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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.

Verified by Dennis M.
Verified by Dennis M.
Paper
Difficulty
Status
Level
Question 1
SL • Paper 1A
Easy
Calculator Permitted

The carbon-carbon bond length decreases as bond order increases.

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

A.

Cāˆ’C<C≔C<C=CC-C < C\equiv C < C=C

B.

Cāˆ’C<C=C<C≔CC-C < C=C < C\equiv C

C.

C=C<Cāˆ’C<C≔CC=C < C-C < C\equiv C

D.

C≔C<C=C<Cāˆ’CC\equiv C < C=C < C-C

Question 2
SL • Paper 1A
Easy
Calculator Permitted

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
Calculator Permitted

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
Easy
Calculator Permitted

In a paper chromatogram, the solvent front travels 8.0Ā cm8.0\ \text{cm} from the baseline. The centre of a dye spot travels 3.2Ā 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

Question 5
HL • Paper 1A
Easy
Calculator Permitted

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 Cāˆ’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
Easy
Calculator Permitted

The structure of ethanoic acid is CH3COOHCH_3COOH.

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

A.

7 σ\sigma bonds and 1 Ļ€\pi bond

B.

8 σ\sigma bonds and 1 Ļ€\pi bond

C.

7 σ\sigma bonds and 2 Ļ€\pi bonds

D.

6 σ\sigma bonds and 1 Ļ€\pi bond

Question 7
SL • Paper 1A
Medium
Calculator Permitted

Methanal, H2COH_2CO, has three electron domains around the carbon atom.

What is the most accurate statement about the Hāˆ’Cāˆ’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 Cāˆ’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
Medium
Calculator Permitted

Each diagram represents a molecule with identical terminal atoms XX bonded to a central atom AA. Each Aāˆ’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.
Question 9
HL • Paper 1A
Medium
Calculator Permitted

The carbonate ion, CO32āˆ’CO_3^{2-}, is represented by three equivalent resonance structures.

What is the bond order of each Cāˆ’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
Calculator Permitted

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
Medium
Calculator Permitted

In one Lewis formula of NO3āˆ’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=VEāˆ’(NBE+12BE)FC = VE - \left(NBE + \frac{1}{2}BE\right), what is the formal charge on nitrogen?

A.

āˆ’1-1

B.

0

C.

+2+2

D.

+1+1

Question 12
SL • Paper 2
Medium
Calculator Permitted

Phosphorus trichloride, PCl3PCl_3, is a covalent molecule.

A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 13
SL • Paper 2
Medium
Calculator Permitted

The table gives information about the carbon-carbon bonds in ethane, ethene and ethyne.

CompoundC–C bond orderC–C bond length / pmC–C bond strength / kJ mol⁻¹
Ethane1154348
Ethene2134612
Ethyne3120837
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 14
SL • Paper 2
Medium
Calculator Permitted

Diamond and graphite are allotropes of carbon with covalent network structures.

A

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

[2]
Write your answer here...
B

Explain why graphite conducts electricity but diamond does not.

[2]
Write your answer here...

0

Question 15
SL • Paper 2
Medium
Calculator Permitted

Propanone, CH3COCH3CH_3COCH_3, and butane, C4H10C_4H_{10}, have similar molar masses but different solubilities in water.

A

State the strongest intermolecular force between propanone molecules.

[1]
Write your answer here...
B

Explain why propanone is much more soluble in water than butane.

[2]
Write your answer here...
C

State why London dispersion forces are present in both substances.

[1]
Write your answer here...

0

Question 16
HL • Paper 2
Medium
Calculator Permitted

Propyne has the structural formula CH3C≔CHCH_3C\equiv CH.

A

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 17
SL • Paper 1B
Medium
Calculator Permitted

A student prepared Lewis formulas for four covalent species using dots and crosses to show valence electrons.

An annotated panel of Lewis formulas for four labelled species including $OF_2$, $BF_3$, $NH_3$ and $NH_4^+$. The display shows bonding pairs, lone pairs, brackets and charge where appropriate, with one species having fewer than an octet around the central atom. The panel should not include total valence electron counts or explanatory notes.
A

State the total number of valence electrons in OF2OF_2.

[1]
Write your answer here...
B

Identify the electron-deficient molecule in the panel.

[1]
Write your answer here...
C

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

[2]
Write your answer here...

0

Question 18
SL • Paper 1B
Medium
Calculator Permitted

Bond length and average bond enthalpy data are shown for carbon-carbon bonds of different bond order.

