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

Your turn: the bond on the continuum

Practice 1 answer: A

B stopped at the metal; S–O inside SO₄²⁻ and Cl–O inside ClO₂⁻ are shared pairs. C missed the chlorite ion: ClO₂⁻ is chlorate, ClO₃⁻, minus one oxygen, held together by shared Cl–O pairs. D kept only the bonds inside the ions; Fe²⁺ attracting SO₄²⁻ is ionic bonding. E read SO₂ as ions; S and O are both nonmetals, so they share: covalent only.

Practice 2 answer: B

A counted the two chlorines as bonded to each other; CaCl₂ is Ca²⁺ with two separate Cl⁻ ions, ionic only. C read the subscripts as a formula unit of ions; P and O are both nonmetals, so P₂O₅ shares: covalent only. D read the OH as hydroxide; CH₃OH has no ions at all, every atom a nonmetal sharing: covalent only.

Practice 3 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 4 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 5 answer: D

A called every large difference ionic: H–F (1.78) and N–F (0.94) join two nonmetals, which share. B called a real difference nonpolar: 0.94 is not zero. C took fluorine's own high electronegativity for polarity; in F–F, 3.98 − 3.98 = 0. E read the 1.7 boundary as a wall: H–F is two nonmetals sharing a pair, polar covalent.

Practice 6 answer: B

A read the shared group as an equal pull: electronegativity climbs up a group, and 0.82 is not zero, so the sharing is unequal. C gave δ− to the larger atom: Cl holds more electrons, but F (3.98) pulls the shared pair harder than Cl (3.16); 3.16 − 3.98 = −0.82 flips only the sign. D read sharing as transfer: two nonmetals 0.82 apart share the pair.

Practice 7 answer: C

A read 1.7 as a wall: P and F are nonmetals, so they share the pair. B ignored the metal: K gives its electron to I. D put δ− on the wrong atom: F (3.98) outpulls P (2.19); 2.19 − 3.98 = −1.79 flips only the sign. E judged by the strongest atom: the difference and the metal test set the type, not F alone.

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.

Practice 3 answer: B

A paired one Al with one S without counting electrons: 3 lost against 2 gained. C wrote each element's own electron count as its own subscript: Al₃S₂ has 3 × 3 = 9 lost against 2 × 2 = 4 gained. D crossed the dot counts, 3 and 6, instead of the electrons that move: Al₆S₃ has 6 × 3 = 18 lost against 3 × 2 = 6 gained.

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 4 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.

Practice 5 answer: D

A counted each element once: 4 + 1 + 7 = 12, leaving 12 − 8 = 4 after the bonds, one pair per Br. B gave hydrogen an octet: 8 + 12 + 12 = 32 drawn from a pool of 20; H takes a duet. C put a lone pair on carbon with none left: 8 + 12 + 2 = 22, and carbon holds 10.

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

Worked example 4: solution

Step 2 · Draw the skeleton Step 3 · Complete the outer octets Step 4 · Place leftovers on the center

Carbon centered, a single bond to each of the three oxygens; each oxygen then takes three lone pairs.

Worked example 4: solution

Step 2 · Draw the skeleton Step 3 · Complete the outer octets Step 4 · Place leftovers on the center

Worked example 4: solution

Step 2 · Draw the skeleton Step 3 · Complete the outer octets Step 4 · Place leftovers on the center

Worked example 4: the electron budget

Practice 6 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.

Practice 7 answer: C

A counted two double bonds as too many: the center N has 4 + 4 = 8. B took the single-bonded O as unfinished: 2 + 6 = 8 with three lone pairs. D took O's triple bond as too many: 6 + 2 = 8 with one lone pair. E stopped before the center atom, where N≡N=O puts 10.

Practice 8 answer: A

B counted the lone-pair electrons, 4 × 2 = 8, not the pairs. C gave each H an octet: 4 lone pairs on the O atoms plus 3 on each H makes 10, which needs 6 + 20 = 26 electrons from a pool of 14. D counted every pair in the pool as a lone pair: 14 ÷ 2 = 7, the 3 bonding pairs included.

