🧪 Contents · Interactive Lab Chapter 7 · Chemical Bonding and Molecular Geometry
Chemistry · Chapter 7

Chemical Bonding and Molecular Geometry

Atoms almost never travel alone. The way they join sets the rules for everything that happens next. This chapter follows two ways of joining. In the first, one atom hands an electron over for good. In the second, two atoms share. Then we get to shape, because a bonded group of atoms is not a shapeless blob. It has angles you can predict, and you predict them by counting. By the end you can take a formula like SF4, draw it, name its shape, and say whether its pull is lopsided or even.

By the end of this chapter you can

  • Show how an ionic bond forms, and say what makes lattice energy large or small
  • Use Pauling electronegativity to sort a bond as nonpolar, polar, or ionic
  • Draw a Lewis structure from a formula, and name the three ways the octet rule breaks
  • Write resonance forms, and use formal charge to pick the best one
  • Count electron domains, then name both the domain shape and the molecular shape
  • Give the bond angle for every common shape, including the squeezed cases
  • Add bond dipoles as arrows to tell a polar molecule from a nonpolar one

Section 1Ionic bonding and lattice energy

One atom gives, the other takes

Picture sodium next to chlorine. Sodium has one electron in its outer shell, while chlorine is one electron short of a full shell. So sodium hands its electron over. Sodium is left with 11 protons and 10 electrons, so it becomes Na+. A positive ion like this is a cation. Chlorine gains the electron and becomes Cl−. A negative ion is an anion. Opposite charges pull on each other, and that pull is an ionic bond.

There are no tidy pairs here. Each Na+ tugs on every Cl− nearby, and each Cl− tugs back on every Na+. The ions stack into a repeating solid called a lattice. In table salt, every sodium ion has six chloride ions around it, and every chloride ion has six sodium ions around it. So NaCl is a ratio, not a tiny two-atom unit you could pull out with tweezers.

The sodium chloride lattice in cross-section Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ r₀ = 283 pm Charges alternate in every row and every column. Six neighbors each Every ion is held by the six ions of opposite charge touching it. r₀ sets the pull r₀ is the gap between two neighboring ion centers.
Figure 7.1 Salt is a repeating grid, not a molecule. The ratio 1:1 in the formula NaCl comes from the pattern, and r0 is the center-to-center gap between neighbors.

Lattice energy measures the glue

Lattice energy (ΔHlattice) is the energy it takes to pull one mole of the solid apart into free gas-phase ions. A large value means a tough solid. For NaCl it is about 769 kJ/mol, while for MgO it is about 3795 kJ/mol. So MgO takes nearly five times as much energy to break up. That shows up in the melting points. Salt melts at 801 °C, while MgO holds out past 2800 °C.

Two things set the size, and both come from Coulomb's law. Charge is the first, and the pull grows with the two charges multiplied together. Distance is the second, and the pull fades as the ion centers move apart.

ΔHlattice  ∝  Q1Q2 / r0

Here Q1 and Q2 are the ion charges, and r0 is the gap between the centers. Double both charges and the energy roughly quadruples, while swapping in a fatter ion makes the energy drop. MgO wins on both counts. Mg2+ and O2− each carry twice the charge of the salt ions, and both are smaller, so r0 is shorter.

Worked example · predict the lattice energy of NaF

Problem. NaCl has a lattice energy near 769 kJ/mol. Use ionic radii to estimate the value for NaF.

Step 1 · get r₀ for each solid Ionic radii: Na+ 102 pm, Cl− 181 pm, F− 133 pm. Add the two radii to get the center-to-center gap.
NaCl: r0 = 102 pm + 181 pm = 283 pm.
NaF: r0 = 102 pm + 133 pm = 235 pm.

Step 2 · check the charges Both solids pair a 1+ ion with a 1− ion. So Q1Q2 is the same for both, and only the distance changes.

Step 3 · scale by the distance Energy goes as 1/r0, so the smaller gap gives the bigger number.
769 kJ/mol × (283 pm / 235 pm) = 769 × 1.204 = 926 kJ/mol.

Step 4 · check it against the real value The measured lattice energy of NaF is about 910 kJ/mol. The estimate is high by 16 kJ/mol, or about 2 percent. Good enough to rank solids, and a reminder that the model is only the leading term.