Bond orderBond length / pmAverage bond enthalpy / kJ mol^-1
1154348
2134614
3120839
A

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

[1]
Write your answer here...
B

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

[1]
Write your answer here...
C

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

[2]
Write your answer here...

0

Question 19
HL • Paper 1A
Medium
Calculator Permitted

In the ethanoate ion, CH3COOāˆ’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
Medium
Calculator Permitted

Methanal, H2COH_2CO, has the Lewis formula shown.

A displayed Lewis formula for methanal showing carbon as the central atom, single bonds from carbon to two hydrogen atoms, a double bond from carbon to oxygen, and two lone pairs on oxygen. The diagram must not include bond angles, molecular geometry labels or dipole arrows.
A

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

[2]
Write your answer here...
B

Suggest why the Hāˆ’Cāˆ’HH-C-H bond angle is slightly less than the ideal angle for this electron domain geometry.

[1]
Write your answer here...
C

Deduce whether methanal is polar.

[1]
Write your answer here...

0

Question 21
SL • Paper 2
Medium
Calculator Permitted

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

A TLC plate diagram with a horizontal pencil baseline near the bottom, a horizontal solvent front near the top, and two labelled spot centres X and Y on the same lane. Spot X must be higher than spot Y. The diagram must show enough scale or distance information from the baseline to allow calculation of the retardation factor, but must not display any calculated $R_F$ values.
A

Calculate the RFR_F value for spot X.

[2]
Write your answer here...
B

Explain which compound, X or Y, is less polar.

[2]
Write your answer here...

0

Question 22
HL • Paper 2
Medium
Calculator Permitted

The carbonate ion, CO32āˆ’CO_3^{2-}, can be represented by resonance structures.

A

State what is meant by resonance structures.

[1]
Write your answer here...
B

Deduce the number of equivalent resonance structures for CO32āˆ’CO_3^{2-}.

[1]
Write your answer here...
C

Explain why the three carbon-oxygen bonds in CO32āˆ’CO_3^{2-} have the same length.

[2]
Write your answer here...

0

Question 23
HL • Paper 2
Medium
Calculator Permitted

Sulfur tetrafluoride, SF4SF_4, and xenon tetrafluoride, XeF4XeF_4, are species with expanded octets around the central atom.

A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...
C

State the molecular geometry of XeF4XeF_4.

[1]
Write your answer here...

0

Question 24
SL • Paper 1B
Medium
Calculator Permitted

The table compares four molecules with four or fewer electron domains around the central atom.

MoleculeCentral atomElectron domains around central atomLone pairs on central atomBond angle(s) / °
CH4C40H-C-H = 109.5
NH3N41H-N-H = 107.0
H2OO42H-O-H = 104.5
H2COC30H-C-H = 116.0; O-C-H = 122.0
A

Deduce the molecular geometry of NH3NH_3 from the data.

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

Suggest why the Hāˆ’Cāˆ’HH-C-H angle in H2COH_2CO is less than the ideal trigonal planar angle.

[1]
Write your answer here...

0

Question 25
SL • Paper 1B
Medium
Calculator Permitted

Selected physical properties of covalent substances are shown.

SubstanceMelting point / °CElectrical conductivityHardness / texture
Diamond3550NoVery hard
Graphite3650Yes, along layersSoft and slippery
Silicon dioxide1710NoHard
Buckminsterfullerene280NoSoft
A

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

[1]
Write your answer here...
B

Explain the high melting point of silicon dioxide.

[2]
Write your answer here...
C

Suggest why buckminsterfullerene has a much lower melting point than diamond.

[1]
Write your answer here...

0

Question 26
SL • Paper 1B
Medium
Calculator Permitted

Boiling point and solubility data are shown for four molecular covalent substances of similar molar mass.

SubstanceBoiling point / °CSolubility in water / g per 100 g water
butane-10.01
methoxyethane77.0
propanal4920
propan-1-ol97100
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

Suggest why the hydrocarbon has the lowest solubility in water.

[1]
Write your answer here...

0

Question 27
HL • Paper 1B
Medium
Calculator Permitted

Experimental bond length data are shown for the carbonate ion and for typical carbon-oxygen bonds.

Bond typeBond length / pm
Typical C–O single bond143
Typical C=O double bond120
C–O bond 1 in CO3^2āˆ’128
C–O bond 2 in CO3^2āˆ’128
C–O bond 3 in CO3^2āˆ’128
A

State what the equal Cāˆ’OC-O bond lengths in CO32āˆ’CO_3^{2-} indicate about the three bonds.

[1]
Write your answer here...
B

Explain why a single Lewis formula with one C=OC=O bond and two Cāˆ’OC-O bonds does not fully represent CO32āˆ’CO_3^{2-}.