Practice 9 answer: D

A stopped at step 4: 7 single bonds and 5 lone pairs place all 24 electrons, but carbon holds 6. B counted each line of C=O as a single bond: 6 + 2 = 8. C counted lone-pair electrons, 4 × 2 = 8, not pairs. E left out the four N–H bonds: 6 − 4 = 2.

Practice 10 answer: B

A stopped at step 4: all 18 electrons are placed, but N holds 2 + 2 + 2 = 6. C put O in the center because it sits in the middle of the formula; O has the highest electronegativity, so it belongs outside. D dropped the 2 leftover electrons: 18 − 2 = 16 drawn, and N holds 6.

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

Practice 1 answer: C

A read the period of the chlorines instead of the center. Boron is in period 2 and stops at six. B took boron's own count, 3, as the molecule's; the total, 24, is even. D called ClO₂ incomplete because one atom falls short of eight; an odd total, 19, puts it in the odd-electron family.

Practice 2 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.

Practice 3 answer: B

A dropped the 1− charge: 5 + 2(6) = 17 looks odd, but the added electron makes 18, and 18 pairs fully. C judged by nitrogen's own count, 5; two nitrogens give 2(5) + 4(6) = 34, even. D took boron's short octet for an unpaired electron; BF₃ holds 24 electrons, all in pairs.

Worked example 2: solution

Step 2 · Build the skeleton, then the outer octets

Worked example 2: solution

Worked example 2: solution

Worked example 2: leave boron at six

The tempting fix Pull a fluorine lone pair into a B=F double bond (Structure 2). Reject it: boron, like beryllium, accepts fewer than eight.

Worked example 2: leave boron at six

The tempting fix Pull a fluorine lone pair into a B=F double bond (Structure 2). Reject it: boron, like beryllium, accepts fewer than eight.

Practice 4 answer: A

Practice 4 answer: A

Practice 5 answer: D

A stopped at single bonds, which leave carbon at six. Carbon is not boron: an O lone pair forms C=O and completes the octet. B skipped the leftovers: 26 − 24 = 2 go on nitrogen, 6 + 2 = 8. C mixed up the families: antimony holds 10, an expanded octet.

Practice 6 answer: B

A kept boron at six, as in BF₃, with the fourth fluorine left as a separate F⁻. The ion bonds all four. C counted bonds, not electrons: 4 bonds hold 4 × 2 = 8. D took boron's own 3 valence electrons as its total; the bonds bring in fluorine's electrons too.

Practice 7 answer: C

A missed the incomplete octet: two Be–H bonds give beryllium 2 + 2 = 4. B stopped silicon at eight as if it were carbon; period-3 silicon holds six bonds, 12 electrons. D added PCl₄⁺'s charge instead of subtracting it: 5 + 4(7) + 1 = 34 leaves 2 for phosphorus, 10 electrons.

Guided example: the Lewis structure

First · Draw the Lewis structure

Oxygen is the central atom. The 4 leftover electrons stay on oxygen as two lone pairs.

Guided example: the Lewis structure

First · Draw the Lewis structure

Two fluorines bond to oxygen: X = 2. Two lone pairs sit on oxygen: E = 2.

Guided example: the Lewis structure

First · Draw the Lewis structure

Two fluorines bond to oxygen: X = 2. Two lone pairs sit on oxygen: E = 2.

Guided example: the shape

Step 1 · Count the electron domains

X + E = 2 + 2 = 4 electron domains.

Guided example: the shape

Four domains spread to 109.5°. The electron-pair geometry is tetrahedral.

Guided example: the shape

Guided example: the shape

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.

Practice 2 answer: B

A named the electron-pair geometry; one corner holds the lone pair, not an atom. C stopped after the outer octets: 26 − 6 − 18 = 2 electrons remain, and they sit on phosphorus. D counted three domains with one a lone pair, which leaves room for only two chlorines. PCl₃ has three bonded Cl plus the lone pair: 3 + 1 = 4 domains.

