Worked example 1: solution

Step 1 · Find the electronegativity difference

Worked example 1: solution

A difference of zero sits at the far left: nonpolar covalent.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the bond on the continuum

Worked example 2: solution

Step 1 · Find the electronegativity difference

Worked example 2: solution

A difference of 0.96 sits between the 0.4 and 1.7 guidelines: polar covalent. The pair is shared, but not equally.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the bond on the continuum

Practice 1 answer: C

B put δ− on the wrong atom: O (3.44) pulls harder than S (2.58), so the pair shifts toward O; subtracting backwards, 2.58 − 3.44 = −0.86, only flips the sign, not the direction. A ignored the difference: 0.86 is not zero, so the sharing is unequal, not nonpolar. D read sharing as transfer: two nonmetals 0.86 apart still share the pair, polar covalent, not ionic.

Worked example 3: solution

Step 1 · Find the electronegativity difference

Worked example 3: solution

1.78 sits just past the 1.7 guideline. The boundary is not a wall: H and F are both nonmetals, and two nonmetals share.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the bond on the continuum

Practice 2 answer: D

A dropped B–F because 1.94 clears 1.7 and settled for Si–Cl at 1.26; a bigger difference only makes a bond more polar, and with no metal in it B–F still shares its pair. B picked the two most electronegative atoms; polarity comes from the difference, and 3.98 − 3.44 = 0.54 is the smallest of the three. C put δ− on the wrong atom: F (3.98) pulls harder than B (2.04); 2.04 − 3.98 = −1.94 flips only the sign.

Practice 1 answer: A

B counted the period: P sits in period 3, but dots come from the group, 15 → 5. C drew the whole group number as 15 dots; only the 5 valence electrons appear. D paired too early: dots fill all four sides singly before any pair forms, so 5 dots make one pair, not two.

Practice 2 answer: A

B drew the dots but dropped the brackets and charge; eight dots on a plain S claims a neutral atom, and neutral sulfur has 6. C pasted the charge onto the original 6 dots; the 2− exists because 2 electrons arrived, so 6 + 2 = 8 dots must show. D ran the transfer backwards: 6 − 2 = 4 dots with 2+ describes a cation, and nonmetals gain.

Worked example 1: solution

Step 2 · Draw the skeleton

Oxygen in the center, one single bond to each H.

Worked example 1: solution

The outer atoms are hydrogen; a single bond already fills each H's duet.

Worked example 1: solution

The outer atoms are hydrogen; a single bond already fills each H's duet.

Worked example 1: the oxygen lone pairs

Four electrons sit in the two O–H bonds; the outer hydrogens are done. Four remain.

Step 4 · Place leftovers on the center

Worked example 1: the oxygen lone pairs

Four electrons sit in the two O–H bonds; the outer hydrogens are done. Four remain.

Step 4 · Place leftovers on the center

Worked example 1: the electron budget

Worked example 2: solution

Step 1 · Count the valence electrons

Worked example 2: solution

Step 1 · Count the valence electrons

Nitrogen is least electronegative, so it takes the center, with a single bond to each F.

Worked example 2: solution

Step 1 · Count the valence electrons

Nitrogen is least electronegative, so it takes the center, with a single bond to each F.

Worked example 2: the fluorine octets

Six electrons sit in the three N–F bonds. Each fluorine still needs three lone pairs.

Step 3 · Complete the outer octets

Worked example 2: the fluorine octets

Six electrons sit in the three N–F bonds. Each fluorine still needs three lone pairs.

Step 3 · Complete the outer octets

Worked example 2: the nitrogen lone pair

Six electrons sit in the three N–F bonds and eighteen fill the fluorine octets. Two remain.

Step 4 · Place leftovers on the center

Worked example 2: the nitrogen lone pair

Six electrons sit in the three N–F bonds and eighteen fill the fluorine octets. Two remain.

Step 4 · Place leftovers on the center

Worked example 2: the electron budget

Practice 3 answer: D

B counted each element once: 6 + 7 = 13. C counted every electron, core included: 8 + 2(9) = 26. A dropped a pair: 20 − 2 = 18.

Worked example 3: solution

Step 2 · Draw the skeleton

Carbon centered, a single bond to each oxygen.

Worked example 3: solution

Step 2 · Draw the skeleton

Carbon centered, a single bond to each oxygen.

Worked example 3: the outer octets

Step 3 · Complete the outer octets

Worked example 3: the outer octets

Step 3 · Complete the outer octets

Step 4 · Place leftovers on the center

None remain: 16 − 4 − 12 = 0. The carbon is still two pairs short of an octet.