Key points
  • Metals lose electrons to form cations. Nonmetals gain them to form anions.
  • An ionic solid is a lattice of ions, so its formula gives a ratio.
  • Lattice energy is the energy to split one mole of solid into gas-phase ions.
  • It rises with bigger charges and falls with a bigger r0.
  • MgO beats NaCl on both counts, so its lattice energy is about five times larger.
Worth knowing. Lattice energy is also why salt dissolves in water but not in oil. Water molecules can surround each ion and pay back most of that 769 kJ/mol. Oil cannot, so the lattice stays put.
Interactive Build a lattice and see what makes lattice energy rise. Open full ↑

Section 2Covalent bonds, electronegativity and polarity

Sharing instead of handing over

Two chlorine atoms cannot both win. Each one needs a single electron to fill its shell, and neither will give one up. So they share a pair instead. A shared pair of electrons holding two atoms together is a covalent bond. The pair sits in the space between the two nuclei, and both nuclei pull on it. That shared pull is the bond.

Sharing is rarely fair, because some atoms tug harder on a shared pair than others. Electronegativity is the number that scores that tug. Linus Pauling built the scale we still use, and he put fluorine at the top at 3.98, which makes it the greediest atom there is. Cesium sits near the bottom at 0.79. Electronegativity rises as you go right across a row and falls as you go down a column. Those two trends track the same cause, because nuclear charge grows to the right and the outer shell drifts farther from the nucleus as you go down.

ElementPauling valueElementPauling value
F3.98Si1.90
O3.44Al1.61
Cl3.16Be1.57
N3.04Mg1.31
Br2.96Li0.98
I2.66Na0.93
S2.58Ca1.00
C2.55K0.82
H2.20Cs0.79
P2.19B2.04

The gap in the numbers tells you the bond type

Subtract the smaller value from the larger one. That difference, written ΔEN, sorts bonds along a smooth range. A tiny gap means the pair sits in the middle, which is a nonpolar covalent bond. A medium gap drags the pair toward the greedier atom, which is a polar covalent bond, and one end picks up a partial negative charge written δ−. The other end goes δ+. A huge gap means the electron is taken, not shared, and the bond counts as ionic.

ΔENBond typeExampleΔEN for it
below 0.4Nonpolar covalentC−H in CH42.55 − 2.20 = 0.35
0.4 to 1.8Polar covalentO−H in H2O3.44 − 2.20 = 1.24
above 1.8IonicNa−Cl in salt3.16 − 0.93 = 2.23

Treat those cutoffs as road signs, not walls. H−F has ΔEN = 1.78 and still behaves as a very polar covalent bond. NaI has ΔEN = 1.73 yet is a true ionic solid. The better habit is to read ΔEN as a dial. Larger gap, more lopsided bond.

Nonpolar, polar and ionic bonding compared Nonpolar covalent Polar covalent Ionic Cl–Cl  ΔEN = 0 H–Cl  ΔEN = 0.96 Na–Cl  ΔEN = 2.23 ClCl HCl δ+δ− dipole points to Cl NaCl 1+1− The cloud sits deadcenter. No dipole. The cloud leans towardthe greedier atom. The electron has movedacross. Full charges.
Figure 7.2 One smooth range, not three separate worlds. As ΔEN grows, the shared cloud slides from the middle toward one atom, and at the far end it stops being shared at all.

Putting a number on the lopsidedness

A polar bond has a dipole moment, written μ. It equals the size of the partial charge times the distance between the charge centers. The unit is the debye (D), where 1 D = 3.336 × 10−30 C·m. Chemists often report the same idea as percent ionic character. To get it you compare the real dipole moment with the one you would see if a whole electron had moved across.

Worked example · how ionic is the H−Cl bond?

Problem. HCl has a measured dipole moment of 1.08 D and a bond length of 127 pm. What fraction of a full charge has shifted?

Step 1 · find μ for a full transfer Use μ = Qr with one whole electron charge moved across the bond.
μ = (1.602 × 10−19 C)(127 × 10−12 m) = 2.03 × 10−29 C·m.

Step 2 · change the unit to debye Divide by the size of one debye.
(2.03 × 10−29 C·m) / (3.336 × 10−30 C·m per D) = 6.10 D.

Step 3 · compare the real value to that limit 1.08 D / 6.10 D = 0.177.

