valence electrons, adjusted for charge.
2Connect atoms with single bonds.
3Fill outer atoms' octets with lone pairs.
4Put leftover electrons on the central atom.
5Central octet short? Turn an outer lone pair into a multiple bond.
Main-group position helps count Lewis electrons. C in group 14 has 4 valence electrons; O in group 16 has 6. CO₂ starts with 4 + 2(6) = 16 valence electrons. Add for negative ion charge; subtract for positive. The periodic table supplies the neutral-atom counts. The formula and charge supply the rest.
The table gives neutral-atom counts; the formula and charge give the rest.
1 · Recognize the clue
Look for this:a formula or ion with a request to count valence electrons or draw its Lewis structure.
2 · Follow the route
  1. Count valence electrons, adjusted for any charge.
  2. Connect the atoms with single bonds.
  3. Fill the outer atoms' octets.
  4. Use multiple bonds if the central atom is short.
  5. Check the total against your starting count.

On the periodic table: The main-group column gives valence electrons: C (Group 14) 4, N (15) 5, O (16) 6, H (1) 1.

On the periodic table: Helium is the exception in Group 18: it has 2 valence electrons, not 8.

Worked example

Build the usual Lewis structure for CO2.

  1. 1
    Count available valence electrons

    On the periodic table: C is in Group 14 (4 valence electrons); O is in Group 16 (6 each).

    C + 2 O = 4 + 2(6) = 16 electrons
  2. 2
    Connect the atoms

    Carbon is the central atom. Two single bonds use four electrons.

    OCO2 bonds use 4 electrons; 12 left
  3. 3
    Complete the outer octets

    The twelve remaining electrons go on the oxygens as 3 lone pairs each.

    OCOEach O has 8; C has only 4
  4. 4
    Share a lone pair from each oxygen

    Turn one lone pair on each O into an extra bonding pair. Carbon now has eight.

    OCOC, O, O: 8 electrons each
  5. 5
    Check the final budget

    The structure uses exactly the electrons we started with.

    PartCountElectrons
    Double bonds28
    Lone pairs48
    Total16
  6. ✓
    Answer

    O=C=O, with two lone pairs on each oxygen.

8.2 Formal charge and resonance

Formal charge is bookkeeping: it tells you how many electrons each atom “owns” in a Lewis structure. It helps you pick the best drawing and recognize resonance, when one drawing is not enough.

What you’ll use from earlier
  • Lewis structures and lone pairs (8.1).
  • Electronegativity (3.4).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
Formal charge = valence − unshared − half of shared
FC = V − N − B2
V: the free atom's valence electrons. N: lone-pair (nonbonding) electrons. B: bonding electrons.
Main-group columns help count valence electrons. White numbers in gold badges: main-group valence electrons. Gray numbers beneath: group numbers. Helium is the exception to the group 18 badge: He has 2 valence electrons, not 8. Carbon (group 14) contributes 4; nitrogen (15) contributes 5; oxygen (16) contributes 6.
V comes from the gold badge above the atom's column.
Choosing the best structure
  • Formal charges must add up to the overall charge of the species.
  • Prefer structures with small formal charges (0 where possible).
  • Put a negative formal charge on the more electronegative atom.
Resonance: one molecule, two drawings
OOO+−OOO+−Same atoms in the same places; only the electrons move.
Ozone, O3. Gold marks show formal charges.
Formal charges in either ozone drawing
Oxygen atomVNB2FC
Central623+1
End, double bond6420
End, single bond661−1
Sum0
The sum matches ozone's overall charge of 0.
What resonance means
Not this
The molecule flips back and forth between drawings.
This
One real structure, a blend of the drawings. Both O–O bonds in O3 are identical.
1 · Recognize the clue
Look for this:"formal charge," counts of bonding and nonbonding electrons, or several drawings of one molecule called resonance structures.
2 · Follow the route
  1. Count each atom's owned electrons: lone-pair electrons plus half its bonding electrons.
  2. Subtract from the valence count to get formal charge.
  3. Check that the formal charges add up to the overall charge.

On the periodic table: The group gives V: N (Group 15) has 5, O (16) has 6, H has 1.

On the periodic table: Electronegativity generally increases up and to the right, toward fluorine (F). Compare positions to see which atom should carry negative charge.

Worked example

In NH4+, nitrogen has four single bonds and no lone pair. Find nitrogen's formal charge.

NHHHH+
Formal charge starts with a valence count. N in group 15 contributes 5 valence electrons; H contributes 1. In NH₄⁺, nitrogen owns half of 8 bonding electrons: formal charge = 5 − 0 − 4 = +1. You also need the Lewis structure to count bonding and nonbonding electrons.
The table gives V; the Lewis structure gives N and B.
  1. 1
    Start with nitrogen's valence count

    On the periodic table: N is in Group 15, so it has 5 valence electrons.

    V = 5
  2. 2
    Count nonbonding electrons

    Nitrogen has no lone pair here.

    N = 0
  3. 3
    Assign half the bonding electrons

    Bonding electrons are split equally between the two bonded atoms.

    B2 = 4 bonds × 2 electrons2 = 4
  4. 4
    Subtract the assigned electrons

    Nitrogen is assigned one fewer electron than a neutral N atom.

    FC = 5 − 0 − 4 = +1
  5. ✓
    Answer

    Nitrogen has formal charge +1.

8.3 Electron regions lead to molecular shape

Molecules are three-dimensional. Count the electron regions around a central atom, and you can predict how they spread out: that gives the shape.

What you’ll use from earlier
  • Lewis structures and lone pairs (8.1).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
What counts as one electron region
  • Each bond is one region: single, double or triple.
  • Each lone pair on the central atom is one region.
  • VSEPR (valence-shell electron-pair repulsion): regions spread as far apart as possible.
Two names for every molecule
Electron geometry
All regions, lone pairs included. Water: tetrahedral.
Molecular shape
Bonded atoms only; lone pairs push but are not named. Water: bent.
The six shapes you will meet most
XXALinear2 regions · 0 lone pairs180°XXXATrigonal planar3 regions · 0 lone pairs120°XXABent3 regions · 1 lone paira little under 120°XXXXATetrahedral4 regions · 0 lone pairs109.5°XXXATrigonal pyramidal4 regions · 1 lone paira little under 109.5°XXABent4 regions · 2 lone pairsabout 104.5° (water)
A = central atom, X = bonded atom, shaded lobe = lone pair. Wedge points toward you; dashes point away.
VSEPR: regions → electron geometry → shape → angle
RegionsLone pairsElectron geometryMolecular shapeIdeal angleExample
20LinearLinear180°CO2
30Trigonal planarTrigonal planar120°BF3
31Trigonal planarBent< 120°SO2
40TetrahedralTetrahedral109.5°CH4
41TetrahedralTrigonal pyramidal< 109.5°NH3
42TetrahedralBent≈ 104.5°H2O
50Trigonal bipyramidalTrigonal bipyramidal90°, 120°PCl5
51Trigonal bipyramidalSeesaw< 90°, < 120°SF4
52Trigonal bipyramidalT-shaped< 90°ClF3
53Trigonal bipyramidalLinear180°XeF2
60OctahedralOctahedral90°SF6
61OctahedralSquare pyramidal< 90°BrF5
62OctahedralSquare planar90°XeF4
5 regions: lone pairs take the middle-ring (equatorial) positions first.

