Lewis DiagramsLewis 结构

Big Idea · Structure and Properties (SAP)AP Science Practice 1 · Models and RepresentationsAP link · Unit 2~40 min

Learning goals

  • Draw Lewis diagrams for molecules and ions step by step, including central atoms with lone pairs, an electron-deficient central atom (B), and two molecules joined through a lone pair
  • Read information from a Lewis diagram: the number of lone pairs, the type and number of bonds, relative bond length and bond energy, and the empirical formula
  • Recognize structures with the same molecular formula but different connections between atoms (isomers)

Before you start

  • Covalent bonds and molecules (2.4.2)
  • Drawing Lewis structures (2.4.4)
  • Types of chemical bonds (U2.1)

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

  1. Count the valence electrons. Add one electron for each negative charge on an anion; subtract one for each positive charge on a cation.
  2. 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.
  3. Connect with single bonds. Each single bond uses 2 electrons.
  4. 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.
  5. 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 HX2O\ce{H2O} and HX2S\ce{H2S}, 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 NHX3\ce{NH3} and PHX3\ce{PH3} each have 1 lone pair, and those of CHX4\ce{CH4} and SiHX4\ce{SiH4} have none.

HOH
H₂O: 2 lone pairs on O
HSH
H₂S: 2 lone pairs on S
NHHH
NH₃: 1 lone pair on N

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 BFX3\ce{BF3} and BHX3\ce{BH3} 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 PClX5\ce{PCl5} (10) and SFX6\ce{SF6} (12). The period 2 elements C, N, O, and F never exceed 8.
  • Odd numbers of electrons: NOX2\ce{NO2} has 17 valence electrons, so some atom always ends up short of 8. This is a limitation of the Lewis model.
BFFF
BF₃: 6 electrons around B
BHHH
BH₃: 6 electrons around B
SFFFFFF
SF₆: 12 electrons around S

An electron-deficient B meets a lone pair

The B in BFX3\ce{BF3} is still one electron pair short, and the N in NHX3\ce{NH3} 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:

NHX3+BFX3→HX3N−BFX3\ce{NH3 + BF3 -> H3N-BF3}
NHHH
NH₃
+
BFFF
BF₃
N+B−HHHFFF
Product H₃N–BF₃: the lone pair on N has become the N–B 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

MoleculeBetween the two C atomsBond length (pm)Bond energy (kJ/mol)Empirical formula
CX2HX6\ce{C2H6}single bond (2 shared electrons)154348CHX3\ce{CH3}
CX2HX4\ce{C2H4}double bond (4 shared electrons)134614CHX2\ce{CH2}
CX2HX2\ce{C2H2}triple bond (6 shared electrons)120839CH
CCHHHHHH
C₂H₆
CCHHHH
C₂H₄
HCCH
C₂H₂

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 CX2HX2\ce{C2H2} and benzene, CX6HX6\ce{C6H6}, share the empirical formula CH. U2.7 uses the same diagrams to predict shapes: all six atoms of CX2HX4\ce{C2H4} lie in one plane, the four atoms of CX2HX2\ce{C2H2} lie on a straight line, and CX2HX6\ce{C2H6} 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 CX2HX6O\ce{C2H6O}: 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.

CCOHHHHHH
Ethanol CH₃CH₂OH
COCHHHHHH
Dimethyl ether CH₃OCH₃

Propanal and acetone are both CX3HX6O\ce{C3H6O}. 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.

CCCOHHHHHH
Propanal CH₃CH₂CHO
CCCOHHHHHH
Acetone CH₃COCH₃

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:

Try it: choose a molecule, click bonds to switch between single, double, and triple, and build a correct Lewis structure

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 PClX3\ce{PCl3} and state how many lone pairs are on the central atom.

