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S2.2 The covalent model

Practice exam-style IB Chemistry questions for The covalent model, aligned with the syllabus and grouped by topic.

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Verified by Dennis M.
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Question 1
SL • Paper 1A
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SL • Paper 1A
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The carbon-carbon bond length decreases as bond order increases.

What is the correct order of increasing carbon-carbon bond length?

A.

C−C<C≡C<C=CC-C < C\equiv C < C=C

B.

C−C<C=C<C≡CC-C < C=C < C\equiv C

C.

C=C<C−C<C≡CC=C < C-C < C\equiv C

D.

C≡C<C=C<C−CC\equiv C < C=C < C-C

Question 2
SL • Paper 1A
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SL • Paper 1A
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The Lewis formula of OF2OF_2 contains two single bonds.

What is the total number of lone pairs in the molecule?

A.

4

B.

8

C.

6

D.

10

Question 3
SL • Paper 1A
Easy
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SL • Paper 1A
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Ammonia reacts with boron trifluoride to form F3B←NH3F_3B\leftarrow NH_3.

What is the donor atom in the coordination bond?

A.

FF

B.

NN

C.

BB

D.

HH

Question 4
SL • Paper 1A
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SL • Paper 1A
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In a paper chromatogram, the solvent front travels 8.0 cm8.0\ \text{cm} from the baseline. The centre of a dye spot travels 3.2 cm3.2\ \text{cm} from the baseline.

What is the RFR_F value of the dye?

A.

2.5

B.

4.8

C.

0.40

D.

0.25

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Question 5
HL • Paper 1A
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HL • Paper 1A
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X-ray diffraction shows that all six carbon-carbon bonds in benzene have the same length.

What conclusion is best supported by this evidence?

A.

Benzene contains three isolated C=CC=C bonds and three isolated C−CC-C bonds.

B.

Benzene is non-planar because the carbon atoms are sp3sp^3 hybridized.

C.

Benzene undergoes addition reactions more readily than alkenes.

D.

Benzene has a delocalized π\pi system with carbon-carbon bond lengths intermediate between single and double bonds.

Question 6
HL • Paper 1A
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HL • Paper 1A
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The structure of ethanoic acid is CH3COOHCH_3COOH.

How many σ\sigma bonds and π\pi bonds are present in one molecule?

A.

7 σ\sigma bonds and 1 π\pi bond

B.

8 σ\sigma bonds and 1 π\pi bond

C.

7 σ\sigma bonds and 2 π\pi bonds

D.

6 σ\sigma bonds and 1 π\pi bond

Question 7
SL • Paper 1A
Medium
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SL • Paper 1A
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Methanal, H2COH_2CO, has three electron domains around the carbon atom.

What is the most accurate statement about the H−C−HH-C-H bond angle?

A.

It is exactly 109.5∘109.5^\circ because carbon forms four bonds.

B.

It is slightly less than 120∘120^\circ because the C=OC=O domain repels more strongly than C−HC-H domains.

C.

It is 180∘180^\circ because the carbon atom has no lone pairs.

D.

It is exactly 120∘120^\circ because all three domains repel equally.

Question 8
SL • Paper 1A
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SL • Paper 1A
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Each diagram represents a molecule with identical terminal atoms XX bonded to a central atom AA. Each A−XA-X bond is polar, with XX more electronegative than AA.

The molecule with a net dipole moment is represented by which diagram?

A.
B.
C.
D.

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Question 9
HL • Paper 1A
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HL • Paper 1A
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The carbonate ion, CO32−CO_3^{2-}, is represented by three equivalent resonance structures.

What is the bond order of each C−OC-O bond in the resonance hybrid?

A.

1121\frac{1}{2}

B.

1

C.

1131\frac{1}{3}

D.

2

Question 10
HL • Paper 1A
Medium
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HL • Paper 1A
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A molecule has five electron domains around the central atom: four bonding domains and one lone pair.

The molecular geometry is represented by which diagram?

A.
B.
C.
D.
Question 11
HL • Paper 1A
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HL • Paper 1A
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In one Lewis formula of NO3−NO_3^-, nitrogen is bonded to one oxygen by a double bond and to two oxygens by single bonds. Nitrogen has no lone pairs.

Using FC=VE−(NBE+12BE)FC = VE - \left(NBE + \frac{1}{2}BE\right), what is the formal charge on nitrogen?