Lone pairs push bonds a little closer, so angles shrink slightly below the ideal value.

1 · Recognize the clue
Look for this:bonds and lone pairs around a central atom, "electron regions," or a question asking for a molecule's shape.
2 · Follow the route
  1. Draw the Lewis structure; pick the central atom.
  2. Count all its regions: bonds plus lone pairs.
  3. Separate lone pairs from bonded atoms.
  4. Name the geometry and shape.

On the periodic table: The group gives main-group valence electrons: C (14) 4, N (15) 5, O (16) 6, H 1. Helium has 2.

On the periodic table: The table builds the Lewis structure; it does not show the shape.

Worked example

Use electron regions to find the shapes of CO2 and H2O.

Main-group columns help count valence electrons. Gold badges give main-group valence electrons; gray numbers beneath are group numbers. Helium has 2 valence electrons, not 8. Carbon (group 14) contributes 4; nitrogen (15) contributes 5; oxygen (16) contributes 6.
Valence electrons for the Lewis structures: C 4, O 6, H 1.
  1. 1
    Count regions around C in CO2

    O=C=O. Each double bond is one region.

    Around CRegions
    C=O double bond1
    C=O double bond1
    Lone pairs0
    Total2
  2. 2
    Name the CO2 shape

    The O atoms sit on opposite sides of C.

    2 regions, 0 lone pairs → linear, 180°
  3. 3
    Count regions around O in H2O

    Lone pairs affect the arrangement, so count them.

    Around ORegions
    O–H bond1
    O–H bond1
    Lone pairs2
    Total4
  4. 4
    Name the H2O shape

    Geometry counts lone pairs; shape names only atoms.

    CountsName
    Electron geometry4 regionsTetrahedral
    Molecular shape2 bonded atomsBent
  5. ✓
    Answer

    CO2 is linear; H2O is bent.

8.3a Bonding models: hybridization, sigma/pi bonds, and bond order

Lewis structures show where bonds are. Bonding models explain how they form: hybrid orbitals, sigma and pi bonds, and bond order. All three start from the Lewis structure.

What you’ll use from earlier
  • Electron regions and shapes (8.3).
  • Resonance (8.2).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
Three molecules drawn with bond lines: H–H labeled single, 1 σ; O=O labeled double, 1 σ + 1 π; and N≡N labeled triple, 1 σ + 2 π. Below: two electron regions give sp, three give sp², four give sp³.
Every bond has one σ part; each extra line is a π bond.
Electron regions set the hybridization
RegionsHybridizationOrbitals mixedArrangementExample
2sp1 s + 1 pLinear, 180°C in CO2
3sp21 s + 2 pTrigonal planar, 120°C in C2H4
4sp31 s + 3 pTetrahedral, 109.5°C in CH4
Each bond (single, double or triple) or lone pair is one region. The superscript counts p orbitals, not electrons.
Two ways orbitals overlap
Sigma (σ) bondOverlap lies ON the linejoining the nucleiPi (π) bondOverlap ABOVE and BELOW the lineTwo lobes, one π bond
Counting sigma and pi bonds
BondσπBond order
Single101
Double112
Triple123
Higher bond order: shorter bond (carbon–carbon)
C–C154 pmC=C134 pmC≡C120 pmTypical bond length (pm)
Approximate values. Compare the same atom pair only; atom size and surroundings also matter.
Bond order in practice
  • Same atom pair: higher bond order means a shorter, stronger bond.
  • Resonance can give an average bond order between whole numbers. Ozone's two O–O bonds are each 1.5.
  • Atoms with 5 or 6 regions (expanded octets, 8.3) sometimes get hybrid labels too. Those labels are simplified models.
Molecular orbital (MO) theory: a second model
  • Atomic orbitals combine into new orbitals that spread over the whole molecule.
  • Bonding orbitals hold electrons between the nuclei and hold the atoms together.
  • Antibonding orbitals (marked with *) hold electrons that weaken the bond.
  • Any unpaired electrons make a substance paramagnetic: pulled into a magnetic field. O2 is the classic example.
MO bond order
bond order = bonding electrons − antibonding electrons2
H2: 2 − 02 = 1, a single bond
1 · Recognize the clue
Look for this:"hybridization," "sigma and pi," double or triple bonds, bond length, or bonding and antibonding electrons for bond order.
2 · Follow the route
  1. Draw the Lewis structure.
  2. Count electron regions on the atom: 2 → sp, 3 → sp2, 4 → sp3.
  3. Count bonds: every bond has 1 σ; each extra line is 1 π.
  4. For MO bond order, subtract antibonding from bonding electrons and divide by 2.

On the periodic table: The group gives the valence electrons for the Lewis structure: C (Group 14) 4, O (Group 16) 6.

On the periodic table: The table does not show hybridization or bond types; those come from the Lewis structure.

Worked example

Give the hybridization of carbon in CO2, and count its sigma and pi bonds.

OCOEach C=O: 1 σ + 1 π
  1. 1
    Count regions on carbon

    Each double bond is one region. Carbon has no lone pairs.

    Around CRegions
    C=O double bond1
    C=O double bond1
    Lone pairs0
    Total2
  2. 2
    Match regions to hybridization

    Two regions sit on opposite sides of carbon.

    2 regions → sp, linear (180°)
  3. 3
    Count sigma and pi bonds

    Each C=O double bond has 1 σ and 1 π.

    Bondσπ
    C=O11
    C=O11
    Total22
  4. ✓
    Answer

    Carbon is sp (linear). CO2 has 2 sigma and 2 pi bonds.

Section practice · 10 questions · tap to open

What this set is for: practicing the method from this section while it’s fresh. Every problem uses only what this section has taught.

Practice 1

A structure contains 1 double bond and no other bonds. Count sigma and pi bonds.

Give me a hint

Count the components of one double bond, then multiply.

Check the answer

Answer: 1 sigma and 1 pi

Explain the worked problem to me

What you know: The structure has 1 double bond and nothing else. What you want: The number of sigma and pi bonds.