Worked example 3: the outer octets

Step 3 · Complete the outer octets

Step 4 · Place leftovers on the center

None remain: 16 − 4 − 12 = 0. The carbon is still two pairs short of an octet.

Worked example 3: two double bonds

Carbon sits two pairs short of an octet. Pull one lone pair from each oxygen into a bond.

Step 5 · Form multiple bonds

Worked example 3: two double bonds

Carbon sits two pairs short of an octet. Pull one lone pair from each oxygen into a bond.

Step 5 · Form multiple bonds

Worked example 3: the electron budget

Practice 4 answer: B

A reported the whole pool of 10, not the lone-pair share. D counted the shared electrons, 2 + 6 = 8, instead of the ones left over. C stopped at single bonds: 10 − 4 = 6 in lone pairs, which leaves carbon with only 4 electrons. E stopped at a double bond: 2 + 4 = 6 shared, 10 − 6 = 4 in lone pairs, which leaves carbon with only 6.

Worked example 1: solution

Step 2 · Count the atom's nonbonding electrons

Single-bonded O: 6. Central O: 2. Double-bonded O: 4.

Worked example 1: solution

Single-bonded O: ½(2) = 1. Central O: ½(6) = 3. Double-bonded O: ½(4) = 2.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the sum check

Step 5 · Check the sum against the overall charge

Worked example 1: the sum check

Step 5 · Check the sum against the overall charge

Worked example 2: the three structures

Worked example 2: the three structures

Worked example 2: the three structures

Worked example 2: formal charges

Step 2 · Count the atom's nonbonding electrons
Step 3 · Count half the atom's bonding electrons

Carbon: no lone pairs, four bonding pairs. Double-bonded O: two lone pairs. Each single-bonded O: three lone pairs.

Worked example 2: formal charges

Worked example 2: formal charges

Worked example 2: the sum check

Step 5 · Check the sum against the overall charge

Worked example 2: the sum check

Step 5 · Check the sum against the overall charge

Practice 1 answer: C

A counted all eight bonding electrons: 5 − 0 − 8 = −3. B gave nitrogen the lone pair it has in ammonia: 5 − 2 − 4 = −1, but here all four pairs sit in bonds. D set every atom to zero; the formal charges must add to the ion's +1, and every hydrogen is 0, so nitrogen carries it.

Practice 1 answer: C

Worked example 3: solution

Worked example 3: solution

Guided example: formal charges

Step 2 · Count the atom's nonbonding electrons
Step 3 · Count half the atom's bonding electrons

Carbon holds four bonds and no lone pair in all three: 4 − 0 − 4 = 0 every time. Only the end atoms change.

Guided example: formal charges

Guided example: the best structure

Rule 1 · Size: charges closest to zero

Structure 3 carries −2 on N. Structures 1 and 2 keep every charge within ±1. Drop structure 3.

Guided example: the best structure

Practice 2 answer: B

A miscounted: the central N holds four bonds and no lone pair in both drawings, 5 − 0 − 4 = +1, so no structure of N₂O is charge-free. C ran the tiebreaker backwards; a negative formal charge belongs on the most electronegative atom, and oxygen (3.44) beats nitrogen (3.04). D treated non-equivalent drawings as equal partners; their charges differ, so formal charge picks the better one.

Practice 3 answer: E

A is [O=C=N]⁻: O 6 − 4 − 2 = 0, N 5 − 4 − 2 = −1. It ties E on size but puts the −1 on N, as in [S=C=N]⁻; oxygen outranks nitrogen. B is [O≡C–N]⁻: O 6 − 2 − 3 = +1, N 5 − 6 − 1 = −2, which fails Rule 1. C gave oxidation numbers: O −2, N −3, so C is −1 + 2 + 3 = +4. D sums to 0; an ion's formal charges must sum to −1.

Worked example 1: solution

Step 2 · Build the skeleton, then the outer octets

Worked example 1: solution

Step 2 · Build the skeleton, then the outer octets

Worked example 1: the expanded octet

Step 3 · Leftovers go on the central atom

Worked example 1: the expanded octet

Step 3 · Leftovers go on the central atom

Worked example 2: bonds and outer octets

Step 2 · Build the skeleton, then the outer octets

Worked example 2: bonds and outer octets

Step 2 · Build the skeleton, then the outer octets

Worked example 2: the expanded center

Step 3 · Leftovers go on the central atom

Worked example 2: the expanded center

Step 3 · Leftovers go on the central atom

Practice 1 answer: D

C added sulfur's own six valence electrons on top of the twelve in its bonds: 12 + 6 = 18; those six are already inside the bonds. A counted the whole pool: 6 + 6(7) = 48. B counted the fluorine lone pairs: 6 × 6 = 36.

