One diagram, a lot of information
Lesson 2.4.4 showed how to draw Lewis structures. AP topic 2.5 asks you to draw them quickly and correctly, and also to read information off them: how many electrons surround each atom, whether the central atom has lone pairs, and whether each bond is single or multiple. Resonance and formal charge (U2.6), molecular shape and hybridization (U2.7), and intermolecular forces (Unit 3) all start from a correct Lewis diagram.
Drawing Lewis diagrams: a quick review
- Count the valence electrons. Add one electron for each negative charge on an anion; subtract one for each positive charge on a cation.
- Sketch the skeleton. The less electronegative atom usually goes in the middle; H can bond to only one atom, so it is always on the outside.
- Connect with single bonds. Each single bond uses 2 electrons.
- Complete the outer atoms. Use the remaining electrons to give each outer atom 8 (H needs only 2), then put any extra electrons on the central atom.
- If the central atom is short of 8, make multiple bonds. Move a lone pair from an outer atom into the space between the two atoms, turning a single bond into a double bond, and a triple bond if needed.
Lone pairs on the central atom
O and S are both in group 16 with 6 valence electrons. In and , the central atom uses 2 of them to bond to the two H atoms, and the remaining 4 form 2 lone pairs孤电子对 on the central atom. Elements in the same group form molecules with matching Lewis diagrams: the central atoms of and each have 1 lone pair, and those of and have none.
A Lewis diagram is a flat sketch: H–O–H is drawn in a straight line, but that doesn’t mean the water molecule is straight. The real shape comes from the VSEPR model (U2.7).
Exceptions to the octet rule
- Electron-deficient atoms: the B in and has only 6 electrons around it. Why not move a lone pair from F to make a B=F double bond? That would give F a positive formal charge (U2.6), which makes no sense for the most electronegative element. So B keeps only 3 electron pairs.
- Expanded octets: central atoms from period 3 and beyond (P, S, Cl, Xe, and so on) are larger and can have more than 8 electrons, as in (10) and (12). The period 2 elements C, N, O, and F never exceed 8.
- Odd numbers of electrons: has 17 valence electrons, so some atom always ends up short of 8. This is a limitation of the Lewis model.
An electron-deficient B meets a lone pair
The B in is still one electron pair short, and the N in has a lone pair. When the two molecules meet, the lone pair on N is shared with B, forming a new N–B covalent bond:
Both electrons in this bond come from N (a coordinate covalent bond). After the reaction, B also has 8 electrons around it. The +1 and −1 on the diagram are formal charges, covered in detail in U2.6.
Reading a Lewis diagram: hydrocarbons
| Molecule | Between the two C atoms | Bond length (pm) | Bond energy (kJ/mol) | Empirical formula |
|---|---|---|---|---|
| single bond (2 shared electrons) | 154 | 348 | ||
| double bond (4 shared electrons) | 134 | 614 | ||
| triple bond (6 shared electrons) | 120 | 839 | CH |
From these diagrams alone: the more electrons the two C atoms share, the shorter and stronger the C–C bond (U2.2). The three empirical formulas all differ (U1.3), while and benzene, , share the empirical formula CH. U2.7 uses the same diagrams to predict shapes: all six atoms of lie in one plane, the four atoms of lie on a straight line, and is not planar.
Isomers
Compounds with the same molecular formula but different connections between atoms are called isomers同分异构体. Ethanol and dimethyl ether are both : in ethanol, O is bonded to one C and one H (an O–H bond), while in dimethyl ether, O is bonded to two C atoms. Different structures give different properties: at room temperature ethanol is a liquid that mixes with water in any proportion, while dimethyl ether is a gas.
Propanal and acetone are both . The difference is where the C=O sits: at the end of the chain in propanal (that C also carries an H), and in the middle in acetone.
To decide whether two diagrams show isomers, first count each kind of atom to confirm the molecular formula is the same, then check whether the atoms are connected differently.
Now draw your own: choose a molecule and click a bond to switch it among single, double, and triple. Lone pairs are filled in automatically by the steps above, and atoms still short of an octet turn orange:
Click a bond to switch it between single, double, and triple. Orange atoms don't have an octet yet; green atoms do.
Example 1: The Lewis diagram of PCl₃
Draw the Lewis diagram of and state how many lone pairs are on the central atom.
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Count the valence electrons. P is in group 15 with 5 valence electrons; Cl is in group 17 with 7:
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Build the skeleton. The less electronegative atom goes in the middle: P (2.19) is less electronegative than Cl (3.16), so P is central, joined to three Cl atoms by single bonds of 2 electrons each:
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Complete the outer atoms first. Each Cl already has one bond (2 electrons) and needs 3 lone pairs (6 electrons) to reach 8:
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Put the rest on the central atom. The last 2 electrons go on P as 1 lone pair.
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Check the octets. P: 3 bonds (6 electrons) + 1 lone pair (2) = 8; each Cl: 1 bond + 3 lone pairs = 8. Every atom has an octet, so no multiple bonds are needed, and 6 + 18 + 2 = 26 electrons are used, matching step 1.
The central P atom has lone pair.
Example 2: Comparing C₂H₄ and C₂H₂
Using their Lewis diagrams, compare the length and strength of the C–C bonds in and , and give the empirical formula of each.
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Identify: draw both Lewis diagrams. Valence electrons:
To give each C an octet, the carbons in share a double bond (H₂C=CH₂, 4 shared electrons), and those in a triple bond (H–C≡C–H, 6 shared electrons).
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Compare bond orders. A double bond has bond order 2; a triple bond has bond order 3.