  1. Recall what one double bond contains: 1 sigma bond (overlap along the line between the nuclei) plus 1 pi bond (overlap above and below that line).
  2. Multiply by the number of double bonds. Sigma: 1 × 1 = 1. Pi: 1 × 1 = 1.
  3. There are no other bonds to add, so the total is 1 sigma and 1 pi.

Check: Each double bond is 2 bonds in all, so the total is 1 × 2 = 2 bonds, and 1 sigma + 1 pi = 2.

Watch out: A double bond is not two sigma bonds. Only the first bond between two atoms is sigma; every extra bond is pi.

That’s right: 1 sigma and 1 pi

Practice 2

A structure contains 2 double bonds and no other bonds. Count sigma and pi bonds.

Give me a hint

Count the components of one double bond, then multiply.

Check the answer

Answer: 2 sigma and 2 pi

Explain the worked problem to me

What you know: The structure has 2 double bonds and nothing else. What you want: The number of sigma and pi bonds.

  1. Recall what one double bond contains: 1 sigma bond (overlap along the line between the nuclei) plus 1 pi bond (overlap above and below that line).
  2. Multiply by the number of double bonds. Sigma: 2 × 1 = 2. Pi: 2 × 1 = 2.
  3. There are no other bonds to add, so the total is 2 sigma and 2 pi.

Check: Each double bond is 2 bonds in all, so the total is 2 × 2 = 4 bonds, and 2 sigma + 2 pi = 4.

Watch out: A double bond is not two sigma bonds. Only the first bond between two atoms is sigma; every extra bond is pi.

That’s right: 2 sigma and 2 pi

Practice 3

A structure contains 3 double bonds and no other bonds. Count sigma and pi bonds.

Give me a hint

Count the components of one double bond, then multiply.

Check the answer

Answer: 3 sigma and 3 pi

Explain the worked problem to me

What you know: The structure has 3 double bonds and nothing else. What you want: The number of sigma and pi bonds.

  1. Recall what one double bond contains: 1 sigma bond (overlap along the line between the nuclei) plus 1 pi bond (overlap above and below that line).
  2. Multiply by the number of double bonds. Sigma: 3 × 1 = 3. Pi: 3 × 1 = 3.
  3. There are no other bonds to add, so the total is 3 sigma and 3 pi.

Check: Each double bond is 2 bonds in all, so the total is 3 × 2 = 6 bonds, and 3 sigma + 3 pi = 6.

Watch out: A double bond is not two sigma bonds. Only the first bond between two atoms is sigma; every extra bond is pi.

That’s right: 3 sigma and 3 pi

Practice 4

A structure contains 4 double bonds and no other bonds. Count sigma and pi bonds.

Give me a hint

Count the components of one double bond, then multiply.

Check the answer

Answer: 4 sigma and 4 pi

Explain the worked problem to me

What you know: The structure has 4 double bonds and nothing else. What you want: The number of sigma and pi bonds.

  1. Recall what one double bond contains: 1 sigma bond (overlap along the line between the nuclei) plus 1 pi bond (overlap above and below that line).
  2. Multiply by the number of double bonds. Sigma: 4 × 1 = 4. Pi: 4 × 1 = 4.
  3. There are no other bonds to add, so the total is 4 sigma and 4 pi.

Check: Each double bond is 2 bonds in all, so the total is 4 × 2 = 8 bonds, and 4 sigma + 4 pi = 8.

Watch out: A double bond is not two sigma bonds. Only the first bond between two atoms is sigma; every extra bond is pi.

That’s right: 4 sigma and 4 pi

Practice 5

A structure contains 5 double bonds and no other bonds. Count sigma and pi bonds.

Give me a hint

Count the components of one double bond, then multiply.

Check the answer

Answer: 5 sigma and 5 pi

Explain the worked problem to me

What you know: The structure has 5 double bonds and nothing else. What you want: The number of sigma and pi bonds.

  1. Recall what one double bond contains: 1 sigma bond (overlap along the line between the nuclei) plus 1 pi bond (overlap above and below that line).
  2. Multiply by the number of double bonds. Sigma: 5 × 1 = 5. Pi: 5 × 1 = 5.
  3. There are no other bonds to add, so the total is 5 sigma and 5 pi.

Check: Each double bond is 2 bonds in all, so the total is 5 × 2 = 10 bonds, and 5 sigma + 5 pi = 10.

Watch out: A double bond is not two sigma bonds. Only the first bond between two atoms is sigma; every extra bond is pi.

That’s right: 5 sigma and 5 pi

Practice 6

Two electron regions around carbon: introductory hybridization?

Give me a hint

2 regions → sp, 3 regions → sp2, 4 regions → sp3.

Check the answer

Answer: sp

Explain the worked problem to me

What you know: The atom has 2 electron regions. What you want: Its introductory hybridization label.

  1. Remember what counts as a region: each single, double, or triple bond is one region, and each lone pair is one region.
  2. Use the matching rule: 2 regions → sp, 3 regions → sp2, 4 regions → sp3.
  3. Two regions match sp.

Check: The label uses one s orbital and one p orbital, which makes 2 hybrid orbitals, one per region. The regions are arranged linearly, 180° apart.

Watch out: A double or triple bond still counts as only one region.

That’s right: sp

Practice 7

Three electron regions: hybridization?

Give me a hint

2 regions → sp, 3 regions → sp2, 4 regions → sp3.

Check the answer

Answer: sp2

Explain the worked problem to me

What you know: The atom has 3 electron regions. What you want: Its introductory hybridization label.

  1. Remember what counts as a region: each single, double, or triple bond is one region, and each lone pair is one region.
  2. Use the matching rule: 2 regions → sp, 3 regions → sp2, 4 regions → sp3.
  3. Three regions match sp2.

Check: The label uses one s orbital and two p orbitals, which makes 3 hybrid orbitals, one per region. The regions are arranged in a trigonal plane, 120° apart (the molecular shape is bent if one region is a lone pair).

Watch out: The superscript counts p orbitals, not electrons. And a double or triple bond still counts as only one region.

That’s right: sp2

Practice 8

Four electron regions: hybridization?

Give me a hint

2 regions → sp, 3 regions → sp2, 4 regions → sp3.

Check the answer

Answer: sp3

Explain the worked problem to me

What you know: The atom has 4 electron regions. What you want: Its introductory hybridization label.

  1. Remember what counts as a region: each single, double, or triple bond is one region, and each lone pair is one region.
  2. Use the matching rule: 2 regions → sp, 3 regions → sp2, 4 regions → sp3.
  3. Four regions match sp3.

Check: The label uses one s orbital and three p orbitals, which makes 4 hybrid orbitals, one per region. The regions are arranged tetrahedrally, 109.5° apart (lone pairs would make the molecular shape trigonal pyramidal or bent).

Watch out: The superscript counts p orbitals, not electrons. And a double or triple bond still counts as only one region.