Step 4 · read the answer The bond is about 18 percent ionic. Only about one fifth of an electron charge has shifted toward chlorine. The ΔEN of 0.96 told us to expect a clearly polar bond, and this number says how polar.

Key points
  • A covalent bond is a shared pair of electrons held by two nuclei.
  • Electronegativity scores how hard an atom tugs on a shared pair.
  • ΔEN below 0.4 is nonpolar, 0.4 to 1.8 is polar, above 1.8 is ionic.
  • A polar bond carries δ+ on one atom and δ− on the other.
  • The dipole moment μ puts a measured number on that split.
Worth knowing. No bond is perfectly ionic. Even in NaCl, a little electron density stays shared. That leftover sharing is why the simple charge-and-distance model misses the measured lattice energy by a few percent.
Interactive Slide the electronegativity difference from nonpolar to ionic. Open full ↑

Section 3Lewis structures

Dots for electrons, lines for bonds

A Lewis structure is a map of where the outer electrons sit. Only valence electrons appear, because only they do the bonding. A shared pair is drawn as a line and called a bonding pair. A pair that belongs to one atom alone is drawn as two dots and called a lone pair. Two lines between the same two atoms make a double bond, and three lines make a triple bond. The number of shared pairs between two atoms is the bond order.

Most atoms stop when their valence shell holds eight electrons. That habit is the octet rule, and it comes straight from the s and p orbitals of the outer shell. Hydrogen is the standing exception, since its shell is full at two.

Five steps that always work

Build every structure the same way. First, add up the valence electrons from all the atoms. Add one more for each unit of negative charge, and take one away for each unit of positive charge. Second, sketch a skeleton with one line per bond. The atom with the lowest electronegativity usually goes in the middle, and hydrogen is never in the middle. Third, give every outer atom a full octet using lone pairs. Fourth, park any electrons that are left on the central atom. Fifth, if the central atom is still short of eight, pull a lone pair in from a neighbor to make a double or triple bond.

Keep a running count the whole time, because every line you draw uses two electrons and every pair of dots uses two. When your total matches the number from step one, the structure is complete. That count is the one thing that catches most mistakes.

Worked example · draw methanal, CH2O

Problem. Give a complete Lewis structure for CH2O.

Step 1 · count valence electrons C has 4, each H has 1, O has 6.
4 + 2(1) + 6 = 12 valence electrons, which is 6 pairs.

Step 2 · draw the skeleton Hydrogen is never in the middle, and carbon is less electronegative than oxygen. So carbon goes in the middle. Both H atoms and the O attach to it. Three single bonds use 3 × 2 = 6 electrons. Six are left.

Step 3 · fill the outer atoms Each H is already full with its two shared electrons. Oxygen needs 6 more, so it gets 3 lone pairs. That uses the last 6 electrons. Running total: 6 + 6 = 12. All 12 are placed.

Step 4 · check the central atom Carbon has only 3 bonds, so it holds 6 electrons. It is 2 short of an octet.

Step 5 · make a multiple bond Shift one lone pair from oxygen into the C−O bond. Now that bond is a double bond. Carbon has 4 bonds, so 8 electrons. Oxygen has 2 bonds plus 2 lone pairs, which is 4 + 4 = 8 electrons. The total is still 12, and both atoms have an octet.

Answer. H2C=O, with two lone pairs on oxygen and a C=O bond order of 2.

Three ways the octet rule breaks

The rule is a strong habit, not a law. It fails in three tidy families.

Some molecules have an odd electron count, so no structure can pair everything up. Nitrogen monoxide, NO, has 5 + 6 = 11 valence electrons. One electron has to stay alone, and the nitrogen ends up with seven. A molecule with an unpaired electron is a free radical, and radicals like this tend to be reactive.

Some central atoms settle for less than eight, and beryllium and boron do this often. BF3 has 3 + 3(7) = 24 valence electrons. Three B−F bonds use 6, and 18 fill the lone pairs on fluorine. Boron is left with only 6 electrons, and the molecule is stable anyway. That gap is why BF3 grabs a lone pair from almost anything.