A.

−1-1

B.

0

C.

+2+2

D.

+1+1

Question 12
SL • Paper 2
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SL • Paper 2
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Phosphorus trichloride, PCl3PCl_3, is a covalent molecule.

A

the total number of valence electrons in one molecule of PCl3PCl_3.

[1]
B

the Lewis formula of PCl3PCl_3, showing all bonding pairs and lone pairs.

[2]
C

whether the phosphorus atom obeys the octet rule in this molecule.

[1]

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Question 13
SL • Paper 2
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SL • Paper 2
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The table gives information about the carbon-carbon bonds in ethane, ethene and ethyne.

Compound

C–C bond order

C–C bond length / pm

C–C bond strength / kJ mol−1^{-1}

Ethane

1

154

348

Ethene

2

134

612

Ethyne

3

120

837

A

the bond order of the carbon-carbon bond in ethyne, HC≡CHHC\equiv CH.

[1]
B

the relationship between bond order, bond length and bond strength for carbon-carbon bonds.

[2]
Question 14
SL • Paper 2
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SL • Paper 2
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Diamond and graphite are allotropes of carbon with covalent network structures.

A

the bonding and arrangement of carbon atoms in diamond and graphite.

[2]
B

why graphite conducts electricity but diamond does not.

[2]
Question 15
SL • Paper 2
Medium
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SL • Paper 2
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Propanone, CH3COCH3CH_3COCH_3, and butane, C4H10C_4H_{10}, have similar molar masses but different solubilities in water.

A

the strongest intermolecular force between propanone molecules.

[1]
B

why propanone is much more soluble in water than butane.

[2]
C

why London dispersion forces are present in both substances.

[1]
Question 16
HL • Paper 2
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HL • Paper 2
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Propyne has the structural formula CH3C≡CHCH_3C\equiv CH.

A

the number of sigma bonds and pi bonds in one molecule of propyne.

[2]
B

why a sigma bond is usually stronger than a pi bond.

[2]

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Question 17
SL • Paper 1B
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SL • Paper 1B
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A student prepared Lewis formulas for four covalent species using dots and crosses to show valence electrons.

A

the total number of valence electrons in OF2OF_2.

[1]
B

Identify the electron-deficient molecule in the panel.

[1]
C

why the central atom in NH4+NH_4^+ has no lone pair, although the central atom in NH3NH_3 does.

[2]
Question 18
SL • Paper 1B
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SL • Paper 1B
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Bond length and average bond enthalpy data are shown for carbon-carbon bonds of different bond order.

Bond order

Bond length / pm

Average bond enthalpy / kJ mol−1^{-1}

1

154

348

2

134

614

3

120

839

A

the bond order of the carbon-carbon bond with the shortest bond length.

[1]
B

the relationship between carbon-carbon bond length and average bond enthalpy shown by the data.

[1]
C

why a carbon-carbon triple bond is stronger than a carbon-carbon single bond.

[2]
Question 19
HL • Paper 1A
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HL • Paper 1A
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In the ethanoate ion, CH3COO−CH_3COO^-, the two carbon-oxygen bonds in the carboxylate group are equivalent because of delocalization.

What is the hybridization of the carbon atom in the carboxylate group and of each oxygen atom in that group?

A.

Carbon sp2sp^2; each oxygen sp2sp^2

B.

Carbon sp3sp^3; each oxygen sp3sp^3

C.

Carbon sp2sp^2; one oxygen sp2sp^2 and one oxygen sp3sp^3

D.

Carbon spsp; each oxygen sp2sp^2

Question 20
SL • Paper 2
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SL • Paper 2
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Methanal, H2COH_2CO, has the Lewis formula shown.

A

the number of electron domains around the carbon atom and the electron domain geometry.

[2]
B

why the H−C−HH-C-H bond angle is slightly less than the ideal angle for this electron domain geometry.

[1]
C

whether methanal is polar.

[1]

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Question 21
SL • Paper 2
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SL • Paper 2
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A mixture of two coloured compounds, X and Y, was separated by thin-layer chromatography on polar silica using a non-polar solvent.

A

the RFR_F value for spot X.

[2]
B

which compound, X or Y, is less polar.

