Practice Topic 4 Test bonding [89 marks] -...

Practice Topic 4 Test bonding [89 marks]

1. [1 mark]Which species contain delocalized electrons?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeB

Examiners report[N/A]

2. [1 mark]What is the correct number of sigma and pi bonds in prop-2-enenitrile, ?

MarkschemeC


(σ) (π) C CHCNH2

3. [1 mark]Which process involves the breaking of hydrogen bonds?

A.

B.

C.

D.

MarkschemeD

2HI(g) → (g) + (g)H2 I2

C (g) → C(g) + 4H(g)H4

(l) → (g)H2 H2

N (l) → N (g)H3 H3

Examiners reportMany considered B, the breaking of C–H bonds, to be correct.

4. [1 mark]How many sigma and pi bonds are there in ?

A. 13 and 5

B. 15 and 2

C. 15 and 3

D. 15 only

MarkschemeC

Examiners reportThis was thought to be a hard example “to test student understanding of hybridization and bond types”. This type ofquestion is not new and was answered correctly by 79% of the candidates. All a candidate needs to do is to count thenumber of -bonds.

(σ) (π) C C CCC COOHH3 H2 H2

σ π

σ π

σ π

σ

π

5. [1 mark]Which compound has the highest boiling point?

A.

B.

C.

D.

MarkschemeC


C CH3 H3

C OHH3

C C OHH3 H2

C C CH3 H2 H3

6. [1 mark]What are the hybridizations of the atoms labelled 1, 2 and 3 in the molecule below?

MarkschemeB


7. [1 mark]Which substance is made up of a lattice of positive ions and free moving electrons?

A. Graphite

B. Sodium chloride

C. Sulfur

D. Sodium

MarkschemeD


8. [1 mark]Which order is correct when the following compounds are arranged in order of increasing melting point?

A.

B.

C.

D.

MarkschemeA

Examiners reportOne respondent stated that it would be best to write from least reactive to most reactive in both of these questions.However, “increasing” is written in bold in both questions and, also, this type of question has been asked extensivelyon previous papers and hence candidates would have understood what was asked for explicitly if they had looked atsome of the previous examination papers. In the case of Q.13 60% of candidates gave the correct answer and in Q.27,68% had the question correct.

C < S < OH4 H2 H2

S < O < CH2 H2 H4

C < O < SH4 H2 H2

S < C < OH2 H4 H2

9. [1 mark]What is the correct order if the compounds are arranged in order of increasing boiling point?

A.

B.

C.

D.

MarkschemeD


C < C Cl < Si < C OHH4 H3 H4 H3

C OH < C < C Cl < SiH3 H4 H3 H4

C OH < C Cl < Si < CH3 H3 H4 H4

C < Si < C Cl < C OHH4 H4 H3 H3

10. [1 mark]What is the type of hybridization of the silicon and oxygen atoms in silicon dioxide?

MarkschemeA

Examiners reportThis question on hybridization was particularly badly answered (Difficulty Index 29%), though it was not clear as towhether this arose from a lack of comprehension of the concept itself or the structure of silicon dioxide.

11. [1 mark]Which compound does not form hydrogen bonds between its molecules?

A.

B.

C.

D.

MarkschemeB


C NH3 H2

C COCH3 H3

C COOHH3

C C OHH3 H2

12. [1 mark]Which substance does not conduct electricity?

A. Solid zinc

B. Molten zinc

C. Solid zinc chloride

D. Molten zinc chloride

MarkschemeC


2

13. [1 mark]Which molecules have hybridization?

I.

II.

III.

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeB


sp2

C2H4

C4H10

C6H6

14. [1 mark]Which combination of shape and bond angle is correct for a molecule of xenon tetrafluoride, ?

MarkschemeB

Examiners reportWhilst 70% gave the correct answer, a significant number (17%) missed the axial lone pairs and thought the moleculeto be tetrahedral.

XeF4

15. [1 mark]Which allotropes of carbon show hybridization?

I. Diamond

II. Graphite

III. fullerene

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeC

sp2

C60


16. [1 mark]Which species have delocalized electrons?

I.

II.