That’s right: sp3

Practice 9

For the same atom pair, is a triple bond usually shorter than a single bond?

Give me a hint

A higher bond order holds the same two atoms closer together.

Check the answer

Answer: Yes

Explain the worked problem to me

What you know: Two bonds join the same pair of atoms; one is triple, one is single. What you want: Whether the triple bond is shorter.

  1. Find each bond order. A single bond has bond order 1. A triple bond has bond order 3.
  2. Apply the rule: for the same pair of atoms, higher bond order means more shared electrons pulling the nuclei together, so the bond is shorter and stronger.
  3. 3 is greater than 1, so the triple bond is shorter: Yes.

Why it matters: Bond order lets you rank bond length and strength without measuring anything.

Watch out: The rule only works for the same atom pair. A C≡C bond and an H–H bond cannot be compared this way.

That’s right: Yes

Practice 10

6 bonding electrons and 2 antibonding electrons: MO bond order?

Give me a hint

Bond order = (bonding electrons − antibonding electrons) ÷ 2.

Check the answer

Answer: 2

Explain the worked problem to me

What you know: 6 bonding electrons and 2 antibonding electrons. What you want: The molecular-orbital (MO) bond order.

  1. Write the formula: bond order = (bonding electrons − antibonding electrons) ÷ 2.
  2. Subtract first, because antibonding electrons cancel some of the bonding: 6 − 2 = 4.
  3. Divide by 2, because a bond is a pair of electrons: 4 ÷ 2 = 2.

Check: A bond order of 2 means a double bond, which fits 4 net bonding electrons (2 pairs).

Watch out: Do not forget to divide by 2. Answering 4 counts electrons, not bonds.

That’s right: 2

8.4 Polarity and attractions between molecules

Whether a molecule is polar depends on two things: its bonds and its shape. Polarity then decides how strongly molecules attract each other, which shows up in properties like boiling point.

What you’ll use from earlier
  • Electronegativity (Chapter 3) and molecular shape (8.3); bonds within a molecule are different from attractions between molecules.

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
From bonds to molecules
  • Polar bond: electrons shared unevenly because one atom pulls harder (higher electronegativity).
  • Polar molecule: the bond pulls do not cancel, leaving a net dipole (a positive end and a negative end).
  • Shape decides whether the pulls cancel.
Linear carbon dioxide has two equal opposing C-to-O bond dipoles that cancel. Bent water has H-to-O bond dipoles that add to a net dipole toward oxygen. Gold arrows show bond dipoles; green shows the net molecular dipole.
CO2: polar bonds, no net dipole. H2O: bent, so the dipoles add. Crossed arrow tails mark the positive end.
Same kind of bonds, different result
CO2: nonpolar
Linear. Two equal pulls point in opposite directions and cancel.
H2O: polar
Bent. The two pulls both lean toward O and add up.
Periodic table with an arrow pointing up the side and an arrow pointing right along the bottom, both marking higher electronegativity. The shaded top-right group of N, O, F, Cl and Br marks the most electronegative elements, with fluorine (F) highest.
Electronegativity rises up a group and to the right, highest at F. Noble gases are usually left out.
Attractions between molecules
AttractionWho has itExample
London dispersionAll atoms and moleculesCH4
Dipole–dipolePolar moleculesNH3, H2O
Hydrogen bondingH bonded to N, O or F, attracted to a lone pair on N, O or F of a neighborNH3, H2O
These act between molecules. They are much weaker than the bonds within a molecule. An H in the formula is not enough for hydrogen bonding.
Stronger attractions, higher boiling point
Methane111.7 K (−161.5 °C)Ammonia239.8 K (−33.3 °C)Water373.2 K (100.0 °C)Boiling point (K)
Similar molar masses: 16.04, 17.03 and 18.02 g/mol. CH4 has dispersion only. NH3 and H2O add dipole–dipole and hydrogen bonding; water hydrogen-bonds the most.
1 · Recognize the clue
Look for this:"polar," "nonpolar," bond dipoles, symmetry, or a question about which attraction (like hydrogen bonding) acts between molecules.
2 · Follow the route
  1. Find each bond's polarity from electronegativity.
  2. Find the molecular shape.
  3. Decide whether the bond dipoles cancel.
  4. Compare the attractions between molecules.

On the periodic table: Electronegativity increases up and to the right, toward fluorine (F).

On the periodic table: O (Group 16) is right of C (Group 14), so O pulls harder in each C=O bond.

On the periodic table: N (Group 15) is more electronegative than H, so each N–H bond is polar.

On the periodic table: Molar masses come from the table: CH4 16.04, NH3 17.03, H2O 18.02 g/mol.

Worked example

Why is CH4 nonpolar overall while NH3 is polar?

Electronegativity helps predict a bond’s uneven pull. N and O attract shared bonding electrons more strongly than H. N–H and O–H bonds are polar. Combine their directional pulls with the molecule’s shape to find overall polarity. Bond polarity alone does not make a polar molecule; symmetric dipoles can cancel.
Bond polarity comes from the table; overall polarity also needs the shape.
  1. 1
    Check methane's shape

    Four identical outer atoms sit in a symmetric arrangement.

    CH4: 4 bonds + 0 lone pairs → tetrahedral
  2. 2
    Combine methane's dipoles

    Equal pulls in symmetric directions balance each other.

    4 equal bond dipoles, symmetric → net dipole = 0
  3. 3
    Check ammonia's shape

    A lone pair takes the fourth region, so the H atoms are not arranged like methane's.

    NH3: 3 bonds + 1 lone pair → trigonal pyramidal
  4. 4
    Combine ammonia's dipoles

    On the periodic table: N (Group 15, near the top right) is more electronegative than H, so each N–H bond is polar. The pyramid does not cancel them.

    Polar bonds?ShapeDipoles cancel?Molecule
    CH4SlightlyTetrahedralYesNonpolar
    NH3YesTrigonal pyramidalNoPolar
  5. ✓
    Answer

    CH4 is nonpolar overall; NH3 is polar. Shape matters as well as bond polarity.

8.5 Alkanes and simple organic molecules

Organic molecules are built on carbon chains. Carbon always makes four bonds, so a few rules give the Lewis structure, formula and bond angles of simple organic molecules.