Atoms in period 3 and below can hold more than eight. This is an expanded valence shell, and it needs a central atom big enough to fit the extra neighbors. PCl5 has 5 + 5(7) = 40 valence electrons, and phosphorus carries 10. SF6 has 6 + 6(7) = 48, and sulfur carries 12. XeF4 has 8 + 4(7) = 36. Four bonds use 8, the fluorines take 24 in lone pairs, and the last 4 sit on xenon as two lone pairs. Xenon ends up with 12 electrons. Second-row atoms such as C, N, O and F never do this, because they have no room for it.

Three families that break the octet rule Too few Expanded to 10 Expanded to 12 B F F F P Cl Cl Cl Cl Cl S F F F F F F BF₃, 24 electrons total. Boron keeps just 6. Still a stable molecule. PCl₅, 40 electrons total. Phosphorus holds 10. Five bonds, no lone pairs. SF₆, 48 electrons total. Sulfur holds 12. Six bonds, no lone pairs.
Figure 7.3 Lone pairs on the outer atoms are left off so the central counts stand out. Boron settles for six electrons. Phosphorus and sulfur take on more than eight, which only period 3 and heavier atoms can do.
Key points
  • Count valence electrons first, adjusting for the charge on an ion.
  • Lines are shared pairs, dots are lone pairs, and both use two electrons.
  • Fill the outer atoms, then the central atom, then add multiple bonds if needed.
  • Odd electron counts, such as NO with 11, cannot pair up at all.
  • B and Be often stop short of eight. Period 3 and below can go past eight.
Interactive Count valence electrons and place bonds and lone pairs. Open full ↑

Section 4Resonance and formal charge

When one drawing is not enough

Ozone, O3, has 18 valence electrons. Follow the five steps and you get a bent molecule with one double bond and one single bond. Now ask which side gets the double bond. Nothing in the rules picks a side, and you can draw it either way. Measurement settles the argument, because both bonds in ozone are 127.8 pm long. A true O=O double bond is about 121 pm and a true O−O single bond is about 148 pm. The real bonds are identical and sit between the two.

So neither drawing is the molecule. The two drawings are called resonance forms, and the real molecule is a single blend of them called the resonance hybrid. The double-headed arrow between forms is important. It does not mean the molecule flips back and forth, but rather that one structure is not enough to hold the truth, so you average the set. In ozone each bond is one and a half bonds, which fits the measured length exactly.

Average the shared pairs to get the real bond order, so ozone, with 3 bonding pairs spread over 2 bond positions, comes out at 3/2 = 1.5. Carbonate, CO32−, has three resonance forms and 4 bonding pairs over 3 positions, so its bond order is 4/3 = 1.33. All three carbonate bonds measure about 129 pm, which again sits between a single bond near 143 pm and a double bond near 123 pm. Spreading electrons out this way also lowers the energy, which is why these ions are so stable.

The two resonance forms of ozone and the blend they average to Resonance form AResonance form B The real molecule O O O +1 0 −1 O O O +1 −1 0 O O O 1.5 1.5 One double bond on theleft, one single on the right. The same drawing with thedouble bond swapped over. Two matching bonds, each127.8 pm, bond order 1.5. Lone pairs are left off to keep the charges readable. The central O carries +1 in both forms.
Figure 7.4 The arrow between forms A and B is not a flip. The real ozone molecule is the average on the right, with two identical bonds of order 1.5.

Formal charge keeps the bookkeeping honest

Formal charge is the charge an atom would carry if every bonding pair were split exactly down the middle. It is a bookkeeping tool rather than a real measured charge, and it is still the fastest way to rank competing structures.

formal charge = valence electrons − lone pair electrons − ½(bonding electrons)

Three habits make the tool reliable. The formal charges must add up to the overall charge on the species. The best structure keeps those charges as close to zero as it can. When a negative charge has to land somewhere, it belongs on the more electronegative atom.

Worked example · which CO2 drawing is better?

Problem. Both O=C=O and O≡C−O use all 16 valence electrons and give every atom an octet. Use formal charge to choose.

Step 1 · score the symmetric form, O=C=O Carbon: 4 − 0 − ½(8) = 4 − 0 − 4 = 0.
Each oxygen: 6 − 4 − ½(4) = 6 − 4 − 2 = 0.
Sum: 0 + 0 + 0 = 0, which matches a neutral molecule.