III.

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeC

Examiners reportAlthough 70% of the candidates gave the expected answer, C, there is minimal delocalization in ethanoic acid, so bothC and D were accepted (giving an 86% success rate on the question).

π

C COCH3 H3

NO−2

CO2−3

17. [1 mark]Which structure has delocalized electrons?

A.

B. CO

C. HCN

D.

MarkschemeA


π

O3

CO2

18. [1 mark]How many bonding pairs and lone pairs of electrons surround the sulfur atom in the molecule?

MarkschemeA

SF4

Examiners reportOne respondent stated that the structure of is not specified in the teachers note corresponding to AS 14.1.1. Thisis a comment that has been made at length in previous subject reports. The AS states that candidates should be ableto determine the shape and bond angles of species with five or six negative charge centres using VSEPR Theory. In theteaching programme, examples such as , , and should be definitely included. However, anyspecies with five or six negative charge centres could be asked in a question and hence examples are not restricted tothese latter four examples.

SF4

PCl5 SF6 XeF4 PF−6

19. [1 mark]What is the hybridization of the carbon atom, and the number of and bonds in the methanal molecule?

MarkschemeA


σ π

20. [1 mark]The Lewis structure of is given below.

What is the shape of the molecule?

A. Bent (V-shaped)

B. Linear

C. T-shaped

D. Triangular planar

MarkschemeA

Examiners reportThere were two G2 comments on this question. One respondent stated that D. should be trigonal planar instead oftriangular planar. Both terms are widely used in fact, though of course the correct answer is A. bent or V-shaped.Another respondent stated that it would have been better to represent the Lewis structure of with valenceexpansion. It is true that could be represented as an alternate Lewis structure. However, the question did notstate what the best Lewis structure representation of was and hence was not basing the representation at anydistinction centred on formal charge differences versus expanded octets. Candidates simply had to look at the threenegative charge centres present which equates to a triangular planar electron-domain geometry and hence a bentmolecular geometry as the final shape giving A as the correct answer.

SO2

SO2

SO2SO2

SO2

21. [1 mark]Which particles are responsible for electrical conductivity in metals?

A. Anions

B. Cations

C. Electrons

D. Protons

MarkschemeC

Examiners reportOne G2 comment stated that the terms cation and anion are not stated on the syllabus. Although strictly correct, itwould be assumed that these terms would be introduced to students in the classroom as they are universally used inchemistry (e.g. even the term carbocation is widely used in explaining certain nucleophilic substitution reactionmechanisms).

22. [1 mark]Which species does not have delocalized electrons?

A.

B.

C.

D.

MarkschemeD

Examiners reportOne respondent mentioned the fact that there is some debate in the literature in relation to possible sigmadelocalization in cyclopropane which is a valid comment and although 63.98% of candidates chose D. as thecorrect answer, it is fair to state that a different example might have been selected where there is no evidence ofdelocalization.

NO−3

NO−2

O3

C3H6

C3H6

23. [1 mark]Which particles are responsible for the conduction of electricity in molten aluminium?

A. Cations

B. Anions

C. Electrons

D. Protons

MarkschemeC

Examiners reportThere were a number of comments on this question and many teachers stated that although they assumed that therequired answer was C. i.e. electrons, many felt that as molten aluminium was involved, the cations are mobile andthus could conduct electricity, so A. could be another answer. Although the correct answer C. (electrons) was given bythe majority of candidates (71.18%), it was decided at Grade Award to also accept A. as clearly some candidates mayhave approached the question in the sense articulated by several teachers.

24. [1 mark]Retinol (vitamin A) contains a total of 5 double bonds and 46 single bonds.

Which statements are correct?

I. There are 51 and 5 bonds.

II. The oxygen atom is hybridized.

III. Retinol is a primary alcohol.

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

MarkschemeD


σ π

sp3

25a. [2 marks]

Airbags are an important safety feature in vehicles. Sodium azide, potassium nitrate and silicon dioxide have beenused in one design of airbag.

Sodium azide, a toxic compound, undergoes the following decomposition reaction under certain conditions.