[2]
Write your answer here...
C

Deduce the average bond order of each carbon-oxygen bond in CO32āˆ’CO_3^{2-}.

[1]
Write your answer here...

0

Question 28
HL • Paper 1B
Medium
Calculator Permitted

Molecular model data are shown for species with expanded octets around the central atom.

Modelelectron domainsbonding domainslone pairslone-pair site90° interactions/LP
TBP axial lone pair5——axial3
TBP equatorial lone pair5——equatorial2
SF4541equatorial2
XeF4642opposite4
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

Deduce the molecular geometry of XeF4XeF_4.

[1]
Write your answer here...

0

Question 29
HL • Paper 2
Medium
Calculator Permitted

Benzene, C6H6C_6H_6, is often represented as a hexagon with a circle inside the ring.

BondLength / pm
Typical C–C single bond154
Typical C=C double bond134
Benzene C1–C2139
Benzene C2–C3139
Benzene C3–C4139
Benzene C4–C5139
Benzene C5–C6139
Benzene C6–C1139
A

State the geometry around each carbon atom in benzene.

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

Suggest why benzene tends to undergo substitution rather than addition reactions.

[1]
Write your answer here...

0

Question 30
HL • Paper 2
Medium
Calculator Permitted

Two possible Lewis formulas, A and B, can be drawn for the cyanate ion, OCNāˆ’OCN^-.

Two labelled Lewis formulas for the cyanate ion with the same atom order O-C-N. Formula A shows a single O-C bond and a triple C-N bond with all lone pairs shown. Formula B shows an O=C=N arrangement with all lone pairs shown. The overall ion charge is shown outside brackets for each formula, but preferred structure and formal charge values are not shown.
A

Calculate the formal charge on oxygen in formula A.

[1]
Write your answer here...
B

Calculate the formal charge on nitrogen in formula B.

[1]
Write your answer here...
C

Deduce which formula is preferred, using formal charge arguments.

[2]
Write your answer here...

0

Question 31
HL • Paper 2
Medium
Calculator Permitted

The ethanoate ion, CH3COOāˆ’CH_3COO^-, contains a carboxylate group in which the two carbon-oxygen bonds are experimentally found to be equal in length.

A displayed structural formula of the ethanoate ion showing the methyl group attached to a carboxylate group with the two oxygen atoms bonded to the same carbon. The ion is enclosed in brackets with an overall negative charge. Do not show resonance arrows, hybridization labels or bond order values.
A

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 32
SL • Paper 1B
Hard
Calculator Permitted

A mixture of three dyes was separated by thin-layer chromatography using a polar silica stationary phase and a non-polar mobile phase.

An annotated TLC plate with a pencil baseline, solvent front, three reference dye lanes and one mixture lane. Distances from the baseline to the solvent front and to the centres of the spots are indicated with a ruler scale. The more polar reference dye remains closer to the baseline. Exact distances should be visible in the final figure, but no $R_F$ values are printed.
A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 33
HL • Paper 1B
Hard
Calculator Permitted

Two possible Lewis formulas for the cyanate ion, OCNāˆ’OCN^-, are shown. In both formulas the atoms are arranged Oāˆ’Cāˆ’NO-C-N.

Two labelled Lewis formulas for $OCN^-$ with the same atom connectivity. Structure A shows a single $O-C$ bond and a triple $C\equiv N$ bond with appropriate lone pairs and overall brackets. Structure B shows double bonds $O=C=N$ with appropriate lone pairs and overall brackets. Formal charge values are not printed.
A

Calculate the formal charge on oxygen in structure A.

[1]
Write your answer here...
B

Calculate the formal charge on nitrogen in structure B.

[1]
Write your answer here...
C

Determine which structure is preferred using formal charge arguments.

[2]
Write your answer here...

0

Question 34
HL • Paper 1B
Hard
Calculator Permitted

A displayed Lewis formula for one resonance structure of the nitrate ion is shown with one N=ON=O bond and two Nāˆ’ON-O bonds.

A bracketed Lewis formula for one resonance structure of $NO_3^-$ showing a central nitrogen atom bonded to three oxygen atoms, with one $N=O$ double bond and two $N-O$ single bonds. Lone pairs and the overall charge are shown. A small table gives measured equal $N-O$ bond lengths in nitrate and typical single and double $N-O$ bond lengths. The visual does not state sigma or pi bond counts.
A

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...
C

Suggest why the measured Nāˆ’ON-O bond length is intermediate between typical single and double Nāˆ’ON-O bond lengths.

[1]
Write your answer here...