Step 2 · score the lopsided form, O≡C−O Carbon: 4 − 0 − ½(8) = 0.
Triple-bonded oxygen: 6 − 2 − ½(6) = 6 − 2 − 3 = +1.
Single-bonded oxygen: 6 − 6 − ½(2) = 6 − 6 − 1 = −1.
Sum: 0 + 1 − 1 = 0, so this one is legal too.

Step 3 · compare the scores Add the sizes of the charges. The symmetric form totals 0. The lopsided form totals 2.

Step 4 · pick the winner O=C=O wins. Its charges are all zero, and the loser puts a positive charge on oxygen, the second greediest atom on the scale. Measured CO2 has two equal bonds of 116 pm, which agrees.

Key points
  • Resonance forms differ only in where electrons sit, never in where atoms sit.
  • The real molecule is one blend of the forms, not a switch between them.
  • Bond order = bonding pairs divided by bond positions, so ozone is 1.5.
  • Formal charge = valence − lone pair electrons − half the bonding electrons.
  • Best structure: charges near zero, and any negative charge on the greedier atom.
Worth knowing. Equal resonance forms share the load equally, as in ozone and carbonate. When the forms are not equal, the better one counts for more. The blend leans toward whichever form scores best on formal charge.
Interactive Compare resonance forms and pick the best by formal charge. Open full ↑

Section 5VSEPR and molecular shapes

Electron groups push each other apart

A Lewis structure is flat on the page. Real molecules are not. To get the shape, use VSEPR, which stands for valence shell electron pair repulsion. The whole idea fits in one line: electron groups around a central atom are all negative, so they spread out as far from each other as they can.

Count groups, not atoms. Each group is an electron domain, and one single bond counts as one domain. A double or triple bond is still one domain, because those pairs sit in the same region. A lone pair is one domain too. Add them up, and the number of domains fixes the electron-domain geometry for you.

Two domains spread to 180°, which is linear. Three settle at 120° in a flat triangle, which is trigonal planar. Four go to 109.5° and point at the corners of a tetrahedron. Five give a trigonal bipyramid, with three equatorial positions 120° apart and two axial positions at 90° to that belt. Six give an octahedron, with every neighbor at 90°.

The five electron-domain geometries and their angles 2 domains3 domains4 domains 5 domains6 domains LinearTrigonal planarTetrahedral TrigonalbipyramidalOctahedral 180°120°109.5° 90° and 120°90° Gold is the central atom. Teal marks each electron domain.
Figure 7.5 Five domain counts, five patterns. These angles hold when every domain is a bond. Lone pairs bend them, as the text below explains.

Two shapes, one molecule

Here is the step that is easiest to skip. The domain geometry counts every domain, including lone pairs. The molecular geometry counts only the atoms, because lone pairs are invisible to a camera. So one molecule gets two names.

Water makes the point. Oxygen has 2 bonds and 2 lone pairs, so 4 domains, and the domain geometry is tetrahedral. The molecular geometry is bent, because you only see the two hydrogen atoms. Ammonia has 3 bonds and 1 lone pair, so also 4 domains. Its domain geometry is tetrahedral and its molecular shape is trigonal pyramidal.

Lone pairs also squeeze the angles. A lone pair is held by one nucleus instead of two, so it spreads wider and pushes harder on its neighbors. Methane has 4 bonds and no lone pairs, and its angle is the ideal 109.5°. Ammonia loses a little, down to 107°, and water loses more, down to 104.5°. Each lone pair shaves off about 2 to 3 degrees. For five domains, lone pairs always take equatorial spots, where there is more elbow room.

Bonds + lone pairsDomain geometryMolecular geometryAnglesExample
2 + 0LinearLinear180°CO2
3 + 0Trigonal planarTrigonal planar120°BF3
2 + 1Trigonal planarBent119°SO2
4 + 0TetrahedralTetrahedral109.5°CH4
3 + 1TetrahedralTrigonal pyramidal107°NH3
2 + 2TetrahedralBent104.5°H2O
5 + 0Trigonal bipyramidalTrigonal bipyramidal90°, 120°PCl5
4 + 1Trigonal bipyramidalSeesaw102°, 173°SF4
3 + 2Trigonal bipyramidalT-shaped87.5°ClF3
2 + 3Trigonal bipyramidalLinear180°XeF2
6 + 0OctahedralOctahedral90°SF6
5 + 1OctahedralSquare pyramidal84.8°BrF5
4 + 2OctahedralSquare planar90°XeF4

Worked example · name both shapes of SF4

Problem. Give the domain geometry, the molecular geometry and the angles for SF4.