Two students looked at data in a simulated computer-based experiment to determine the volume of nitrogengenerated in an airbag.

Sodium azide involves ionic bonding, and metallic bonding is present in sodium. Describe ionic and metallicbonding.

MarkschemeIonic:

(electrostatic) attraction between oppositely charged ions/cations and anions/positive and negative ions;

Do not accept answers such as compounds containing metal and non-metal are ionic.

Metallic:

(electrostatic attraction between lattice of) positive ions/cations/nuclei and delocalized electrons / (bed of) positiveions/cations/nuclei in sea of electrons / OWTTE;

Examiners reportQuestion 1 tested a number of concepts and very few students were able to gain all the marks available. Part (a) wasfairly well done and students could explain ionic and metallic bonding although weak students did not explain thebonding but simply stated that ionic was between metal and non metal etc.

2Na (s) → 2Na(s) + 3 (g)N3 N2

25b. [1 mark]

Using the simulation programme, the students entered the following data into the computer.

State the number of significant figures for the temperature, mass and pressure data.

T:

m:

p:

MarkschemeT: 4 and m: 3 and p: 3;

Examiners reportSurprisingly in part (b) (i) a number of students could not state the number of significant figures and many stated that25.00 was 2 SF instead of 4.

25c. [1 mark]Calculate the amount, in mol, of sodium azide present.

Markscheme;

No penalty for using whole number atomic masses.

Examiners reportPart (b) (ii) required the calculation of the amount of substance in moles, and was generally well done although somedid not realise the value was in kg and so had a value 1000 times too small.

n = (65.0/65.02) = 1.00 (mol)

d 3

25d. [4 marks]Determine the volume of nitrogen gas, in , produced under these conditions based on this reaction.

Markscheme;

;

;

;

Award [4] for correct final answer.

Award [3 max] for 0.0341 (dm ) or 22.7 (dm ).

Award [3 max] for 34.4 (dm ).




Examiners reportIn part (b) (iii) a number of students lost marks for forgetting to convert temperature or pressure and also to multiplythe amount by 1.5. Also many forgot to convert the pressure into kPa if they wanted their answer in . However,most students could obtain at least one of the marks available.

dm3

n( ) = ( × 1.00 =) 1.50 (mol)N232

T = ((25.00 + 273.15) =) 298.15 K/(25.00 + 273) = 298 K

p = 1.08 × 1.01 × Pa/1.08 × 1.01 × kPa/1.09 × Pa/1.09 × kPa105 102 105 102

V = = = 34.1 (d )nRTp

( )(1.50)(8.31)(298.15/298)103

(1.08×1.01× )105 m3

3 3

3

3

3

3

dm3

25e. [1 mark]

The chemistry of the airbag was found to involve three reactions. The first reaction involves the decomposition ofsodium azide to form sodium and nitrogen. In the second reaction, potassium nitrate reacts with sodium.

Suggest why it is necessary for sodium to be removed by this reaction.

Markschemesodium could react violently with any moisture present / sodium is (potentially) explosive / sodium (is dangerous sinceit is flammable when it) forms hydrogen on contact with water / OWTTE;

Do not accept answers such as sodium is dangerous or sodium is too reactive.

Examiners reportIn part (c) (i) many did not relate the removal of sodium to the potential for it to react with water and instead gave a fartoo vague of answer that it was reactive. However, the very best students were able to answer this hypothesis typequestion and stated that sodium reacts with water. This proved a good discriminator at the top end of the candidature.

2KN (s) + 10Na(s) → O(s) + 5N O(s) + (g)O3 K2 a2 N2

25f. [2 marks]The metal oxides from the second reaction then react with silicon dioxide to form a silicate in the third reaction.

Draw the structure of silicon dioxide and state the type of bonding present.

Structure:

Bonding:

O(s) + N O(s) + Si (s) → N Si (s)K2 a2 O2 a2K2 O4

MarkschemeStructure:

drawing of giant structure showing tetrahedrally arranged silicon;

Minimum information required for mark is Si and 4 O atoms, in a tetrahedral arrangement (not 90° bond angles) butwith each of the 4 O atoms showing an extension bond.