Worked example 3: the fix fails

Worked example 3: the fix fails

F: 7 − 4 − 2 = +1. Fluorine, the most electronegative element, goes positive.

Worked example 3: the fix fails

Worked example 3: the fix fails

Practice 2 answer: C

A dropped the 1− charge: 7 + 2(7) = 21 leaves 5, two pairs and a stray electron; the added electron makes 22, an even count that pairs fully. B subtracted the charge instead: 7 + 2(7) − 1 = 20 leaves 4, two pairs and an octet; a negative charge adds an electron. D placed the right 22 but parked the leftovers on the chlorines; every outer atom already has its octet, so leftovers go on the center.

Worked example 1: solution

Step 1 · Count the electron domains

Four bonded hydrogens and no lone pairs on carbon. Together that is 4 + 0 = 4 electron domains.

Worked example 1: solution

Four domains spread as far apart as possible, to 109.5°. The electron-pair geometry is tetrahedral.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the row in the table

Worked example 2: solution

A common first attempt

A double bond is one region of electrons, so it counts once. Two double bonds are two domains, not four.

Worked example 2: solution

Two bonded oxygens and no lone pairs on carbon: 1 + 1 = 2 electron domains.

Worked example 2: solution

Two domains point straight apart, to 180°. The electron-pair geometry is linear.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the row in the table

Practice 1 answer: A

B named the electron-pair geometry, not the shape: the lone pair takes a corner, so the atoms are not tetrahedral. C ignored the lone pair; three bonded atoms alone would be trigonal planar, but the fourth corner is filled. D miscounts: four domains, not three, so the base is a tetrahedron, not a triangle.

Worked example 3: solution

Step 1 · Count the electron domains

Two bonded oxygens count as two domains: the double bond counts once. Add the lone pair: 2 + 1 = 3 electron domains.

Worked example 3: solution

Three domains spread to 120°. The electron-pair geometry is trigonal planar.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the row in the table

Worked example 4: solution

Step 1 · Count the electron domains

Two bonded hydrogens and two lone pairs on oxygen: 2 + 2 = 4 electron domains.

Worked example 4: solution

Four domains land on the AX₂E₂ row. The electron-pair geometry is tetrahedral.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the row in the table

Practice 2 answer: C

A dropped the lone pairs: only 4 of the 8 electrons sit in the two bonds, and the other 4 stay on selenium as two more domains. B named the electron-pair geometry; two of the four corners hold lone pairs, so the atoms are bent, not tetrahedral. D miscounted the leftovers as one pair: 8 − 4 = 4 electrons make two pairs, four domains, not three.

Worked example 1: solution

Step 1 · Recall the shape

Carbon has two bonding groups and no lone pairs. VSEPR gives a linear molecule: the two oxygens sit 180° apart.

Worked example 1: solution

Both arrows point outward, away from carbon, toward the oxygens.

Worked example 1: solution

The two arrows pull in exactly opposite directions and cancel. No net arrow. Nonpolar.

Worked example 1: solution

Worked example 1: the route on the flowchart

Worked example 2: solution

Step 1 · Recall the shape

Oxygen has two bonding groups and two lone pairs. VSEPR gives a bent molecule; the two O–H bonds meet at about 104.5°.

Worked example 2: solution

Both arrows point from the hydrogens up toward the oxygen.

Worked example 2: solution

The two arrows point the same way, so they reinforce instead of cancel. A net arrow runs through the oxygen. Polar.

Worked example 2: solution

Worked example 2: the route on the flowchart

Practice 1 answer: C

B took polar bonds as proof of a polar molecule and skipped the shape: the symmetric triangle cancels the three equal arrows. A called the symmetric shape lopsided, but three identical arrows 120° apart balance exactly. D denied the bonds are polar. ΔEN(B–Cl) = 3.16 − 2.04 = 1.12, so each bond is polar, and the molecule is nonpolar because of the shape, not because the bonds are nonpolar.

Worked example 3: shape and dipoles

Step 1 · Recall the shape

Both are tetrahedral: four bonding groups on carbon, no lone pairs, the outer atoms 109.5° apart.

Worked example 3: shape and dipoles

Each C–Cl arrow points toward its chlorine; the weaker C–H arrow points toward carbon.