Step-by-step solution
  1. Count the valence electrons. P is in group 15 with 5 valence electrons; Cl is in group 17 with 7:

    5+3×7=265 + 3 \times 7 = 26
  2. 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:

    26−3×2=2026 - 3 \times 2 = 20
  3. Complete the outer atoms first. Each Cl already has one bond (2 electrons) and needs 3 lone pairs (6 electrons) to reach 8:

    20−3×6=220 - 3 \times 6 = 2
  4. Put the rest on the central atom. The last 2 electrons go on P as 1 lone pair.

  5. 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 1\boxed{1} lone pair.

Example 2: Comparing C₂H₄ and C₂H₂

Using their Lewis diagrams, compare the length and strength of the C–C bonds in CX2HX4\ce{C2H4} and CX2HX2\ce{C2H2}, and give the empirical formula of each.

Step-by-step solution
  1. Identify: draw both Lewis diagrams. Valence electrons:

    CX2HX4:2×4+4×1=12CX2HX2:2×4+2×1=10\begin{aligned} \ce{C2H4} &: 2 \times 4 + 4 \times 1 = 12 \\ \ce{C2H2} &: 2 \times 4 + 2 \times 1 = 10 \end{aligned}

    To give each C an octet, the carbons in CX2HX4\ce{C2H4} share a double bond (H₂C=CH₂, 4 shared electrons), and those in CX2HX2\ce{C2H2} a triple bond (H–C≡C–H, 6 shared electrons).

  2. Compare bond orders. A double bond has bond order 2; a triple bond has bond order 3.

  3. 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 CX2HX2\ce{C2H2} is shorter and stronger.

  4. Empirical formulas. An empirical formula is the lowest whole-number ratio:

    CX2HX4:2:4=1:2→CHX2CX2HX2:2:2=1:1→CH\begin{aligned} \ce{C2H4} &: 2 : 4 = 1 : 2 \to \boxed{\ce{CH2}} \\ \ce{C2H2} &: 2 : 2 = 1 : 1 \to \boxed{\ce{CH}} \end{aligned}
Practice: how many lone pairs are on the central atom?
  1. S in HX2S\ce{H2S}
  2. P in PClX3\ce{PCl3}
  3. B in BFX3\ce{BF3}
  4. C in CHX4\ce{CH4}
  5. O in OFX2\ce{OF2}
  6. C in HCN
  7. O in HX3OX+\ce{H3O+}
  8. N in NHX4X+\ce{NH4+}
  9. S in SClX2\ce{SCl2}
  10. Xe in XeFX2\ce{XeF2}

Common mistakes

Mistake 1: “Every atom must have 8 electrons.” H needs only 2; B in BFX3\ce{BF3} has only 6; central atoms from period 3 on can exceed 8, like the 12 around S in SFX6\ce{SF6}.

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 HX2S\ce{H2S}, and why?

In which molecule does the central atom have fewer than 8 electrons around it?

NHX3\ce{NH3} reacts with BFX3\ce{BF3} to form HX3N−BFX3\ce{H3N-BF3}. Which statement about the new N–B bond is correct?

Based on the Lewis diagrams of CX2HX6\ce{C2H6}, CX2HX4\ce{C2H4}, and CX2HX2\ce{C2H2}, which claim is supported?

How are CHX3CHX2OH\ce{CH3CH2OH} and CHX3OCHX3\ce{CH3OCH3} related?

Which molecule contains only single bonds?

AP-style free response

AP-style free response

(a) Find the total number of valence electrons in BFX3\ce{BF3}, draw its Lewis diagram, and state how many electrons surround B.

(b) A student draws BFX3\ce{BF3} 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) BFX3\ce{BF3} reacts with NHX3\ce{NH3} to form HX3N−BFX3\ce{H3N-BF3}. 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 NHX3\ce{NH3}; 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 HX2O\ce{H2O} and HX2S\ce{H2S}, 1 in NHX3\ce{NH3}.
  • 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 HX3N−BFX3\ce{H3N-BF3}.
  • 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?

Practice this chapter: Unit 2 · Compound Structure and Properties · 12 questionsPractice this topic: U2.5 · Lewis Diagrams · 5 questions

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