Bonding:

(giant/network/3D) covalent;

Examiners reportPart (c)(ii) was very poorly answered and the majority of students believed that had a similar structure to .The very few students that drew a giant structure often did not then show a tetrahedral arrangement of the atoms,however most did realise that the bonding was covalent.

SiO2 CO2

25g. [1 mark]

An airbag inflates very quickly.

It takes just 0.0400 seconds to produce nitrogen gas in the simulation. Calculate the average rate of formation ofnitrogen in (b) (iii) and state its units.

Markscheme;

Accept 851 dm s .

Units required for mark.

Examiners reportPart (d) was generally well answered and most students calculated a rate from their results although some lost themark for incorrect or absent units.

( ) = 853 d / ( ) = 37.5 mol34.10.0400

m3s−1 1.500.0400

s−1

3 –1

25h. [1 mark]The students also discovered that a small increase in temperature (e.g. 10 °C) causes a large increase (e.g.doubling) in the rate of this reaction. State one reason for this.

Markschememore energetic collisions / more species have energy ;

Allow more frequent collisions / species collide more often.

Examiners reportMost students could then successfully explain why the rate increased with temperature. However a minority forgot torefer to time (i.e. more frequent) in relation to collisions.

⩾ Ea

26a. [2 marks]

Calcium nitrate contains both covalent and ionic bonds.

State the formula of both ions present and the nature of the force between these ions.

Ions:

Nature of force:

Markscheme and ;

electrostatic (attraction);

Do not accept ionic.

Ca2+ NO−3

Examiners reportIt was distressing how many students taking HL Chemistry (over 50%?) do not know the formula of the nitrate ion!Many students also gave the type of bonding present between the ions, rather than the nature of the force asked for,though almost all could correctly identify the covalently bonded atoms. Hardly any could explain delocalization interms of the overlap of p-orbitals, or the extension of a -bond, over more than two atoms, though its effect onstructure and stability were better known. In part (c), which tested Aim 8 of the syllabus, most managed to gain someof the marks available for atmospheric pollution from oxides of nitrogen. Inevitably, owing to some overlap inassessment statements these concepts would be more familiar to those studying the Environmental Chemistry option,but undoubtedly studying other options assists in other areas, such as organic chemistry.

π

26b. [1 mark]State which atoms are covalently bonded.

Markschemenitrogen/N and oxygen/O;

Do not accept nitrate/NO .

Accept atoms in nitrate/NO


3–

3–

π

26c. [2 marks]Bonding in the nitrate ion involves electron delocalization. Explain the meaning of electron delocalization andhow it affects the ion.

Markschemepi/ -electrons shared by more than two atoms/nuclei / a pi/

-bond/overlapping p-orbitals that extends over more than two atoms/nuclei;

all (N–O) bonds equal length/strength/bond-order / charge on all oxygen/O atoms equal / increases stability/lowers PE(of the ion);

Accept a diagram that clearly shows one or both points.


π−−π−−

π

26d. [1 mark]

Nitrogen also forms oxides, which are atmospheric pollutants.

Outline the source of these oxides.

Markschemeproduced by high temperature combustion;

Accept combustion/jet/car engines / car exhaust/emissions / lightning / action of bacteria/microorganisms.

Do not accept combustion/burning, cars, planes, jets, factories, power plants etc.


π

26e. [1 mark]State one product formed from their reaction with water.

Markschemenitric acid/ / nitrous acid/nitric(III) acid/ ;

Accept “form acidic solutions / acid rain”.


HNO3 HNO2

π

26f. [1 mark]State one environmental problem caused by these atmospheric pollutants.

Markschemeacid deposition/rain / respiratory problems / corrosion problems / decomposition of ozone layer / photochemical smog/ acidification/pollution of lakes / damage to plants/ trees;

Accept “acid rain” in either part (ii) or part (iii) but not both.

Do not accept air pollution.


π

27a. [4 marks]

Two groups of students (Group A and Group B) carried out a project* on the chemistry of some group 7 elements (thehalogens) and their compounds.

* Adapted from J Derek Woollins, (2009), Inorganic Experiments and Open University, (2008), Exploring the MolecularWorld.