What you’ll use from earlier
  • Lewis structures (8.1).
  • Electron regions, shapes, and bond angles (8.3).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea
The building rules
  • Carbon has 4 valence electrons, so it forms 4 bonds. Hydrogen forms 1.
  • Alkane: only carbon and hydrogen, joined by single bonds.
  • The name's start gives the number of carbons; the ending “-ane” means all single bonds.
Alkane formula
CnH2n+2
propane, n = 3: H = 2(3) + 2 = 8 → C3H8
The first ten alkanes
CarbonsNameFormula
1methaneCH4
2ethaneC2H6
3propaneC3H8
4butaneC4H10
5pentaneC5H12
6hexaneC6H14
7heptaneC7H16
8octaneC8H18
9nonaneC9H20
10decaneC10H22
Lewis structures of methane, ethane and propane, each carbon with four single bonds and about 109.5 degree angles; ethene with a C=C double bond and 120 degree angles; ethyne with a C≡C triple bond and a 180 degree angle.
Four regions: about 109.5°. Three (a double bond): about 120°. Two (a triple bond): 180°.
Bonds on carbon set the shape
Carbon hasRegionsShapeAngleExample
4 single bonds4Tetrahedralabout 109.5° (H–C–H, C–C–C)Alkanes, CH4
1 double + 2 single3Trigonal planarabout 120°Ethene, C2H4
1 triple + 1 single2Linear180°Ethyne, C2H2
1 · Recognize the clue
Look for this:alkane names ending in -ane, a carbon count, or a simple organic molecule with a question about formula or bond angle.
2 · Follow the route
  1. Count valence electrons: 4 per C, 1 per H.
  2. Connect the carbons in a chain.
  3. Add hydrogens so every carbon has 4 bonds.
  4. Count regions on each carbon and name the shape.

On the periodic table: Carbon is in Group 14 (4 valence electrons, 4 bonds) and hydrogen in Group 1 (1 valence electron, 1 bond).

Worked example

Draw the Lewis structure of propane, C3H8, and give its bond angles.

  1. 1
    Count valence electrons

    On the periodic table: C (Group 14) has 4; H (Group 1) has 1.

    3(4) + 8(1) = 20 electrons
  2. 2
    Connect the carbons

    Carbons form the chain. Hydrogen makes only one bond, so it goes on the outside.

    C–C–C
  3. 3
    Add hydrogens

    Give each carbon 4 bonds. Ten bonds use all 20 electrons.

    CCCHHHHHHHH2 C–C + 8 C–H = 10 bonds, no lone pairs
    10 bonds × 2 electrons = 20 electrons    CH3–CH2–CH3
  4. 4
    Name the shape

    Every carbon has 4 bonding regions and no lone pairs.

    Each CRegionsLone pairsShapeAngle
    CH3–, –CH2–, –CH340Tetrahedralabout 109.5°
  5. ✓
    Answer

    CH3–CH2–CH3, with angles of about 109.5° at every carbon.

Section practice · 5 questions · tap to open

What this set is for: practicing the method from this section while it’s fresh. Every problem uses only what this section has taught.

Practice 1

What is the formula of heptane?

Give me a hint

Hepta- means 7. Use CnH2n+2.

Check the answer

Answer: C7H16

Explain the worked problem to me

What you know: Heptane is an alkane, and hepta- means 7 carbons. What you want: Its formula.

  1. Use the alkane formula CnH2n+2 with n = 7.
  2. Hydrogens: 2n + 2 = 2(7) + 2 = 14 + 2 = 16.
  3. So the formula is C7H16.

Check: Each end carbon holds 3 H and each of the 5 middle carbons holds 2 H: 3 + 3 + 5(2) = 16.

Watch out: Do the multiplying before adding 2. 2(7 + 2) = 18 is wrong.

That’s right: C7H16

Practice 2

An alkane has 5 carbons. Name it and give its formula.

Give me a hint

Pent- means 5. Use CnH2n+2.

Check the answer

Answer: Pentane, C5H12

Explain the worked problem to me

What you know: An alkane with 5 carbons. What you want: Its name and formula.

  1. Name it from the prefix: 5 carbons is pentane (penta- means 5).
  2. Use CnH2n+2 with n = 5. Hydrogens: 2(5) + 2 = 12.
  3. The answer is Pentane, C5H12.

Check: Two end carbons with 3 H each and three middle carbons with 2 H each: 3 + 3 + 3(2) = 12.

That’s right: Pentane, C5H12

Practice 3

What is the H–C–H bond angle in methane, CH4?

Give me a hint

Count the bonding regions on carbon.

Check the answer

Answer: About 109.5°

Explain the worked problem to me

What you know: CH4 has carbon bonded to 4 H atoms. What you want: The H–C–H bond angle.

  1. Count carbon's electron regions: 4 single bonds and no lone pairs, so 4 regions.
  2. Four regions spread as far apart as possible in 3D, giving a tetrahedral shape.
  3. The ideal tetrahedral angle is about 109.5°.

Check: All four regions are identical bonds with no lone pairs, so nothing squeezes the angle below 109.5°.

Watch out: A flat drawing on paper suggests 90°. Methane is 3D, so the angle is 109.5°.

That’s right: About 109.5°

Practice 4

Ethene, C2H4, has a C=C double bond. What is the shape around each carbon, and the H–C–H angle?

Give me a hint

A double bond counts as one region.

Check the answer

Answer: Trigonal planar, about 120°

Explain the worked problem to me

What you know: In C2H4 each carbon has a C=C double bond and two C–H bonds. What you want: The shape at each carbon and the H–C–H angle.

  1. Count regions on one carbon: the double bond counts as 1 region, plus 2 C–H bonds, giving 3 regions.
  2. Carbon has no lone pairs (4 bonds use all 4 valence electrons).
  3. Three regions with no lone pairs spread out in a flat triangle: Trigonal planar, about 120°.

Check: Bonds to carbon: 2 (double) + 1 + 1 = 4, so carbon's octet is complete.

Watch out: Do not count the double bond as 2 regions. That would give 4 regions and the wrong angle, 109.5°.

That’s right: Trigonal planar, about 120°

Practice 5

Methanol, CH3OH: how many lone pairs are on oxygen, and what is the shape at oxygen?

Give me a hint

Oxygen has 6 valence electrons and makes 2 bonds here.

Check the answer

Answer: 2 lone pairs; bent, a little less than 109.5°

Explain the worked problem to me

What you know: In CH3OH, oxygen bonds to C and to H. What you want: Oxygen's lone pairs and the shape at oxygen.

  1. Oxygen has 6 valence electrons. It uses 2 of them in its 2 bonds, leaving 6 − 2 = 4 electrons, which is 2 lone pairs.
  2. Count regions on O: 2 bonds + 2 lone pairs = 4 regions, arranged tetrahedrally.
  3. Only 2 of the 4 positions hold atoms, so the C–O–H shape is bent. Lone pairs push the bonds closer, so the angle shrinks: 2 lone pairs; bent, a little less than 109.5°.

Check: Oxygen's octet: 2 bonds × 2 + 2 lone pairs × 2 = 8 electrons. It matches water, which is also bent.

Watch out: Lone pairs set the angle but are not named in the shape. Do not call it tetrahedral.