Step 1 · count valence electrons S has 6 and each F has 7.
6 + 4(7) = 34 valence electrons, which is 17 pairs.

Step 2 · place them Four S−F bonds use 8 electrons. Giving each fluorine 3 lone pairs uses 24 more. Running total: 8 + 24 = 32. That leaves 34 − 32 = 2 electrons, so sulfur gets 1 lone pair.

Step 3 · count domains Sulfur has 4 bonds plus 1 lone pair, so 5 domains. Five domains means a trigonal bipyramidal domain geometry. Sulfur is in period 3, so holding 10 electrons is allowed.

Step 4 · drop the lone pair and name the shape The lone pair takes an equatorial spot. That leaves 2 axial and 2 equatorial fluorines. The molecular geometry is a seesaw.

Step 5 · adjust the angles Start from the ideal 120° and 90°, then let the lone pair push. The measured equatorial angle is about 102° and the axial pair closes from 180° to about 173°.

Key points
  • Count domains. A multiple bond is one domain, and a lone pair is one domain.
  • 2, 3, 4, 5 and 6 domains give linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
  • The ideal angles are 180°, 120°, 109.5°, then 90° with 120°, then 90°.
  • Domain geometry counts lone pairs. Molecular geometry counts only atoms.
  • Lone pairs push harder, so each one trims the angle by a few degrees.
  • With five domains, lone pairs always sit in the equatorial belt.
Worth knowing. Shape decides function. Water is bent, so it clings to its neighbors and stays liquid at room temperature. If it were linear, the pulls would cancel, and water would boil far below 0 °C.
Interactive Add bonding and lone pairs and watch the shape change. Open full ↑

Section 6Molecular polarity

Arrows that can cancel

Every polar bond has a little arrow on it. The arrow points toward the greedier atom, and its length tracks ΔEN. That arrow is the bond dipole, and a whole molecule is polar only if those arrows fail to cancel. Arrows are vectors, so you add them head to tail, the same way you add forces. The leftover arrow is the molecular dipole moment.

Carbon dioxide shows the catch. Both C=O bonds are strongly polar, since ΔEN = 3.44 − 2.55 = 0.89. But CO2 is linear, so the two arrows point in exactly opposite directions, which means they cancel and the measured dipole moment is 0 D. A molecule can be built entirely from polar bonds and still be nonpolar overall.

Water has the same two polar bonds and a bigger ΔEN of 1.24. It is bent at 104.5°, so the two arrows do not oppose each other, and they add to a strong net arrow pointing from the hydrogen side toward the oxygen. Water's dipole moment is 1.85 D, and that one number explains much of its behavior.

Bond dipoles cancelling in carbon dioxide and adding in water CO₂ · linear H₂O · bent C O O ΔEN = 0.89 on each bond Equal arrows, opposite ways. They add to nothing. μ = 0 D · nonpolar O H H 104.5° net Bent, so the arrows add. μ = 1.85 D · polar
Figure 7.6 Same kind of polar bond, opposite outcome. Shape decides whether the bond dipoles cancel. The gold arrows are bond dipoles and the teal arrow is the sum.

A two-question test

You can usually judge polarity without any math. Ask two questions. Are all the outer atoms the same? Are they placed symmetrically around the center, with no lone pairs on the central atom? If both answers are yes, the arrows cancel and the molecule is nonpolar. Any no, and it is polar.

That test sorts the common cases fast. CH4, CCl4, BF3, CO2, PCl5 and SF6 all pass, so all are nonpolar. Swap one atom and symmetry dies. CHCl3 has a dipole moment of 1.04 D because one corner is now hydrogen, and a lone pair usually breaks symmetry just as well. SO2 is bent, and its dipole moment is 1.63 D, while linear CO2 sits at zero. XeF2 and XeF4 are the exception. Their lone pairs sit exactly opposite each other, so the bond arrows still cancel.

Worked example · how big is one O−H bond dipole?

Problem. Water has μ = 1.85 D and a bond angle of 104.5°. Find the dipole of a single O−H bond.

Step 1 · set up the vector sum The two bond arrows are equal in size. Each leans half the bond angle away from the net arrow. Only the part along the net direction survives, so the sum is
μnet = 2 μbond cos(θ/2).