When iodine reacts with excess chlorine, can form. Deduce the Lewis (electron dot) structure of and and state the name of the shape of each species.

Markscheme

No ECF for shape if Lewis structure is incorrect.

ICl3 ICl3ICl−

2

Examiners reportPart (e) was by far one of the most disappointing questions on the entire paper with only the top-end candidatesscoring all four marks. Many mistakes were seen, such as the usual mistakes of omitting lone pairs on terminal atomsand not including square brackets and the negative charge for the Lewis structure of the anion. The biggest problemhowever for candidates was failing to realise that for Lewis structures based on five negative charge centres or fiveelectron domains, the lone pairs are inserted in the equatorial position and not the axial position, resulting in a T-shaped molecular geometry for and a linear shape for . Candidates may benefit in class from a carefuldiscussion of the various angles resulting from LP-LP, LP-BP and BP-BP repulsions for such structures emanating fromfive electron domains. As a result of poor comprehension of this aspect of VSEPR Theory, a common incorrectmolecular geometry of trigonal planar was often cited for the molecular geometry for .

ICl3 ICl−2

ICl3

27b. [1 mark]

In this project the students explored several aspects of the chemistry of the halogens. In the original preparation ofICl(l), they observed the yellow-green colour of chlorine gas, Cl (g), reacting with solid iodine, I (s).

State the full electron configuration of iodine .

Markscheme;

No mark for 2,8,18,18,7 or [Kr] 5s 4d 5p .

Allow electron configurations with order of sublevels interchanged.

Electrons must be represented as superscript to award mark.

Examiners reportIn part (f), the better candidates gave the correct full electron configuration for iodine. Surprisingly some of the weakercandidates gave electron arrangements which scored no marks and a few candidates gave rather sloppyconfigurations, either putting subscripts instead of superscripts or not putting the number of electrons as superscripts,which was rather disconcerting to see at HL.

2 2

(Z = 53)

1 2 2 3 3 4 3 4 5 4 5 /1 2 2 3 3 3 4 4 4 5 5s2 s2 p6 s2 p6 s2 d10 p6 s2 d10 p5 s2 s2 p6 s2 p6 d10 s2 p6 d10 s2 p5

2 10 5

27c. [2 marks]One important use of chlorine is in the synthesis of poly(chloroethene), PVC. Identify the monomer used to makePVC and state one of the uses of PVC.

Monomer:

Use:

MarkschemeMonomer:

/ chloroethene / ;

Accept vinyl chloride or chloroethylene.

Allow C H Cl.

Use:

raincoats / packaging / window frames / pipes / carpets / gutters / electrical cable sheathing / covers for electricalwires / rope / bottles;

Accept suitable alternatives.

Do not allow glue.

Do not allow just plastic(s) or just windows.

Allow plastic bag.

Examiners reportIn part (iii), a large number of candidates stated chloroethane instead of chloroethene for the monomer. Plastic wasoften given as a use of PVC. This however was not allowed for M2 and a more precise answer was required.

C CHClH2

2 3

28a. [1 mark]

Lithium and boron are elements in period 2 of the periodic table. Lithium occurs in group 1 (the alkali metals) andboron occurs in group 3. Isotopes exist for both elements.

The electron configuration of boron is . Draw the shape of an s orbital and a orbital on the axesbelow.

Markschemesymmetrical shape of s orbital and dumbbell-shaped p orbital with electron density along x-axis;

1 2 2s2 s2 p1 px

Examiners reportIn part (iii), a common mistake involved candidates drawing the lobe of electron density around the y or z axes for the

orbital. Some candidates drew three dumbells for the s-orbital. Other candidates incorrectly drew hybrid orbitals.px

28b. [5 marks](ii) Cobalt is a transition metal. One common ion of cobalt is . Draw the orbital diagram (using the arrow-in-box notation) for the ion.

(iii) State the other most common ion of cobalt.

(iv) Explain why the complex is coloured.

Markscheme(ii)

Allow full arrows instead of half-arrows for example .

Do not allow arrows with the same spin for example or in the same orbital.