That’s right: 2 lone pairs; bent, a little less than 109.5°

Mixed chapter review

Answer in your notebook before revealing help. Keep your first attempt separate from an answer corrected with help.

Mixed chapter review · 10 questions · tap to open

What this set is for: problems from every section of the chapter, mixed together and not labeled, the way a test asks them. Your job is to pick the right method yourself. Do this before the skill check.

Practice 1

Count the valence electrons for NO3−. Use H = 1, N = 5, O = 6 valence electrons.

Give me a hint

Add the valence electrons of every atom, then add 1 electron for the negative charge.

Check the answer

Answer: 24

Explain the worked problem to me

What you know: NO3− has 1 N, 3 O and a 1− charge. What you want: Total valence electrons.

  1. Add every atom's valence electrons: N gives 5; three O give 3 × 6 = 18. 5 + 18 = 23.
  2. The 1− charge means one extra electron: 23 + 1 = 24.

Check: Recount atom by atom: 5 (N) + 6 + 6 + 6 (three O) + 1 (charge) = 24.

Watch out: A negative charge adds electrons; a positive charge removes them. Subtracting here would give 22.

Your route: Add every atom’s valence count → Add for negative charge or subtract for positive.

On the periodic table: Nitrogen (N) is in Group 15, so it has 5 valence electrons. Oxygen (O) is in Group 16, so it has 6.

That’s right: 24

Practice 2

A central atom has one double bond, one single bond, and one lone pair. How many VSEPR electron regions does it have?

Give me a hint

Count each bond once, even a double bond, then add one region for each lone pair.

Check the answer

Answer: 3

Explain the worked problem to me

What you know: The central atom has 1 double bond, 1 single bond and 1 lone pair. What you want: The number of VSEPR electron regions.

  1. Count each bonded neighbor once, whatever the bond type: double bond = 1, single bond = 1. That is 2 regions.
  2. Add the lone pair as 1 more region: 2 + 1 = 3.

Check: 3 regions with 1 lone pair gives a bent shape, like the central O in O3.

Watch out: Counting the double bond as 2 gives 4, which is wrong. Regions count directions, not electron pairs in bonds.

Your route: Count each bonded group once → Add lone pairs.

That’s right: 3

Practice 3

A central atom has 6 bonding regions and no lone pairs. What is its ideal molecular shape?

Give me a hint

Bond-region and lone-pair counts are supplied.

Check the answer

Answer: Octahedral

Explain the worked problem to me

What you know: 6 bonding regions, 0 lone pairs. What you want: The molecular shape.

  1. Count regions: 6 bonds + 0 lone pairs = 6 regions.
  2. Six regions point to the corners of an octahedron (up, down, and four around the middle), neighboring corners 90° apart.
  3. With no lone pairs, every position holds an atom, so the molecular shape equals the electron arrangement: Octahedral.

Check: For 6 regions, the shapes are octahedral (0 lone pairs), square pyramidal (1) and square planar (2). Zero lone pairs picks octahedral.

Your route: Count regions → Verify no lone pairs → Name region arrangement.

That’s right: Octahedral

Practice 4

A central atom has 2 bonds and 3 lone pairs, all counted as separate regions. What molecular shape results?

Give me a hint

Both bond and lone-pair counts are given; each one is a separate region.

Check the answer

Answer: Linear

Explain the worked problem to me

What you know: 2 bonds and 3 lone pairs on the central atom. What you want: The molecular shape.

  1. Count regions: 2 + 3 = 5, so the regions form a trigonal bipyramid.
  2. Lone pairs take the middle ring (equatorial) positions first. All 3 middle positions fill with lone pairs.
  3. The 2 bonded atoms sit at the top and bottom, 180° apart. Naming only the atoms gives Linear.

Check: For 5 regions, 0, 1, 2 or 3 lone pairs give trigonal bipyramidal, seesaw, T-shaped or linear. 3 lone pairs → linear.

Watch out: 2 bonded atoms does not mean you can skip the region count. Here the lone pairs decide where the atoms go.

Your route: Count bonds plus lone pairs → Arrange regions → Name bonded-atom shape.

That’s right: Linear

Practice 5

Why is an ideal symmetric BeCl2 molecule nonpolar even though its bonds are polar?

Give me a hint

Molecular polarity is asked from a known shape.

Check the answer

Answer: Its bond dipoles cancel as vectors

Explain the worked problem to me

What you know: BeCl2 has two polar Be–Cl bonds. What you want: Why the molecule is still nonpolar.

  1. Find the shape: Be has 2 bonds and no lone pairs, so 2 regions, which is linear (180°).
  2. Each Be–Cl bond dipole points from Be toward Cl. The two dipoles are equal in size and point in exactly opposite directions.
  3. Equal, opposite pulls add to zero, like a balanced tug-of-war: Its bond dipoles cancel as vectors.

Why it matters: CO2 works the same way. A bent molecule like H2O cannot cancel, so it is polar.

Watch out: Polar bonds do not guarantee a polar molecule. Always check the shape.

Your route: Identify molecular symmetry → Add equal bond dipoles as vectors.

On the periodic table: Chlorine (Cl) in Group 17 is far to the right of beryllium (Be) in Group 2, so Cl is more electronegative and each Be–Cl bond is polar.

That’s right: Its bond dipoles cancel as vectors

YOUR PERIODIC-TABLE REFERENCE

17ClChlorine35.45Group 17 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 6

Among CH3OH, CH4, H2S, and Ar, which pure substance can hydrogen-bond with its own particles?

Give me a hint

Look for an H bonded directly to N, O, or F.

Check the answer

Answer: CH3OH

Explain the worked problem to me

What you know: Four substances: CH3OH, CH4, H2S and Ar. What you want: Which can hydrogen-bond with itself.

  1. Look for an H bonded directly to N, O or F. CH4 has H only on C. H2S has H on S, which is not N, O or F. Ar has no H at all.
  2. CH3OH has an O–H bond, so it qualifies.
  3. Check for a partner: the O in a neighboring CH3OH has 2 lone pairs to accept that H. So the answer is CH3OH.

Check: Both conditions are met only by methanol: the H–O donor and the O lone-pair acceptor.

Watch out: The three H atoms in the CH3 group do not count. Only the H on O can hydrogen-bond.

Your route: Check H bonded to N O or F → Check lone-pair partner.

That’s right: CH3OH

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1
6CCarbon12.01Group 14 · Period 2
7NNitrogen14.01Group 15 · Period 2
8OOxygen16.00Group 16 · Period 2
9FFluorine19.00Group 17 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 7

In NH3, N has 2 nonbonding electrons and 6 bonding electrons. What is its formal charge? Use 5 valence electrons.

Give me a hint

Formal charge = valence − nonbonding electrons − half the bonding electrons.