[2]
Question 22
HL • Paper 2
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HL • Paper 2
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The carbonate ion, CO32−CO_3^{2-}, can be represented by resonance structures.

A

what is meant by resonance structures.

[1]
B

the number of equivalent resonance structures for CO32−CO_3^{2-}.

[1]
C

why the three carbon-oxygen bonds in CO32−CO_3^{2-} have the same length.

[2]
Question 23
HL • Paper 2
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HL • Paper 2
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Sulfur tetrafluoride, SF4SF_4, and xenon tetrafluoride, XeF4XeF_4, are species with expanded octets around the central atom.

A

the electron domain geometry and molecular geometry around sulfur in SF4SF_4.

[2]
B

why the lone pair in SF4SF_4 is placed in an equatorial position in the VSEPR model.

[1]
C

the molecular geometry of XeF4XeF_4.

[1]
Question 24
SL • Paper 1B
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SL • Paper 1B
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The table compares four molecules with four or fewer electron domains around the central atom.

Molecule

Central atom

Electron domains around central atom

Lone pairs on central atom

Bond angle(s) / ∘^\circ

CH4CH_4

C

4

0

H-C-H = 109.5

NH3NH_3

N

4

1

H-N-H = 107.0

H2OH_2O

O

4

2

H-O-H = 104.5

H2COH_2CO

C

3

0

H-C-H = 116.0; O-C-H = 122.0

A

the molecular geometry of NH3NH_3 from the data.

[1]
B

the bond angles in CH4CH_4, NH3NH_3 and H2OH_2O using the VSEPR model.

[2]
C

why the H−C−HH-C-H angle in H2COH_2CO is less than the ideal trigonal planar angle.

[1]

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Question 25
SL • Paper 1B
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SL • Paper 1B
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Selected physical properties of covalent substances are shown.

Substance

Melting point / ∘C^\circ\text{C}

Electrical conductivity

Hardness / texture

Diamond

3550

No

Very hard

Graphite

3650

Yes, along layers

Soft and slippery

Silicon dioxide

1710

No

Hard

Buckminsterfullerene

280

No

Soft

A

Identify the substance that conducts electricity well along layers and can act as a lubricant.

[1]
B

the high melting point of silicon dioxide.

[2]
C

why buckminsterfullerene has a much lower melting point than diamond.

[1]
Question 26
SL • Paper 1B
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SL • Paper 1B
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Boiling point and solubility data are shown for four molecular covalent substances of similar molar mass.

Substance

Boiling point / ∘C^\circ\text{C}

Solubility in water / g per 100 g water

butane

-1

0.01

methoxyethane

7

7.0

propanal

49

20

propan-1-ol

97

100

A

Identify the substance most likely to form hydrogen bonds between its own molecules.

[1]
B

why the alcohol has a higher boiling point than the ether, although their molar masses are similar.

[2]
C

why the hydrocarbon has the lowest solubility in water.

[1]
Question 27
HL • Paper 1B
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HL • Paper 1B
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Experimental bond length data are shown for the carbonate ion and for typical carbon-oxygen bonds.

Bond type

Bond length / pm

Typical C–O single bond

143

Typical C=O double bond

120

C–O bond 1 in CO32−CO_3^{2-}

128

C–O bond 2 in CO32−CO_3^{2-}

128

C–O bond 3 in CO32−CO_3^{2-}

128

A

what the equal C−OC-O bond lengths in CO32−CO_3^{2-} indicate about the three bonds.

[1]
B

why a single Lewis formula with one C=OC=O bond and two C−OC-O bonds does not fully represent CO32−CO_3^{2-}.

[2]
C

the average bond order of each carbon-oxygen bond in CO32−CO_3^{2-}.

[1]
Question 28
HL • Paper 1B
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HL • Paper 1B
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Molecular model data are shown for species with expanded octets around the central atom.

Model

electron domains

bonding domains

lone pairs

lone-pair site

90∘90^\circ interactions/LP

TBP axial lone pair

5

—

—

axial

3

TBP equatorial lone pair

5

—

—

equatorial

2

SF4SF_4

5

4

1

equatorial

2

XeF4XeF_4

6

4

2

opposite

4

A

the molecular geometry of a species with five electron domains, four bonding domains and one lone pair.

[1]
B

why the lone pair in SF4SF_4 occupies an equatorial position in a trigonal bipyramidal electron domain geometry.