Do not allow an orbital diagram with a configuration.

(iii) ;

Accept +2, 2+, cobalt(II), II.

(iv) partially filled/incomplete d subshell/sub-level/orbitals;

d orbitals split (into two sets of different energies);

(colour due to) electron transition between (split) d orbitals / d to d transitions / frequencies of visible light absorbed byelectrons moving from lower to higher d levels ;

colour due to remaining frequencies / complementary colour seen;

Allow wavelength as well as frequency.

Co3+

Co3+

[Co(N ]CH3)6 l3

↑↓

↑↑ ↓↓

4 3s1 d5

Co2+

Examiners reportThe orbital diagram in (ii) also proved to be quite a good discriminating question. Many candidates failed to realisethat the electrons are removed from the 4s level before the 3d for a first-row transition metal ion. In addition, asignificant number of candidates showed poor understanding of Hund‟s Rule of Maximum Multiplicity which states thatwhen degenerate orbitals are available, electrons fill the orbitals singly before filling them in pairs. Hence, in manycases incorrect representations were seen for the 3d which involved three pairs of electrons of opposite spin beinginserted in three 3d orbitals. Most candidates stated the ion, though a common incorrect answer was . Part(iv) involved candidates having to explain why the complex is coloured. This question was asked anumber of times in previous examinations and previously was typically really very poorly answered. In N12, theexplanations certainly were better though some candidates mixed up the principles of the line emission spectrum ofhydrogen with the d to d transitions involved in the explanation of colour pertaining to a transition metal complex.

Co2+ Co4+

[Co(N ]CH3)6 l3

29a. [7 marks]

Ethene, C H , and hydrazine, N H , are hydrides of adjacent elements in the periodic table.

(i) Draw Lewis (electron dot) structures for and showing all valence electrons.

(ii) State and explain the H–C–H bond angle in ethene and the H–N–H bond angle in hydrazine.

Markscheme

(i)

Accept x’s, dots or lines for electron pairs

(ii) H–C–H:

any angle between 118° and 122°;

due to three negative charge centres/electron domains/electron pairs;

H–N–H:

any angle between 104° and 108°;

due to four negative charge centres/electron domains/electron pairs;

extra repulsion due to lone electron pairs;

Do not allow ECF for wrong Lewis structures.

Examiners reportThis was a popular question and was answered quite successfully. The Lewis structure for ethene was given correctlyby the great majority of the candidates, but that of hydrazine by only about half of them. Incorrect answers had doublebonds appearing between the 2 nitrogen atoms and lone pairs on nitrogen atoms not shown. Those who could drawthe correct structure in (i) gave the correct bond angle, but the explanation was not given correctly by many. Only veryfew scored the five marks as many failed to mention the extra repulsion of the lone pair.

2 4 2 4

C2H4 N2H4

29b. [4 marks]

The polarity of a molecule can be explained in terms of electronegativity.

(i) Define the term electronegativity.

(ii) Compare the relative polarities of the C–H bond in ethene and the N–H bond in hydrazine.

(iii) Hydrazine is a polar molecule and ethene is non-polar. Explain why ethene is non-polar.

Markscheme(i) (relative) measure of an atoms attraction for electrons;

in a covalent bond / shared pair;

(ii) C–H is less polar as C is less electronegative / N–H bond is more polar as N is more electronegative / difference inelectronegativity is greater for N-H than C-H;

(iii) bond polarities cancel in / OWTTE;

Examiners reportThe definition of electronegativity was not well known and many forgot to mention covalent bond or got confused withionization and electron affinity and talked about a mole of gaseous atoms.

C2H4

29c. [2 marks]The boiling point of hydrazine is much higher than that of ethene. Explain this difference in terms of theintermolecular forces in each compound.

Markschemeweaker van der Waals’/London/dispersion/intermolecular forces in ethene;

stronger (intermolecular) hydrogen bonding in hydrazine;

If no comparison between strengths then [1 max].

Examiners reportIn part (c) most knew that hydrogen bonding in hydrazine was stronger than the van der Waals’ forces in ethene andexplained its higher boiling point. However, some candidates described hydrogen bonding as the bond between N andH in the molecule, and some omitted a comparison of the relative strengths.