Check the answer

Answer: 0

Explain the worked problem to me

What you know: N has 5 valence electrons, 2 nonbonding and 6 bonding electrons. What you want: N's formal charge.

  1. Use formal charge = valence − nonbonding − half of bonding.
  2. Half the bonding electrons: 6 ÷ 2 = 3, because each bond is shared with an H.
  3. Formal charge = 5 − 2 − 3 = 0.

Check: Each H is also 0 (1 − 0 − 1). The total 0 matches neutral NH3.

Watch out: Subtract only half of the bonding electrons. Subtracting all 6 gives −3.

Your route: Valence minus nonbonding minus half bonding.

On the periodic table: Nitrogen (N) is in Group 15, so it has 5 valence electrons.

That’s right: 0

Practice 8

Two proposed drawings of O3 are called resonance contributors. Which feature must stay unchanged?

Give me a hint

Alternative Lewis drawings are being compared.

Check the answer

Answer: The atom connections and total charge

Explain the worked problem to me

What you know: Two drawings of O3 are resonance contributors. What you want: What cannot change between them.

  1. Recall the rule: resonance drawings keep the same atom connections but place electrons differently.
  2. In O3, the double bond moves from the left O–O to the right O–O: one end O turns a lone pair into a bond pair, and the other end O turns a bond pair back into a lone pair. The O–O–O skeleton stays.
  3. The electrons are only moved, never added or removed, so the total charge stays too: The atom connections and total charge.

Check: Both O3 drawings use 3 × 6 = 18 valence electrons and have formal charges adding to 0.

Watch out: Moving an atom to a new position makes a different molecule, not a resonance form.

Your route: Keep atoms connected → Redistribute electrons.

That’s right: The atom connections and total charge

Practice 9

How many valence electrons are available for NH4+? N contributes 5 and each H contributes 1.

Give me a hint

Add the valence electrons of every atom, then subtract 1 electron for the positive charge.

Check the answer

Answer: 8

Explain the worked problem to me

What you know: NH4+ has 1 N, 4 H and a 1+ charge. What you want: Total valence electrons.

  1. Add the atoms: N gives 5; four H give 4 × 1 = 4. 5 + 4 = 9.
  2. The 1+ charge means one electron is missing: 9 − 1 = 8.

Check: 8 electrons make exactly 4 N–H bonds of 2 electrons each, with no lone pair left.

Watch out: Adding 1 for a positive charge gives 10. Positive means fewer electrons.

Your route: Add valence counts → Subtract positive charge.

On the periodic table: Nitrogen (N) is in Group 15, so it has 5 valence electrons. Hydrogen (H) is in Group 1, so it has 1.

That’s right: 8

Practice 10

A central atom has two double bonds and no lone pairs. How many electron regions surround it?

Give me a hint

Each double bond counts as one region; then add any lone pairs.

Check the answer

Answer: 2

Explain the worked problem to me

What you know: The central atom has 2 double bonds and 0 lone pairs. What you want: The number of electron regions.

  1. Count each bonded group once: each double bond is 1 region, so 2 double bonds give 2 regions.
  2. Add lone pairs: 2 + 0 = 2.

Check: This is the carbon in CO2 (O=C=O): 2 regions, linear, 180°.

Watch out: Counting each double bond as 2 gives 4 regions and a wrong tetrahedral shape.

Your route: Count each bonded group once → Add lone pairs.

That’s right: 2

Skill check: the chapter’s key skills

Try each fresh question without the worked examples. Record your first answer and whether you used a hint. Explain the route as well as the answer.

Skill check · 7 questions · tap to open

What this set is for: one question on each key skill from this chapter. Try each on your own, without hints, then mark how it went in the tracker below. A miss tells you which section to review.

Skill 1: Count valence electrons and draw a simple Lewis structure.

How many valence electrons in H2O? Describe its Lewis structure.

Give me a hint

Count valence electrons, draw bonds and lone pairs, then count electron regions around the central atom. Use those regions to identify shape.

Check the answer

2(1) + 6 = 8; two O–H bonds and two lone pairs on O.

On the periodic table: Hydrogen (H) is in Group 1, so it has 1 valence electron. Oxygen (O) is in Group 16, so it has 6.

Explain the worked problem to me

What you know: H2O has 2 H and 1 O. What you want: The valence electron count and the Lewis structure.

  1. Add valence electrons: 2 H × 1 = 2, plus O = 6. Total 2 + 6 = 8.
  2. Put O in the center (H can make only one bond). Draw two O–H single bonds: that uses 2 × 2 = 4 electrons.
  3. Place the remaining 8 − 4 = 4 electrons on O as 2 lone pairs. Result: 2(1) + 6 = 8; two O–H bonds and two lone pairs on O.

Check: O has 4 + 4 = 8 electrons (octet); each H has 2. 4 regions on O give a bent shape.

Watch out: Do not forget the lone pairs. Without them O would have only 4 electrons.

That’s right: 2(1) + 6 = 8; two O–H bonds and two lone pairs on O.

On the periodic table: Hydrogen (H) is in Group 1, so it has 1 valence electron. Oxygen (O) is in Group 16, so it has 6.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1
8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 2: Find molecular shape from bonds and lone pairs.

Three bonds and one lone pair around the central atom: shape?

Give me a hint

Add the bonds and lone pairs to get the electron regions around the central atom. Then name only the positions of the bonded atoms.

Check the answer

Trigonal pyramidal; four electron regions.

Explain the worked problem to me

What you know: The central atom has 3 bonds and 1 lone pair. What you want: The molecular shape.

  1. Count regions: 3 bonds + 1 lone pair = 4 regions, arranged tetrahedrally.
  2. Atoms sit at only 3 of the 4 corners. The lone pair sits on top and pushes the 3 bonds down.
  3. Naming only the atom positions gives a pyramid with a triangle base: Trigonal pyramidal; four electron regions.

Check: NH3 fits this pattern: three N–H bonds and one lone pair on N make it trigonal pyramidal.

Watch out: Trigonal planar has 3 regions and no lone pair. One lone pair adds a 4th region and lifts the shape out of the plane.

That’s right: Trigonal pyramidal; four electron regions.

Skill 3: Use electron-region counts for introductory hybridization and distinguish sigma and pi bonds.

For CO2, give carbon’s hybridization and count the sigma and pi bonds.

Give me a hint

Count carbon’s electron regions (a double bond counts as one region). Every bond has one sigma bond; each double bond adds one pi bond.

Check the answer

CO2 carbon has two regions: sp, linear. It has two C=O double bonds, so the molecule contains 2 sigma and 2 pi bonds.

Explain the worked problem to me

What you know: CO2 is O=C=O. What you want: Carbon's hybridization and the sigma and pi bond counts.