[2]
C

the molecular geometry of XeF4XeF_4.

[1]

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Question 29
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HL • Paper 2
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Benzene, C6H6C_6H_6, is often represented as a hexagon with a circle inside the ring.

Bond

Length / pm

Typical C−CC-C single bond

154

Typical C=CC=C double bond

134

Benzene C1–C2

139

Benzene C2–C3

139

Benzene C3–C4

139

Benzene C4–C5

139

Benzene C5–C6

139

Benzene C6–C1

139

A

the geometry around each carbon atom in benzene.

[1]
B

how carbon-carbon bond length data support the delocalized model of benzene.

[2]
C

why benzene tends to undergo substitution rather than addition reactions.

[1]
Question 30
HL • Paper 2
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HL • Paper 2
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Two possible Lewis formulas, A and B, can be drawn for the cyanate ion, OCN−OCN^-.

A

the formal charge on oxygen in formula A.

[1]
B

the formal charge on nitrogen in formula B.

[1]
C

which formula is preferred, using formal charge arguments.

[2]
Question 31
HL • Paper 2
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HL • Paper 2
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The ethanoate ion, CH3COO−CH_3COO^-, contains a carboxylate group in which the two carbon-oxygen bonds are experimentally found to be equal in length.

A

the hybridization and electron domain geometry of the carbon atom in the carboxylate group.

[2]
B

why the two carbon-oxygen bonds in the carboxylate group are equal in length.

[2]
Question 32
SL • Paper 1B
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SL • Paper 1B
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A mixture of three dyes was separated by thin-layer chromatography using a polar silica stationary phase and a non-polar mobile phase.

A

the RFR_F value of the dye spot in the mixture that travelled 3.2 cm3.2\ \text{cm} when the solvent front travelled 8.0 cm8.0\ \text{cm}.

[2]
B

Identify which reference dye matches the mixture spot with RF=0.40R_F=0.40.

[1]
C

why the most polar dye travels the shortest distance on this TLC plate.

[1]

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Question 33
HL • Paper 1B
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HL • Paper 1B
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Two possible Lewis formulas for the cyanate ion, OCN−OCN^-, are shown. In both formulas the atoms are arranged O−C−NO-C-N.

A

the formal charge on oxygen in structure A.

[1]
B

the formal charge on nitrogen in structure B.

[1]
C

which structure is preferred using formal charge arguments.

[2]
Question 34
HL • Paper 1B
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HL • Paper 1B
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A displayed Lewis formula for one resonance structure of the nitrate ion is shown with one N=ON=O bond and two N−ON-O bonds.

A

the number of sigma bonds and pi bonds in the displayed resonance structure.

[2]
B

why the measured N−ON-O bonds in nitrate are equal, although the displayed structure contains one double bond and two single bonds.

[2]
C

why the measured N−ON-O bond length is intermediate between typical single and double N−ON-O bond lengths.

[1]
Question 35
HL • Paper 1B
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HL • Paper 1B
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The structural formula of acrylonitrile, CH2=CH−C≡NCH_2=CH-C\equiv N, is shown with three carbon atoms labelled.

A

the hybridization of C1C_1 and C3C_3.

[2]
B

the linear geometry around C3C_3 in terms of hybridization.

[1]
C

Analyse the pi bonding in acrylonitrile.

[2]
Question 36
SL • Paper 2
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SL • Paper 2
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Methanal, H2COH_2CO, is a small covalent molecule. Carbon is the central atom.

A

The Lewis formula of methanal is used to its shape.

I.

the Lewis formula of methanal, showing all bonding and non-bonding electron pairs.

[2]
II.

the electron domain geometry and the molecular geometry around the carbon atom.

[2]
B

why the H-C-H bond angle in methanal is slightly less than the ideal angle for its electron domain geometry.

[2]
C

Methanal contains polar C=O bonds and polar C-H bonds. whether methanal is polar overall.

[1]

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Question 37
SL • Paper 2
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SL • Paper 2
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Ammonia reacts with boron trifluoride to form the adduct H3NBF3H_3NBF_3. Boron trifluoride is an electron-deficient molecule.

A

The reaction can be described using Lewis acid-base theory.

I.

Identify the Lewis base and the Lewis acid in the reaction.