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Connect to length and strength. The more electrons shared between two nuclei, the more strongly both nuclei are attracted, so they are pulled closer and are harder to separate:
- Length: C≡C 120 pm < C=C 134 pm
- Bond energy: C≡C 839 > C=C 614 (kJ/mol)
The C–C bond in is shorter and stronger.
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Empirical formulas. An empirical formula is the lowest whole-number ratio:
- S in
S has 6 valence electrons; 2 are used in bonds, leaving 4: 2 lone pairs.
- P in
P has 5 valence electrons; 3 are used in bonds, leaving 2: 1 lone pair.
- B in
All 3 valence electrons of B are used in bonds, so B has only 6 electrons around it.
- C in
All 4 valence electrons of C are used in bonds.
- O in
As in , O uses 2 valence electrons in bonds and keeps 2 lone pairs.
- C in HCN
C has one single bond and one triple bond, using all 4 valence electrons.
- O in
8 valence electrons in all (6 + 3 − 1); the three O–H bonds use 6, leaving 1 lone pair on O.
- N in
8 valence electrons in all (5 + 4 − 1), all in the four N–H bonds.
- S in
As in , S has 2 lone pairs.
- Xe in
22 valence electrons in all: the two Xe–F bonds use 4, each F gets 6 more, and 6 are left on Xe, which is 3 lone pairs (an expanded octet).
Common mistakes
Mistake 1: “Every atom must have 8 electrons.” H needs only 2; B in has only 6; central atoms from period 3 on can exceed 8, like the 12 around S in .
Mistake 2: “A molecule has the shape its Lewis diagram is drawn in.” A Lewis diagram shows only how atoms are connected and where electrons are. H–O–H is drawn straight, yet water is bent. Shape comes from the VSEPR model (U2.7).
Mistake 3: “Isomers are just different ways of drawing the same substance.” Isomers are different compounds, with atoms connected differently and different properties. Structures that differ only in where the electrons are, with the atoms unmoved, are different drawings of the same substance, called resonance structures (U2.6).
Quick check
How many lone pairs are on the central S atom in , and why?
S is in the same group as O, so the Lewis diagram of matches that of . B: each S–H bond uses only 1 electron from S. D: S can expand its octet, but has no extra electrons to place there.
In which molecule does the central atom have fewer than 8 electrons around it?
B has only 3 valence electrons, so after forming 3 bonds it has just 6 electrons around it. N, C, and O each have 8 as central atoms.
reacts with to form . Which statement about the new N–B bond is correct?
B starts with only 6 electrons and has no lone pair to offer; N has one lone pair. N and B are both nonmetals, so the bond is covalent, not ionic.
Based on the Lewis diagrams of , , and , which claim is supported?
More shared electrons mean a shorter, stronger bond. A: the empirical formulas are , , and CH. C: the number of H atoms doesn't decide C–C bond strength. D: a double bond shares 4 electrons.
How are and related?
Both are , but ethanol has an O–H bond while the O in dimethyl ether is bonded to two C atoms: they are isomers, with different properties. Resonance structures only move electrons and never change which atoms are connected.
Which molecule contains only single bonds?
In , C is bonded to 3 H atoms and 1 O, and O to 1 H, and every atom already has an octet (2 for H). HCN has C≡N, has C=O, and has C=C.
AP-style free response
(a) Find the total number of valence electrons in , draw its Lewis diagram, and state how many electrons surround B.
(b) A student draws with one B=F double bond so that B has an octet. Calculate the formal charges on B and on the double-bonded F in this structure (formal charge = valence electrons − lone-pair electrons − bonds), and explain why it is not the best structure.
(c) reacts with to form . Identify the source of the two electrons in the new N–B bond.
(d) In the reaction, the number of electron pairs around B changes from 3 to 4. Predict how the F–B–F bond angle changes.
Model answer and scoring
(a) 3 + 3 × 7 = 24 valence electrons. B is in the middle, joined to three F atoms by single bonds, with 3 lone pairs on each F. B has only 6 electrons around it.
(b) Double-bonded F: 7 − 4 − 2 = +1; B: 3 − 0 − 4 = −1. F, the most electronegative element, would carry a positive formal charge while B carries a negative one, whereas in the all-single-bond structure every formal charge is 0. So the single-bond structure is better.
(c) Both electrons come from the lone pair on the N atom of ; B, which was one electron pair short, accepts this pair.
(d) The angle decreases: B goes from 3 electron pairs (trigonal planar, 120°) to 4 (tetrahedral), so the F–B–F angle becomes close to 109.5°.
Scoring: (a) 24 electrons, a correct diagram, and “6 electrons”; (b) both formal charges correct, plus the point that the positive charge sits on F, the most electronegative element; (c) “the lone pair on N”; (d) predicts a decrease and cites 4 electron pairs and about 109.5°.
Summary
- Drawing a Lewis diagram: count valence electrons (adjusting for charge), sketch the skeleton, connect with single bonds, complete the outer atoms and then the central atom, and form multiple bonds if needed.
- Show the lone pairs on the central atom: 2 each in and , 1 in .
- Exceptions: B can have only 6 electrons; central atoms from period 3 on can exceed 8; molecules with an odd number of electrons always leave some atom short.
- A lone pair can form a new covalent bond with an electron-deficient atom, as in .
- A Lewis diagram shows bond types, relative bond lengths and energies, and the empirical formula; compounds with the same molecular formula but different connections are isomers.
AP tip: Questions often list several molecules and ask which has two lone pairs on the central atom, which doesn’t follow the octet rule, or which claim a set of Lewis diagrams supports. Count the total valence electrons before you draw, and check at the end: is the electron total right, and how many electrons surround each atom?