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 3 answer: D

B named the electron-pair geometry; the lone pair takes a corner, so the atoms are bent. C counted the N=O double bond as two domains: 2 + 1 + 1 = 4, a tetrahedral base. A made the same count and then named that geometry. E subtracted the charge: 5 + 2(6) − 1 = 16 leaves 16 − 4 − 12 = 0 for N, so O=N=O would be linear.

Practice 4 answer: C

A dropped the leftover pair: the charge's electron makes 26, and 2 of them stay on chlorine. B named the electron-pair geometry; one corner holds the lone pair, not an atom. D took the water row: bent needs two lone pairs and two atoms, and ClO₃⁻ has three bonded O and one lone pair.

Extra practice 1 answer: C

D forced boron's octet with a lone pair. No electrons remain for one, so the 3 + 1 = 4 domains of a 107° pyramid never form. A added the same lone pair and took the tetrahedral angle, 109.5°. B read the angle off the flat drawing; three domains spread to 120°, not 90°.

Extra practice 2 answer: E

B added the charge: 7 + 2(6) + 1 = 20 leaves 20 − 4 − 12 = 4, two lone pairs and the 2 + 2 = 4 domains of chlorite, ClO₂⁻; a 1+ ion has one electron fewer. A dropped the leftover pair: 2 of the 18 electrons stay on chlorine as a third domain. C named the electron-pair geometry twice; the lone pair takes a corner, so the atoms are bent. D counted the Cl=O double bond as two domains: 2 + 1 + 1 = 4, then read a pyramid.

Extra practice 3 answer: D

A gave nitrogen a lone pair, as in NH₃. NO₃⁻ has 5 + 3(6) + 1 = 24 electrons, and 24 − 6 − 18 = 0 left; N at 6 → one O shares a pair, N=O: 3 domains, trigonal planar. B counted the S=O double bond twice: 3 + 1 = 4 domains around three atoms. SO₃ has 6 + 3(6) = 24, none left on S; S at 6 → one O shares a pair, S=O: 3 + 0 = 3 domains, trigonal planar. C matched the elements, not the counts. NH₄⁺ has 5 + 4(1) − 1 = 8 electrons, all in four N–H bonds: tetrahedral.

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.

Practice 2 answer: A

B skipped the difference: 0.03 is far under 0.4, and even polar bonds would cancel in a straight line. C gave carbon a lone pair; its four valence electrons all go into the two double bonds. D bent the molecule with the outer atoms' lone pairs; only lone pairs on the central atom set the shape.

Practice 3 answer: D

A missed the two lone pairs and drew H₂S straight, like CS₂. B skipped the first question: 2.58 − 2.20 = 0.38 is under 0.4, but a central lone pair makes the molecule polar anyway. C read the flat drawing as the shape; four domains spread in three dimensions, and both lone pairs sit on one side of sulfur.

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 4 answer: C

A stopped at the lone-pair question; a tetrahedral shape cancels only when all four outer atoms are identical, as in CCl₄. B trusted the flat drawing; in three dimensions the two C–Cl bonds sit 109.5° apart, never opposite, so their arrows add. D used a rule CCl₄ breaks: four polar bonds, nonpolar molecule.

Practice 5 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.

Practice 6 answer: A

B stopped at three identical fluorines; identical outer atoms cancel only around a symmetric center, and the lone pair on nitrogen pushes all three N–F arrows to one side. C drew SO₂ straight like CO₂; the lone pair on sulfur bends it, and the two sulfur–oxygen arrows add. D stopped at the tetrahedral shape; one Br among three H leaves the C–Br arrow with no equal partner.

Practice 7 answer: D

A stopped at the shape; trigonal planar cancels only when all three outer atoms are identical, as in BCl₃. B let the outer atom's lone pairs set the shape; only lone pairs on the central atom do, and carbon has none. C has the right verdict for a wrong reason: CO₂ has polar bonds and is nonpolar.

Practice 8 answer: B

A skipped the lone-pair question for NCl₃: 0.12 is under 0.4, but the lone pair on nitrogen makes the molecule pyramidal and polar anyway. C judged by the bonds alone: SiCl₄ has the most polar bonds in the set, 1.26, yet its four identical arrows cancel. D read HCN as symmetric like CO₂; a linear shape cancels only when both outer atoms are identical.