[2]
II.

the feature of the ammonia molecule that allows it to form a coordination bond.

[1]
B

the Lewis formula of H3NBF3H_3NBF_3, using an arrow to show the coordination bond.

[2]
C

why BF3BF_3 can accept an electron pair even though each fluorine atom has an octet.

[1]
Question 38
SL • Paper 2
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SL • Paper 2
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The table gives information about bonds between nitrogen atoms in gaseous covalent species.

Species

N–N bond order

N–N bond length / pm

N–N bond enthalpy / kJ mol−1^{-1}

Hydrazine, N2H4N_2H_4

1

145

163

Diazene, N2H2N_2H_2

2

125

418

Nitrogen, N2N_2

3

110

945

A

Use the data to analyse the relationship between bond order, bond length and bond strength.

[3]
B

Hydrazine, N2H4N_2H_4, contains an N-N single bond. Nitrogen, N2N_2, contains an N≡NN\equiv N triple bond.

I.

the number of shared electron pairs in the bond between the nitrogen atoms in hydrazine and in nitrogen.

[1]
II.

which N-N bond is more difficult to break, referring to the data.

[1]
C

the statement: “A molecule containing a triple bond is always less reactive than a molecule containing a single bond because the triple bond is stronger.”

[2]
Question 39
SL • Paper 2
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SL • Paper 2
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A student separates three coloured compounds, X, Y and Z, by thin layer chromatography. The stationary phase is polar silica and the mobile phase is a relatively non-polar solvent.

A

The retardation factor is calculated from the distances on the chromatogram.

I.

the equation used to RFR_F.

[1]
II.

the RFR_F value of compound Y using the distances shown on the chromatogram. The distance from the baseline to Y is 6.0 cm6.0\ \text{cm} and the distance from the baseline to the solvent front is 10.0 cm10.0\ \text{cm}.

[2]
B

which compound is most strongly attracted to the stationary phase.

[2]
C

why RFR_F values from this experiment should not be used to identify compounds unless the chromatography conditions are specified.

[2]
Question 40
HL • Paper 1B
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HL • Paper 1B
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The hydrogenation enthalpy of benzene can be compared with that expected for a hypothetical molecule containing three isolated carbon-carbon double bonds.

Species / bond

Hydrogenation enthalpy / kJ mol−1^{-1}

C–C bond length / pm

Hypothetical molecule with three isolated C=C bonds

-360

—

Benzene (all six C–C bonds)

-208

139

Typical C–C single bond

—

154

Typical C=C double bond

—

134

A

the resonance energy of benzene using a hypothetical hydrogenation enthalpy of −360 kJ mol−1-360\ \text{kJ mol}^{-1} and an experimental value for benzene of −208 kJ mol−1-208\ \text{kJ mol}^{-1}.

[2]
B

how the carbon-carbon bond length data support delocalization in benzene.

[1]
C

why benzene tends to undergo substitution rather than addition reactions.

[2]

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Question 41
SL • Paper 2
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SL • Paper 2
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Diamond, graphite and silicon dioxide are covalent network substances with different structures and properties.

A

the bonding around carbon atoms in diamond and graphite.

[3]
B

The properties of these covalent networks can be explained by their structures.

I.

why diamond and silicon dioxide have high melting points.

[2]
II.

why graphite can be used as a lubricant.

[1]
C

why graphite conducts electricity but silicon dioxide does not.

[2]
Question 42
SL • Paper 2
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SL • Paper 2
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Propanone, propan-1-ol and butane have similar molar masses but different physical properties.

Substance

Molar mass / g mol−1^{-1}

Boiling point / ∘C^\circ\text{C}

Water solubility in water at 25∘C25^\circ\text{C}

Butane

58.1

-0.5

very low

Propanone

58.1

56.0

miscible

Propan-1-ol

60.1

97.2

miscible

A

the strongest type of intermolecular force between molecules of each substance.

[3]
B

Use intermolecular forces to two trends shown by the data.

I.

why propan-1-ol has a higher boiling point than propanone.

[1]
II.

why butane has poor solubility in water.

[1]
C

why molar mass alone is not sufficient to the volatility of these compounds.

[2]
Question 43
HL • Paper 2
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HL • Paper 2
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Ozone, O3O_3, is an allotrope of oxygen. Its bonding is described using resonance.