Worked example 3: shape and dipoles

Worked example 3: adding the arrows

Step 3 · Add the arrows

Worked example 3: the route on the flowchart

Practice 2 answer: B

A read OCS as symmetric like CO₂; linear cancels only when both outer atoms are identical, and C=O out-pulls C=S. C missed the lone pair on phosphorus and treated PCl₃ as a flat triangle; three bonds plus one lone pair make it pyramidal and polar. D took polar bonds as proof of a polar molecule; SiF₄ and CF₄ have polar bonds and cancel by shape.

Worked example 1: solution

Step 1 · Count the domains

Five bonded fluorines and no lone pairs on phosphorus: 5 + 0 = 5 electron domains.

Worked example 1: solution

Five domains spread into a trigonal bipyramid: three equatorial corners at 120°, two axial corners at 90° to them.

Worked example 1: solution

There are none. Every corner holds a fluorine.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

Step 1 · Count the domains

Four bonded fluorines plus one lone pair: 4 + 1 = 5 electron domains.

Worked example 2: solution

Five domains arrange as a trigonal bipyramid.

Worked example 2: solution

An axial lone pair would press on three bonding pairs at 90°. An equatorial corner has only two such neighbors, so the wide pair takes it.

Worked example 2: solution

The lone pair takes an equatorial corner. Two axial fluorines and two equatorial fluorines remain.

Worked example 2: solution

Worked example 2: solution

Guided example: the lone pairs

Step 1 · Count the domains

Every fluorine has its octet. The 6 leftover electrons stay on krypton as 3 lone pairs.

Guided example: the lone pairs

Guided example: the shape

Step 2 · Name the geometry

Five domains spread into a trigonal bipyramid.

Guided example: the shape

Lone pairs go equatorial first. Three lone pairs fill all three equatorial corners, so both fluorines sit on the axis.

Guided example: the shape

Guided example: the shape

Practice 1 answer: B

A named the electron-pair geometry; two of the five corners hold lone pairs, so the atoms are not a bipyramid. C seats both lone pairs axial to leave a flat triangle, but an axial lone pair presses on three bonding pairs at 90°; the lone pairs go equatorial. D forces an octet with 3 + 1 = 4 domains; bromine holds an expanded octet, 3 + 2 = 5.

Practice 1 answer: B

Worked example 3: the shape

Step 1 · Count the domains

Four bonded fluorines plus two lone pairs: 4 + 2 = 6 electron domains.

Worked example 3: the shape

Six domains spread into an octahedron. Every corner sits at 90° to its four neighbors.

Worked example 3: the shape

All six corners are identical, so no corner is roomier. The two wide pairs get as far from each other as possible: opposite corners, 180° apart.

Worked example 3: the shape

Worked example 3: the shape

Worked example 3: the polarity

Polarity · Add the bond dipoles

Each Xe–F arrow has a partner pointing exactly the other way, so the four cancel in pairs. The two lone pairs sit opposite each other and balance as well.

Worked example 3: the polarity

Polarity · Add the bond dipoles

Each Xe–F arrow has a partner pointing exactly the other way, so the four cancel in pairs. The two lone pairs sit opposite each other and balance as well.

Practice 2 answer: A

B assumes identical atoms always cancel; they cancel only when every arrow has a partner straight across, and here the fifth arrow faces a lone pair instead. C denies the bonds are polar; fluorine out-pulls bromine, so every bond is polar. D overgeneralizes; polar bonds alone decide nothing, since SF₆ holds six polar bonds and is nonpolar.

Practice 2 answer: A

Practice 3 answer: E

A stopped after the outer octets; the 2 leftover electrons are a lone pair on iodine, so 4 + 0 = 4 undercounts. B named the electron-pair geometry; one corner holds the lone pair. C added an electron for the 1+ charge: 7 + 4(7) + 1 = 36 and 36 − 8 − 24 = 4, two lone pairs and 4 + 2 = 6 domains. A positive charge removes electrons. D seated the lone pair axial; an equatorial corner has fewer neighbors at 90°.

Practice 3 answer: E

Guided example: solution

Step 1 · Charges first

Both are 2+/2−. The charge lever cannot decide.

Guided example: solution

On a charge tie, the shorter d wins. Compare both ions.

Guided example: solution

Mg sits above Ca in group 2. O sits above S in group 16. Both MgO ions are smaller.

Guided example: solution

The larger lattice energy holds the solid harder, so MgO melts higher.