  1. Count regions on C: 2 double bonds, no lone pairs, so 2 regions. Two regions → sp, linear.
  2. Each double bond has 1 sigma and 1 pi bond.
  3. Two double bonds: 2 × 1 = 2 sigma and 2 × 1 = 2 pi. Result: CO2 carbon has two regions: sp, linear. It has two C=O double bonds, so the molecule contains 2 sigma and 2 pi bonds.

Check: Carbon makes 2 + 2 = 4 bonds in all, and 2 sigma + 2 pi = 4.

Watch out: Count regions, not bonds, for hybridization. 4 bonds would wrongly suggest sp3.

That’s right: CO2 carbon has two regions: sp, linear. It has two C=O double bonds, so the molecule contains 2 sigma and 2 pi bonds.

YOUR PERIODIC-TABLE REFERENCE

6CCarbon12.01Group 14 · Period 2
8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 4: Choose the strongest relevant intermolecular attraction.

Pure H2O: which strong intermolecular attraction applies?

Give me a hint

Look for H bonded directly to N, O, or F, and a lone pair on N, O, or F of a neighboring molecule.

Check the answer

Hydrogen bonding, because H is bonded to O.

Explain the worked problem to me

What you know: The substance is pure H2O. What you want: The strongest relevant intermolecular attraction.

  1. Check for H bonded to N, O or F: water has two O–H bonds, so yes.
  2. Check for a partner: the O on a neighboring water molecule has 2 lone pairs to accept the H.
  3. Both conditions are met, so the answer is Hydrogen bonding, because H is bonded to O.

Why it matters: Water also has dispersion and dipole–dipole attractions, but hydrogen bonding is the strongest of these, which explains water's high boiling point.

Watch out: A hydrogen bond is an attraction between molecules, not the O–H covalent bond inside one molecule.

That’s right: Hydrogen bonding, because H is bonded to O.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1
8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 5: Use bond polarity and shape to decide molecular polarity.

CO2 has polar bonds and is linear/symmetric. Polar overall?

Give me a hint

Consider bond polarity and molecular shape together. Check whether the bond dipoles cancel.

Check the answer

No: its equal opposing bond dipoles cancel.

Explain the worked problem to me

What you know: CO2 has polar C=O bonds and a linear, symmetric shape. What you want: Whether the molecule is polar overall.

  1. Each C=O bond dipole points from C toward the more electronegative O.
  2. Linear means the two O atoms sit 180° apart, so the two dipoles point in opposite directions.
  3. The two bonds are identical, so the pulls are equal. Equal and opposite pulls cancel: No: its equal opposing bond dipoles cancel.

Why it matters: Compare bent H2O: its dipoles do not point in opposite directions, so they add to a net dipole and water is polar.

Watch out: Polar bonds alone do not make a polar molecule. Shape decides.

That’s right: No: its equal opposing bond dipoles cancel.

Skill 6: Draw Lewis structures and give bond angles for the first ten alkanes and a few simple organic molecules.

Give the formula of butane and its C–C–C bond angle.

Give me a hint

Four carbons; use CnH2n+2.

Check the answer

Answer: C4H10; about 109.5°

Explain the worked problem to me

What you know: Butane is an alkane with 4 carbons. What you want: Its formula and C–C–C angle.

  1. Use CnH2n+2 with n = 4. Hydrogens: 2(4) + 2 = 10.
  2. Each carbon has 4 single bonds and no lone pairs, so 4 regions: tetrahedral.
  3. Tetrahedral angle is about 109.5°, so the answer is C4H10; about 109.5°.

Check: End carbons hold 3 H each and middle carbons 2 H each: 3 + 3 + 2 + 2 = 10.

Watch out: The chain is drawn as a straight line on paper, but it is zigzag. The angle is 109.5°, not 180°.

That’s right: C4H10; about 109.5°

Skill 7: Use formal charge and resonance to choose the best Lewis structure.

Two Lewis structures for CO2 each have 16 valence electrons. (A) O=C=O with two lone pairs on each O. (B) O≡C–O with one lone pair on the triple-bonded O and three lone pairs on the single-bonded O. Find each atom’s formal charge and choose the better structure.

Give me a hint

For each atom, start from its valence count, subtract every lone-pair electron, then subtract half of its bonding electrons. A triple bond counts as 6 bonding electrons.

Check the answer

Structure A; every atom in A has formal charge 0, while B puts +1 and −1 on the two O atoms.

On the periodic table: Carbon (C) is in Group 14, so it has 4 valence electrons. Oxygen (O) is in Group 16, so it has 6.

Explain the worked problem to me

What you know: CO2 has 16 valence electrons. Structure A is O=C=O with two lone pairs on each O. Structure B is O≡C–O with one lone pair on the triple-bonded O and three lone pairs on the single-bonded O. What you want: The formal charges in each structure and the better structure.

  1. Use FC = V − N − B/2 for each atom, where V is valence electrons, N is nonbonding electrons and B is bonding electrons.
  2. Structure A: each O has FC = 6 − 4 − 4/2 = 0. Carbon has FC = 4 − 0 − 8/2 = 0.
  3. Structure B: the triple-bonded O has FC = 6 − 2 − 6/2 = +1. Carbon has FC = 4 − 0 − 8/2 = 0. The single-bonded O has FC = 6 − 6 − 2/2 = −1.
  4. Prefer the structure with the smallest formal charges. A has all zeros, so the answer is Structure A.

Check: The formal charges add to the overall charge in both structures: A gives 0 + 0 + 0 = 0, and B gives +1 + 0 + (−1) = 0. CO2 is neutral, so both totals fit.

Watch out: B and its mirror drawing O–C≡O keep the same atom connections and move only electrons, so they are resonance drawings of each other. Being a resonance drawing does not make B better. Its +1 sits on oxygen, a very electronegative atom, which is unfavorable.

That’s right: Structure A; every atom in A has formal charge 0, while B puts +1 and −1 on the two O atoms.

On the periodic table: Carbon (C) is in Group 14, so it has 4 valence electrons. Oxygen (O) is in Group 16, so it has 6.

YOUR PERIODIC-TABLE REFERENCE

6CCarbon12.01Group 14 · Period 2
8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

What to do next

Mark each skill: “I can do it independently,” “I needed help,” or “I need another explanation.” Revisit the section with the same goal for any skill needing help, then try a different question. A successful first attempt is evidence of progress; one answer alone does not establish lasting mastery.

Practice with instant feedbackNext: Chapter 9

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Take the 70-question practice final

Modeled on standardized first-term general chemistry finals: 70 questions, 110 minutes. Timed or untimed. Full feedback on every question, plus your score by topic. The first 20 are free; the full final is $7.99 for 2 weeks.

Start the 20 free questionsGet the full final · $7.99

Original ChemGem questions.