0

Question 35
HL • Paper 1B
Hard
Calculator Permitted

The structural formula of acrylonitrile, CH2=CHāˆ’C≔NCH_2=CH-C\equiv N, is shown with three carbon atoms labelled.

A displayed structural formula of acrylonitrile, $CH_2=CH-C\equiv N$, with the terminal alkene carbon labelled $C_1$, the central alkene carbon labelled $C_2$, and the nitrile carbon labelled $C_3$. The diagram includes approximate bond angles around $C_1$ and $C_3$ and indicates the $C=C$ and $C\equiv N$ multiple bonds, but it does not name hybridizations or orbital types.
A

Deduce the hybridization of C1C_1 and C3C_3.

[2]
Write your answer here...
B

Explain the linear geometry around C3C_3 in terms of hybridization.

[1]
Write your answer here...
C

Analyse the pi bonding in acrylonitrile.

[2]
Write your answer here...

0

Question 36
SL • Paper 2
Hard
Calculator Permitted

Methanal, H2COH_2CO, is a small covalent molecule. Carbon is the central atom.

A

The Lewis formula of methanal is used to predict its shape.

I.

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

[2]
Write your answer here...
II.

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 37
SL • Paper 2
Hard
Calculator Permitted

Ammonia reacts with boron trifluoride to form the adduct H3NBF3H_3NBF_3. Boron trifluoride is an electron-deficient molecule.

A Lewis-style diagram showing separate molecules of ammonia and boron trifluoride before reaction. Ammonia has nitrogen bonded to three hydrogens with one lone pair on nitrogen. Boron trifluoride has boron bonded to three fluorine atoms in a trigonal planar arrangement; lone pairs on fluorine atoms may be indicated, and boron is shown with fewer than eight electrons. No product structure or coordination arrow is shown.
A

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

I.

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

[2]
Write your answer here...
II.

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 38
SL • Paper 2
Hard
Calculator Permitted

The table gives information about bonds between nitrogen atoms in gaseous covalent species.

SpeciesN–N bond orderN–N bond length / pmN–N bond enthalpy / kJ mol⁻¹
Hydrazine, Nā‚‚Hā‚„1145163
Diazene, Nā‚‚Hā‚‚2125418
Nitrogen, Nā‚‚3110945
A

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

[3]
Write your answer here...
B

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

I.

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

[1]
Write your answer here...
II.

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

[1]
Write your answer here...
C

Evaluate the statement: ā€œA molecule containing a triple bond is always less reactive than a molecule containing a single bond because the triple bond is stronger.ā€

[2]
Write your answer here...

0

Question 39
SL • Paper 2
Hard
Calculator Permitted

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 TLC plate diagram with a pencil baseline near the bottom, a solvent front near the top, and three spots labelled X, Y and Z at different heights above the baseline. X is closest to the baseline, Y is intermediate and Z is closest to the solvent front. Distances from the baseline to each spot centre and to the solvent front are indicated with measuring arrows, but exact numerical values are not included in this placeholder.
A

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

I.

State the equation used to calculate RFR_F.

[1]
Write your answer here...
II.

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

[2]
Write your answer here...
B

Explain which compound is most strongly attracted to the stationary phase.

[2]
Write your answer here...
C

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

[2]
Write your answer here...

0

Question 40
HL • Paper 1B
Hard
Calculator Permitted

The hydrogenation enthalpy of benzene can be compared with that expected for a hypothetical molecule containing three isolated carbon-carbon double bonds.

Species / bondHydrogenation enthalpy / kJ mol^-1C–C bond length / pm
Hypothetical molecule with three isolated C=C bonds-360—
Benzene (all six C–C bonds)-208139
Typical C–C single bond—154
Typical C=C double bond—134
A

Calculate the resonance energy of benzene using a hypothetical hydrogenation enthalpy of āˆ’360Ā kJĀ molāˆ’1-360\ \text{kJ mol}^{-1} and an experimental value for benzene of āˆ’208Ā kJĀ molāˆ’1-208\ \text{kJ mol}^{-1}.

[2]
Write your answer here...
B

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

[1]
Write your answer here...
C

Discuss why benzene tends to undergo substitution rather than addition reactions.

[2]
Write your answer here...

0

Question 41
SL • Paper 2
Hard
Calculator Permitted

Diamond, graphite and silicon dioxide are covalent network substances with different structures and properties.