29d. [3 marks]Hydrazine is a valuable rocket fuel.

The equation for the reaction between hydrazine and oxygen is given below.

Use the bond enthalpy values from Table 10 of the Data Booklet to determine the enthalpy change for this reaction.

Markschemebonds broken: 4 N–H, N–N, O=O / ;

bonds formed: N N, 4O–H / ;

;

Award [3] for correct final answer.

Examiners reportThe calculation for the enthalpy change produced some completely correct calculations but many candidates lostmarks here for using the wrong bond energies, although ECF was applied to the structures drawn in part (a).

(g) + (g) → (g) + 2 O(g)N2H4 O2 N2 H2

+2220 (kJ mo )l−1

≡ −2801 (kJ mo )l−1

−581 (kJ mo )l−1

29e. [2 marks]State the name of the product and identify the type of reaction which occurs between ethene and hydrogenchloride.

Markschemechloroethane;

(electrophilic) addition;

Do not accept free radical/nucleophilic addition.

Examiners reportIn (e) ‘addition’ was correctly identified as the reaction type by most but when asked in (f) to identify the final reactiontype few recognised it as an acid-base reaction, however, the bond angle was given correctly by many.

29f. [2 marks]

The reaction between N H (aq) and HCl (aq) can be represented by the following equation.

(i) Identify the type of reaction that occurs.

(ii) Predict the value of the H–N–H bond angle in .

Markscheme(i) acid-base/neutralization;

(ii) 109°/109.5°;

Examiners reportIn (e) ‘addition’ was correctly identified as the reaction type by most but when asked in (f) to identify the final reactiontype few recognised it as an acid-base reaction, however, the bond angle was given correctly by many.

2 4

(aq) + 2HCl(aq) → (aq) + 2C (aq)N2H4 N2H2+6 l−

N2H2+6

30. [6 marks]SF , SF and SF have different shapes. Draw their Lewis structures and use the VSEPR theory to predict the nameof the shape of each molecule.

2 4 6

Markscheme

Penalise missing lone pairs on fluorine atoms once in correct structures only.

For Lewis structures candidates are not expected to draw exact shapes of molecules.

Do not allow ECF for wrong Lewis structures.

Accept dots or crosses instead of lines.

Examiners reportThis question was generally well answered and many high scores were seen. Common errors included the omission ofnon-bonding pairs of electrons on fluorine atoms and sometimes on sulfur. The shape of SF was not well known butcandidates were able to write correctly the name of the shape for SF and SF .

4 2 6

31a. [3 marks]

Consider the structure and bonding in and .

State and explain the difference in the electrical conductivity in the liquid state of the two chlorides.

Markscheme conducts and does not;

ionic and covalent/molecular/(consists of) molecules;

Award [1 max] for MgCl conducts and ionic.

Award [1 max] for PCl does not conduct and covalent/molecular/(consists of molecules).

ions can move in liquid (in ) / OWTTE;

Examiners reportThis was usually well answered.

MgCl2 PCl5

MgCl2 PCl5

MgCl2 PCl5

2

5

MgCl2

Printed for Southeast High School

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31b. [2 marks]Compare the formation of sigma ( ) and pi ( ) bonds between the carbon atoms in a molecule of ethyne.

Markscheme bond:

end-on/axial overlap with electron density between the two carbon atoms/nuclei / end-on/axial overlap of orbitals soshared electrons are between atoms / OWTTE;

bond:

sideways/parallel overlap of p orbitals with electron density above and below internuclear axis/ bond /sideways/parallel overlap of p orbitals so shared electrons are above and below internuclear axis/ bond / OWTTE;

Marks can be scored from a suitable diagram.

Award [1 max] for stating end-on/axial overlap for and sideways/parallel overlap for only i.e. without mentioning electron density OR stating electron density between the two atoms/nuclei for and

above and below internuclear axis for .

Examiners reportVery few candidates scored both marks on sigma and pi bonds.

σ π

σ

π

σσ

σπ σ

π

Practice Topic 4 Test bonding [89 marks] -...

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