Step 2 · put in the angle θ/2 = 104.5° / 2 = 52.25°, and cos 52.25° = 0.6124.

Step 3 · solve for the bond dipole μbond = μnet / (2 cos 52.25°)
μbond = 1.85 D / (2 × 0.6124) = 1.85 / 1.2248 = 1.51 D.

Step 4 · sanity check A single O−H bond is more polar than the whole water molecule, which sounds odd until you look at the arrows. The two bonds point partly against each other, so some of each one is lost. Compare 1.51 D with 1.08 D for H−Cl. The O−H bond has the larger ΔEN, 1.24 against 0.96, so the larger dipole fits.

Key points
  • A bond dipole is an arrow pointing at the greedier atom, sized by ΔEN.
  • Add the arrows as vectors. The leftover is the molecular dipole moment.
  • Polar bonds in a symmetric shape cancel, as in CO2 and CCl4.
  • Matching outer atoms and no lone pairs on the center means nonpolar.
  • For a bent AX2 molecule, μnet = 2μbond cos(θ/2).
Worth knowing. NH3 and NF3 have the same shape, yet their dipole moments are 1.47 D and 0.23 D. In ammonia the bond arrows point at nitrogen, the same way the lone pair leans. In NF3 the arrows point out at the fluorines, against the lone pair, so most of the pull cancels.
Interactive Add bond dipoles as vectors to find the molecule's polarity. Open full ↑

The chapter in one breath

  • Ionic bonds move an electron across. The solid is a lattice, so the formula is a ratio.
  • Lattice energy climbs with bigger ion charges and shrinks as r0 grows.
  • Covalent bonds share a pair, and ΔEN scores how fairly: under 0.4 nonpolar, 0.4 to 1.8 polar, over 1.8 ionic.
  • Lewis structures start with a valence electron count and end with every octet checked.
  • The octet rule bends three ways: odd counts, too few electrons, and expanded shells.
  • Resonance blends two or more drawings, which is why ozone has two 1.5 bonds.
  • Formal charge ranks competing structures, and charges near zero win.
  • VSEPR turns a domain count into two shape names and a bond angle.
  • A molecule is polar only when its bond dipoles fail to cancel.

ReferenceWords worth knowing

Twenty terms carry this chapter. Read the list once now, then come back to it when you are drawing structures and naming shapes.

anion
An ion with a negative charge, made when an atom gains one or more electrons.
bond dipole
The arrow on one bond, pointing to the more electronegative atom. Its length tracks ΔEN.
bond order
The number of shared pairs between two atoms. Resonance can make it a fraction, such as 1.5 in ozone.
cation
An ion with a positive charge, made when an atom loses one or more electrons.
covalent bond
A bond made of one or more electron pairs shared between two atoms.
dipole moment (μ)
The measured size of a charge separation, in debye. 1 D = 3.336 × 10−30 C·m.
electron domain
One group of valence electrons around a central atom. A single, double or triple bond counts as one, and so does a lone pair.
electronegativity
A number for how strongly an atom pulls on a shared pair. On the Pauling scale F is 3.98 and Cs is 0.79.
expanded valence shell
A central atom holding more than eight valence electrons, as phosphorus does in PCl5. Only period 3 and heavier atoms can.
formal charge
Valence electrons minus lone pair electrons minus half the bonding electrons. A bookkeeping score used to rank structures.
ionic bond
The pull between oppositely charged ions after an electron has moved from one atom to another.
lattice energy (ΔHlattice)
The energy needed to pull one mole of an ionic solid apart into gas-phase ions. About 769 kJ/mol for NaCl.
Lewis structure
A drawing of the valence electrons, with lines for shared pairs and dots for lone pairs.
lone pair
A valence pair that belongs to one atom and is not shared.
molecular geometry
The shape traced by the atoms alone, with lone pairs left out. Water is bent.
nonpolar covalent bond
A bond with ΔEN under about 0.4, where the shared pair sits near the middle.
octet rule
The habit of ending up with eight valence electrons. Hydrogen stops at two.
polar covalent bond
A bond with ΔEN of about 0.4 to 1.8, giving δ+ on one atom and δ− on the other.
resonance
Two or more valid Lewis structures for one species. The real molecule is their blend, not a switch between them.
VSEPR
Valence shell electron pair repulsion. The model that predicts shape by spreading electron domains as far apart as possible.