Worked example 1: solution

Step 1 · Charges first

MgO has Q₁Q₂ = 2 × 2 = 4; NaF and NaBr each have 1 × 1 = 1. The high charge product puts MgO far in front.

Worked example 1: solution

Step 1 · Charges first
Step 2 · Then size

NaF and NaBr tie on charge, so compare d. F⁻ is smaller than Br⁻ (d ≈ 235 vs 298 pm), so NaF > NaBr.

Worked example 1: solution

Step 1 · Charges first
Step 2 · Then size

Worked example 2: solution

Step 1 · Charges first

LiF is 1+/1−, so Q₁Q₂ = 1. CaO is 2+/2−, so Q₁Q₂ = 4: four times larger.

Worked example 2: solution

Step 1 · Charges first
Size does not apply

LiF's ions are smaller (d ≈ 209 vs 240 pm), but size only breaks a charge tie. Here 4 beats 1 decisively, so the route stops at Step 1.

Practice 1 answer: A

B let one 2− ion carry it: Na₂O is Na⁺ with O²⁻, so Q₁Q₂ = 1 × 2 = 2, half of BaO's 4. C let size beat charge: LiCl's ions sit closer than BaO's (d ≈ 257 vs 275 pm), but its product is 1. D ran size backward: KI has the largest ions here, and a longer d lowers the lattice energy.

Practice 2 answer: C

A ran size backward: a larger cation means a longer d, and d sits in the denominator. B misread the charge: every group 1 ion is 1+, whatever its row. D stopped at the tie: equal charges hand the decision to size. E credited mass: formula mass is not in Q₁Q₂/d.

Practice 3 answer: D

A added the charges: (2 + 2)/(1 + 1) × 319/284 = 2.25. B flipped the distance ratio: 4 × 284/319 = 3.56, which rewards the longer d. C stopped at the charge product: 4/1 = 4.00, with size never applied. E skipped the charge lever: 319/284 = 1.12, the distance ratio alone.

Worked example 1: solution

Step 1 · Draw every Lewis structure

Three diatomic molecules, one single bond each: H–H, Cl–Cl, and H–Cl twice.

Worked example 1: solution

One H–H and one Cl–Cl: Σ broken = 436 + 243 = 679 kJ.

Worked example 1: solution

Two H–Cl bonds, from the coefficient 2: Σ formed = 2 × 431 = 862 kJ.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

Step 1 · Draw every Lewis structure

H–H twice, one O=O (one double bond, its own D), and two H–O–H: each water holds two O–H bonds.

Worked example 2: solution

Two H–H (2 × 436 = 872) plus one O=O (498): Σ broken = 872 + 498 = 1370 kJ.

Worked example 2: solution

Two H₂O × two O–H each = 4 O–H bonds: Σ formed = 4 × 467 = 1868 kJ.

Worked example 2: solution

Worked example 2: solution

Practice 1 answer: B

A (−760.) summed only the bonds formed; the 808 kJ of breaking was never charged. C (+428) forgot the coefficient 2 on OF₂: 808 − 380. = 428. D (−48) reversed the formula: 760. − 808 = −48, the right size with the wrong sign.

Practice 2 answer: A

B (+1307) reversed the formula: 5064 − 3757. C (+1225) ignored both product coefficients, one CO₂ and one H₂O: 1598 + 2 × 467 = 2532, and 3757 − 2532 = +1225. D (−5064) summed the products only. E (+3757) stopped at Σ broken and never subtracted.

Guided example: solution

Step 1 · Draw every Lewis structure

Per mole of NH₃ puts a 1 in front of NH₃; fractions balance the rest. NH₃ holds three N–H, O₂ is O=O, N₂ is N≡N, and each H₂O holds two O–H.

Guided example: solution

Three N–H (3 × 391 = 1173) plus 3/4 mol O=O (3/4 × 498 = 373.5): Σ broken = 1546.5 kJ.

Guided example: solution

1/2 mol N≡N (1/2 × 941 = 470.5) plus 3/2 H₂O × two O–H = 3 O–H (3 × 467 = 1401): Σ formed = 1871.5 kJ.

Guided example: solution

Guided example: solution

Practice 3 answer: C

A (−3662) left O₂ at 1, so one O=O broke: 3436 + 498 = 3934, and 3934 − 7596 = −3662. B (−3838) is ΔH for 2 CH₂CHCH₃ + 9 O₂ → 6 CO₂ + 6 H₂O, two moles of propene. D (−4160.) never broke the O=O bonds: 3436 − 7596.