A three-panel schematic of covalent network structures. One panel shows a three-dimensional tetrahedral carbon network for diamond. One panel shows graphite as stacked hexagonal layers with weak attractions between layers and delocalized electrons within layers. One panel shows silicon dioxide as a continuous network with silicon atoms bonded to four oxygens and oxygen atoms bridging two silicon atoms. Labels identify only the substances and general structural features, not properties.
A

Compare the bonding around carbon atoms in diamond and graphite.

[3]
Write your answer here...
B

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

I.

Explain why diamond and silicon dioxide have high melting points.

[2]
Write your answer here...
II.

Explain why graphite can be used as a lubricant.

[1]
Write your answer here...
C

Explain why graphite conducts electricity but silicon dioxide does not.

[2]
Write your answer here...

0

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

SubstanceMolar mass / g mol^-1Boiling point / °CWater solubility in water at 25 °C
Butane58.1-0.5very low
Propanone58.156.0miscible
Propan-1-ol60.197.2miscible
A

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

[3]
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B

Use intermolecular forces to explain two trends shown by the data.

I.

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

[1]
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II.

Explain why butane has poor solubility in water.

[1]
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C

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

[2]
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Question 43
HL • Paper 2
Hard
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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.

Draw two resonance structures for ozone.

[2]
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II.

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

[2]
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B

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

[1]
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C

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

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

An unlabeled VSEPR reference diagram showing the positions available in trigonal bipyramidal and octahedral electron-domain arrangements. The trigonal bipyramidal arrangement distinguishes axial and equatorial positions. The octahedral arrangement shows six equivalent positions before lone pairs are considered. No molecular geometries for the named species are shown.
A

Analyse the VSEPR model for PF5PF_5.

I.

State the number of electron domains around phosphorus in PF5PF_5.

[1]
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II.

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

[2]
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B

Sulfur tetrafluoride has five electron domains around sulfur.

I.

Deduce the molecular geometry of SF4SF_4.

[1]
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II.

Explain why the lone pair in SF4SF_4 occupies an equatorial position.

[2]
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C

Deduce the molecular geometry of XeF4XeF_4 and explain your answer.

[2]
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Question 45
HL • Paper 2
Hard
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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).

Two skeletal Lewis formula candidates for sulfur dioxide labelled A and B. Formula A shows O-S=O with the left S-O bond single and the right S=O bond double; lone pairs are indicated sufficiently for formal charge calculation. Formula B shows O=S=O with two double bonds, one lone pair on sulfur, and lone pairs on both oxygen atoms; sulfur has an expanded octet (10 electrons total). Formal charges are not shown.
A

Formal charge can be used to compare the two Lewis formulas.

I.

State the formula used to calculate formal charge.

[1]
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II.

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

[2]
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III.

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

[1]
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B

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

[2]
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C

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

[1]
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Question 46
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 displayed structural formula of acrylonitrile, H2C=CH-C≔N, with all atoms labelled and all bonds shown clearly. The C=C double bond and C≔N triple bond are visible. No sigma or pi labels are shown.
A

Analyse the bonding shown in acrylonitrile.

I.

State the number of σ\sigma bonds and Ļ€\pi bonds in one molecule of acrylonitrile.

[2]
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II.

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

[2]
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B

Explain why the C≔\equivN triple bond contains two Ļ€\pi bonds but only one σ\sigma bond.

[2]
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C

Suggest why a Ļ€\pi bond is usually weaker than a σ\sigma bond.

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

ObservationBond length / pmHydrogenation ΔH / kJ mol^-1
All six C–C bonds in benzene139, 139, 139, 139, 139, 139—
Typical C–C bond154—
Typical C=C bond134—
Hydrogenation of cyclohexene—-120
Hydrogenation of benzene—-208
A

Use physical evidence to discuss the bonding in benzene.

[3]
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B

The hydrogenation data provide evidence for resonance energy.

I.

State what is meant by resonance energy in benzene.

[1]
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II.

Explain how the hydrogenation data support the existence of resonance energy.

[2]
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C

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

[2]
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Question 48
HL • Paper 2
Hard
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The ethanoate ion, CH3COOāˆ’CH_3COO^-, can be represented by two equivalent resonance structures. The two C-O bonds in the carboxylate group have the same length.

A structural diagram of the ethanoate ion showing the CH3 group attached to a carboxylate carbon bonded to two oxygen atoms. The diagram indicates that the two C-O bonds are equivalent, but does not show the resonance structures or hybridization labels.
A

Analyse the resonance description of the carboxylate group.

I.

Draw the two equivalent resonance structures of the ethanoate ion.

[2]
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II.

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

[1]
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B

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

I.

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

[2]
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II.

Explain how unhybridized p orbitals allow delocalization in the carboxylate group.

[1]
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C

Explain 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