Check yourself

1. How many valence electrons go into the Lewis structure of the chlorate ion, ClO3−?
26 electrons. Cl brings 7 and each O brings 6, so 7 + 3(6) = 25. The 1− charge adds one more electron: 25 + 1 = 26, which is 13 pairs.
2. NaCl has a lattice energy near 769 kJ/mol. Estimate the value for NaBr, given radii Na+ 102 pm, Cl− 181 pm and Br− 196 pm.
About 730 kJ/mol. r0(NaCl) = 102 + 181 = 283 pm and r0(NaBr) = 102 + 196 = 298 pm. The charges match, so scale by distance: 769 × (283/298) = 769 × 0.9497 = 730 kJ/mol. The measured value is about 732 kJ/mol.
3. Find ΔEN for the Si−O bond and classify it. Use Si 1.90 and O 3.44.
ΔEN = 1.54, a polar covalent bond. 3.44 − 1.90 = 1.54, which lands between 0.4 and 1.8. Oxygen takes the δ− end. This is the bond that builds sand and glass.
4. Work out the formal charge on the central oxygen in ozone. It has one double bond, one single bond and one lone pair.
+1. Oxygen has 6 valence electrons. The lone pair is 2 nonbonding electrons. Three bonds mean 6 bonding electrons, and half of that is 3. So 6 − 2 − 3 = +1. The single-bonded outer oxygen comes out at 6 − 6 − 1 = −1, so the molecule is neutral overall.
5. What is the bond order in the nitrate ion, NO3−?
1.33. Nitrate has 5 + 3(6) + 1 = 24 valence electrons. Each resonance form has one N=O and two N−O bonds, so 4 bonding pairs are spread over 3 bond positions: 4/3 = 1.33. All three bonds are the same length.
6. SO2 has μ = 1.63 D and a bond angle of 119°. How big is one S−O bond dipole?
About 1.61 D. Use μnet = 2μbond cos(θ/2). Here θ/2 = 59.5° and cos 59.5° = 0.5075. So μbond = 1.63 / (2 × 0.5075) = 1.63 / 1.015 = 1.61 D.
7. Give both geometry names and the approximate bond angle for ClF3.
Trigonal bipyramidal domains, T-shaped molecule, about 87.5°. ClF3 has 7 + 3(7) = 28 valence electrons. Three bonds use 6 and the fluorine lone pairs use 18, leaving 4 for two lone pairs on chlorine. So 3 bonds + 2 lone pairs = 5 domains, and both lone pairs take equatorial spots.
8. Both CO2 and H2O are built from polar bonds. Why is only one of them polar?
Shape decides it. CO2 is linear, so its two bond dipoles point exactly opposite and cancel to μ = 0 D. Water is bent at 104.5°, so its two arrows add to 1.85 D pointing toward oxygen.
9. Which of NO, CO2 and CH4 cannot possibly follow the octet rule, and why?
NO. It has 5 + 6 = 11 valence electrons, an odd number, so one electron cannot be paired. Nitrogen ends up with seven, which makes NO a reactive free radical.
10. Name the electron-domain geometry and the molecular geometry of water, and say why they differ.
Tetrahedral domains, bent molecule. Oxygen has 2 bonds and 2 lone pairs, so 4 domains spread tetrahedrally. The domain name counts lone pairs and the molecular name does not, so you see only the bent H−O−H frame at 104.5°.
11. MgO has a lattice energy near 3795 kJ/mol against 769 kJ/mol for NaCl. Account for the difference using charges and radii. Use Mg2+ 72 pm, O2− 140 pm.
Charge does most of it, and the shorter distance adds the rest. The charge product goes from (1)(1) to (2)(2), a factor of 4. r0 falls from 283 pm to 72 + 140 = 212 pm, a factor of 283/212 = 1.33. Together the model predicts 4 × 1.33 = 5.3 times, and the real ratio is 3795/769 = 4.9 times.
12. A classmate says ozone spends half its time as form A and half as form B. What is wrong with that?
There is no switching. Ozone is one molecule with one structure, the blend of both forms. Both bonds are 127.8 pm at all times, with a bond order of 1.5. If it really switched, you would catch one short bond and one long bond.