A

Resonance structures can be used to model the bonding in ozone.

I.

two resonance structures for ozone.

[2]
II.

why the two O-O bonds in ozone have the same length.

[2]
B

the bond order of each O-O bond in ozone using the resonance model.

[1]
C

Oxygen, O2O_2, and ozone absorb different wavelengths of ultraviolet radiation in the atmosphere. why different photon energies are needed to break bonds in O2O_2 and O3O_3.

[2]
Question 44
HL • Paper 2
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HL • Paper 2
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Phosphorus pentafluoride, PF5PF_5, sulfur tetrafluoride, SF4SF_4, and xenon tetrafluoride, XeF4XeF_4, contain central atoms with expanded octets.

A

Analyse the VSEPR model for PF5PF_5.

I.

the number of electron domains around phosphorus in PF5PF_5.

[1]
II.

the electron domain geometry, molecular geometry and approximate F-P-F bond angles in PF5PF_5.

[2]
B

Sulfur tetrafluoride has five electron domains around sulfur.

I.

the molecular geometry of SF4SF_4.

[1]
II.

why the lone pair in SF4SF_4 occupies an equatorial position.

[2]
C

the molecular geometry of XeF4XeF_4 and your answer.

[2]

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Question 45
HL • Paper 2
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HL • Paper 2
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Two possible Lewis formulas can be drawn for sulfur dioxide, SO2SO_2. Formula A has one S-O single bond and one S=O double bond. Formula B has two S=O double bonds and one lone pair on sulfur, giving sulfur an expanded octet (10 electrons total).

A

Formal charge can be used to the two Lewis formulas.

I.

the formula used to formal charge.

[1]
II.

the formal charge on sulfur in Formula A and in Formula B.

[2]
III.

the sum of all formal charges in any valid Lewis formula of SO2SO_2.

[1]
B

which of Formula A and Formula B is preferred using formal charge.

[2]
C

Distinguish formal charge from oxidation in terms of the assumption made about bonding electrons.

[1]
Question 46
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Acrylonitrile, CH2=CHC≡NCH_2=CHC\equiv N, is used to make polymers. Its displayed structure contains single, double and triple covalent bonds.

A

Analyse the bonding shown in acrylonitrile.

I.

the number of σ\sigma bonds and π\pi bonds in one molecule of acrylonitrile.

[2]
II.

how a σ\sigma bond differs from a π\pi bond in terms of orbital overlap and electron density.

[2]
B

why the C≡\equivN triple bond contains two π\pi bonds but only one σ\sigma bond.

[2]
C

why a π\pi bond is usually weaker than a σ\sigma bond.

[1]
Question 47
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Benzene, C6H6C_6H_6, is often represented as a hexagon with a circle inside the ring.

Observation

Bond length / pm

Hydrogenation ΔH\Delta H / kJ mol−1^{-1}

All six C–C bonds in benzene

139, 139, 139, 139, 139, 139

—

Typical C–C bond

154

—

Typical C=C bond

134

—

Hydrogenation of cyclohexene

—

−120-120

Hydrogenation of benzene

—

−208-208

A

Use physical evidence to the bonding in benzene.

[3]
B

The hydrogenation data provide evidence for resonance energy.

I.

what is meant by resonance energy in benzene.

[1]
II.

how the hydrogenation data support the existence of resonance energy.

[2]
C

why benzene tends to undergo substitution reactions rather than addition reactions typical of alkenes.

[2]
Question 48
HL • Paper 2
Hard
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HL • Paper 2
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The ethanoate ion, CH3COO−CH_3COO^-, can be represented by two equivalent resonance structures. The two C-O bonds in the carboxylate group have the same length.

A

Analyse the resonance description of the carboxylate group.

I.

the two equivalent resonance structures of the ethanoate ion.

[2]
II.

the average C-O bond order in the carboxylate group.

[1]
B

Use hybridization to the geometry and delocalization in the carboxylate group.

I.

the hybridization of the carboxylate carbon atom and the electron domain geometry around it.

[2]
II.

how unhybridized p orbitals allow delocalization in the carboxylate group.

[1]
C

why the two C-O bonds in the ethanoate ion have the same length and why this length is intermediate between typical C-O and C=O bonds.

[2]

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S2.1 The ionic model

S2.3 The metallic model